TITLE: Can we find a lower bound for $P(V>1)$ when we know $EV=1$ and $P(V \le 1- \eta) \ge \delta? V \ge 0$ is non-constant.
QUESTION [0 upvotes]: Let $V \ge 0$ be a non-negative, non-constant, random variable and assume that we know $EV=1$ and $P(V \le 1- \eta) \ge \delta.$ I'm trying to see if I can find a lower bound for $P(V>1),$ in terms of $\eta, \delta?$
Intuitively, speaking, since $EV=1,$ it stands to reason that $P(V>1)$ and $P(V<1)$ both must be strictly positive as $V$ is non-constant. Now in this problem: we've a lower bound on $P(V \le 1- \eta),$ but does this guarantee a lower bound for the probability of the "other side", namely $P(V>1),$ which intuitively seems correct as this probability has to be big enough to balance the expectation of $V$ to be equal to $1.$ Intuitively, if this probability is too small, then $P(V>1)$ is close to $0,$ so $P(V<1)$ is close to $1,$ and as a result, it seems the $EV$ will be strictly less than $1.$
EDIT: Is it possible to have this kind of uniform lower bound if we instead work with a uniformly bounded or uniformly integrable family of random variables $\{V_{\epsilon}\}?$ If yes, what'd be the bound?

REPLY [1 votes]: Unless $V$ is bounded, $P(V > 1)$ can still be arbitrarily small.  For fixed $0 < \eta, \delta < 1$ let $0 < \varepsilon < 1-\delta$.  Then define $V = \frac{1}{\varepsilon}$ with probability $\varepsilon$ and $V = 0$ with probability $1-\varepsilon$.  Then $\mathbb{E}[V] = 1$ and $P(V \le 1-\eta) = 1-\varepsilon > \delta$, but $P(V > 1) = \varepsilon$.


TITLE: Some equations similar to Goormaghtigh problem
QUESTION [3 upvotes]: Let $m,n,t$ be some primes and $x$ and $y$ be some prime powers. Also suppose that $mn\mid(t-1)$. Which one of the following is correct?
1- If $t=\dfrac{x^{n}-1}{(x-1)(x-1,n)}=\dfrac{y^{m}-1}{(y-1)(y-1,m)}$, then $x=y$ and $m=n$.
2- If $t=\dfrac{x^{n}+1}{(x+1)(x+1,n)}=\dfrac{y^{m}-1}{(y-1)(y-1,m)}$, then there is no solution for $x,y,m,n$.
3- If $t=\dfrac{x^{n}+1}{(x+1)(x+1,n)}=\dfrac{y^{m}+1}{(y+1)(y+1,m)}$, then $x=y$ and $m=n$.

REPLY [6 votes]: There are counter-examples to all three claims in the question above, so none of the claims are correct.
1- A counter example is {x,n,y,m,t}={2,5,5,3,31} as was indicated in a comment above.
Another counter example is {x,n,y,m,t}={2,3,13,2,7}.
2- A counter example is {x,n,y,m,t}={3,3,13,2,7}.
Another counter example is {x,n,y,m,t}={5,3,13,2,7}.
3- A counter example is {x,n,y,m,t}={2,7,7,3,43}.
For those that enjoy searching for patterns in data, the table below for $(x^n,y^m)\le 10,000$ provides a bit more to contemplate. The table includes cases that violate the assumption $m\,n|(t-1)$ and trivial cases of Claims 1 and 2. The non-trivial cases that satisfy the assumption $m\,n|(t-1)$ are contained in rows 38 to 56 below. I've currently generated results for $(x^n,y^m)\le 100,000,000$ which is too much data to post here.
$\begin{array}{ccccccccccccccccc}
 \# & \text{mn$|$(t-1)} & \text{Trivial} & \text{Claim} & \text{x=y} & \text{n=m} & \text{x} & \text{n} & \text{y} & \text{m} & \text{t} & \text{$x^n$} & \text{$y^m$} & \{\text{Factor(x)}\} & \{\text{Factor(y)}\} & \{\text{Factor($x^n\pm$1)}\} & \{\text{Factor($y^m\pm$1 )}\} \\
 1 & \text{False} & \text{False} & 1 & \text{False} & \text{True} & 2 & 2 & 5 & 2 & 3 & 4 & 25 & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 5 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
 3 & 1 \\
\end{array}
\right) \\
 2 & \text{False} & \text{False} & 1 & \text{False} & \text{True} & 2 & 3 & 4 & 3 & 7 & 8 & 64 & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 7 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 2 \\
 7 & 1 \\
\end{array}
\right) \\
 3 & \text{False} & \text{False} & 2 & \text{False} & \text{True} & 3 & 3 & 2 & 3 & 7 & 27 & 8 & \left(
\begin{array}{cc}
 3 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
 7 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 7 & 1 \\
\end{array}
\right) \\
 4 & \text{False} & \text{False} & 2 & \text{False} & \text{True} & 3 & 3 & 4 & 3 & 7 & 27 & 64 & \left(
\begin{array}{cc}
 3 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
 7 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 2 \\
 7 & 1 \\
\end{array}
\right) \\
 5 & \text{False} & \text{False} & 2 & \text{False} & \text{True} & 4 & 3 & 3 & 3 & 13 & 64 & 27 & \left(
\begin{array}{cc}
 2 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 5 & 1 \\
 13 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 13 & 1 \\
\end{array}
\right) \\
 6 & \text{False} & \text{False} & 2 & \text{False} & \text{True} & 5 & 3 & 2 & 3 & 7 & 125 & 8 & \left(
\begin{array}{cc}
 5 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 3 & 2 \\
 7 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 7 & 1 \\
\end{array}
\right) \\
 7 & \text{False} & \text{False} & 2 & \text{False} & \text{True} & 5 & 3 & 4 & 3 & 7 & 125 & 64 & \left(
\begin{array}{cc}
 5 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 3 & 2 \\
 7 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 2 \\
 7 & 1 \\
\end{array}
\right) \\
 8 & \text{False} & \text{False} & 3 & \text{False} & \text{True} & 3 & 3 & 5 & 3 & 7 & 27 & 125 & \left(
\begin{array}{cc}
 3 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 5 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
 7 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 3 & 2 \\
 7 & 1 \\
\end{array}
\right) \\
 9 & \text{False} & \text{True} & 1 & \text{True} & \text{True} & 2 & 2 & 2 & 2 & 3 & 4 & 4 & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 1 \\
\end{array}
\right) \\
 10 & \text{False} & \text{True} & 1 & \text{True} & \text{True} & 2 & 3 & 2 & 3 & 7 & 8 & 8 & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 7 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 7 & 1 \\
\end{array}
\right) \\
 11 & \text{False} & \text{True} & 1 & \text{True} & \text{True} & 2 & 5 & 2 & 5 & 31 & 32 & 32 & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 31 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 31 & 1 \\
\end{array}
\right) \\
 12 & \text{False} & \text{True} & 1 & \text{True} & \text{True} & 2 & 7 & 2 & 7 & 127 & 128 & 128 & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 127 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 127 & 1 \\
\end{array}
\right) \\
 13 & \text{False} & \text{True} & 1 & \text{True} & \text{True} & 2 & 13 & 2 & 13 & 8191 & 8192 & 8192 & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 8191 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 8191 & 1 \\
\end{array}
\right) \\
 14 & \text{False} & \text{True} & 1 & \text{True} & \text{True} & 3 & 2 & 3 & 2 & 2 & 9 & 9 & \left(
\begin{array}{cc}
 3 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
\end{array}
\right) \\
 15 & \text{False} & \text{True} & 1 & \text{True} & \text{True} & 3 & 3 & 3 & 3 & 13 & 27 & 27 & \left(
\begin{array}{cc}
 3 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 13 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 13 & 1 \\
\end{array}
\right) \\
 16 & \text{False} & \text{True} & 1 & \text{True} & \text{True} & 3 & 7 & 3 & 7 & 1093 & 2187 & 2187 & \left(
\begin{array}{cc}
 3 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 1093 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 1093 & 1 \\
\end{array}
\right) \\
 17 & \text{False} & \text{True} & 1 & \text{True} & \text{True} & 4 & 3 & 4 & 3 & 7 & 64 & 64 & \left(
\begin{array}{cc}
 2 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 2 \\
 7 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 2 \\
 7 & 1 \\
\end{array}
\right) \\
 18 & \text{False} & \text{True} & 1 & \text{True} & \text{True} & 5 & 2 & 5 & 2 & 3 & 25 & 25 & \left(
\begin{array}{cc}
 5 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 5 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
 3 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
 3 & 1 \\
\end{array}
\right) \\
 19 & \text{False} & \text{True} & 1 & \text{True} & \text{True} & 5 & 3 & 5 & 3 & 31 & 125 & 125 & \left(
\begin{array}{cc}
 5 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 5 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
 31 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
 31 & 1 \\
\end{array}
\right) \\
 20 & \text{False} & \text{True} & 1 & \text{True} & \text{True} & 13 & 2 & 13 & 2 & 7 & 169 & 169 & \left(
\begin{array}{cc}
 13 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 13 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
 3 & 1 \\
 7 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
 3 & 1 \\
 7 & 1 \\
\end{array}
\right) \\
 21 & \text{False} & \text{True} & 1 & \text{True} & \text{True} & 13 & 3 & 13 & 3 & 61 & 2197 & 2197 & \left(
\begin{array}{cc}
 13 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 13 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
 3 & 2 \\
 61 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
 3 & 2 \\
 61 & 1 \\
\end{array}
\right) \\
 22 & \text{False} & \text{True} & 1 & \text{True} & \text{True} & 37 & 2 & 37 & 2 & 19 & 1369 & 1369 & \left(
\begin{array}{cc}
 37 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 37 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
 3 & 2 \\
 19 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
 3 & 2 \\
 19 & 1 \\
\end{array}
\right) \\
 23 & \text{False} & \text{True} & 1 & \text{True} & \text{True} & 61 & 2 & 61 & 2 & 31 & 3721 & 3721 & \left(
\begin{array}{cc}
 61 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 61 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
 3 & 1 \\
 5 & 1 \\
 31 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
 3 & 1 \\
 5 & 1 \\
 31 & 1 \\
\end{array}
\right) \\
 24 & \text{False} & \text{True} & 3 & \text{True} & \text{True} & 2 & 5 & 2 & 5 & 11 & 32 & 32 & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 1 \\
 11 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 1 \\
 11 & 1 \\
\end{array}
\right) \\
 25 & \text{False} & \text{True} & 3 & \text{True} & \text{True} & 2 & 7 & 2 & 7 & 43 & 128 & 128 & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 1 \\
 43 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 1 \\
 43 & 1 \\
\end{array}
\right) \\
 26 & \text{False} & \text{True} & 3 & \text{True} & \text{True} & 2 & 11 & 2 & 11 & 683 & 2048 & 2048 & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 1 \\
 683 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 1 \\
 683 & 1 \\
\end{array}
\right) \\
 27 & \text{False} & \text{True} & 3 & \text{True} & \text{True} & 2 & 13 & 2 & 13 & 2731 & 8192 & 8192 & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 1 \\
 2731 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 1 \\
 2731 & 1 \\
\end{array}
\right) \\
 28 & \text{False} & \text{True} & 3 & \text{True} & \text{True} & 3 & 3 & 3 & 3 & 7 & 27 & 27 & \left(
\begin{array}{cc}
 3 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
 7 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
 7 & 1 \\
\end{array}
\right) \\
 29 & \text{False} & \text{True} & 3 & \text{True} & \text{True} & 3 & 5 & 3 & 5 & 61 & 243 & 243 & \left(
\begin{array}{cc}
 3 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
 61 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
 61 & 1 \\
\end{array}
\right) \\
 30 & \text{False} & \text{True} & 3 & \text{True} & \text{True} & 3 & 7 & 3 & 7 & 547 & 2187 & 2187 & \left(
\begin{array}{cc}
 3 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
 547 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
 547 & 1 \\
\end{array}
\right) \\
 31 & \text{False} & \text{True} & 3 & \text{True} & \text{True} & 4 & 3 & 4 & 3 & 13 & 64 & 64 & \left(
\begin{array}{cc}
 2 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 5 & 1 \\
 13 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 5 & 1 \\
 13 & 1 \\
\end{array}
\right) \\
 32 & \text{False} & \text{True} & 3 & \text{True} & \text{True} & 4 & 5 & 4 & 5 & 41 & 1024 & 1024 & \left(
\begin{array}{cc}
 2 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 5 & 2 \\
 41 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 5 & 2 \\
 41 & 1 \\
\end{array}
\right) \\
 33 & \text{False} & \text{True} & 3 & \text{True} & \text{True} & 5 & 3 & 5 & 3 & 7 & 125 & 125 & \left(
\begin{array}{cc}
 5 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 5 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 3 & 2 \\
 7 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 3 & 2 \\
 7 & 1 \\
\end{array}
\right) \\
 34 & \text{False} & \text{True} & 3 & \text{True} & \text{True} & 5 & 5 & 5 & 5 & 521 & 3125 & 3125 & \left(
\begin{array}{cc}
 5 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 5 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 3 & 1 \\
 521 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 3 & 1 \\
 521 & 1 \\
\end{array}
\right) \\
 35 & \text{False} & \text{True} & 3 & \text{True} & \text{True} & 7 & 3 & 7 & 3 & 43 & 343 & 343 & \left(
\begin{array}{cc}
 7 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 7 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
 43 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
 43 & 1 \\
\end{array}
\right) \\
 36 & \text{False} & \text{True} & 3 & \text{True} & \text{True} & 13 & 3 & 13 & 3 & 157 & 2197 & 2197 & \left(
\begin{array}{cc}
 13 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 13 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 7 & 1 \\
 157 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 7 & 1 \\
 157 & 1 \\
\end{array}
\right) \\
 37 & \text{False} & \text{True} & 3 & \text{True} & \text{True} & 16 & 3 & 16 & 3 & 241 & 4096 & 4096 & \left(
\begin{array}{cc}
 2 & 4 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 4 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 17 & 1 \\
 241 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 17 & 1 \\
 241 & 1 \\
\end{array}
\right) \\
 38 & \text{True} & \text{False} & 1 & \text{False} & \text{False} & 2 & 3 & 13 & 2 & 7 & 8 & 169 & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 13 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 7 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
 3 & 1 \\
 7 & 1 \\
\end{array}
\right) \\
 39 & \text{True} & \text{False} & 1 & \text{False} & \text{False} & 2 & 5 & 5 & 3 & 31 & 32 & 125 & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 5 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 31 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
 31 & 1 \\
\end{array}
\right) \\
 40 & \text{True} & \text{False} & 1 & \text{False} & \text{False} & 2 & 5 & 61 & 2 & 31 & 32 & 3721 & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 61 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 31 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
 3 & 1 \\
 5 & 1 \\
 31 & 1 \\
\end{array}
\right) \\
 41 & \text{True} & \text{False} & 1 & \text{False} & \text{False} & 2 & 7 & 19 & 3 & 127 & 128 & 6859 & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 19 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 127 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 3 & 3 \\
 127 & 1 \\
\end{array}
\right) \\
 42 & \text{True} & \text{False} & 1 & \text{False} & \text{False} & 3 & 3 & 25 & 2 & 13 & 27 & 625 & \left(
\begin{array}{cc}
 3 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 5 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 13 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 4 \\
 3 & 1 \\
 13 & 1 \\
\end{array}
\right) \\
 43 & \text{True} & \text{False} & 1 & \text{False} & \text{False} & 4 & 3 & 13 & 2 & 7 & 64 & 169 & \left(
\begin{array}{cc}
 2 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 13 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 2 \\
 7 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
 3 & 1 \\
 7 & 1 \\
\end{array}
\right) \\
 44 & \text{True} & \text{False} & 1 & \text{False} & \text{False} & 5 & 3 & 61 & 2 & 31 & 125 & 3721 & \left(
\begin{array}{cc}
 5 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 61 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
 31 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
 3 & 1 \\
 5 & 1 \\
 31 & 1 \\
\end{array}
\right) \\
 45 & \text{True} & \text{False} & 1 & \text{False} & \text{False} & 7 & 3 & 37 & 2 & 19 & 343 & 1369 & \left(
\begin{array}{cc}
 7 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 37 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 3 & 2 \\
 19 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
 3 & 2 \\
 19 & 1 \\
\end{array}
\right) \\
 46 & \text{True} & \text{False} & 1 & \text{False} & \text{True} & 4 & 2 & 9 & 2 & 5 & 16 & 81 & \left(
\begin{array}{cc}
 2 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 1 \\
 5 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 4 \\
 5 & 1 \\
\end{array}
\right) \\
 47 & \text{True} & \text{False} & 2 & \text{False} & \text{False} & 3 & 3 & 13 & 2 & 7 & 27 & 169 & \left(
\begin{array}{cc}
 3 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 13 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
 7 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
 3 & 1 \\
 7 & 1 \\
\end{array}
\right) \\
 48 & \text{True} & \text{False} & 2 & \text{False} & \text{False} & 3 & 5 & 13 & 3 & 61 & 243 & 2197 & \left(
\begin{array}{cc}
 3 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 13 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
 61 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
 3 & 2 \\
 61 & 1 \\
\end{array}
\right) \\
 49 & \text{True} & \text{False} & 2 & \text{False} & \text{False} & 4 & 3 & 25 & 2 & 13 & 64 & 625 & \left(
\begin{array}{cc}
 2 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 5 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 5 & 1 \\
 13 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 4 \\
 3 & 1 \\
 13 & 1 \\
\end{array}
\right) \\
 50 & \text{True} & \text{False} & 2 & \text{False} & \text{False} & 4 & 5 & 81 & 2 & 41 & 1024 & 6561 & \left(
\begin{array}{cc}
 2 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 4 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 5 & 2 \\
 41 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 5 \\
 5 & 1 \\
 41 & 1 \\
\end{array}
\right) \\
 51 & \text{True} & \text{False} & 2 & \text{False} & \text{False} & 5 & 3 & 13 & 2 & 7 & 125 & 169 & \left(
\begin{array}{cc}
 5 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 13 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 3 & 2 \\
 7 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
 3 & 1 \\
 7 & 1 \\
\end{array}
\right) \\
 52 & \text{True} & \text{False} & 2 & \text{False} & \text{False} & 8 & 3 & 37 & 2 & 19 & 512 & 1369 & \left(
\begin{array}{cc}
 2 & 3 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 37 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 3 \\
 19 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
 3 & 2 \\
 19 & 1 \\
\end{array}
\right) \\
 53 & \text{True} & \text{False} & 2 & \text{False} & \text{False} & 11 & 3 & 73 & 2 & 37 & 1331 & 5329 & \left(
\begin{array}{cc}
 11 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 73 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
 3 & 2 \\
 37 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 4 \\
 3 & 2 \\
 37 & 1 \\
\end{array}
\right) \\
 54 & \text{True} & \text{False} & 2 & \text{False} & \text{True} & 8 & 3 & 7 & 3 & 19 & 512 & 343 & \left(
\begin{array}{cc}
 2 & 3 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 7 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 3 \\
 19 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 3 & 2 \\
 19 & 1 \\
\end{array}
\right) \\
 55 & \text{True} & \text{False} & 2 & \text{False} & \text{True} & 9 & 3 & 8 & 3 & 73 & 729 & 512 & \left(
\begin{array}{cc}
 3 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 5 & 1 \\
 73 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 7 & 1 \\
 73 & 1 \\
\end{array}
\right) \\
 56 & \text{True} & \text{False} & 3 & \text{False} & \text{False} & 2 & 7 & 7 & 3 & 43 & 128 & 343 & \left(
\begin{array}{cc}
 2 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 7 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 1 \\
 43 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
 43 & 1 \\
\end{array}
\right) \\
 57 & \text{True} & \text{True} & 1 & \text{True} & \text{True} & 4 & 2 & 4 & 2 & 5 & 16 & 16 & \left(
\begin{array}{cc}
 2 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 1 \\
 5 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 1 \\
 5 & 1 \\
\end{array}
\right) \\
 58 & \text{True} & \text{True} & 1 & \text{True} & \text{True} & 7 & 3 & 7 & 3 & 19 & 343 & 343 & \left(
\begin{array}{cc}
 7 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 7 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 3 & 2 \\
 19 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 3 & 2 \\
 19 & 1 \\
\end{array}
\right) \\
 59 & \text{True} & \text{True} & 1 & \text{True} & \text{True} & 8 & 3 & 8 & 3 & 73 & 512 & 512 & \left(
\begin{array}{cc}
 2 & 3 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 7 & 1 \\
 73 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 7 & 1 \\
 73 & 1 \\
\end{array}
\right) \\
 60 & \text{True} & \text{True} & 1 & \text{True} & \text{True} & 9 & 2 & 9 & 2 & 5 & 81 & 81 & \left(
\begin{array}{cc}
 3 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 4 \\
 5 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 4 \\
 5 & 1 \\
\end{array}
\right) \\
 61 & \text{True} & \text{True} & 1 & \text{True} & \text{True} & 16 & 2 & 16 & 2 & 17 & 256 & 256 & \left(
\begin{array}{cc}
 2 & 4 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 4 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 1 \\
 5 & 1 \\
 17 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 1 \\
 5 & 1 \\
 17 & 1 \\
\end{array}
\right) \\
 62 & \text{True} & \text{True} & 1 & \text{True} & \text{True} & 17 & 3 & 17 & 3 & 307 & 4913 & 4913 & \left(
\begin{array}{cc}
 17 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 17 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 4 \\
 307 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 4 \\
 307 & 1 \\
\end{array}
\right) \\
 63 & \text{True} & \text{True} & 1 & \text{True} & \text{True} & 19 & 3 & 19 & 3 & 127 & 6859 & 6859 & \left(
\begin{array}{cc}
 19 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 19 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 3 & 3 \\
 127 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 3 & 3 \\
 127 & 1 \\
\end{array}
\right) \\
 64 & \text{True} & \text{True} & 1 & \text{True} & \text{True} & 25 & 2 & 25 & 2 & 13 & 625 & 625 & \left(
\begin{array}{cc}
 5 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 5 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 4 \\
 3 & 1 \\
 13 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 4 \\
 3 & 1 \\
 13 & 1 \\
\end{array}
\right) \\
 65 & \text{True} & \text{True} & 1 & \text{True} & \text{True} & 73 & 2 & 73 & 2 & 37 & 5329 & 5329 & \left(
\begin{array}{cc}
 73 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 73 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 4 \\
 3 & 2 \\
 37 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 4 \\
 3 & 2 \\
 37 & 1 \\
\end{array}
\right) \\
 66 & \text{True} & \text{True} & 1 & \text{True} & \text{True} & 81 & 2 & 81 & 2 & 41 & 6561 & 6561 & \left(
\begin{array}{cc}
 3 & 4 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 4 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 5 \\
 5 & 1 \\
 41 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 5 \\
 5 & 1 \\
 41 & 1 \\
\end{array}
\right) \\
 67 & \text{True} & \text{True} & 3 & \text{True} & \text{True} & 8 & 3 & 8 & 3 & 19 & 512 & 512 & \left(
\begin{array}{cc}
 2 & 3 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 3 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 3 \\
 19 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 3 \\
 19 & 1 \\
\end{array}
\right) \\
 68 & \text{True} & \text{True} & 3 & \text{True} & \text{True} & 9 & 3 & 9 & 3 & 73 & 729 & 729 & \left(
\begin{array}{cc}
 3 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 3 & 2 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 5 & 1 \\
 73 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 1 \\
 5 & 1 \\
 73 & 1 \\
\end{array}
\right) \\
 69 & \text{True} & \text{True} & 3 & \text{True} & \text{True} & 11 & 3 & 11 & 3 & 37 & 1331 & 1331 & \left(
\begin{array}{cc}
 11 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 11 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
 3 & 2 \\
 37 & 1 \\
\end{array}
\right) & \left(
\begin{array}{cc}
 2 & 2 \\
 3 & 2 \\
 37 & 1 \\
\end{array}
\right) \\
\end{array}$


TITLE: Evaluate the integral with and without Green's theorem
QUESTION [2 upvotes]: Evaluate the line integral $\oint_C y^2dx + xdy$ when $C$ has the vector equation $\alpha(t)=(2\cos^3t)i+(2\sin^3t)j$, $0 \leq t \leq 2\pi$.
My attempt
BY USING GREEN's THEOREM,
i.e., $P=y^2$, $Q=x$
we get $\iint_c 1-2y dxdy$
Now putting $x=r\cos^3t$ and $y=r\sin^3t$
we get $dxdy=Jdrdt$, where $J=3r \sin^2t \cos^2t$
hence $dxdy=3r \sin^2t \cos^2t drdt$
Hence the required integral is,
$\int_0^{2\pi} \int_0^2 (1-2r \sin^3t)(3r \sin^2t \cos^2t)drdt$
which is equal to $\pi /4$
But the answer given in APOSTOL's is $3 \pi /2$
NOW BY NORMAL METHOD
put $x=2 \cos^3t$ and $y=2 \sin^3t$
$dx=-6 \cos^2t \sin t dt$ and $dy=6 \sin^2t cost dt$,
Hence the integral becomes $-24 \int_0^{2 \pi} \sin^7t \cos^2t dt + 12 \int_0^{2 \pi} \cos^4t \sin^2t dt$ which is equal to ZERO.
WHAT MISTAKE AM I DOING AND THE CORRECT ANSWER IS $3 \pi /2$  ?

REPLY [3 votes]: Your second method
Use $$\int_0^{2\pi}\sin^n x\ dx=\dfrac{2\pi}{2^{2k}}{2k\choose k}$$
for even $n=2k$ and for odd $n$ it is zero, then
$$-24 \int_0^{2 \pi} \sin^7t \cos^2t dt + 12 \int_0^{2 \pi} \cos^4t \sin^2t dt$$
$$-24\int_0^{2 \pi} \sin^7t (1-\sin^2t) dt + 12 \int_0^{2 \pi} (1-\sin^2t)^2\sin^2t dt$$
$$-24\int_0^{2 \pi} \sin^7t\ dt + 24\int_0^{2 \pi} \sin^9t\ dt + 12\int_0^{2 \pi} \sin^2t\ dt -24 \int_0^{2 \pi} \sin^4t\ dt + 12\int_0^{2 \pi} \sin^6t\ dt $$
$$0+0+12\dfrac{2\pi}{2^{2}}{2\choose 1} -24\dfrac{2\pi}{2^{4}}{4\choose 2} + 12\dfrac{2\pi}{2^{6}}{6\choose 3}=\color{blue}{\dfrac{3\pi}{2}}$$


text:
\begin{document}

\title{CSI-based Outdoor Localization for Massive MIMO: Experiments with a Learning Approach}

\author{Alexis~Decurninge, Luis~Garc\'ia~Ord\'o\~nez,  Paul Ferrand, He Gaoning, Li Bojie, Zhang Wei, Maxime~Guillaud\\
Huawei Technologies}

\maketitle

\begin{abstract}
We report on experimental results on the use of a learning-based approach to infer the location of a mobile user of a cellular network within a cell, for a 5G-type Massive multiple input, multiple output (MIMO) system.
We describe how the sample spatial covariance matrix computed from the CSI can be used as the input to a learning algorithm which attempts to relate it to user location.
We discuss several learning approaches, and analyze in depth the application of extreme learning machines, for which theoretical approximate performance benchmarks are available, to the localization problem.
We validate the proposed approach using experimental data collected on a Huawei 5G testbed, provide some performance and robustness benchmarks, and discuss practical issues related to the deployment of such a technique in 5G networks.
\end{abstract}

\section{Introduction}

\noindent Future 5G systems are expected to incorporate base stations (BSs) equipped with massive multiple-input multiple-output (MIMO) technology, which uses arrays with the number of antenna elements being some orders of magnitude larger than current state-of-the-art MIMO BSs \cite{Lar14}.
Beyond the expected network-level performance improvements, the presence of excess antennas substantially increases the spatial resolution of the antenna array.
This opens the door to localization methods of increased accuracy based on measuring the propagation characteristics of the wireless channel.
In this article, we investigate to which extent the uplink channel state information (CSI), as estimated by the base station for communications purposes,
can be exploited to infer the position of a user equipment (UE)\footnote{Note that we focus on network-side localization here, i.e., the ability for the network to localize a UE, as opposed to user-side localization such as the one enabled by the GPS system, whereby the location information is directly obtained at the UE, but not necessarily shared with the network operator.}.

There is a rich literature on localization strategies based on estimating the state of the wireless propagation channel.
Most of the existing techniques can be understood as follows: the network measures or estimates some propagation parameter---e.g. channel impulse response, propagation delay, angle of arrival or received signal strength---at one or multiple positions in the network, typically BSs or access points (APs).
This \emph{fingerprint} of the user location can be  then either compared to existing databases and maps \cite{Triki_etal_PDP_fingerprinting_VTC2006,Khalajmehrabadi_wlan_fingerprinting_2017}, or subsequently intepreted in a geometric sense.
The assumption of the first approach is that there should be a bijective relationship between the user position and the fingerprint, although this relationship is usually too rich to be modeled and must thus be learned.
A variety of learning algorithms can be applied in this context, among which $K$-nearest neighbors ($K$-nNs), neural network (NNs), support vector machines (SVMs) and Bayesian learning are the most popular.
In contrast, for the geometric approach, the fingerprints are translated into distances and angles from which the user location is triangulated or trilaterated.
In both approaches, the  dimension of the fingerprint is usually the parameter limiting the localization accuracy.
For example, it was shown in~\cite{Khalajmehrabadi_wlan_fingerprinting_2017} that received signal strength indicator (RSSI)-based approaches become extremely accurate for indoor environments as the number of reference measurements increases.

This limitation can be overcome through the use of multiple antennas at the APs or BSs, as proposed in~\cite{Dayekh_localization_underground_mines_2014,Chapre_CSI_MIMO_2014}.
A similar approach has also been applied in~\cite{Savic_fingerprinting_positioning_distributed_MIMO_2015} to outdoor localization in the so-called \emph{distributed} MIMO case, in which the antenna elements belonging to the same (virtual) BS are not collocated in a single antenna array but distributed within the cell.

A common way to obtain spatial fingerprints is to extract the cluster of scatterers angle and delay from the raw CSI~\cite{Shutin_angle_delay_2004}.
This fingerprinting method has been studied both in~\cite{Vieira_etal_DCNN_positioning_PIMRC17} and~\cite{Sun_single_site_localization_2018}.
In the former, the authors applied deep neural networks to the obtained fingerprints, while in the latter $K$-nNs type algorithms were considered for their affordable complexity.

In this article, we focus on fingerprinting at a single BS equipped with a large (massive MIMO) antenna array.
We adopt a NN-based learning approach and compare it with a more classical $K$-nN learning algorithm.
In particular, we choose a single-layer random feed-forward neural network, usually referred to as an extreme learning machine (ELM) \cite{huang06,Huang_trends15}.
Beyond its simplicity, the ELM has been proven to exhibit universal approximation capabilities and good generalization performance for a wide range of problems.
In the context of localization, the authors of~\cite{Zou_positioning_procrustes_ELM_TWC16,Zhang_etal_device_free_localization_Sensors17,ELM_positioning_Lu_TransCybernetics17} use RSSI measurements coupled with extreme learning machines for indoor localization.
They however only consider single-antenna elements at the APs.
The rest of the paper is outlined as follows. We introduce our approach in Section~\ref{section_CSI_based_localization}; the experimental results are presented in Section~\ref{section_experimental}. Finally, in Section~\ref{section_practical_issues} we discuss some general implementation issues that need to be considered before deployment.

\section{Proposed CSI-Based Localization Approach}
\label{section_CSI_based_localization}

\subsection{Input Data Pre-Processing}

\noindent When multiple antennas are used in a single BS at a single cell, some additional pre-processing of the raw channel data is needed to form a useful input for any learning algorithm.
We consider a system where CSI acquisition takes place in the frequency domain, consistent with the OFDM/OFDMA approach being standardized for 5G systems and we let $\h(s,t) \in \mathbb{C}^{N}$ denote the complex vector representing the instantaneous uplink propagation channel between the single-antenna user equipment (UE) and the $N$ BS antennas on subcarrier $s$ at time $t$. Even if the UE position in the horizontal plane (denoted by $\y \in \mathbb{R}^{2}$) is completely static, the instantaneous channel as measured by the baseband processor typically shows some variations due to small-scale fading and to the residual carrier frequency offset resulting from the impossibility of perfectly synchronizing  the oscillators of the transmitter and the receiver.
In order to reduce the noise present in the input data and simultaneously remove the common phase component due to the clock offset, we compute the BS-side channel covariance as
\begin{equation} \label{eq_covariance}
\C (\y) = \mathop{\mathbb{E}}\left[ \h(s,t) \h(s,t)^\dagger \right]
\end{equation}
for some fixed reference subcarrier $s$ and where $(\cdot)^\dagger$ denotes the Hermitian transposition. We implicitly assume that the expectation is computed at time $t$ over a small time horizon $\Delta T$ while the UE remains fixed in a given location $\y$, so that the fading process can be assumed stationary. In practice, the expectation in \eqref{eq_covariance} will be replaced by a sample covariance estimator (see Section~\ref{section_experimental} for more details about our experimental set-up).

In the sequel, we consider as the input to the learning algorithm a vectorized and normalized\footnote{Here we remove the pathloss information (see Section \ref{section_experimental}). Note that there could be alternative approaches to taking the normalized covariance to pre-process the CSI; we did not attempt to exhaustively research this question.} version of the covariance $\C(\y)$, denoted by $ \x \in\mathbb{R}^{N^2}$ and obtained as
\begin{equation} \label{eq:input}
\x = \Big(  \mathrm{vec_{ut}}\big( \mathrm{Re}\big\{\tfrac{\C(\y) }{\text{Tr}(\C(\y))}\big\}\big)^{\mathrm{T}},  \mathrm{vec_{ut}}\big( \mathrm{Im}\big\{\tfrac{\C(\y) }{\text{Tr}(\C(\y))}\big\}\big)^{\mathrm{T}}\Big)^{\mathrm{T}}
\end{equation}
where $\mathrm{vec_{ut}}(\cdot)$ stands for the operator stacking in a vector the upper triangular entries of a $N\times N$ matrix as long as they are non-zero and, hence, removes the inherent redundancy present in $\C(\y)$. This input choice comes from the intuition that the second-order channel statistics capture most of the location-related characteristics of the CSI and, consequently, $\C(\y)$ can be directly related to the UE location $\y$. However, the complexity of the relationship between the spatial scattering environment surrounding a user and the CSI as measured by the BS makes it extremely difficult to express or even model; therefore, we propose to infer it through statistical learning.

\subsection{Learning approaches}

\noindent It is convenient to distinguish between the following two phases of the use of a learning algorithm:
\begin{itemize}
\item[$(\mathrm{i})$] \emph{the training phase}, in which the CSI related input and the UE positions are acquired, i.e., the \emph{training} data set $(\x_j,\y_j)_{j\in\mathcal{T}}$ consisting of $T=|\mathcal{T}|$ pairs $\x_j\in\mathbb{R}^M$, with $M=N^2$, as defined in \eqref{eq:input} and the corresponding user location $\y_j\in\mathbb{R}^2$; this training set is used to infer the mapping relating $\x_j$ to $\y_j$, and
\item[$(\mathrm{ii})$] \emph{the exploitation phase}, in which the mapping obtained in $(\mathrm{i})$ is used to infer the UE position from the observed CSI; the performance of this step is evaluated on a separate \emph{testing} data set  $(\x_j,\y_j)_{j \in \mathcal{T}^{\prime}}$ with $T^{\prime} = |\mathcal{T}^{\prime}|$ data pairs.
\end{itemize}

The adopted learning approaches are detailed below.
\medskip
\subsubsection{K-nearest neighbors}
\label{sec:NN}
As a baseline method for comparison purposes, we consider a regression mapping using $K$-nearest neighbors. In this case the training phase only consists in acquiring the training data set. Then, for any observed input data $\x$ during the exploitation phase, we find the $K$ nearest neighbors of $\x$ in the training set, i.e., the $K$ indices $j_1, \ldots, j_K \in \mathcal{T}$  such that, for some distance $d( \cdot ,\cdot )$ defined on the input space (e.g., $d(\x_1,\x_2)=\|\x_1-\x_2\|_2$), we have $d(\x_{j_1},\x) \leq  \cdots \leq  d(\x_{j_K},\x) \leq d(\x_{j},\x)$ for any other $j \in \mathcal{T}$. Finally, the estimated UE location $\hat{\y}$ is computed as the barycenter of the locations of the $K$ nearest neighbors:
\begin{equation}
\hat{\y} = \frac{1}{K} \sum_{k=1}^K\y_{j_k}.
\end{equation}

\smallskip
\subsubsection{Extreme Learning Machine} \label{section_ELM}

As the main learning method we consider ELMs, which are feedforward neural networks consisting of a random single-hidden layer of neurons (see an example in Fig.~\ref{fig:elm}). An ELM can be understood as a non-linear regression function of the form
\begin{equation}
f :  \x\in\mathbb{R}^{M} \mapsto \sum_{i=1}^n \varphi(\w_i^{\mathrm{T}} \x)\dbeta_i\in\mathbb{R}^2
\end{equation}
where $\w_1,\dots,\w_n\in\mathbb{R}^{M}$ are weights of the $n$ neurons, $\varphi(\cdot)$ is an activation function (see Table~\ref{table:activation} for examples), and  $\dbeta_1,\dots,\dbeta_n\in\mathbb{R}^2$ are the output weights. In ELMs the entries of the weight matrix $\W = (\w_1,\dots\w_n)$ are independently drawn, e.g., from a Gaussian standard distribution and only the output weights $\dbeta=(\dbeta_i)_{i=1\dots n}$ need to be optimized.

During the training phase, $\dbeta$ is obtained using a simple linear ridge regression between the augmented version of the input, $(\varphi(\w_1^{\mathrm{T}} \x),\dots,\varphi(\w_n^{\mathrm{T}} \x))^{\mathrm{T}} \in \mathbb{R}^n$, and the output $\y$:
\begin{equation}
\label{eq:elm2}
 {\dbeta} = \Y_{\mathcal{T}}(\bSigma_{\mathcal{T}}^{\mathrm{T}}\bSigma_{\mathcal{T}}+\gamma\I_{\mathcal{T}})^{-1}\bSigma_{\mathcal{T}}^{\mathrm{T}}
\end{equation}
where we denote $\bSigma_{\mathcal{T}}=(\varphi(\w_i^{\mathrm{T}} \x_j))_{i=1,\ldots n, \ j \in \mathcal{T}}\in\mathbb{R}^{n\times T}$ and $\Y_{\mathcal{T}}=(\y_j)_{j \in \mathcal{T}}\in\mathbb{R}^{2\times T}$,
and we have introduced an $L_2$ regularization parameter $\gamma$, which makes the regression more robust without adding complexity.

In the exploitation phase, the UE location for a new observed input $\x$ is estimated as
\begin{equation}
\label{eq:elm}
\hat{\y} = \sum_{i=1}^n \varphi(\w_i^{\mathrm{T}} \x){\dbeta}_i={\dbeta}^{\mathrm{T}}\varphi(\W^{\mathrm{T}} \x).
\end{equation}

\begin{table}[b!]
\centering
\begin{tabular}{l|c}
  \hline
  ReLu   & $\varphi(t) = \max(0,t)$\\
  Step   & $\varphi(t) = \mathds{1}_{t\geq 0}$\\
  Sign  & $\varphi(t) = \mathds{1}_{t\geq 0} - \mathds{1}_{t<0}$\\
  \hline
\end{tabular}
\caption{Examples of activation functions.}\label{table:activation}
\end{table}

\begin{figure}[t!]
\centering

\begin{tikzpicture}[
transform shape,
scale=0.8,
y=0.7cm,
>=latex,
every node/.style={circle},
y=0.7cm
]
\def\nI{4}
\def\nC{7}
\def\nO{2}
\pgfmathsetmacro\Ishift{(\nI+1)/2}
\pgfmathsetmacro\Cshift{(\nC+1)/2}
\pgfmathsetmacro\Oshift{(\nO+1)/2}
\foreach \x in {1,...,\nI}{

	\draw (-4,\x-\Ishift) node[draw] (I\x) {\phantom{1}};
}
\foreach \x in {1,...,\nC}{
	\draw (0,\x-\Cshift) node[draw] (C\x) {\phantom{1}};
}
\foreach \x in {1,...,\nO}{
	\draw (4,\x-\Oshift) node[draw] (O\x) {\phantom{1}};
}

\foreach \i in {1,...,\nI}{
	\foreach \c in {1,...,\nC}{
	\draw[->] (I\i) -- (C\c);
}
}
\foreach \c in {1,...,\nC}{
	\foreach \o in {1,...,\nO}{
	\draw[->] (C\c) -- (O\o);
}
}
\node at (-2, \Ishift+1) {\large $\mat \W$};
\node at (2, \Ishift+1) {\large $\mat \dbeta$};

\node at (0, -\Cshift) {\large $\varphi(\W^{\mathrm{T}}\x)$};
\node at (-4, -\Cshift) {\large $\x$};
\node at (4, -\Cshift) {\large $\y$};
\end{tikzpicture}
\vspace{-4mm}
\caption{A typical extreme learning machine with $M = 4$ inputs, $n = 8$ neurons and $2$ outputs.}
\label{fig:elm}
\vspace{-2mm}
\end{figure}

The choice of ELM as a learning approach is motivated by its theoretical capacity to learn any non-linear mapping, as well as its simplicity \cite{huang06,Huang_trends15} and amenability to theoretical analysis \cite{Louart_etal_RMT_neural_networks_AP17}.

\subsection{Localization Performance}

\noindent In order to analyze the performances of the considered localization approaches, we measure the localization error as
\begin{equation} \label{eq:error}
e(\y_j,\hat{\y}_j) = \|\y_j - \hat{\y}_j\|
\end{equation}
where $\hat{\y}_j$ is the estimated position for the input $\x_j$. We define the corresponding average training and testing errors as

$E  = \frac{1}{T} \sum_{j \in \mathcal{T}} \|\hat{\y}_j - \y_j\|$ and
$E^{\prime} = \frac{1}{T^{\prime}} \sum_{j \in \mathcal{T}^{\prime}} \|\hat{\y}_j - \y_j\|$.

In the case of the ELM approach we also introduce the training and testing mean square localization errors:
\begin{align} \label{eq:mse_i}
\mathsf{mse}(\W) & = \frac{1}{T} \sum_{j \in \mathcal{T}} \|\hat{\y}_j - \y_j\|^2 \\
\label{eq:mse_ii}
\mathsf{mse}^{\prime}(\W) & = \frac{1}{T^{\prime}} \sum_{j \in \mathcal{T}^{\prime}} \|\hat{\y}_j - \y_j\|^2
\end{align}
where we have made explicit their randomness through the dependency on the random weight matrix $\W$.  The authors of \cite{Louart_etal_RMT_neural_networks_AP17} show that for the regression in \eqref{eq:elm}, both training and testing mean square errors almost surely converge to some deterministic limit, i.e.,
\begin{align}
\label{eq:det_eq1}
\mathsf{mse}(\W) - \overline{\mathsf{mse}} &\xrightarrow[ n,M,T \rightarrow \infty]{\mathrm{a.s.}} 0 \\%\qquad \text{and} \qquad
\mathsf{mse}^{\prime}(\W) - \overline{\mathsf{mse}}^{\prime} &\xrightarrow[ n,M,T^{\prime}\rightarrow \infty]{\mathrm{a.s.}} 0 \label{eq:det_eq2}
\end{align}
where the expressions of the deterministic equivalents $\overline{\mathsf{mse}}$ and $\overline{\mathsf{mse}}^{\prime}$ are given in \cite[Conject.~1 and Thm.~3]{Louart_etal_RMT_neural_networks_AP17}.
Note that $\overline{\mathsf{mse}}$ and $\overline{\mathsf{mse}}^{\prime}$ still depend on the training and testing data sets in order to avoid assuming a generative model of the entries.

The computation of these deterministic values allow us to choose the best hyperparameters, i.e., the regularization parameter $\gamma$, the activation function and the number of neurons independently of the random matrix $\W$.

\begin{figure}[t!]
\centering

 \includegraphics[width=.9\columnwidth]{fig_error_map.pdf}\\

\caption{Map of the measurement area with the locations of the training set (black dots) and testing data sets (colored dots) with testing location error (using the scale on the left) plotted for the true UE location.
Imagery \textcopyright \ Google, DigitalGlobe.}
\label{fig:error_map}
\vspace{-2mm}
\end{figure}

\section{Experimental Results}
\label{section_experimental}
\noindent Data was collected on the Huawei 5G testbed on the campus of the University of the Chinese Academy of Science in Huairou, P.R. China. CSI data was measured for a single-antenna UE on the uplink of a massive MIMO system using a 32 dual-polarized antenna array at the BS, which is located on the roof of a five-floor building. Due to some malfunctioning RF chains, the measurements from 8 antennas had to be discarded and, hence, unless otherwise stated, $N=56$.   In total, 90 minutes of outdoors measurements were acquired, covering a maximum extent of $196\times303$ meters, depicted on Fig.~\ref{fig:error_map}. Since the mobile UE was moved on a cart, the paths for which data could be recorded was limited to roads and pedestrian walkways. The scenario includes a majority of line-of-sight (LOS) locations, as well as some non-LOS locations. The UE position (latitude/longitude) was recorded using a commercial GPS with 1 Hz sampling rate; it was transformed to a local coordinate system to have dimensions in meters, and subsequently linearly interpolated in time when required. The CSI sampling rate was 10 Hz. Path-loss information was not deemed reliable due to automatic gain control, and therefore all covariance matrices were normalized as in \eqref{eq:input}. Gaps in the trace are due to the UE being out of range from the BS.

Clearly, when the UE is moving, $\h(s,t)$ can not be considered to be stationary and the statistics cannot be related to the position. This makes it difficult to evaluate the expectation when computing in practice the covariance as defined in \eqref{eq_covariance}. In order to escape this apparent contradiction and to be able to process the experimental data acquired for a moving user, we apply a sample covariance estimator over a 1-second horizon,\footnote{Note that at pedestrian speeds, $1$ s translates into approximately $1$ m, which is small with respect to the size of the considered area, and roughly commensurate with the expected accuracy of the GPS location data.} where the statistics can be assumed approximately constant. In the experiments presented here, a single subcarrier (always the same) is used, since including multiple ones have shown little or no gain, even when the selected subcarriers were spaced apart within the available 10 MHz band. We speculate that this is due to the lack of channel frequency selectivity in the considered scenario.

\begin{figure} [t!]
\centering
{\scriptsize \input{fig_histogram.tikz}}
\vspace{-1mm}
\caption{Localization error distribution. } \label{fig:error_distribution}
\end{figure}

\begin{figure} [t!]
\vspace{-2mm}
\centering
{\scriptsize \input{fig_accuracy_vs_antennas.tikz}}
\vspace{-1mm}
\caption{Average localization error vs. number of antennas}  \label{fig:error_vs_Nantennas}
\vspace{-2mm}
\end{figure}

In Fig.~\ref{fig:error_map} we present the resulting localization error as defined in \eqref{eq:error} for the ELM approach with $n=10^4$, a ReLu activation function, and $\W$ with i.i.d.~standard normal entries. The black dots depict the training data set of size $T=3500$, while the colored dots show the localization error for each location within the testing set of size $T^{\prime}=550$, using the scale on the left. Observe that the accuracy lies mostly below 10 m for most locations, while some loss of accuracy is noticeable for locations far away from the BS (where the CSI estimates are more noisy due to a lower SNR) and for a set of NLOS locations where the BS is shadowed by a tall (six floors) building. This can be further confirmed by the testing error distribution estimated over 1000 realizations of $\W$ in Fig.~\ref{fig:error_distribution}.

\subsection{Influence of the number of antennas}

\noindent In order to evaluate the sensitivity of the localization error to the number of antennas, we ran simulations with varying number of BS antennas by discarding part of the data. For each value of $N$, the subset of antenna elements in use was chosen so as to heuristically maximize the inter-element spacing, and the polarization diversity. The result is shown in Fig.~\ref{fig:error_vs_Nantennas}. Our experiments indicate that the localization error is low and exhibits little sensitivity to $N$ as long as $N\geq 12$, while for $N \leq 8$ the localization accuracy decreases significantly. This indicates that the potential applications of the proposed method are not limited to the massive MIMO regime, but also extends to the typical number of antennas found in 4.5G MIMO systems.

\subsection{Optimization of the ELM hyperparameters}
\noindent The theoretical framework of \cite{Louart_etal_RMT_neural_networks_AP17} was used to optimize certain parameters of the ELM (number of neurons $n$, activation function, and regularization parameter $\gamma$) based on the training and testing data sets. The results are presented in Fig.~\ref{fig:error_rmt}, where it can be noticed that the ReLu activation function uniformly outperforms the sign and step functions.

The dependency on the number of neurons $n$ is shown in Fig.~\ref{fig_mse_vs_neurons}. Both deterministic equivalents for the training and testing MSEs decrease when $n$ grows. The sharp drop in the training MSE that can be observed for $n\geq 10^4$ corresponds to $n\geq T$. Note, however, that no adverse over-fitting effect is observed for large $n$ (the testing MSEs keep monotonically decreasing towards the asymptotic values of $\overline{\mathsf{mse}}^{\prime}$ plotted in dash-dotted lines). The curves of Fig.~\ref{fig_mse_vs_neurons} have been generated by computing the optimum $\gamma$ minimizing $\overline{\mathsf{mse}}^{\prime}$ for each $n$.

Fig.~\ref{fig_mse_vs_gamma} shows the influence of the regularization parameter $\gamma$ on the localization MSE, for the three considered activation functions. Here, we fixed $n=10^4$ since it provides a good tradeoff between complexity and performance. The dots correspond to the experimental values of $\mathbb{E}_{\W}\{\mathsf{mse}(\W)\}$ and  $\mathbb{E}_{\W}\{\mathsf{mse}^{\prime}(\W)\}$  directly computed from running the trained ELM on the training and testing data for different realizations of $\W$, while the solid and dashed lines show the corresponding deterministic equivalents (see~\eqref{eq:det_eq1} and \eqref{eq:det_eq2}). It can be seen that there is an excellent match between the deterministic equivalents and the experimental values.

\subsection{Comparison of ELM with $K$-nN approach}
\noindent  In order to benchmark our results, we compare in Table~\ref{table:NNELM} the ELM learning approach with the $K$-nN approach presented in Section~\ref{sec:NN} wherein the chosen distance in the input space is the $2$-norm. As we see in Table~\ref{table:NNELM}, $K$-nN outperforms the ELM-based approach in terms of the average accuracy, while at the same time being more prone to catastrophic errors. However, at this stage we refrain from drawing conclusions from this comparison, since the set-up is unfavorable to the ELM approach due to the small size of the training dataset (with respect to the dimension of the data). Furthermore, the dataset is very structured in the sense  that it consists only of pathways, and this guarantees that for every data point in the testing data set there are always a couple of ``good'' nearest neighbors, which results in the best situation for the $K$-nN algorithm. Finally, we would like also to remark that the computational complexity for the $K$-nN localization strategy grows with the size of the training data set, whereas in the ELM case is only determined by the number of neurons. This can render the $K$-nN approach unfeasible is practice.

\begin{table}[b!]
\centering
\begin{tabular}{l |c c}
  \hline
Statistic & ELM (ReLu) & 3-nN\\
  \hline
  Average localization error [in m] & 9.60 &  8.16\\
  Median localization error [in m] & 7.56 & 4.29\\
  Maximum localization error [in m] & 73.26 & 108.25\\
  \hline
\end{tabular}
\caption{Comparison of statistics computed on the testing set on localization error between $3$-nNs and ELM.} \label{table:NNELM}
\end{table}

\begin{figure*}[t!]
\begin{subfigure}[t!]{0.47\textwidth}
                {\scriptsize \input{fig_mse_vs_neurons.tikz}}
								\subcaption{$\sqrt{\overline{\mathsf{mse}}}$ (solid) and $\sqrt{\overline{\mathsf{mse}}^{\prime}}$ (dashed) vs number of neurons.}\label{fig_mse_vs_neurons}
        \end{subfigure} \hspace{0.5cm}
        \begin{subfigure}[t!]{0.47\textwidth}
               {\scriptsize \input{fig_mse_vs_gamma.tikz}}
							\subcaption{$\sqrt{\overline{\mathsf{mse}}}$ (solid) and $\sqrt{\overline{\mathsf{mse}}^{\prime}}$ (dashed) vs. regularization parameter.} \label{fig_mse_vs_gamma}
        \end{subfigure}
        \caption{Optimization of the neural network parameters based on the deterministic equivalent analysis of the ELM.} \label{fig:error_rmt}
 \vspace{-2mm}
\end{figure*}

\section{Practical implementation issues}
\label{section_practical_issues}
\noindent  Here we discuss the practical issues related to the implementation of a learning-based localization approach in a cellular network. In particular, the availability of training data, and potentially the necessity to periodically update this training, are important questions in the context of the proposed application.

\subsection{Potential training approaches}
\noindent The method outlined in Section~\ref{section_CSI_based_localization} relies on the availability of a training dataset for which CSI has been obtained for a large number of reference positions, i.e., we need to assume that a reference position information is available for this data set.

Since the mapping between CSI and location is strongly dependent on the geometric properties of the scattering environment which surrounds the BS (i.e. shape and location of buildings and other macroscopic obstacles), it is expected that a dedicated training must be performed for each BS site. This also indicates that additional training might be required if significant changes occur in the scattering environment (new buildings...).
Therefore, it is important to identify practically feasible strategies to deal with the acquisition (and potentially, with the updates) of the training data set. Among the possible approaches, we distinguish between operator-based  and crowdsourcing-based training.

\paragraph{Operator-based training}
In this strategy, the training data is acquired through dedicated measurements performed by the operator. The location of the reference UE is determined either through GPS, or through other kinds of geometric measures (distance to a reference point in the horizontal plane, elevation based on the floor number in case of 3-D localization in indoor environments...); this operation can be performed by a human or robotic operator.
Although it is clearly a costly approach, it comes with the guarantee that the training data set is accurate and exhaustive.

\paragraph{Crowdsourcing-based training}
In this approach, the location data is collected by a subset of the end users of the cellular network. Practically, this can take the form of an operator-provided application which runs on the users' mobile device and relies on its existing localization service (GPS-based or other) to determine its position, and feed it back to the operator; the operator can provide a reward (monetary, or through a discount or additional services) in exchange for the voluntary participation of the user to this crowdsourcing data collection. For a given user participating in the crowdsourcing, this data collection does not have to be permanent, so that the associated impact on power and data consumption is minimized.
In that scenario, the operator has no control over the geographical area covered by the training data, since it merely follows the end user's behavior; thus, it can not be ensured that the training data covers uniformly and/or exhaustively the geographical area covered by the cell. On the other hand, it can be argued that, if sufficiently many users can be enrolled in the crowdsourcing data collection, the distribution of user locations in the training data will match the real user distribution, ensuring a faithful location estimate during the exploitation phase.

\subsection{Training Update}
\noindent In the context of the considered application, the mapping between CSI and localization within the cell is clearly a function of the scattering environment, which can not always be assumed constant over long periods of time (due e.g. to new buildings...). Hardware updates, or the aging of the electronic components can also make this mapping vary over time. It is therefore necessary to take into account these slow-scale time variations to properly assess the accuracy of learning-based localization. This calls for updates of the learned function over time, and thus the collection of new data.

If the old training data is not completely stale, it might be useful to re-train the neural network using the merged old and new training datasets. In order to reduce the complexity of this re-training, update mechanisms (such as \cite{Liang06_sequential_learning} for ELMs) can readily be used to provide an online solution.

The question of when to trigger an update (through the collection of new training data) can be answered by using the theoretical framework of \cite{Louart_etal_RMT_neural_networks_AP17}, e.g., by checking how good the new training data can be predicted by the old one.
In the case of crowdsourcing, the system can include on-demand acquisition of training data through a mechanism whereby data collection is enabled if the measured CSI for a crowdsourcing user can not be accurately matched to its location.

\subsection{Channel frequency selectivity}
\noindent Modern cellular networks rely on some form of frequency-division multiple access (FDMA) to multiplex the users in a cell. As a consequence, for a given user, the channel estimate is available only for a fraction of the spectrum (typically, a group of OFDM subcarriers, or \emph{resource block group}). The consequence is that the channel covariance matrix associated to a given user can only be estimated for a fraction of the spectrum. If the total system bandwidth is large, this covariance matrix can not be assumed constant in the frequency domain. This complicates the learning process, since the mapping between channel covariance and UE location will depend on the considered frequency. One approach is to fractionate the system bandwidth into subbands in which the channel covariance can be assumed constant, and repeat the learning process independently for each subband. Another solution is to add a pre-processing step consisting in transposing (in the frequency domain) the measured covariance matrix to a reference frequency (see e.g. \cite{YantaoHan_FDD_covariance_reciprocity_CHINACOM10,FDD_covariance_interpolation_Globecom2015,Haghighatshoar_covariance_interpolation_2018,Miretti_etal_FDD_covariance_conversion_ICASSP18}), and to let the neural network operate solely on the reference frequency.

\subsection{Channel covariance estimation}
\noindent A full estimate of the uplink channel matrix is not always available at the BS; in particular this is the case for multiple-antenna UEs when precoded reference signals are in use---in this case, only the product between the precoder and the instantaneous channel can be estimated. Another case is when channel estimation is performed for multiple users simultaneously using non-orthogonal reference symbols \cite{Efficient_CSI_acquisition_WSA2017}. However, even in these cases where the complete instantaneous CSI matrix is not estimated, it is generally possible to obtain unbiased estimates of the covariance matrix of each user, at the cost of more complex processing \cite{covariance_estimation_EUSIPCO17}. \\

\bibliographystyle{IEEEtran}

\bibliography{refs}
\end{document}

config: arxiv


TITLE: Martingales and different definitions
QUESTION [3 upvotes]: Are there any differences between the following definitions of Martingales and if so what are they?

Let $(X_{i})_{i=1}^{n}$ and $(Z_{i})_{i=1}^{n}$ be sequences of random variables then $(X_{i})$ is a martingale if $\mathbb{E}(X_{i}|Z_{1}...Z_{i})=X_{i-1}$ for all $i$.
Let $\mathcal {F}_{0} \subseteq \mathcal{F}_{1} \subseteq...\subseteq \mathcal{F}_{n}$ be a sequence of $\sigma$-subfields. Then $(X_{i})$ is a martingale with respect to $\mathcal{F}_{i}$ if $\mathbb{E}(X_{i}|\mathcal{F}_{i})=X_{i-1}$ for all $i$

In addition i was wondering what $\mathbb{E}(X_{i}|\mathcal{F}_{i})$ actually means, as it combines two notions of probability that are usually distinct from my previous experience in the sense that a random variable is a function defined on a probability space. I appreciate any help thanks!

REPLY [2 votes]: (1) is the definition of the martingale in the special case where the filtration $\mathscr{F}_i$ is the $\sigma$-algebra generated by the random variables $Z_1,\ldots,Z_i$. (2) is the "general" case.


TITLE: Choose $k$ numbers from $n$ numbers and what is the probability that they are divisible by given number $q$?
QUESTION [1 upvotes]: I have $n$ numbers (from $1$ to $n$). I choose $k$ numbers from these. So I need to calculate the probability that these $k$ numbers are divisible by given number $q$.

REPLY [4 votes]: Without loss of generality we may assume that $q$ is positive.
There are $\dbinom{n}{k}$ equally likely ways to choose $k$ numbers from $n$.
There are $m=\lfloor n/q\rfloor$ of our numbers that are divisible by $q$. (Here $\lfloor x\rfloor$ denotes the greatest integer which is $\le x$.)
So there are $\dbinom{m}{k}$ "favourable" outcomes.
Our probability is therefore $\dfrac{\binom{m}{k}}{\binom{n}{k}}$.


TITLE: Induction proof for $V - E + F = 2$ (Connected planar graphs)
QUESTION [3 upvotes]: This might be a bit of a silly question, but I'm struggling to understand this proof for $V - E + F = 2$, where $V$ = number of vertices, $E =$ number of edges and $F = $ number of faces.
I understand the first part, but I'm confused as to why we can assume true for a number of faces but then subtract from $F$, if we've proved it for $F = 1$.

An explanation would be much appreciated. Thanks.

REPLY [4 votes]: For an easy to picture example, consider the complete graph on 4 vertices. Initially $F=3, V=4, E=6.$ Notice that $V-E+F=2$. Now imagine removing the edge joining the bottom-left vertex to the center vertex (this edge is appropriate because the proof requires us to remove an edge that belongs to a cycle.) How has the picture changed? We haven't removed any vertices, so $V=4$ still. We have removed one edge, so our new $E=5$. We have also combined two faces into 1 new face, so now there are two remaining, so our new  $F=2$. Notice that it is still the case that $V-E+F=2$.
The idea is that this is what will always happen when you remove an edge that belongs to a cycle in a planar graph - doing so will reduce the number of edges and the number of faces both by 1, so the value of $V-E+F$ will not change. In this way, the author is reducing the problem for a graph with $n$ faces to a graph with $n-1$ faces. They are saying that because the value of $V-E+F$ is the same in each case, and we have assumed in the $n-1$ case that $V-E+F=2$, we have shown that the same is true in the case with $n$ faces.


text:
\begin{document}

\title{Extended Generalized DoF Optimality Regime of Treating Interference as Noise in the X Channel}

\author{\IEEEauthorblockN{Soheil Gherekhloo, Anas Chaaban, and Aydin Sezgin}\thanks{This work was supported by the DFG grant SE 1697/10.
}
\IEEEauthorblockA{Institute of Digital Communication Systems\\
Ruhr-Universit\"at Bochum\\
Email: soheyl.gherekhloo,anas.chaaban,aydin.sezgin@rub.de}
}

\maketitle

\begin{abstract}
The simple scheme of treating interference as noise (TIN) is studied in this paper for the $3\times2$ X channel. A new sum-capacity upper bound is derived. This upper bound is transformed into a generalized degrees-of-freedom (GDoF) upper bound, and is shown to coincide with the achievable GDoF of a scheme that combines TDMA and TIN for some conditions on the channel parameters. These conditions specify a noisy interference regime which extends noisy interference regimes available in literature. As a by-product, the sum-capacity of the $3\times 2$ X channel is characterized within a constant gap in the given noisy interference regime.
\end{abstract}

\IEEEpeerreviewmaketitle

\section{Introduction}

Dealing with interference is a main challenge in wireless communications.
Compared with noise, interference contains information. Using this
property, some techniques have been investigated which decode the interference~\cite{HanKobayashi1981}
in order to have a cleaner version of the received signal. This was  shown to be optimal in some cases such as 2-user interference channel (IC) with strong interference~\cite{Sato, Carleial_vsi}. On the other hand,
there is another extreme case in which the interference is so weak that
the undesired receiver is not able to decode it. Ignoring the interference
completely at the undesired receiver is a common way to deal with interference in such cases.
This technique is known as treating interference as noise
(TIN)~\cite{CharafeddineSezginPaulraj}.

TIN is simple from a computational point of view, and is not demanding in terms of channel state information and coordination between different nodes.
This simplicity makes TIN an appropriate choice for practical communication scenarios.
The practical advantages of TIN makes it interesting to identify cases where TIN is capacity-optimal~\cite{ShangKramerChen}. In~\cite{EtkinTseWang}, it is shown
that TIN is capacity optimal within a gap of $1$ bit in a 2-user IC which satisfies $\sqrt{\text{INR}}< \text{SNR}$. TIN is constant-gap optimal in the symmetric $K$-user IC ($K>2$) under the same condition~\cite{JafarVishwanath}. This result has been extended for asymmetric $K$-user IC in~\cite{GengNaderializadehAvestimehrJafar}.
The optimality of TIN has also been studied for scenarios in which numbers of transmitters and receivers are not equal~\cite{ChaabanSezgin_SubOptTIN}. In such cases, it turns out that the constant-gap optimality regime of TIN is increased~\cite{GherekhlooDiChaabanSezgin2014} by switching some users off.
For a more general $M\times N$ X channel, conditions for \textcolor{comm_Ay}{constant gap optimality in the} noisy
interference regime were identified in~\cite{GengSunJafar2014}. \textcolor{comm_Ay}{The X channel models a cellular network, in which a user is communicating with multiple base stations in order to achieve a soft hand-over.}

In this paper, we show that the noisy interference regime of the X channel \textcolor{comm_Ay}{for which a constant gap to capacity is achieved} is in fact larger than the one given in~\cite{GengSunJafar2014}. To do this, we consider a $3\times2$ X channel for simplicity, and we derive a noisy interference regime where a TDMA-TIN scheme (which combines TDMA and TIN) is constant-gap optimal. The resulting noisy
interference regime identified in our work not only subsumes the
regime in~\cite{GengSunJafar2014}, but also extends it. This is mainly due to a novel
upper bound that we establish in this paper.

Throughout the paper, we use $C(x)=\log_2(1+x)$ for $x>0$, $\bar{x}=1-x$ for $x\in[0,1]$, and $x^n=(x_1,\cdots,x_n)$.

\section{System Model}
\label{Model}
The system we consider is a $3\times 2$ Gaussian X channel which consists of three senders and two receivers (Figure \ref{sysmod}). Each sender wants to communicate with each receiver. Namely, transmitter $i$ (Tx$i$) wants to send the messages $W_{ji}$ to receiver $j$ (Rx$j$), where $i\in\{1,2,3\}$ and $j\in\{1,2\}$. The message $W_{ji}$ has a rate $R_{ji}$. Tx$i$ encodes $(W_{1i},W_{2,i})$ into a codeword $x_i^n\in\mathbb{C}^n$ of $n$ symbols. The transmitters have power constraints $\rho$ which must be satisfied by their transmitted signals.

\newcommand{\GaussianSystemmodel}[0]{
\node (t3) at (0,0) [inner sep=0] {};
\node (t1) at (0,-2) [inner sep=0] {};
\node (t2) at (0,-4) [inner sep=0] {};
\node (r1) at (3.5,-1) [inner sep=0] {};
\node (r2) at (3.5,-3) [inner sep=0] {};
\node (Z1U) at (3.6,-.3) [inner sep=0] {};
\node (Z1D) at (3.6,-.9) [inner sep=0] {};
\node (Z2D) at (3.6,-3.7) [inner sep=0] {};
\node (Z2U) at (3.6,-3.1) [inner sep=0] {};
\draw[->] (t3) to (r1);
\draw[->] (t1) to (r1);
\draw[->] (t2) to (r2);
\draw[->] (Z1U) to (Z1D);
\draw[->] (Z2D) to (Z2U);
\draw[->] (t3) to (r2);
\draw[->] (t1) to (r2);
\draw[->] (t2) to (r1);
\draw [fill=white, white] (0.6,-4) rectangle (1.3,0);
\node (h13) at (1,-0.3) [inner sep=0] {$h_{11}$};
\node (h23) at (1,-0.9) [inner sep=0] {{$h_{21}$}};
\node (h11) at (1,-1.7) [inner sep=0] {{$h_{12}$}};
\node (h22) at (1,-3.8) [inner sep=0] {{$h_{23}$}};
\node (h21) at (1,-2.3) [inner sep=0] {{$h_{22}$}};
\node (h22) at (1,-3.2) [inner sep=0] {{$h_{13}$}};
\node (tx3) at (-1.5,0) [inner sep=0] {\small{$\begin{pmatrix}W_{11}\\W_{21}\end{pmatrix}\rightarrow x_1^n$}};
\node (tx1) at (-1.5,-2) [inner sep=0] {\small{$\begin{pmatrix}W_{12}\\W_{22} \end{pmatrix}\rightarrow x_2^n$}};
\node (tx2) at (-1.5,-4) [inner sep=0] {\small{$\begin{pmatrix}
 W_{13} \\ W_{23}
\end{pmatrix}\rightarrow x_3^n$}};
\node (W1Rl) at (3.6,-1) [inner sep=0] {{$\oplus$}};
\node (W2Rl) at (3.6,-3) [inner sep=0] {{$\oplus$}};
\node (Z_1) at (3.75,0) [inner sep=0] {{$z_1^n$}};
\node (Z_2) at (3.75,-4) [inner sep=0] {{$z_2^n$}};
\node (Rx1) at (5.2,-1) [inner sep=0] {\small{  $\rightarrow y_1^n\rightarrow \begin{pmatrix}
\hat{W}_{11}\\ \hat{W}_{12}\\ \hat{W}_{13}\end{pmatrix}$}};
\node (Rx2) at (5.2,-3) [inner sep=0] {\small{$\rightarrow y_2^n\rightarrow \begin{pmatrix}
\hat{W}_{21}\\ \hat{W}_{22}\\ \hat{W}_{23}
\end{pmatrix}$}};
}
\begin{figure}[t]
\centering
\textcolor{comm_Ay}{
\begin{tikzpicture}[scale=0.8]
\GaussianSystemmodel
\end{tikzpicture}}
\caption{System model of the $3\times2$ Gaussian X channel.}
\label{sysmod}
\end{figure}

At time instant $t\in\{1,\cdots,n\}$, Rx$j$ receives\footnote{The time index $t$ will be suppressed henceforth for clarity.}
\begin{align}
\label{recsig1}
y_j[t]&= h_{j1} x_1[t]+h_{j2} x_2[t]+h_{j3} x_3[t]+z_j[t],
\end{align}
where $z_j[t]$, $j\in\{1,2\}$, is a complex-valued Gaussian noise with zero mean and unit variance, and the constant $h_{ji}$ represents the complex (static) channel coefficient between Tx$i$ and Rx$j$. \textcolor{comm_Ay}{Since noise variance is unit, the transmit signal-to-noise ratio is given by $\rho$.} The noises $z_1$ and $z_2$ are independent of each other and are both independent and identically distributed (i.i.d.) over time.

After $n$ transmissions, Rx$j$ has $y_j^n$ and decodes $W_{ji}$, $i\in\{1,2,3\}$. The probability of error, achievable rates $R_{ji}$, capacity region $\mathcal{C}$ are defined in the standard Shannon sense \cite{CoverThomas}. The sum-capacity $C_\Sigma$ is the maximum achievable sum-rate $R_\Sigma=\sum_{i=1}^3\sum_{j=1}^2R_{ji}$ for all rate tuples in the capacity region $\mathcal{C}$.

In this work,

we focus on the interference limited scenario, and hence, we assume that all signal-to-noise and interference-to-noise ratios are larger than 1, i.e., $\rho|h_{ji}|^2>1$ for all $j\in\{1,2\}$ and $i\in\{1,2,3\}$. Defining
\begin{align}
\alpha_{ji}=\frac{\log_2(\rho|h_{ji}|^2)}{\log_2(\rho)}, \label{eq:def_alpha}
\end{align} we define the generalized degrees-of-freedom (GDoF) of the channel as in \cite{GengNaderializadehAvestimehrJafar}
\begin{align}
\label{GDoFdef}
d_\Sigma(\boldsymbol{\alpha})=\lim_{\rho\to\infty}\frac{C_{\Sigma}(\boldsymbol{\alpha})}{\log_2(\rho)},
\end{align}
where $\boldsymbol{\alpha}$ is a vector which contains all $\alpha_{ji}$.

The focus of this work is studying constant gap optimality of TIN for the $3\times 2$ Gaussian X channel. Next, we introduce the transmission strategy we propose in this paper.

\section{TDMA-TIN}
\label{sec:TDMA-TIN}
In this scheme, we allow only two transmitters to be active simultaneously. In addition to this, for each active transmitter only one dedicated receiver is considered. \textcolor{comm_Ay}{Thus, we decompose the X channel into its underlying interference channels (IC).} In total, we have six  2-user IC's in the $3\times2$ X channel. Using TDMA, we assign a $\tau_s>0$ fraction of time to each of those six 2-user IC's, with $\sum_{s=1}^6 \tau_s = 1$. If the achievable sum-rate using TIN for one of those IC's is $R_s$, then the achievable sum-rate of TDMA-TIN is given by
\begin{align}
R_{TT}= \max_{\tau_1,\ldots,\tau_6}\sum_{s=1}^6 \tau_s R_s.
\end{align}
This optimization problem is linear in $\tau_s$ and is solved by setting $\tau_s=1$ for some $s\in\{1,\cdots, 6\}$ and setting the remaining $\tau_{s'}=0$. Namely, the maximization above is achieved by activating the 2-user IC which yields the highest sum-rate. Without loss of generality, suppose that the 2-user IC with maximum TIN sum-rate is the one in which Tx$i_1$ and Tx$i_2$ want to send messages $W_{j_1i_1}$ and $W_{j_2i_2}$ to Rx$j_1$ and Rx$j_2$, respectively. The transmitters encode their message into a codeword with power $\rho$. This causes interference at undesired receivers. Therefore, the receivers decode their desired messages using TIN. Using this scheme, the following sum-rate is achievable
\begin{align*}
R_{j_1i_1} + R_{j_2i_2} = C\left(\frac{\rho^{\alpha_{j_1i_1}}}{1+\rho^{\alpha_{j_1i_2}}}\right)+ C\left(\frac{\rho^{\alpha_{j_2i_2}}}{1+\rho^{\alpha_{j_2i_1}}}\right) .
\end{align*}
In general, the achievable sum-rate by using TDMA-TIN is presented
in the following proposition.

\begin{Prop}
\label{Pro:Ach_TDMA-TIN_Gaussian_32X}
The achievable sum-rate of TDMA-TIN in the $3\times 2$ Gaussian X channel is given by
\begin{align}
\label{eq:GausAchTDMATIN}
R_{TT}  =  &\max_{\boldsymbol{p}_{TT}} R_{TT}(\boldsymbol{p}_{TT})
\end{align}
where $\boldsymbol{p}_{TT}=(i_1,i_2,j_1,j_2)$ with $i_1,i_2 \in \{1,2,3\}$, $j_1,j_2\in\{1,2\}$, $i_1\neq i_2$, $j_1\neq j_2$, and where $$R_{TT}(\boldsymbol{p}_{TT})=C\left(\frac{\rho^{\alpha_{j_1i_1}}}{1+\rho^{\alpha_{j_1i_2}}}\right) + C\left(\frac{\rho^{\alpha_{j_2i_2}}}{1+\rho^{\alpha_{j_2i_1}}}\right).$$
\end{Prop}
Let us transform this achievable rate expression to an achievable GDoF expression. We first bound $R_{TT}(\boldsymbol{p}_{TT})$ as follows
\begin{align*}
&R_{TT}(\boldsymbol{p}_{TT})\\
&>  \left[(\alpha_{j_1i_1}-\alpha_{j_1i_2})^+
+(\alpha_{j_2i_2}-\alpha_{j_2i_1})^+\right]
\log_2(\rho)-2.
\end{align*}
Therefore, for this particular set of transmitters and receivers,  TDMA-TIN achieves a GDoF of
\begin{align*}
D_{TT}(\boldsymbol{p}_{TT})=(\alpha_{j_1i_1}-\alpha_{j_1i_2})^+
+(\alpha_{j_2i_2}-\alpha_{j_2i_1})^+.
\end{align*}
As a result, TDMA-TIN achieves the following GDoF
\begin{align}
\label{eq:GDoFTT}
d_{TT}(\boldsymbol{\alpha}) =  &\max_{\boldsymbol{p}_{TT}} D_{TT}(\boldsymbol{p}_{TT}).
\end{align}
Despite the simplicity of TDMA-TIN, this scheme is constant-gap optimal in some cases as we shall see next.

\section{Constant-gap optimality of TDMA-TIN}
Here, we want to introduce a noisy interference regime in which TDMA-TIN achieves the GDoF of the $3\times 2$ Gaussian X channel, and \textcolor{comm_Ay}{moreover}, achieves its sum-capacity within a constant gap. The following theorem characterizes the GDoF of the channel in such a noisy interference regime.
\begin{Theorem}
\label{Th:TIN_con_gap_32X_Gauss}
If there exist distinct $i_1,i_2,i_3\in\{1,2,3\}$ and distinct $j_1,j_2\in\{1,2\}$ such that the following noisy interference regime conditions are satisfied:
\begin{align}
\label{eq:condition_Opt_TDMA_TIN_Gauss1}
\alpha_{j_1i_1}-\alpha_{j_2i_1}&\geq \psi\\
\label{eq:condition_Opt_TDMA_TIN_Gauss2}
\alpha_{j_2i_2}-\alpha_{j_1i_2}&\geq \max\{\alpha_{j_2i_1},\alpha_{j_2i_3}\},
\end{align}
where $\psi=\max\{\alpha_{j_1i_3}-(\alpha_{j_2i_3}-\alpha_{j_2i_1})^+,\alpha_{j_1i_2}\}$, then TDMA-TIN achieves the GDoF of the $3\times2$ Gaussian X channel given by
\begin{align}
\label{Thm:GDoFExp}
d_\Sigma(\boldsymbol{\alpha})&\leq \alpha_{j_1i_1}-\alpha_{j_2i_1}+\alpha_{j_2i_2}-\alpha_{j_1i_2}.
\end{align}
\end{Theorem}
In other words, if there exists a permutation of the transmitters and receivers such that conditions \eqref{eq:condition_Opt_TDMA_TIN_Gauss1} and \eqref{eq:condition_Opt_TDMA_TIN_Gauss2} hold, then TDMA-TIN is GDoF-optimal. Using this theorem, we can show that TDMA-TIN achieves the sum-capacity of the channel within a constant gap. It can be shown that this gap can be upper bounded by 7 bits as long as the conditions in \eqref{eq:condition_Opt_TDMA_TIN_Gauss1} and  \eqref{eq:condition_Opt_TDMA_TIN_Gauss2} are satisfied. Due to space limitations, the gap analysis is not included.

Note that conditions \eqref{eq:condition_Opt_TDMA_TIN_Gauss1} and \eqref{eq:condition_Opt_TDMA_TIN_Gauss2} specify a larger noisy interference regime than that identified in \cite{GengSunJafar2014}. This is true since $\psi=\max\{\alpha_{j_1i_3}-(\alpha_{j_2i_3}-\alpha_{j_2i_1})^+,\alpha_{j_1i_2}\}$ is smaller than $\max\{\alpha_{j_1i_3},\alpha_{j_1i_2}\}$ as identified in \cite{GengSunJafar2014}, specifically, if $\alpha_{j_2i_3}-\alpha_{j_2i_1}>0$. \textcolor{Comm_An}{In Fig.~\ref{fig:GDoF_opt_TIN}, the GDoF optimality regime of TDMA-TIN is illustrated for $3\times 2$ X channel with $\alpha_{11} = \alpha_{22}=1$, $\alpha_{13}= \alpha_{23}=\beta$, where $\beta$ is larger than $0.5$ and smaller than $1$.
The noisy interference regime obtained from Theorem \ref{Th:TIN_con_gap_32X_Gauss} is given by the union of the rectangle defined by $(\alpha_{21},\alpha_{12})\in[0, 0.5]\times[0,1-\beta]$ and the rectangle defined by $(\alpha_{21},\alpha_{12})\in[0,1-\beta]\times[0, 0.5]$. On the other hand, the noisy interference regime obtained from \cite{GengSunJafar2014} is given by the intersection of these two rectangles.
Obviously, the new noisy interference regime does not only subsume the previously known regime from \cite{GengSunJafar2014} but also extends it. Note that if the channel gains from Tx$i_3$ to the receivers decrease, then the intersection of the two rectangles increases. At the point $\beta=1/2$, both regimes will coincide and the noisy interference regime becomes a rectangle with width and height of $1/2$.}
\begin{figure}
\begin{tikzpicture}

\begin{axis}[
width=6cm,
height=6cm,
scale only axis,
xmin=0,
xmax=0.75,
xtick={0,0.25,0.5},
xticklabels={{0},{$1-\beta$},{0.5}},
xlabel={$\alpha_{21}$},
xmajorgrids,
ymin=0,
ymax=0.75,
ytick={0,0.25,0.5},
yticklabels={{0},{$1-\beta$},{0.5}},
ylabel={$\alpha_{12}$},
ymajorgrids,
legend style={at={(0.97,0.8)},anchor=south east,draw=black,fill=white,legend cell align=left}
]
\addplot [color=red,dash pattern=on 1pt off 3pt on 3pt off 3pt,line width=1pt]
  table[row sep=crcr]{0	0.25\\
0.25	0.25\\
0.25	 0\\
};
\addlegendentry{Noisy interf. reg. \cite{GengSunJafar2014}};

\addplot [color=blue,solid,line width=1.0pt]
  table[row sep=crcr]{0	0.5\\
0.25 	0.5\\
0.25	 0.25\\
0.5 0.25\\
0.5 0.0\\
};
\addlegendentry{Noisy interf. reg. (Theorem \ref{Th:TIN_con_gap_32X_Gauss})};

\addplot[area legend,solid,fill=red,opacity=0.3,draw=red,forget plot]
table[row sep=crcr] {
x	y\\
0	0\\
0	0.25\\
0.25	 0.25\\
0.25	 0\\
};

\addplot[area legend,solid,fill=blue,opacity=0.3,draw=blue,forget plot]
table[row sep=crcr] {
x	y\\
0	0\\
0	0.5\\
0.25	0.5\\
0.25	0\\
};
\addplot[area legend,solid,fill=blue,opacity=0.3,draw=blue,forget plot]
table[row sep=crcr] {
x	y\\
0	0\\
0.5	0\\
0.5	0.25\\
0	0.25\\
};

\end{axis}
\end{tikzpicture}
\caption{The GDoF optimality regime of TDMA-TIN for $3\times 2$ X channel with $\alpha_{11} = \alpha_{22}=1$, and $\alpha_{13}=\alpha_{23}=\beta$, where $0.5<\beta<1$.}
\label{fig:GDoF_opt_TIN}
\end{figure}

The GDoF expression given in Theorem \ref{Th:TIN_con_gap_32X_Gauss} is clearly achievable by TDMA-TIN. Namely, consider a permutation of transmitters and receivers given by distinct $i_1,i_2,i_3\in\{1,2,3\}$ and distinct $j_1,j_2\in\{1,2\}$. Then, \eqref{eq:GDoFTT} leads to
\begin{align}
d_{TT}(\boldsymbol{\alpha}) &=  \max_{\boldsymbol{p}_{TT}} D_{TT}(\boldsymbol{p}_{TT}) >  D_{TT}(i_1,i_2,j_1,j_2)\\

&= \alpha_{j_1i_1}-\alpha_{j_1i_2}+\alpha_{j_2i_2}
-\alpha_{j_2i_1}
\end{align}
where the last step follows since the conditions in Theorem \ref{Th:TIN_con_gap_32X_Gauss} dictate that $\alpha_{j_1j_1}\geq\alpha_{j_1i_2}$ and $\alpha_{j_2j_2}\geq\alpha_{j_2i_1}$. This achievable GDoF coincides with \eqref{Thm:GDoFExp}.

To prove Theorem \ref{Th:TIN_con_gap_32X_Gauss}, it remains to prove the converse. In other words, we still need to establish an upper bound on the GDoF which coincides with \eqref{Thm:GDoFExp} under the conditions \eqref{eq:condition_Opt_TDMA_TIN_Gauss1} and \eqref{eq:condition_Opt_TDMA_TIN_Gauss2}. The converse is provided in the next section.

\section{Converse \textcolor{comm_Ay}{for Theorem \ref{Th:TIN_con_gap_32X_Gauss}}}
Here, we derive an upper bound on the sum-capacity of the $3\times 2$ Gaussian X channel which proves the converse of Theorem \ref{Th:TIN_con_gap_32X_Gauss}. The upper bound is given in the following lemma.
\begin{Lemma}
\label{Th:UB_Det_32X}
The sum-capacity of the $3\times 2$ Gaussian X channel is upper bounded by
\begin{align}
\label{eq:UB_32X}
C_\Sigma\leq &\min_{\boldsymbol{p}} B(\boldsymbol{p})
\end{align}
where $\boldsymbol{p}=(i_1,i_2,i_3,j_1,j_2)$ for distinct $i_1,i_2,i_3 \in\{1,2,3\}$ and distinct $j_1,j_2\in\{1,2\}$, and where $B(\boldsymbol{p})$ is as given in \eqref{eq:Bij} \textcolor{comm_Ay}{on the top of the next page}.
\begin{figure*}
\begin{align}
\label{eq:Bij}
&B(\boldsymbol{p})= C\left(\rho^{\alpha_{j_1i_2}} + \bar{d} \rho^{\alpha_{j_1i_3}}+ \frac{\rho^{\alpha_{j_1i_1}} + d \rho^{\alpha_{j_1i_3}}}{1+c^2(\rho^{\alpha_{j_1i_1}} +d \rho^{\alpha_{j_1i_3}})}\right) + C\left(\rho^{\alpha_{j_2i_1}} + \rho^{\alpha_{j_2i_3}} +\frac{\rho^{\alpha_{j_2i_2}} }{1+\rho^{\alpha_{j_1i_2}}}\right)+1\\
&(c^2,d) = \begin{cases}
(\rho^{\alpha_{j_2i_1}-\alpha_{j_1i_1}}, 0)& \text{if}\quad  \alpha_{j_2i_3}\leq \alpha_{j_2i_1},
\\
(\rho^{\alpha_{j_2i_1}-\alpha_{j_1i_1}}, 1)& \text{if}\quad  \alpha_{j_2i_3}> \alpha_{j_2i_1}\text{ and }
\alpha_{j_2i_1} - \alpha_{j_1i_1} \leq \alpha_{j_2i_3}-\alpha_{j_1i_3} - \alpha_{j_2i_1}, \\
(\rho^{\alpha_{j_2i_3}-\alpha_{j_2i_1}-\alpha_{j_1i_3}},1)& \text{otherwise.} \end{cases}
\label{eq:param_def_c_d}
\end{align}
\hrule
\end{figure*}
\end{Lemma}
\begin{proof}

We consider the following permutation of transmitters and receivers: $\boldsymbol{p}=(i_1,i_2,i_3,j_1,j_2)=(1,2,3,1,2)$; the other cases can be proved similarly.
We give \textcolor{comm_Ay}{$V_{j_1}=V_1 = \{W_{21},W_{23}\}$} and $S_{j_1}^n=S_1^n=c(h_{11}X_1^n+ d\cdot h_{13}X_3^n)+N_1^n$ to Rx1 as side information,\footnote{The capital letter notation is used for random variables.} where\textcolor{comm_Ay}{
\begin{align}
(c,d) =
\begin{cases}
(\frac{h_{21}}{h_{11}},0)                 & \frac{|h_{23}|}{|h_{21}|}\leq 1 \\
(\frac{h_{21}}{h_{11}},1)                 & \frac{|h_{23}|}{|h_{21}|}> 1,\,  \frac{\rho|h_{21}|^4}{|h_{11}|^2}\leq \frac{|h_{23}|^2}{|h_{13}|^2}\\
(\frac{h_{23}}{h_{21}\sqrt{\rho}h_{13}},1) & \frac{|h_{23}|}{|h_{21}|}> 1,\, \frac{\rho|h_{21}|^4}{|h_{11}|^2}> \frac{|h_{23}|^2}{|h_{13}|^2},
\end{cases}\label{eq:c_param}
\end{align}}
and where $N_1^n$ is a zero-mean unit-variance Gaussian noise independent of $Z_1$ and $Z_2$ and i.i.d. over time.

We also give $V_{j_2}=V_2=W_{12}$ and $S_{j_2}^n=S_2^n = h_{12}X_2^n+N_2^n$ to Rx2 as side information, where $N_2^n=Z_1^n$.

Using Fano's inequality, and defining $\tilde{W}_j$ as the set of messages desired at Rx$j$, i.e. $\{W_{j1},W_{j2},W_{j3}\}$, we obtain
\begin{align}
nR_\Sigma\leq & \sum_{j=1}^2 I(\tilde{W}_j;Y_j^n,S_j^n,V_j)+n\epsilon_n, \notag
\end{align}
where $\epsilon_n\to 0$ as $n\to\infty$. Then, using the chain rule, and since all messages are independent, we can write
\begin{align*}
nR_\Sigma&\leq  \sum_{j=1}^2 \left[I(\tilde{W}_j;S_j^n|V_j) + I(\tilde{W}_j;Y_j^n|S_j^n,V_j)\right]+n\epsilon_n.
\end{align*}
Now by using $h(S_j^n|\tilde{W}_j,V_j)=h(N_j^n)$ we get
\begin{align}
nR_\Sigma&\leq

 \sum_{j=1}^2 \left[h(S_j^n|V_j) - h(N_j^n)+ h(Y_j^n|S_j^n,V_j)\right.\nonumber\\
&\qquad\qquad \left.- h(Y_j^n|\tilde{W}_j,V_j)\right]+n\epsilon_n\label{RsSum4}
\end{align}
Now, defining $\tilde{S}_1^n = h_{21}X_1^n + h_{23} X_3^n + Z_{2}^n$ and $\tilde{S}_2^n = h_{12}X_2^n + Z_{1}^n$
and using the fact that $(X_1^n,X_3^n)$ and $X_2^n$ can be reconstructed from $(\tilde{W_1},V_1)$ and $(\tilde{W_2},V_2)$, respectively, we obtain $h(Y_j^n|\tilde{W}_j,V_j)=h(\tilde{S}_i^n|V_i)$, where $i\neq j$, $i,j\in\{1,2\}$. Furthermore, since conditioning does not increase entropy, we have $h(\tilde{S}_1^n|V_1)\geq h(\tilde{S}_1^n|V_1,\bar{d}X_3^n)$. Substituting in \eqref{RsSum4} yields
\begin{align}
nR_\Sigma&\leq \sum_{j=1}^2\left[h(S_j^n|V_j) - h(N_j^n)+ h(Y_j^n|S_j^n,V_j)\right]  \nonumber\\
&\quad - h(\tilde{S}_{2}^n|V_{2}) - h(\tilde{S}_{1}^n|V_{1},\bar{d} X_3^n)+n\epsilon_n\notag \\
\label{Rsigm4t}
&\overset{(c)}{\leq} \sum_{j=1}^2 \left[h(Y_j^n|S_j^n,V_j)- h(N_j^n)\right]+ n +n\epsilon_n
\end{align}
where $(c)$ follows since $h(S_2^n|V_2) = h(\tilde{S}_2^n|V_2)$, and since  $h(S_1^n|V_1) = h(\tilde{S}_1^n|V_1,X_3^n)$ if $d=0$, and $h(S_1^n|V_1) - h(\tilde{S}_1^n|V_1)\leq n$ if $d=1$ as shown in Lemma~\ref{Lemma:Entropy_diff_Gaussian} in Appendix~\ref{Proof_Gaus_UB_32X}. By dropping the conditioning on $V_1$ and $V_2$, using Lemma~1 in~\cite{AnnapureddyVeeravalli} which shows that a circularly symmetric complex Gaussian distribution maximizes the
conditional differential entropy for a given covariance constraint,

dividing by $n$, letting $n\to\infty$, and using \eqref{eq:def_alpha}, we obtain
\begin{align}
R_\Sigma
&\leq
C\left(\rho^{\alpha_{12}} + \bar{d} \rho^{\alpha_{13}}+ \frac{\rho^{\alpha_{11}} + d \rho^{\alpha_{13}}}{1+c^2(\rho^{\alpha_{11}} +d  \rho^{\alpha_{13}})}\right) \notag\\ &\quad + C\left(\rho^{\alpha_{21}} + \rho^{\alpha_{23}} +\frac{\rho^{\alpha_{22}} }{1+\rho^{\alpha_{12}}}\right)+1 \label{eq:UB_32X_end_proof}
\end{align}
which is equal to the desired bound $B(\boldsymbol{p})$ \eqref{eq:Bij} for this specific permutation $\boldsymbol{p}=(i_1,i_2,i_3,j_1,j_2)=(1,2,3,1,2)$ of transmitters and receivers. By rewriting the parameters $c$, $d$ as a function of $\rho$, we obtain \eqref{eq:param_def_c_d} for this permutation.

Writing the upper bound in \eqref{eq:UB_32X_end_proof} for all permutations of $i_1,i_2,i_3\in\{1,2,3\}$ and $j_1,j_2\in\{1,2\}$, we obtain the upper bound in \eqref{eq:UB_32X}.
\end{proof}

With this, we obtain a sum-capacity upper bound. We can use the definition of the GDoF in \eqref{GDoFdef} to write this upper bound as a GDoF upper bound. For this purpose, we divide $B(\boldsymbol{p})$ by $\log(\rho)$ and we let $\rho\to\infty$ to obtain a GDoF upper bound for each of the cases in \eqref{eq:param_def_c_d}. By combining the resulting GDoF upper bounds, we get the GDoF upper bound for a specific permutation $\boldsymbol{p}$ (details can be found in Appendix \ref{app:Constant Gap Analysis for TDMA-TIN})
\begin{align}
\label{GDoFUpperBound}
&D(\boldsymbol{p})<\max\{\alpha_{j_2i_1},\alpha_{j_2i_3},\alpha_{j_2i_2}-\alpha_{j_1i_2}\}\\
&+ \max\{\alpha_{j_1i_2},\alpha_{j_1i_1}-\alpha_{j_2i_1},\alpha_{j_1i_3}-(\alpha_{j_2i_3}-\alpha_{j_2i_1})^+\}
\nonumber.
\end{align}
Therefore $d_\Sigma(\boldsymbol{\alpha})\leq \min_{\boldsymbol{p}} D(\boldsymbol{p}).$

Now, we have a general GDoF upper bound. Let us specialize this bound to the noisy interference regime of Theorem \ref{Th:TIN_con_gap_32X_Gauss}. Suppose that the conditions in Theorem \ref{Th:TIN_con_gap_32X_Gauss} are satisfied for some permutation of transmitters and receivers $\hat{\boldsymbol{p}}=(t_1,t_2,t_3,r_1,r_2)$. By using these conditions we get
\begin{align*}
d_\Sigma(\boldsymbol{\alpha})&\leq \min_{\boldsymbol{p}} D(\boldsymbol{p})\nonumber\\
&\leq D(\hat{\boldsymbol{p}})=\alpha_{r_1t_1}-\alpha_{r_2t_1}+\alpha_{r_2t_2}-\alpha_{r_1t_2},
\end{align*}
which proves the converse of Theorem \ref{Th:TIN_con_gap_32X_Gauss}.

\appendices

\section{}
\label{Proof_Gaus_UB_32X}
In this appendix, we introduce a lemma which is necessary for proving the bound \eqref{eq:UB_32X}. Let $W_A$ and $W_B$ be two independent messages, and let $X_A$ (independent of $W_B$) and $X_B$ (independent of $W_A$) be two independent complex-valued signals satisfying a power constraint $\rho$. Define $Y_A$ and $Y_B$ as noisy channel outputs given by
 \begin{align}
   Y_A & = h_1 X_A + h_2 X_B+Z_A\\
   Y_B & = h_3 X_A + h_4 X_B+Z_B,
   \end{align}
where $Z_A$ and $Z_B$ are zero-mean unit-variance Gaussian noises, and are independent of each other and of all other random variables, and where the constants $h_1$, $h_2$, $h_3$ and $h_4$ are complex-valued and satisfy
\begin{align}
|h_1|^2  \leq  |h_3|^2  \leq \frac{|h_4|^2}{\rho |h_2|^2}\quad \text{and} \quad 1  < \rho |h_3|^2 \label{eq:ExtCondition3}.
\end{align}
Let $Y_A^{n}$ and $Y_B^{n}$ be the outputs corresponding to inputs $X_A^{n}$ and $X_B^{n}$ of length $n$, and define $W_C=(W_A,W_B)$. Then,  we have the following lemma.
\begin{Lemma}
\label{Lemma:Entropy_diff_Gaussian}
If conditions \eqref{eq:ExtCondition3} are satisfied, then we have
\begin{align}
\label{lem2}
h(Y_A^n|W_C)-h(Y_B^n|W_C)\leq n.
\end{align}
\end{Lemma}
\begin{Rem}
The parameter $c$ in \eqref{eq:c_param} is chosen such that the conditions in \textcolor{comm_Ay}{\eqref{eq:ExtCondition3}} are satisfied, and therefore in converse for Theorem \ref{Th:TIN_con_gap_32X_Gauss}, $h(S_1^n|V_1) - h(\tilde{S}_1^n|V_1)\leq n$.
\end{Rem}
\begin{proof}
We start by upper bounding the difference as follows
\begin{align*}
h(Y_A^n&|W_C)-h(Y_B^n|W_C)\nonumber\\
&=I(X_A^n,X_B^n;Y_A^n|W_C) - I(X_A^n,X_B^n;Y_B^n|W_C)\\
&\overset{(a)}{\leq} I(X_A^n,X_B^n;Y_A^n|W_C) - I(X_A^n,X_B^n;Y_B^n|W_C)\nonumber\\
&\quad + I(X_A^n;X_B^n|Y_A^n,W_C)\\
&\overset{(b)}{\leq}I(X_A^n;Y_A^n,X_B^n|W_C) + I(X_B^n;Y_A^n|X_A^n,W_C)\nonumber\\
&\quad - I(X_B^n;Y_B^n|W_C) - I(X_A^n;Y_B^n|X_B^n,W_C),
\end{align*}
where $(a)$ follows from the non-negativity of mutual information and $(b)$ follows by using chain rule. Note that $I(X_A^n;X_B^n|W_C)=0$, and hence, $I(X_A^n;Y_A^n,X_B^n|W_C)=I(X_A^n;Y_A^n|X_B^n,W_C)$. Using some standard steps, we get
\begin{align}
&h(Y_A^n|W_C)-h(Y_B^n|W_C)  \nonumber\\
&\leq I(X_A^n;h_1 X_A^n + Z_A^n|W_C) + I(X_B^n;h_2 X_B^n+Z_A^n|W_C) \notag \\ &\quad- I(X_B^n;Y_B^n|W_C) - I(X_A^n; h_1 X_A^n +\frac{h_1}{h_3}Z_B^n|W_C)\nonumber\\
&\stackrel{(c)}{\leq} I(X_B^n;h_2 X_B^n+Z_A^n|W_C) - I(X_B^n;Y_B^n|W_C),\label{StB}
\end{align}
where $(c)$ follows since $I(X_A^n;h_1 X_A^n + Z_A^n|W_C)\leq I(X_A^n; h_1 X_A^n +\frac{h_1}{h_3}Z_B^n|W_C)$ since $Z_A$ and $Z_B$ have the same distribution and $|h_1|^2\leq |h_3|^2$. Next, we proceed by bounding $T=I(X_B^n;Y_B^n|W_C)$. First, we write
\begin{align}
T&= I(X_B^n; \tilde{X}^n_A + \tilde{X}^n_B+\tilde{Z}_B^n|W_C)
\end{align}
where we define $\tilde{X}_A= \frac{X_A}{\sqrt{\rho}}$,  $\tilde{X}_B=h_4\frac{X_B}{\sqrt{\rho}h_3}$, and  $\tilde{Z}_B=\frac{Z_B}{\sqrt{\rho}h_3}$. Note that $I(X_B^n; \tilde{X}_A^{n} + \tilde{X}_B^n+\tilde{Z}_B^n|W_C)\geq I(X_B^n; \tilde{X}_A^{n} + \tilde{X}_B^n+{Z}_B^n|W_C)$ since increasing the noise variance (by $1-\frac{1}{\rho h_3^2}$) leads to a degraded channel, and hence, decreases the mutual information. This leads to $T\geq I(X_B^n; \tilde{X}_A^{n} + \tilde{X}_B^n+Z_B^n|W_C)$. Now, observe that $ I(X_B^n; \tilde{X}_A^{n} + \tilde{X}_B^n+Z_B^n|W_C)$ is larger than $h(\tilde{X}_B^n+Z_B^n|\tilde{X}_A^{n},W_C) - h ( \tilde{X}_A^{n} + Z_B^n|X_B^n,W_C)$ since conditioning reduces entropy. As a result,
\begin{align*}
T &\geq h(\tilde{X}_B^n+Z_B^n|\tilde{X}_A^{n},W_C) - h ( \tilde{X}_A^{n} + Z_B^n|\tilde{X}_B^n,W_C) \\
&= h(\tilde{X}_B^n+Z_B^n|W_C) - h ( \tilde{X}_A^{n} + Z_B^n|W_C),
\end{align*}
since $h(\tilde{X}_B^n+Z_B^n|\tilde{X}_A^{n},W_C)=h(\tilde{X}_B^n+Z_B^n|W_B)=h(\tilde{X}_B^n+Z_B^n|W_A,W_B)$ because $(\tilde{X}_A^n,W_A)$ is independent of $\tilde{X}_B^n$ and $W_B$, and similarly $h(\tilde{X}_A^n+Z_B^n|\tilde{X}_B^{n},W_C)=h(\tilde{X}_A^n+Z_B^n|W_A,W_B)$. Thus,
\begin{align*}
T &\geq  I(X_B^n;\tilde{X}_B^n+Z_B^n|W_C)  - I (\tilde{X}_A^{n}; \tilde{X}_A^{n} + Z_B^n|W_C)\\
&= I(X_B^n; X_B^n+\widehat Z_B^n|W_C)- I (\tilde{X}_A^{n}; \tilde{X}_A^{n} + Z_B^n|W_C)\\
&\geq I(X_B^n; X_B^n+\frac{1}{h_2}Z_B^n|W_C) - I (\tilde{X}_A^{n}; \tilde{X}_A^{n} + Z_B^n|W_C),
\end{align*}
where \textcolor{comm_Ay}{$\widehat Z_B^n = \frac{\sqrt{\rho}h_3}{h_4}Z_B^n$} and the last step follows by increasing the noise variance by $\frac{1}{|h_2|^2} - \frac{\rho|h_3|^2}{|h_4|^2}\geq 0$ (cf. \eqref{eq:ExtCondition3}). Now, we plug in \eqref{StB} to obtain
\begin{align}
&h(Y_A^n|W_C)-h(Y_B^n|W_C)  \nonumber\\
&\leq I(X_B^n;h_2 X_B^n+Z_A^n|W_C) - I(X_B^n; X_B^n+\frac{1}{h_2}Z_B^n|W_C)\nonumber\\
&\quad  + I (\tilde{X}_A^{n}; \tilde{X}_A^{n} + Z_B^n|W_C) \\
&= h (\tilde{X}_A^{n} + Z_B^n|W_C) - h(Z_B^n|W_C) \\
&\leq h ( \tilde{X}_A^{n} + Z_B^n) - h(Z_B^n),
\end{align}
which follows since conditioning reduces entropy and since $Z_B^n$ is independent of $W_C$.  Finally, $h ( \tilde{X}_A^{n} + Z_B^n) - h(Z_B^n)\leq nC(1)=n$ where $C(1)$ is the capacity of a Gaussian channel with input $\tilde{X}_A$ and noise $Z_B^n$, both of unit-power. This concludes the proof of Lemma \ref{Lemma:Entropy_diff_Gaussian}.
\end{proof}

\section{}
\label{app:Constant Gap Analysis for TDMA-TIN}
In this appendix, we transform the upper bound given in Lemma \ref{Th:UB_Det_32X} into a GDoF upper bound. We distinguish between two cases: $\alpha_{j_2i_3}>\alpha_{j_2i_1}$ and $\alpha_{j_2i_3}\leq\alpha_{j_2i_1}$.

\noindent{\bf Case $\alpha_{j_2i_3}>\alpha_{j_2i_1}$:}
In this case, $d=1$. If $\alpha_{j_2i_1}-\alpha_{j_1i_1}\leq \alpha_{j_2i_3}-\alpha_{j_1i_3}-\alpha_{j_2i_1}$, then $c^2=\rho^{\alpha_{j_2i_1}-\alpha_{j_1i_1}}$. By substituting in $B(\boldsymbol{p})$, we can obtain
\begin{align*}
B(\boldsymbol{p})
&< C\left(\rho^{\alpha_{j_1i_2}} + \frac{\rho^{\alpha_{j_1i_1}}}{\rho^{\alpha_{j_2i_1}}}\right)\notag\\
&\quad + C\left(\rho^{\alpha_{j_2i_1}} + \rho^{\alpha_{j_2i_3}} +\frac{\rho^{\alpha_{j_2i_2}} }{\rho^{\alpha_{j_1i_2}}}\right)+1\\

&< \max\{\alpha_{j_1i_2},\alpha_{j_1i_1}-\alpha_{j_2i_1}\}\log_2(\rho)
\\
&\quad+ \max\{\alpha_{j_2i_1},\alpha_{j_2i_3},\alpha_{j_2i_2}-\alpha_{j_1i_2}\}\log_2(\rho)+5\nonumber.
\end{align*}
For the other case where $\alpha_{j_2i_1}-\alpha_{j_1i_1}> \alpha_{j_2i_3}-\alpha_{j_1i_3}-\alpha_{j_2i_1}$, we have $c^2=\rho^{\alpha_{j_2i_3}-\alpha_{j_2i_1}-\alpha_{j_1i_3}}$. Using  similar steps as above, we can get
\begin{align*}
B(\boldsymbol{p})
&< \max\{\alpha_{j_1i_2},\alpha_{j_1i_3}-(\alpha_{j_2i_3}-\alpha_{j_2i_1})\}\log_2(\rho)
\\
&\quad+ \max\{\alpha_{j_2i_1},\alpha_{j_2i_3},\alpha_{j_2i_2}-\alpha_{j_1i_2}\}\log_2(\rho)+5\nonumber.
\end{align*}
By combining both cases, dividing by $\log(\rho)$, and letting $\rho\to\infty$, we get the following GDoF bound
\begin{align}
&d_{\Sigma,1}(\boldsymbol{p})< \max\{\alpha_{j_2i_1},\alpha_{j_2i_3},\alpha_{j_2i_2}-\alpha_{j_1i_2}\}\label{eq:UB_3x2_appG5}\\
&\quad +\max\{\alpha_{j_1i_2},\alpha_{j_1i_1}-\alpha_{j_2i_1},\alpha_{j_1i_3}-(\alpha_{j_2i_3}-\alpha_{j_2i_1})\}
\nonumber.
\end{align}

\noindent{\bf Case $\alpha_{j_2i_3}\leq \alpha_{j_2i_1}$:}
In this case, $d=0$ and $c^2=\rho^{\alpha_{j_2i_1}-\alpha_{j_1i_1}}$. Similar to the previous case, we can obtain
\begin{align}
\label{eq:UB_3x2_appG6}
d_{\Sigma,2}(\boldsymbol{p})
&< \max\{\alpha_{j_2i_1},\alpha_{j_2i_3},\alpha_{j_2i_2}-\alpha_{j_1i_2}\}\\
&\ +\max\{\alpha_{j_1i_2},\alpha_{j_1i_3},\alpha_{j_1i_1}-\alpha_{j_2i_1}\}
\nonumber.
\end{align}

Now, by combining the results in \eqref{eq:UB_3x2_appG5} and \eqref{eq:UB_3x2_appG6}, we obtain
\begin{align*}
&d_\Sigma(\boldsymbol{p})< \max\{\alpha_{j_2i_1},\alpha_{j_2i_3},\alpha_{j_2i_2}-\alpha_{j_1i_2}\}\nonumber\\
&\ \  +\max\{\alpha_{j_1i_2},\alpha_{j_1i_1}-\alpha_{j_2i_1},\alpha_{j_1i_3}-(\alpha_{j_2i_3}-\alpha_{j_2i_1})^+\}
\end{align*}
and as a result, we get $D(\boldsymbol{p})$ as given in \eqref{GDoFUpperBound}.

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