text:
\begin{proof}[Proof of Lemma \ref{lem:weighted}]
Using the re-parameterization $\vbeta =  \Sigma^{-q} \vtheta$ , the weighted min-norm estimator is $\hbeta_T := \tFst^{\dagger} \vy,  \hbeta_{T^c}:= \v0$, where $\vy = \tFst \bt + \tFstc \btc$ and $\tF = \mF \Sigma^q$.

Since $\tFst$ has full rank, the
matrix $\tFst \tFst^* = \Fst \sTqs \Fst^*$ is invertible and $\tFst^{\dagger} = \tFst^*(\tFst \tFst^*)^{-1}$.

The estimation error satisfies
\begin{align} \label{eq:risk_orig}
    \begin{split}
        \|\vtheta  - \hat{\vtheta}\|_2^2  & = \| \sTq (\bt -\hbeta_{T})\|^2 + \|\sTq \sTc(\btc -\hbeta_{T^c})\|^2 \\
        & = \norms{ \sTq \bt - \sTq \dtFst \p{ \tFst \bt + \tFstc \btc} } + \|\sTcq \btc \|^2 \\
        & = \norms{ \sTq  \p{\mI - \dtFst \tFst } \bt - \sTq \dtFst \tFstc \btc } + \norms{\sTcq \btc} \\
        & =  \norms{ \sTq \p{\mI - \dtFst\tFst } \bt } + \norms{\sTq \dtFst \tFstc \btc} +  \norms{\sTcq \btc} \\
        & \quad -  \underbrace{
        2 \operatorname{Re} \left( \bt ^* \p{ \mI - \dtFst \tFst } \sTqs \dtFst \tFstc \btc \right) }_{ =:\gC_1}.\\
    \end{split}
\end{align}
Since $ \dtFst\tFst$ is Hermitian, we have
\begin{align}\label{eq:proj}
        \norms{ \sTq \p{ \mI - \dtFst\tFst } \bt } & = \norms{\sTq \bt} + \norms{\sTq \dtFst\tFst \bt }
         - \underbrace{
         2 \left( \bt^* \sT^{2q} \dtFst\tFst \bt \right) }_{=:\gC_2}.
\end{align}
Combining (\ref{eq:risk_orig}) and (\ref{eq:proj}) and taking expectation yields
\begin{align} \label{eq:simple}
    \begin{split}
    \ex [ \| \Sigmaq (\vbeta -\hbeta) \|^2 ] & =  \ex [  \norms{\Sigmaq \vbeta} ] + \ex [ \| \sTq \dtFst\tFst \bt \|^2 ] +  \ex [ \|\sTq \dtFst \tFstc \btc\|^2 ]  -  \ex \bs{\gC_1}  -  \ex \bs{\gC_2} .
    \end{split}
\end{align}
The “trace trick” and $ \dtFst\tFst = \tFst^*(\tFst \tFst^*)^{-1}\tFst$ give
\begin{align*}
    \begin{split}
        \ex  & \bs{ \norms{  \sTq \dtFst\tFst \bt } }
         = \ex \bs{ \tr{\bt^* \dtFst\tFst \sTqs \dtFst\tFst \bt} }
         = \tr{ \dtFst\tFst \sTqs \dtFst\tFst \ex \bs{ \bt \bt^* } } \\
        & = \tr{ \sTqs \dtFst \tFst \sT^{-q} \kt \sT^{-q} }
         = \tr{ \tFst \kt \tFst^*(\tFst \tFst^*)^{-1} }  = \tr(K_T \dtFst\tFst).
    \end{split}
\end{align*}
Moreover,
\begin{align*}
    \begin{split}
    \ex \bs{ \norms{ \sTq \dtFst\tFstc \btc } }
    & = \tr{ \tFstc^* (\dtFst)^* \sTqs \dtFst \tFstc \ex[ \btc \btc^* ] } \\
    & =  \tr{  \sTqs \dtFst \tFstc \sTc^{-q} \ktc \sTc^{-q}}  \tFstc^* (\dtFst)^*   \\
    & = \tr{  \sTqs \tFst^*(\tFst \tFst^*)^{-1}\Fstc \ktc \Fstc^* (\tFst \tFst^*)^{-1}\tFst } .
    \end{split}
\end{align*}
Since $\ex \bs{ \btc \bt^*} = \sTc^{-q} \ex  \bs{ \vtheta_{T^c} \vtheta_T^*}  \sT^{-q} = 0$ we have $\ex \bs{\gC_1} = 0$. Furthermore, since $\mK$ commutes with $\Sigma^{-q}$ by diagonality, we have
$\sTqs \ex \left[ \bt \bt ^* \right] = \sTqs \ex \bs{ \Sigma^{-q} \vtheta \vtheta ^* \Sigma^{-q} } = \mK $ so that
\begin{align*}
    \begin{split}
        \ex \bs{\gC_2} & = 2 \, \tr{  \dtFst\tFst \sTqs \ex\left[ \bt \bt^*\right] } = 2\, \tr{ \kt  \dtFst\tFst} .
    \end{split}
\end{align*}
Plugging all terms into (\ref{eq:simple}), we have
\begin{align*}
        \risk_q = \ex & \bs{ \norms{  \Sigmaq (\vbeta -\hbeta)} }
           = \tr{\mK } - \tr{ \Fst  \sTqs \kt \Fst^* (\Fst \sTqs \Fst^*)^{-1} } \\
         & \quad +  \tr{ \Fst \sT^{4q} \Fst^*(\Fst \sTqs \Fst^*)^{-1}\Fstc \ktc \Fstc^* (\Fst \sTqs \Fst^*)^{-1} } .
\end{align*}
The risk of the plain min-norm estimator corresponds to $q=0$, which gives
\begin{align*}

       \risk_0 & = \ex  \bs{  \| \vtheta - \hat{\vtheta}\|^2 }  = \tr{\mK } - \tr{ \Fst \kt \Fst^* (\Fst \Fst^*)^{-1} }  +  \tr{\Fstc \ktc \Fstc^* (\Fst  \Fst^*)^{-1} } .

\end{align*}
\end{proof}

\subsection{Proof of Lemma \ref{lem:equi-fst}}
\begin{proof} [Proof of Lemma \ref{lem:equi-fst}]

For $ u \geq 0$, we set $ \mau = \Fst \stu \Fst^*$ and $\mcu = \Fstc \stcu \Fstc^*$, and define $\omn = \exp(-\frac{2\pi i}{n})$. Since $p=n l$, we have, for  $j_1, j_2 \in [n]$,
\begin{align*}
\begin{split}
    (\mau)_{j_1,j_2} & = (\Fst \stu \Fst^*)_{j_1,j_2} = \sum_{k=0}^{p-1} t_k^u \exp \p{ \frac{-2\pi i}{n} (j_2 -j_1)  \cdot k }
    = \sum_{\nu=0}^{l-1} \sum_{k=0}^{n-1} t_{k+n\nu}^u  \omn^{(j_2-j_1)k} ,\\
    (\mcu)_{j_1,j_2} & = (\Fstc \stcu \Fstc^*)_{j_1,j_2} =  \sum_{k=p}^{D-1} t_{k}^u \exp \p{\frac{-2\pi i}{n}  (j_2 -j_1)  \cdot k }
    = \sum_{\nu=l}^{\tau -1}\sum_{k=0}^{n-1} t_{k+n\nu}^u \omn^{(j_2-j_1)k} .
\end{split}
\end{align*}
In the above equations, we use $\omn^{k+n\nu} = \omn^k$ for $\nu \in \sN_+$.

For $j \in [n]$,
let $a_{j} = \aj $ and
$c_j = \cj $. Then
\begin{align} \label{eq:prod}
\begin{split}
    (\mau)_{j_1,j_2} & =  \sum_{\nu=0}^{l-1} \sum_{k=0}^{n-1} t_{k+n\nu}^u  \omn^{(j_2-j_1)k} = a_{j_2-j_1 \pmod n} , \\
    (\mcu)_{j_1,j_2} & =   \sum_{\nu=l}^{\tau-1}\sum_{k=0}^{n-1} t_{k+n\nu}^u \omn^{(j_2-j_1)k}  = c_{j_2-j_1 \pmod n} .
\end{split}
\end{align}
Hence, for any $u \geq 0$, $\mau$ and $\mcu$ are circulant matrices.

For $u=0$ we use again $p = nl, l\in \sN_+$ to obtain that, for $j_1,j_2 \in [n]$,

\begin{align*}

  (\Fst & \Fst^*)_{j_1, j_2} =  \sum_{k=0}^{p-1}  \omn^{(j_2-j_1)k}
   =
  \begin{cases}
   p, & \text{if} \quad j_1 = j_2, \\

    0  , & \text{if} \quad j_1 \neq j_2 .
  \end{cases}

\end{align*}
Hence, $\Fst \Fst^* = p\mI_n$ as claimed.

\end{proof}

config: arxiv


TITLE: Kovacic's algorithm
QUESTION [2 upvotes]: Is there any reference with some example, about how to solve a "riccati" equation in this (below) form :$$y'(x)+a(x)y^2(x)+b(x)y(x)+c(x)=0$$ by Kovacic's algorithm?
Or can anybody help me to understand how Kovacic's algorithm works?
(In Kovacic's 1985 article, the algorithm is described only for the case of $y''=ry$ with some complicated examples. I am searching for some easier examples. I am new to this study.)
Thanks in advance.

REPLY [4 votes]: In order to use Kovacic's result, we shall transform your equation into precisely the one he studies. All the coefficients will be functions of $x$, so for an easier typing I shall omit the argument. Keep in mind that Kovacic's algorithm applies to equations the coefficients of which are rational functions of a complex variable.
1) First, note that your equation is not linear, so Kovacic's algorithm cannot be applied to it. Fortunately, if $a \ne 0$, the usual change of unknown $w = -a y$ transforms your equation into another Riccati equation:
$$ w' = w^2 + (\frac {a'} a - b) w + a c .$$
2) The above is still not linear, so we next perform the usual change of unknown $w = - \frac {v'} v$, obtaining
$$v'' + (b - \frac {a'} a) v' + ac = 0 .$$
3) Let $f = b - \frac {a'} a$. In order to get rid of the term containing $v'$, make the last substitution $v = \Bbb e ^{\frac f 2} u$, obtaining
$$u'' = -(\frac {(f')^2} 4 + ac) u .$$
Let $r = -(\frac {(f')^2} 4 + ac) = \frac {a'' a - a^2 b' - (a')^2 - 4 a^3 c} {4 a^2}$ in order for your equation to become $u'' = r u$.
Now you may apply Kovacic's algorithm if $r$ is a fraction of polynomial functions of a complex variable. Unfortunately, there is nothing to be done in the general case if you do not provide the concrete expressions of $a,b,c$.


TITLE: Selecting limits of volume integral when the limits are dependent on another variable.
QUESTION [2 upvotes]: Evaluate $\int\int\int_v (2x+y)dV$ where V is the closed region bounded by the parabolic cylinder $z=4-x^2$ and the planes x=0,y=0,y=2 and z=0.
I computed the integral to get a final answer of $-\frac{80}{3}$ but the expected answer is $\frac{80}{3}$.Here's what I did:
$\int\int\int_v (2x+y)dV$
$=\int_0^2\int_{4-x^2}^0\int_0^2 (2x+y) dy dz dx$
$=\int_0^2\int_{4-x^2}^0 [(2xy+\frac{y^2}{2})]_0^2 dz dx$
$=\int_0^2\int_{4-x^2}^0 (4x+2) dz dx$
$=\int_0^2 [4xz+2z]_{4-x^2}^0 dx$
$=\int_0^2 {-4x(4-x^2)-2(4-x^2)} dx$
$=\int_0^2 {4x(x^2-4)+2(x^2-4)} dx$
$=-\frac{80}{3}$
I have a feeling I did something wrong with the limits and probably those of z. But since the parabolic cylinder is closed ABOVE by z=0,shouldn't the limits I chose be correct?

REPLY [1 votes]: Note that the region $V$ is above the plane $z=0$:
$$x\in[0,2] \implies z=4-x^2\ge 0.$$
Thus, the limits for $z$ are from $0$ to $4-x^2$ and not otherwise:
$$\int_0^2\int_0^{4-x^2}\int_0^2 (2x+y) dy dz dx.$$