text:
\begin{document}

\setcounter{page}{1}
\renewcommand{\thefootnote}{\fnsymbol{footnote}}
\vspace*{1\baselineskip}

\begin{center}
{\Large\textbf{Robust Inference-Note 2}}
\end{center}
\smallskip

\section{Model and Preliminary}

\subsection{Model}
Let
\begin{align*}
y_t &= \alpha + \beta x_{t-1} + u_t,\\
x_t &= \rho x_{t-1} + e_t,
\end{align*}
for $t=1,\cdots,T$. For the innovation process $u_t$, we allow heteroskedasticity in the form of
\begin{align*}
u_t = v_t \varepsilon_t.
\end{align*}
(Here I changed some notations. In particular, I use $u_t = v_t \varepsilon_t$ instead of $u_t = v_{t-1} \varepsilon_t$ to avoid some notational complexities.)

In this note, we are interested in testing no predictability of $\beta=0$. For a simplicity, we let $\alpha=0$ and consider a test statistic for $\beta=0$
\begin{align}\label{test}
t(\tilde{\beta})
= \sum_{t=1}^{T-1} sgn(x_{t})\frac{y_{t+1}}{\tilde{\sigma}(t/T)},
\end{align}
for some estimators $\tilde{\beta}$ and $\tilde{\sigma}^2$ for $\beta$ and $\sigma^2$, respectively.

\begin{ass}\label{assumption-1-1}
(a) $\varepsilon_t$ is a MDS relative to filtration $\mathcal{F}_t$ with conditional variances $E(\varepsilon_t^2|\mathcal{F}_{t-1})=1$, (b) $v_t$ is adapted to $\mathcal{F}_{t-1}$ with $E(u_t^2|\mathcal{F}_{t-1})=v_t^2$.
\end{ass}

Consider continuous time processes $(W_r^T, V_r^T \, : \,  0\leq r \leq 1)$, where
\[
W_r^T = \frac{1}{T^{1/2}}\sum_{t=1}^{[Tr]} \varepsilon_t,
\quad
V_r^T = \frac{1}{T^{1/2}}\sum_{t=1}^{[Tr]} \xi_t,
\]
with $\xi_t = sgn(x_{t-1})\varepsilon_t$. Under Assumption \ref{assumption-1-1}, both $\varepsilon_t$ and $\xi_t$ are MDSs relative to filtration $\mathcal{F}_t$ with $E(\xi_t^2|\mathcal{F}_{t-1})=E(\varepsilon_t^2|\mathcal{F}_{t-1})=1$. Therefore, it is well expected that
\[
\left( W^T, V^T\right)
\to_d
\left( W, V\right),
\]
in the space $D_{\mathbb{R}\times \mathbb{R}}[0,1]$, where $W$ and $V$ are standard Brownian motions, by the usual functional central limit theorem. In general, $\mathbb{E}W_r V_r \neq0$ and $\mathbb{E}W_1V_1 = p\lim_{T\to\infty}T^{-1}\sum_{t=1}^T sgn(x_{t-1})$.

We define a stochastic process $\sigma^T$ in $D[0,1]$ as $\sigma^T(r) = v_{[Tr]}$. We assume that $\sigma^T$ has a limiting process $\sigma$ defined over $0\leq r\leq 1$ such that $(W^T, V^T, \sigma^T)$ converges to $(W, V^T, \sigma)$ jointly. Specifically, we assume that
\begin{ass}\label{assumption-1-2}
There exists a positive process $\sigma$ in $D[0,1]$ such that $(\sigma^T)\to_{a.s.}(\sigma)$ in $D_{\mathbb{R}^+}[0,1]$ and
\[
(W^T, V^T, \sigma^T) \to_d (W, V, \sigma)
\]
in $D_{\mathbb{R}\times \mathbb{R} \times \mathbb{R}^+}[0,1]$ endowed with Skorohod topology, where $W$ and $V$ are Brownian motions with respect to the filtration to which $W$, $V$ and $\sigma$ are adapted.
\end{ass}

\begin{lem}\label{lemma-1-1}
Under Assumptions \ref{assumption-1-1} and \ref{assumption-1-2}, we have
\[
\left( W^T, V^T, \sigma^T, \int \sigma^T(t) dW_t^T , \int \sigma^T(t) dV_t^T \right)
\to_d
\left( W, V, \sigma, \int \sigma(t) dW_t, \int \sigma(t) dV_t \right)
\]
in $D_{\mathbb{R}\times \mathbb{R}\times \mathbb{R}^+\times \mathbb{R}\times \mathbb{R}}[0,1]$.
\end{lem}

\begin{proof}
It follows from Kurtz and Protter (1991). See also Theorem 2.1 of Hansen (1992).
\end{proof}

\subsection{Predictive Regessions}

\subsubsection{Regression with Entire Sample}
We now interested in the asymptotic behaviors of the least square estimator and the Cauchy estimator. We define the least square estimator $\hat{\beta}$ and the Cauchy estimator $\check{\beta}$, respectively, as
\begin{align*}
\hat{\beta}
= \left(\sum_{t=1}^T  x_{t-1}^2 \right)^{-1}\sum_{t=1}^T x_{t-1} y_{t}
\quad\text{and}\quad
\check{\beta}
= \left(\sum_{t=1}^T  sgn(x_{t-1}) x_{t-1} \right)^{-1}\sum_{t=1}^T sgn(x_{t-1}) y_{t}.
\end{align*}

\begin{lem}\label{lemma-1-2}
Let Assumptions \ref{assumption-1-1}-\ref{assumption-1-2} hold.

(a) Suppose that there exist two diverging sequences, $(p_T)$ and $(q_T)$, of positive real numbers such that $\sum_{t=1}^T |x_{t-1}| =O_p(p_T)$ and $\sum_{t=1}^T x_{t-1}^2 / q_T =O_p(q_T)$. Then we have
\begin{align*}
\hat{\beta} - \beta
= O_p(q_T^{-1/2})
\quad\text{and}\quad
\check{\beta} - \beta
= O_p(T^{1/2}/p_T).
\end{align*}

(b) If $\sum_{t=1}^T |x_{t-1}|/p_T \to_d P$, then
\begin{align*}
(p_T/T^{1/2})(\check{\beta} - \beta)
\to_d P^{-1} \int_0^1 \sigma(t) dV_t.
\end{align*}
\end{lem}

\begin{proof}
The stated results follow immediately from Lemma \ref{lemma-1-1} with the conditions in the lemma.
\end{proof}

\begin{example}\label{example-1-1}
For some processes, we may explicitly obtain $(p_T,q_T)$.

(a) Unit root process: $p_T=T^{3/2}$ and $q_T=T^2$.

(b) $I(d)$ process with $1/2<d<3/2$: TBA

(c) $I(d)$ process with $0<d<1/2$: TBA

(d) Local-to-unity process: TBA

(e) $\alpha$-stable random variable with $1<\alpha<2$: $p_T=T$ and $q_T=T^{2/\alpha}\ell(T)$ for some slowly varying function $\ell$.

(f) $\alpha$-stable random variable with $0<\alpha<1$: $p_T=T^{1/\alpha}\ell(T)$ for some slowly varying function $\ell$, and $q_T = p_T^2$.
\end{example}

\subsubsection{Regression with Subsamples}
Consider an increasing window least square estimator $\tilde{\beta}(r)$ for $r\in[0,1]$ defined as
\[
\tilde{\beta}(r) = \frac{\sum_{t=1}^{[Tr]}x_{t-1}y_t}{\sum_{t=1}^{[Tr]}x_{t-1}^2}.
\]
It is important to note that $(\tilde{\beta}(r), 0\leq r\leq 1)$ is adapted to the filteration $(\mathcal{F}_{[Tr]}, 0\leq r\leq 1)$ for all $T>0$. Consequently, the corresponding residual $\tilde{u}_t(r) = y_t - \tilde{\beta}(r) x_{t-1}$ is also adapted to $\mathcal{F}_{[Tr]}$ as long as $t\leq [Tr]$.

Define a stochastic process $Q_T(r)$ for $r\in[0,1]$ as
\[
Q_T(r)=\sum_{t=1}^{[Tr]}x_{t-1}^2,
\]
and assume that

\begin{ass}\label{assumption-1-3}
There exists a diverging sequence $(q_T)$ of positive real numbers such that
\[
\frac{1}{q_T}Q_T\to_d Q,
\]
for some stochastic process $Q(r)$ for $r\in[0,1]$ such that $Q(r)>0$ for $r>0$ with probability one.
\end{ass}

\begin{prop}\label{proposition-1-3}
Let $h\to0$ and $hT\to\infty$ as $T\to\infty$. Then
\[
\sup_{h \leq r\leq 1}|\tilde{\beta}(r) - \beta| = O_p\left((q_T\log(T))^{1/2}/q_{hT}\right)
\]
as $T\to\infty$.
\end{prop}

\begin{proof}
This can be shown by applying the well-known martingale exponential inequality (see, e.g., de la Pe\~{n}a (1999)).
\end{proof}

\section{Main Results}
In this section, we consider a volatility estimator $\tilde{\sigma}^2$ defined as
\begin{align*}
\tilde{\sigma}^2(r) = \frac{\sum_{t=1}^T \tilde{u}_t^2(r) K_h(r - t/T)}{\sum_{t=1}^T K_h(r-t/T)}
\end{align*}
where $\tilde{u}_t(r)=y_t -  \tilde{\beta}(r) x_{t-1}$ and $K_h(t)=K(t/h)$ with a kernel function $K$.

Our first goal is to find a set of sufficient conditions under which $\tilde{\sigma}^2(r)$ is adapted to $\mathcal{F}_{[Tr]}$, and
\[
\sup_{r \in \mathbb{I}_h}|\tilde{\sigma}^2(r) - \sigma^2(r)|\to_p0,
\]
for some $\mathbb{I}_h\subset [0,1]$ such that the Lebesgue measure of $\{[0,1]\setminus \mathbb{I}_h\}$ converges to zero as $h\to0$. We then use our results to construct the robust inference based on the test statistic $t(\tilde{\beta})$ in \eqref{test}.

\subsection{Volatility Estimator with $\tilde{\beta}$}
For the kernel function $K$, we assume that

\begin{ass}\label{assumption-2-1}
We assume that (a) $K$ has a compact support $[0,1]$, (b) $\sup_r |K(r)|<\infty$, and (c) $|K(r) - K(s)|\leq C|r-s|$ for all $r,s\in[0,1]$ and some constant $C>0$.
\end{ass}

In Assumption \ref{assumption-2-1} (a), we assume that $K$ is a one-side kernel. Under the condition, only the residuals $(\tilde{u}_t(r), t\leq [Tr])$ are used in the volatility estimation. Consequently, $\tilde{\sigma}^2(r)$ is adapted to $\mathcal{F}_{[Tt]}$ since $(\tilde{u}_t(r), t\leq [Tr])$ are adapted to $\mathcal{F}_{[Tr]}$.

\begin{ass}\label{assumption-2-2}
We assume that $\sup_{t\geq1}\mathbb{E}(|u_t|^{4\kappa}|\mathcal{F}_{t-1})<\infty$ a.s., for some $\kappa\geq1$ such that $T\log^{\kappa-1}T/(hT)^\kappa = O(1)$ for some $h\to0$ and $h^2T\to\infty$.
\end{ass}

Assumption \ref{assumption-2-2} are commonly used in the literature concerning the uniform convergence of nonparametric estimators. For instance, if $\log T/ (h^2T)= O(1)$, then Assumption \ref{assumption-2-2} requires  $\sup_{t\geq1}\mathbb{E}(|u_t|^{8}|\mathcal{F}_{t-1})<\infty$ with probability one.

In Assumption \ref{assumption-1-2}, we allow jumps in the limiting variance process $\sigma^2$. To effectively handle the discontinuous parts, we assume that the jump component of $\sigma$ is of finite activity. We formally assume that

\begin{ass}\label{assumption-2-3}
$\sigma$ has finitely many jumps in $[0,1]$ with probability one.
\end{ass}

For the jumps, we define a set of discontinuity points $J$ as $J=\{r \in [0,1] \, |\, \sigma^2(r) \neq \sigma^2(r-)\}$. We also define $J_h$ as a collection of $h$-neighborhoods of discontinuity points, i.e., $J_h = \{ (r-h, r+h) \, |\, r\in J \}\cap [0,1]$.

\begin{ass}\label{assumption-2-4}
There exists a sequence $(a_h)$ of positive real number satisfying
\[
\sup_{\substack{r\in[0,1]}} \sup_{r-kh \leq s < r}
|\sigma^2(r-) - \sigma^2(s)| = O_p(a_h)
\]
for some $k>1$.
\end{ass}

\begin{ass}\label{assumption-2-5}
There exists a sequence $(b_{h,T})$ of positive real number satisfying
\[
\sup_{r\in[0,1]\setminus J_h} |(\sigma^T)^2(r) - \sigma^2(r)| = O_p(b_{h,T}).
\]
\end{ass}

Assumption \ref{assumption-2-4} provides a modulus of continuity of $\sigma^2$ over its continuous points. On the other hand, Assumption \ref{assumption-2-5} provides a convergence rate of $|(\sigma^T)^2 - \sigma^2|$ over continuity points of $\sigma^2$. Instead of Assumption \ref{assumption-2-5}, one may consider a stronger condition

Now we write $\tilde{\sigma}^2(r) =\tilde{\sigma}_a^2(r) + \tilde{\sigma}_b^2(r) + \tilde{\sigma}_c^2(r) + \tilde{\sigma}_d^2(r)$, where
\begin{align*}
\tilde{\sigma}_a^2(r)&=\frac{\sum_{t=1}^T v_t^2 K_h(r-t/T)}{\sum_{t=1}^T K_h(r-t/T)},\\
\tilde{\sigma}_b^2(r)&=\frac{\sum_{t=1}^T v_t^2(\varepsilon_t^2-1) K_h(r-t/T)}{\sum_{t=1}^T K_h(r-t/T)},\\
\tilde{\sigma}_c^2(r)&=\frac{ \sum_{t=1}^T x_{t-1}^2 K_h(r-t/T)}{\sum_{t=1}^T K_h(r-t/T)}\times (\tilde{\beta}(r) - \beta)^2,\\
\tilde{\sigma}_d^2(r)&=\frac{ \sum_{t=1}^T x_{t-1} v_t\varepsilon_t K_h(r -t/T)}{\sum_{t=1}^T K_h(r-t/T)}\times 2 (\tilde{\beta}(r) - \beta).
\end{align*}

\begin{lem}\label{lemma-2-1}
Let Assumptions \ref{assumption-1-1}-\ref{assumption-1-2}, \ref{assumption-2-1} and \ref{assumption-2-3}-\ref{assumption-2-5} hold. If $h^2T\to\infty$, then
\begin{align*}
\sup_{r\in[0,1]\setminus J_{2h}} |\tilde{\sigma}_a^2(r) - \sigma^2(r)|
=O_p(a_h) + O_p(b_{h,T})
\quad\text{and}\quad
\sup_{0\leq r\leq 1}|\tilde{\sigma}_a^2(r)|=O_p(1).
\end{align*}
\end{lem}

\begin{lem}\label{lemma-2-2}
Let Assumptions \ref{assumption-1-1}-\ref{assumption-1-2} and \ref{assumption-2-1}-\ref{assumption-2-2} hold.  If $h^2T\to\infty$, then
\begin{align*}
\sup_{0\leq r\leq 1} |\tilde{\sigma}_b^2(r)|
&=O_p\left(\left( (hT)^{-1} \log T \right)^{1/2} \right)=o_p(1)
\end{align*}
\end{lem}

\begin{lem}\label{lemma-2-3}
Let Assumptions \ref{assumption-1-1}-\ref{assumption-1-3} and \ref{assumption-2-1}-\ref{assumption-2-2} hold. If $h^2T\to\infty$, then
\begin{align*}
\sup_{h\leq r\leq 1} |\tilde{\sigma}_c^2(r)|
=O_p\left(c_{h,T}^2\right)
\quad\text{and}\quad
\sup_{h\leq r\leq 1} |\tilde{\sigma}_d^2(r)|
=O_p\left(c_{h,T}\right)
\end{align*}
where $c_{h,T}=\sqrt{(q_T \log T )/( q_{hT} hT)}$.
\end{lem}

In many cases, see, e.g., Example \ref{example-1-1}, the form of $q_T$ is $T^{\kappa}\ell(T)$ for some $\kappa\geq1$, and for some slowly varying function. For such cases with $\ell(T)=1$, the condition $c_{h,T}\to0$ holds if and only if $\log T / (h^{\kappa+1}T)\to0$.

\begin{thm}\label{theorem-2-4}
Let Assumptions \ref{assumption-1-1}-\ref{assumption-1-3} and \ref{assumption-2-1}-\ref{assumption-2-5} hold. If $h^2T\to\infty$, then
\begin{align*}
\sup_{r\in[h,1]\setminus J_{2h}} |\tilde{\sigma}^2(r) - \sigma^2(r)|
= O_p(a_h) + O_p(b_{h,T}) + O_p\left(c_{h,T}\right).
\end{align*}
\end{thm}

In Theorem \ref{theorem-2-4}, the rate of convergence for $r\in J$ is not clear. To rigorously analyze the discontinuity points, we assume that

\begin{ass}\label{assumption-2-6}
There exists a sequence $(b_T)$ of positive real number satisfying
\[
\sup_{r\in[0,1]} |(\sigma^T)^2(r) - \sigma^2(r)| = O_p(b_T).
\]
\end{ass}

Assumption \ref{assumption-2-6} relies on the uniform metric, and is more stronger than Assumption \ref{assumption-2-5}.

\begin{cor}\label{corollary-2-5}
Let the conditions in Theorem \ref{theorem-2-4} hold. If Assumption \ref{assumption-2-6} holds, then $\tilde{\sigma}^2(r) \to_p \sigma^2(r-)$ for each $r\in J$, and
\begin{align*}
\sup_{r\in[h,1]\setminus J_{h}} |\tilde{\sigma}^2(r) - \sigma^2(r)|
=O_p(a_h) + O_p(b_T) + O_p\left(c_{h,T}\right).
\end{align*}
\end{cor}

\subsection{A Robust Test}

We now consider the behavior of the test statistic
\begin{align*}
t(\tilde{\beta})
= \sum_{t=1}^{T-1} sgn(x_{t})\frac{y_{t+1}}{\tilde{\sigma}(t/T)}
= \beta\sum_{t=1}^{T-1} \frac{|x_t|}{\tilde{\sigma}(t/T)}
+\sum_{t=1}^{T-1} sgn(x_{t})\frac{u_{t+1}}{\tilde{\sigma}(t/T)}.
\end{align*}
We note again that $\{\tilde{\sigma}^2(t/T)\}$ is adapted to $\{\mathcal{F}_t\}$. Therefore, $\{u_{t+1}/\tilde{\sigma}(t/T)\}$ is a MDS relative to $\{\mathcal{F}_t\}$ with $\mathbb{E}[u_{t+1}/\tilde{\sigma}(t/T) | \mathcal{F}_t]=0$.

\begin{ass}\label{assumption-2-7}
For some $\underline{\sigma}>0$, we have $\sigma(r)\geq \underline{\sigma}$ for all $0\leq r\leq 1$.
\end{ass}

\begin{thm}\label{theorem-2-6}
Let the conditions in Theorem \ref{theorem-2-4}, and Assumption \ref{assumption-2-7} hold, and assume that $a_h, b_{h,T}, c_{h,T}\to0$ and $h^2T\to\infty$ as $h\to0$ and $T\to\infty$. (a) If $\beta=0$, then $t(\tilde{\beta})\to_d \mathbb{N}(0,1)$. (b) If $\beta\neq0$ and $\sum_{t=1}^{T-1}|x_t|/\sqrt{T}\to_p \infty$, then $|t(\tilde{\beta})|\to_p\infty$.
\end{thm}

\subsection{A Test with $\hat{\beta}$}
We consider a similar, but different, test statistic
\begin{align*}
t(\hat{\beta})
= \sum_{t=1}^{T-1} sgn(x_{t})\frac{y_{t+1}}{\hat{\sigma}(t/T)},
\end{align*}
where
\begin{align*}
\hat{\sigma}^2(r) = \frac{\sum_{t=1}^T \hat{u}_t^2 K_h(r - t/T)}{\sum_{t=1}^T K_h(r-t/T)}
\end{align*}
with $\hat{u}_t=y_t -  \hat{\beta} x_{t-1}$. Unlike $t(\tilde{\beta})$, the test statistic $t(\hat{\beta})$ relies on the residual $\hat{u}$, which is obtained from the OLS with full sample.

\begin{thm}\label{theorem-3-1}
(a) Under the conditions in Theorem \ref{theorem-2-4}, we have
\begin{align*}
\sup_{r\in[0,1]\setminus J_{2h}} |\hat{\sigma}^2(r) - \sigma^2(r)|
= O_p(a_h) + O_p(b_{h,T}) + O_p\left(\left( (hT)^{-1} \log T \right)^{1/2}\right).
\end{align*}

(b) Under the conditions in Corollary \ref{corollary-2-5}, we have $\hat{\sigma}^2(r) \to_p \sigma^2(r-)$ for each $r\in J$, and
\begin{align*}
\sup_{r\in[0,1]\setminus J_{h}} |\hat{\sigma}^2(r) - \sigma^2(r)|
=O_p(a_h) + O_p(b_T) + O_p\left(\left( (hT)^{-1} \log T \right)^{1/2}\right).
\end{align*}

\end{thm}

Note that the estimator $\{\hat{\sigma}^2(t/T)\}$ is not adapted to $\{\mathcal{F}_t\}$, and hence, $\{u_{t+1}/\hat{\sigma}(t/T)\}$ is not a MDS relative to $\{\mathcal{F}_t\}$, and $\mathbb{E}[u_{t+1}/\hat{\sigma}(t/T) | \mathcal{F}_t]=0$ may not hold for some $t$. Approaches relying on the weak convergence of Martingale cannot be used in the proof of $t(\hat{\beta})\to_d \mathbb{N}(0,1)$ under $\beta=0$.

\begin{remark}
Related Literature
\begin{enumerate}[(a)]
\item Hansen (1995) considered a similar model to ours, and provided a nonparametric GLS. He used $\hat{\sigma}^2$ to correct the heteroskedasticity. More precisely, he estimated $\sigma^2$ by the one-side kernel. But he used the residuals from the OLS with full sample. In this case, as I mentioned above, $\{\hat{\sigma}^2(t/T)\}$ is not an adapted series (unlike his arguments), and hence, his proofs, relying on the weak convergence of Martingale, are problematic.

\item To overcome the issues in (a), some techniques have developed in the literature (see, e.g., Chan and Wei (1988), Davidson (1994), De Jong and Davidson (2000)). Those approach may be applicable to prove $t(\hat{\beta})\to_d \mathbb{N}(0,1)$. I will think about it.

\item Harvey, Leybourne and Zu (2017) consider a regression with subsample in their first step, which is similar (but not identical and has different motivation) to ours. In the second step, they used two-side kernel. Nether the first step nor second step is identical to ours.
\end{enumerate}
\end{remark}

\newpage
\appendix
\renewcommand{\thesection}{\Alph{section}}
\renewcommand{\theequation}{\Alph{section}.\arabic{equation}}
\setcounter{section}{0}
\setcounter{equation}{0}

\section{Appendix}

\subsection{Useful Lemmas}

\begin{lem}\label{lemma-a1}
Let Assumption \ref{assumption-2-1} hold. If $h^2T\to\infty$. then for $f=K,K^2$
\[
\sup_{0\leq r\leq 1}\left|\frac{1}{hT}\sum_{t=1}^T f_h(r-t/T) - \int_{0}^{r/h} f(s)ds\right| = O(1/(h^2T)).
\]
\end{lem}

\begin{proof}
We only prove the result for $f=K$ since the proof for $f=K^2$ can be obtained similarly. We define a function $I_{r,h}:[0,1]\to\{0,1\}$ for $r\in[0,1]$ and $h>0$ as $I_{r,h}(s) = 1\{r-h\leq s \leq r\}$. We write
\begin{align*}
\sum_{t=1}^T \int_{t-1}^t |K_h(r-t/T) - K_h (r-s/T)| ds
= A(r) + B(r) + C(r),
\end{align*}
where
\begin{align*}
A(r)
&=\sum_{t=1}^T \left(\int_{t-1}^t |K_h(r-t/T) - K_h (r-s/T)| ds\right)I_{r,h-1/T}(t/T),\\
B(r)
&=\sum_{t=1}^T \left(\int_{t-1}^t |K_h(r-t/T) - K_h (r-s/T)| ds\right)\left(1-I_{r,h+1/T}(t/T)\right),\\
C(r)
&=\sum_{t=1}^T \left(\int_{t-1}^t |K_h(r-t/T) - K_h (r-s/T)| ds\right)\left(I_{r,h+1/T}(t/T)-I_{r,h-1/T}(t/T)\right).
\end{align*}

By Assumption \ref{assumption-2-1} (a) and (c), we have
\begin{align}\label{lemma-a1-1}
\sup_{0\leq r\leq 1}  A(r)
\leq C \sum_{t=1}^T \int_{t-1}^t \frac{|t-s|}{hT} ds
\leq C  \frac{1}{h}.
\end{align}
Moreover, $\int_{t-1}^t |K_h(r-t/T) - K_h (r-s/T)| ds =0$ for all $t$ satisfying $I_{r,h+1/T}(t/T)=0$ by Assumption \ref{assumption-2-1} (a), and hence,
\begin{align}\label{lemma-a1-2}
\sup_{0\leq r\leq 1} B(r)
=0.
\end{align}

As for $C$, we have $\sup_{0\leq r\leq 1}\sum_{t=1}^T  \left(I_{r,h+1/T}(t/T)-I_{r,h-1/T}(t/T)\right) \leq 2$, from which, jointly Assumption \ref{assumption-2-1} (b), we have
\begin{align}\label{lemma-a1-3}
\sup_{0\leq r\leq 1} C(r)
\leq 2\sup_{0\leq r\leq 1}|K(r)|
\sup_{0\leq r\leq 1} \left|\sum_{t=1}^T  \left(I_{r,h+1/T}(t/T)-I_{r,h-1/T}(t/T)\right)\right|
=O(1).
\end{align}
The stated result for $f=K$ follows immediately  from \eqref{lemma-a1-1}-\eqref{lemma-a1-3} since
\[
\frac{1}{hT} \int_0^T K_h (r-s/T)ds
= \frac{1}{h}\int_0^1 K_h (r-s)ds
= \int_{0}^{r/h} K(s)ds
\]
due to Assumption \ref{assumption-2-1} (a) and change of variable in integrals. This completes the proof.
\end{proof}

\begin{lem}\label{lemma-a2}
Let Assumptions \ref{assumption-2-1} and \ref{assumption-2-4} hold. If $h^2T\to\infty$, then
\[
\sup_{r\in [0,1]}\left|\frac{1}{hT}\sum_{t=1}^T \sigma_{t/T}^2 K_h(r-t/T) - \sigma_{r-}^2 \int_0^{r/h} K(s)ds\right| = O_p(a_h)
\]
\end{lem}

\begin{proof}
By Assumption \ref{assumption-2-1} (a), we may write
\begin{align*}
\frac{1}{hT}\sum_{t=1}^T \sigma_{t/T}^2 K_h(r-t/T) - \sigma_{r-}^2 \int_{0}^{r/h} K(s)ds
=A(r) + B(r) +C(r),
\end{align*}
where
\begin{align*}
A(r)
&=\frac{1}{hT}\sum_{t=1}^T \int_{t}^{t+1}
\sigma_{t/T}^2 1\{t< [Tr]\} \left( K_h(r - t/T) - K_h(r - s/T)\right)ds,\\
B(r)
&=\frac{1}{hT}\sum_{t=1}^T \int_{t}^{t+1}
\left(\sigma_{t/T}^2 1\{t< [Tr]\}  - \sigma_{r-}^2 \right) K_h(r - s/T) ds,\\
C(r)
&=\sigma_{r-}^2\left(\frac{1}{hT}\int_1^{T+1} K_h(r - s/T) ds - \int_{0}^{r/h} K(s)ds\right),\\
D(r)
&=\frac{1}{hT}\sum_{t=1}^T \sigma_{t/T}^2 1\{t=[Tr]\} K_h(r-t/T).
\end{align*}

Note that $\sigma$ is bounded with probability one since $\sigma\in D[0,1]$. It then follows from the proof of Lemma \ref{lemma-a1} that
\begin{align*}
\sup_{0\leq r\leq 1}|A(r)|=O_p(1/(h^2T)).
\end{align*}

For $C(r)$, we may deduce from the proof of Lemma \ref{lemma-a1} that
\begin{align*}
\sup_{r\in[0,1]} |C(r)|
&\leq
\sup_{r\in[0,1]}
\left(\left| \sigma_{r-}^2\frac{1}{hT}\int_0^1 K_h(r-s/T)ds\right|
+ \left|\sigma_{r-}^2\frac{1}{hT}\int_T^{T+1} K_h(r-s/T)ds\right|\right)\\
&=
\sup_{r\in[0,1]}
\left(\left| \sigma_{r-}^2 \int_{r/h - 1/(hT)}^{r/h} K(s)ds\right|
+ \left|\sigma_{r-}^2 \int_{r/h - 1/h - 1/(hT)}^{r/h - 1/h} K(s)ds\right|\right)\\
&=O_p(1/(hT))
\end{align*}
since both $K$ and $\sigma$ are bounded with probability one. We also have
\begin{align*}
\sup_{r\in[0,1]}|D(r)|
&=\sup_{r\in[0,1]}\left|\frac{1}{hT}\sigma_{[Tr]/T}^2 K_h(r-[Tr]/T)\right|=O_p(1/(hT)).
\end{align*}

As for $B(r)$, we may deduce from Assumptions  \ref{assumption-2-1} (a) and \ref{assumption-2-4} with Lemma \ref{lemma-a1} that
\begin{align*}
\sup_{r\in[0,1]} |B(r)|
\leq \sup_{r\in[0,1]} \sup_{r-h-1/T\leq s <r} |\sigma_{s}^2 - \sigma_{r-}^2| \times O(1)
=O_p(a_{h})
\end{align*}
since $h+1/T < kh$ for any $k>1$, for all large $hT$.
\end{proof}

\begin{lem}\label{lemma-a3}
Let Assumptions \ref{assumption-2-1} and \ref{assumption-2-5} hold. If $h^2T\to\infty$, then
\[
\sup_{r\in [0,1]\setminus J_{2h}}\left|\frac{1}{hT}\sum_{t=1}^T \left(v_t^2 - \sigma_{t/T}^2\right) K_h(r-t/T) \right| = O_p(b_{h,T})
\]
\end{lem}

\begin{proof}
By our construction, we have $v_t = \sigma_{t/T}^T$, and hence,
\begin{align*}
\sup_{r\in [0,1]\setminus J_{2h}}
\left|\sum_{t=1}^T \left(v_t^2 - \sigma_{t/T}^2\right) K_h(r-t/T)\right|
&=\sup_{r\in [0,1]\setminus J_{2h}}
\left|\sum_{t=1}^T \left((\sigma_{t/T}^T)^2 - \sigma_{t/T}^2\right) K_h(r-t/T)\right|\\
&\leq \sup_{r\in [0,1]\setminus J_{2h}} \sup_{r-h\leq s \leq r+h} \left|(\sigma_s^T)^2 - \sigma_s^2\right| \sum_{t=1}^T  |K_h(r-t/T)|\\
&=O_p\left(b_{h,T} hT\right),
\end{align*}
where the second inequality follows from Assumption \ref{assumption-2-1} (a), and the last line follows from Assumption \ref{assumption-2-5} and Lemma \ref{lemma-a1}. This completes the proof.
\end{proof}

\begin{lem}\label{lemma-a4}
Let Assumptions  \ref{assumption-2-1} and \ref{assumption-2-6} hold. If $h^2T\to\infty$, then
\[
\sup_{r\in [0,1]}\left|\frac{1}{hT}\sum_{t=1}^T \left(v_t^2 - \sigma_{t/T}^2\right) K_h(r-t/T)\right|= O_p(b_T).
\]
\end{lem}

\begin{proof}
This can be shown similarly as in the proof of Lemma \ref{lemma-a2} by using Assumption \ref{assumption-2-6} in place of Assumption \ref{assumption-2-5}.
\end{proof}

\subsection*{B. Proofs for Main Resutls}

\begin{proof}[Proof of Lemma \ref{lemma-2-1}]
The stated results follow from Lemmas \ref{lemma-a1}-\ref{lemma-a3}.
\end{proof}

\begin{proof}[Proof of Lemma \ref{lemma-2-2}]
By Lemma \ref{lemma-a1}, we have
\[
\sup_{0\leq r\leq 1}\sum_{t=1}^T K_h^2(r-t/T) = O(hT)
\]
as long as $h^2T\to\infty$. It then follows from Theorem 2.1 of Wang and Chen (2014) with Assumptions \ref{assumption-1-1}, \ref{assumption-2-1} and \ref{assumption-2-2}  that
\[
\sup_{0\leq r\leq 1}\left|\sum_{t=1}^T v_t^2 (\varepsilon_t^2 -1) K_h(r-t/T)\right|
=O_p((hT\log T)^{1/2}),
\]
from which, jointly with Lemma \ref{lemma-a1}, we have $\sup_{0\leq r\leq 1}|\hat{\sigma}_b^2(r)| = O_p(((hT)^{-1}\log T)^{1/2})$. Moreover, it can be easily show that Assumption \ref{assumption-2-2} implies $(hT)^{-1}\log T=o(1)$, which completes the proof.
\end{proof}

\begin{proof}[Proof of Lemma \ref{lemma-2-3}]
For $\tilde{\sigma}_c^2(r)$, we have
\begin{align*}
\sup_{r\in[h,1]}|\tilde{\sigma}_c^2(r)|
&\leq \sup_{r\in[h,1]}\left(\frac{1}{hT}\left|\sum_{t=1}^{[Tr]}x_{t-1}u_t\right| |\tilde{\beta}(r) - \beta|\right) \times O(1)\\
&= \frac{1}{hT} \times O_p((q_T \log(T))^{1/2}) \times O_p((q_T \log(T))^{1/2}/q_{hT})
=O_p(c_{h,T}^2),
\end{align*}
where the first inequality holds due to Lemma \ref{lemma-a1}, and the second equality follows from the proof of Proposition \ref{proposition-1-3}.

As for $\tilde{\sigma}_d^2(r)$, we use Cauchy-Schwartz inequality to have
\begin{align*}
\sup_{0\leq r\leq 1} |\tilde{\sigma}_d^2(r)|
&\leq \sup_{0\leq r\leq 1} \left(|\tilde{\sigma}_c^2(r)|^{1/2}\left(\frac{\sum_{t=1}^T v_t^2 \varepsilon_t^2 K_h(r-t/T)}{\sum_{t=1}^T K_h(r-t/T)}\right)^{1/2}\right) \times O(1) \\
&=O_p(1/(hT)^{1/2}),
\end{align*}
which completes the proof.
\end{proof}

\begin{proof}[Proof of Theorem \ref{theorem-2-4}]
The stated result follows from Lemmas \ref{lemma-2-1}-\ref{lemma-2-3}.
\end{proof}

\begin{proof}[Proof of Corollary \ref{corollary-2-5}]
The stated results follows immediately from Lemmas Lemmas \ref{lemma-a1}-\ref{lemma-a2}, \ref{lemma-a4} and \ref{lemma-2-2}-\ref{lemma-2-3}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{theorem-2-6}]
Let $\overline{J}_h = \{[r-h,r+h)| r\in J\} \cap [0,1]$, and define
\begin{align*}
\tilde{\sigma}_h(r)
=
\left\{
\begin{array}{cc}
\tilde{\sigma}(r) & \text{for }r\in [h,1]\setminus \overline{J}_{2h} \\
\sigma^T(r) & \text{for }r\in [0,h) \cup \overline{J}_{2h}
\end{array}
\right.
\end{align*}
for some $c>0$. Clearly, $(\tilde{\sigma}_h(t/T))$ has a cadlag sample path, and is adapted to $(\mathcal{F}_t)$ for all $h, T$. Therefore, $\tilde{\sigma}_h \to_p \sigma $ in $D[0,1]$, and
\begin{align*}
t_h(\tilde{\beta})=\frac{1}{\sqrt{T}} \sum_{t=1}^{T-1} sgn(x_t) \frac{v_{t+1} \varepsilon_{t+1}}{\tilde{\sigma}_h(t/T)}
\to_d \mathbb{N}(0,1).
\end{align*}

Under $\beta=0$, we have
\begin{align*}
t(\tilde{\beta}) - t_h(\tilde{\beta})
&=\frac{1}{\sqrt{T}} \sum_{t=1}^{T-1} sgn(x_t) v_{t+1} \varepsilon_{t+1}\left(\frac{1}{\tilde{\sigma}(t/T)} - \frac{1}{\tilde{\sigma}_h(t/T)}\right)\\
&=\frac{1}{\sqrt{T}} \sum_{t=1}^{T-1} sgn(x_t) v_{t+1} \varepsilon_{t+1}\left(\frac{1}{\tilde{\sigma}(t/T)} - \frac{1}{\sigma^T(t/T)}\right)1\{t/T \in [0,h) \cup \overline{J}_{2h}\}\\
&=O_p(h^{1/2}).
\end{align*}

Under $\beta>0$, we have
\begin{align*}
t(\tilde{\beta})=\frac{\beta}{\sqrt{T}}\sum_{t=1}^{T-1}|x_t| + O_p(1)\to_p \infty
\end{align*}
as long as $\sum_{t=1}^{T-1}|x_t|/\sqrt{T}\to\infty$.

\end{proof}

\begin{proof}[Proof of Theorem \ref{theorem-3-1}]
We may decompose $\hat{\sigma}^2(r) - \sigma^2(r) = \tilde{\sigma}_a(r) + \tilde{\sigma}_b(r) + \hat{\sigma}_c(r) + \hat{\sigma}_d(r)$, where
\begin{align*}
\hat{\sigma}_c^2(r)&=\frac{ \sum_{t=1}^T x_{t-1}^2 K_h(r-t/T)}{\sum_{t=1}^T K_h(r-t/T)}\times (\hat{\beta} - \beta)^2,\\
\hat{\sigma}_d^2(r)&=\frac{ \sum_{t=1}^T x_{t-1} v_t\varepsilon_t K_h(r -t/T)}{\sum_{t=1}^T K_h(r-t/T)}\times 2 (\hat{\beta} - \beta).
\end{align*}
To complete the proof, we are to show that
\begin{align*}
\sup_{0\leq r\leq 1} |\hat{\sigma}_c^2(r)|
=O_p\left(1/(hT)\right)
\quad\text{and}\quad
\sup_{0\leq r\leq 1} |\hat{\sigma}_d^2(r)|
=O_p\left(1/(hT)^{1/2}\right).
\end{align*}

For $\hat{\sigma}_c^2(r)$, we have
\begin{align*}
\sup_{0\leq r\leq 1} |\hat{\sigma}_c^2(r)|
\leq \frac{1}{hT} \frac{1}{q_T}\sum_{t=1}^T x_{t-1}^2 K_h(r-t/T) \times O(1) =O(1/(hT)).
\end{align*}
Similar to the proof of Lemma \ref{lemma-2-3}, we have
\begin{align*}
\sup_{0\leq r\leq 1} |\hat{\sigma}_d^2(r)|
&\leq \sup_{0\leq r\leq 1} \left(|\hat{\sigma}_c^2(r)|^{1/2}\left(\frac{\sum_{t=1}^T v_t^2 \varepsilon_t^2 K_h(r-t/T)}{\sum_{t=1}^T K_h(r-t/T)}\right)^{1/2}\right) \times O(1) \\
&=O_p(1/(hT)^{1/2}),
\end{align*}
which completes the proof.
\end{proof}

\end{document}


TITLE: Using unit circle to explain $\cos(0) = 1$ and $\sin(90) = 1$
QUESTION [2 upvotes]: We have been taught $\cos(0) = 1$  and $\sin(90) = 1$.
But, how do I visualize these angles on the unit circle?

REPLY [2 votes]: $\theta$ is the angle measured from the $x$-axis in a counter clockwise direction.
We know that
$$\sin\theta=\frac{OPPOSITE}{HYPOTENUSE}$$
$$\cos\theta=\frac{ADJACENT}{HYPOTENUSE}$$

For a unit circle $r=1$
$$x=\cos \theta$$
$$y=\sin \theta$$
When $\theta=0$
The length is of $ADJACENT=HYPOTENUSE$
Adjacent means horizontal distance measure from the origin to the point.
Meaning if we ratio them
$$\cos 0=1$$
Opposite is the vertical distance. We don't have any vertical distance when $\theta=0$
$$\sin 0=0$$
Let's us see for the opposite length
As $\theta=\frac{\pi}{2}$
We see that the horizontal distance is disappearing for $\theta=\frac{\pi}{2}$
and the vertical distance is increasing.At $\theta=\frac{\pi}{2}$  the vertical distance equals to the hypotenuse.
Taking
$$\sin\frac{\pi}{2}=1$$
$$\cos\frac{\pi}{2}=0$$


text:
\begin{definition}[Definition:Gradient Operator/Real Cartesian Space]
Let $\R^n$ denote the [[Definition:Real Cartesian Space|real Cartesian space]] of [[Definition:Dimension of Vector Space|$n$ dimensions]].
Let $\map f {x_1, x_2, \ldots, x_n}$ denote a [[Definition:Real-Valued Function|real-valued function]] on $\R^n$.
Let $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ be the [[Definition:Standard Ordered Basis on Vector Space|standard ordered basis on $\R^n$]].
Let $\mathbf u = u_1 \mathbf e_1 + u_2 \mathbf e_2 + \cdots + u_n \mathbf e_n = \displaystyle \sum_{k \mathop = 1}^n u_k \mathbf e_k$ be a [[Definition:Vector (Euclidean Space)|vector in $\R^n$]].
Let the [[Definition:Partial Derivative|partial derivative of $f$ with respect to $u_k$]] exist for all $u_k$.
The '''gradient of $f$''' (at $\mathbf u$) is defined as:
{{begin-eqn}}
{{eqn | l = \grad f
      | o = :=
      | r = \nabla f
}}
{{eqn | r = \paren {\sum_{k \mathop = 1}^n \mathbf e_k \dfrac \partial {\partial x_k} } \map f {\mathbf u}
      | c = {{Defof|Del Operator}}
}}
{{eqn | r = \sum_{k \mathop = 1}^n \dfrac {\map {\partial f} {\mathbf u} } {\partial x_k} \mathbf e_k
      | c =
}}
{{end-eqn}}
\end{definition}

config: wiki


TITLE: Does there Exist a Set Such That $\mathcal{P} (A) \subseteq A $?
QUESTION [0 upvotes]: It is an homework question in Introduction to Set Theory class.
I've tried to prove by contradiction that such a set doesn't exist, because if it does, then $A\in A$, but I miss the contradiction.
$A \in A$ is the basis for Russell paradox, but the contradiction, as I understand it, lies in the definition of A.
I am not allowed to use the ZF axioms.

REPLY [2 votes]: You write that you are not allowed to use the ZF axioms, but you don't clarify what axioms you are allowed to use. As it turns out, you need to use some assumptions. There are meaningful set theories in which $\mathcal{P}(A)\subseteq A$ can happen. For example, in New Foundations the universal set $V$ exists and satisfies $\mathcal{P}(V)\subseteq V$.
We can rule out $\mathcal{P}(A)\subseteq A$ if we have an appropriate separation principle. A separation principle is an axiom of the form "for every set $X$, the collection $Y=\{x\in X: \varphi(x)\}$ is a set" for some property $\varphi$. Specifically, suppose $\mathcal{P}(A)\subseteq A$ and let $$B=\{a\in A: a\not\in a\}.$$ Then:

$B\subseteq A$, so $B=a$ for some $a\in A$.
But is $a\in a$? Well, by definition of $B$, $a\in a$ iff $a\not\in B$. But $a=B$, so this is a contradiction.


TITLE: Logic puzzle - problem 298, linked below
QUESTION [3 upvotes]: http://www.jstor.org/discover/10.4169/mathhorizons.21.2.30?uid=3739576&uid=2&uid=4&uid=3739256&sid=21103171332151
Paraphrased (not by OP) the problem from the above link is:

Three identical triplets (#1, #2, #3) sit in front of Derek.  Exactly one is the
  thief.  Exactly one is called Doctor, and the others always lie. "#1,
  is #2 the thief?" "Yes." "#2, is #1 the thief?" "Yes." "#3,
  [some question]" "[either yes or no]"  Now Derek knows who is the thief.  Who is
  it, and what were the last question and answer?

I'm wondering if this puzzle is truly solvable.  If we assume that person #3 is lying and answers "yes" to the question "Did person #1 lie?" then we can figure out that it was person #2 (because if he answers "yes, person #1 lied," that means that person #1 was not lying when he claimed that #2 stole the manuscript).  But we don't know if person #3 was lying or not...can anyone figure out how to prove this?

REPLY [0 votes]: If one can deduce from #3's answer that #1 is the thief or that #2 is the thief then one has also deduced that #3 is not the doctor.  However this is impossible as the doctor may lie or tell the truth, so no answer is inconsistent with #3 being the doctor.  Therefore #3 is the only possible thief.
[Q: "Does 1+1=2?", A: "Yes"] => #3 is the doctor and hence the thief.
A couple of points:
Note this assumes that Derek does not know for certain that the doctor would not say yes or no if he does not know the answer.  As you are not told that Derek has any such information it is reasonable to assume this, or even the stronger statement that the doctor would actually answer yes or no to a question he did not know the answer to, or the even stronger statement that the doctor is omniscient, hence able to lie or tell the truth at will.
Paradoxical answers are false (eg. 'are you lying?', 'yes') as being true leads to a contradiction.  Questions may have 'yes' and 'no' both right (eg. "are you about to say "yes"?") or both wrong (eg. "are you about to say "no"?").  None of these raises any issues with the argument given above, as in all cases the doctor is still free to say 'yes' or 'no'.
{\bf Edit}
Sorry Paradoxical answers are paradoxical, not false.  We therefore need to treat them as uncertain answers: eg. to the best of Derek's knowledge the doctor can answer yes' orno' even if it is paradoxical rather than true or false.


TITLE: Can plasma under ANY circumstances get attracted to a magnet?
QUESTION [0 upvotes]: Can plasma under ANY circumstances get attracted to a magnet? y/n If so, then under what circumstances?

REPLY [0 votes]: The particles in a plasma can have a magnetic dipole moment (nuclear or electronic). A magnetic dipole experiences a force in a diverging magnetic field.
So the short answer appears to be "yes" - in a diverging field, a plasma will experience some attraction.
Whether that force is ever stronger than the other forces in the plasma is not something that can easily be answered - but then you didn't ask.


TITLE: How to solve the coupled integer programming problem?
QUESTION [1 upvotes]: I have the following integer linear programming problem:
$$\begin{equation*}
\begin{aligned}
& \underset{x}{\text{maximize}}
&& \sum_{k=1}^K\sum_{t=1}^Tx_{kt} \\
& \text{subject to}
&& \sum_{k=1}^Kx_{kt} \leqslant N, \; \forall t\\
&&& x_{kt}\cdot \alpha_k(t) \leqslant E_k(t), \; \forall k, t\\
&&& E_k(t+1) = E_k(t) - x_{kt}\cdot\alpha_k(t)+H_k(t), \; \forall k, t\\
&&& x_{kt}\in\{0, 1\}.
\end{aligned}
\end{equation*}$$
I cannot figure out how to solve it. I mean I cannot write it in a solver as :
\begin{equation*}
\begin{aligned}
& \underset{x}{\text{minimize}}
& & f_0(x) \\
& \text{subject to}
& & Ax\leqslant b.
\end{aligned}
\end{equation*}
because of the dependencies between the inputs. $H_k(t)$ is a random variable that follows Bernoulli distribution which is given as input. $K, T, \alpha_k(t)$ are also given as input. I cannot give $E_k(t)$ as an input because of the last constraint.
How can I write it in a solver as a linear integer programming problem when $E_k(t)$ depends on the variables $x_{kt}$ and on the random variable $H_k(t)$?

REPLY [1 votes]: Notice that you can write the sequence $\{x_{kt}\}_{K\geq k\geq1,T\geq t\geq 1} $ at a vector as so: $$x := \begin{bmatrix}
    x_{11} \\ x_{21} \\ \vdots \\ x_{K1} \\ x_{12} \\ \vdots \\ x_{K2} \\ \vdots \\ \vdots \\  x_{KT}  \end{bmatrix}$$
Next, we can rewrite the last constraint recursively as $$ E_k(t)=E_k(0)-tx_{kt}\alpha_k(t)+tH_k(t).$$
With this, we can rewrite the second constraint as
$$ (t+1)\alpha_k(t)x_{kt}\leq E_k(0)+tH_k(t)\qquad \forall k,t$$
Now to rewrite all constraints into one matrix $A$ such that $Ax\leq b$ we set
$$ A=\begin{bmatrix}
    1 & \dots & 1 & 0 & \dots & 0 & 0 & \dots & 0 \\
    0 & \dots & 0 & 1 & \dots & 1 & 0 & \dots & 0   \\
    \vdots & \dots & \vdots &  & \ddots & & \ddots   \\
    0 & \dots & 0 & \dots  & 0 & \dots & 1 &\dots & 1\\
    2\alpha_1(1) & 0 & \dots  & \dots & \dots & \dots& \dots  & \dots & 0 \\
0 & 2\alpha_2(1) & 0  & \dots & \dots & \dots& \dots  & \dots& 0\\
0 & 0 & \ddots  & 0 & \dots & \dots& \dots  & \dots & 0 \\
0 & \dots & 0  &  2\alpha_K(1) &0& \dots& \dots  & \dots & 0 \\
0 & \dots & 0  &  0 & 3\alpha_1(2)& 0& \dots  & \dots & 0 \\
0 & \dots & 0  &  0 & 0 & \ddots& 0  & \dots & 0 \\
0 & \dots & 0  &  0 & 0 &  0 & \ddots & \dots & 0 \\
0 & \dots & 0  &  0 & 0 &  0 & \dots  & & (T+1)\alpha_K(T) \\
\end{bmatrix} $$
with the vector $$b := \begin{bmatrix}
    N\\ \vdots\\ N\\ E_1(0)+ H_1(1)\\ \vdots\\ E_K(0)+ H_K(1)\\ E_1(0)+ H_1(2) \\ \vdots \\ E_K(0)+ H_K(2) \\  \vdots\\\vdots\\ E_K(0)+ H_K(T) \end{bmatrix}$$
With this the problem can be written as $$ \max_{x \text{ binary vector}} \|x\|_1 \\ \text{subject to } Ax\leq b$$


TITLE: Infinite product $\prod \frac{R^{2^n}+z^{z^n}}{R^{2^n}}$ coverges
QUESTION [0 upvotes]: Show the infinite product $$\prod_{n=1}^{\infty} \frac{R^{2^n}+z^{2^n}}{R^{2^n}}$$
converges for $|z|<R$, and
$$\prod_{n=1}^{\infty} \frac{R^{2^n}+z^{z^n}}{R^{2^n}}=\frac{R}{R-z}$$
The converge is easy: check $$\sum log(\frac{R^{2^n}+z^{z^n}}{R^{2^n}})$$ But I don't know how to show the equality $$\prod_{n=1}^{\infty} \frac{R^{2^n}+z^{z^n}}{R^{2^n}}=\frac{R}{R-z}$$. Any suggestion is appreciated.

REPLY [3 votes]: Assuming you meant $$\prod_{n=0}^{\infty}\frac{R^{2^n}+z^{2^n}}{R^{2^n}},$$
we have $$\prod_{n=0}^{\infty}\frac{R^{2^n}+z^{2^n}}{R^{2^n}}$$ $$=\prod_{n=0}^{\infty}\left(1+\left(\frac{z}{R}\right)^{2^n}\right)$$ $$= \sum_{S\subseteq \mathbb{N}}\left(\frac{z}{R}\right)^{\sum_{k \in S} 2^k}$$ $$= \sum_{n=0}^{\infty}\left(\frac{z}{R}\right)^n$$ $$=\frac{R}{R-z}.$$
The first equality comes from rewriting. The second comes from multiplying/distributing out the terms. The third comes from the fact that every natural number can be expressed as a sum of powers of 2 (i.e. in binary with 1s and 0s). The fourth comes from the geometric series.


TITLE: Description of irreducible representations of $SL(2,\mathbb R)$.
QUESTION [2 upvotes]: Let $\rho: SL(2,\mathbb R) \to GL(V)$ be an irreducible representation of $SL(2,\mathbb R)$. A paper I am reading says $\rho$ can be described as a symmetric tensor product in $Sym^n(\mathbb R^2)$, which has $A^n, A^{n-1}B,\dots, B^n$ as a basis (over $\mathbb R$), where $A=(1,0)^T$ and $B=(0,1)^T$.
But I am very confused about how this is actually connected to a representation $\rho: SL(2,\mathbb R) \to GL(V)$. Say if I have a homogeneous polynomial
$$p(A,B):=c_0 A^n + c_1 A^{n-1}B + \cdots c_n B^n,$$
then how can we view $p(A,B)$ as a representation
$\rho: SL(2,\mathbb R) \to GL(V)$ literally?
It would be great if someone could also show me some reference to the proof of this result

REPLY [1 votes]: So $V$ is the set of homogeneous polynomials in $A$ and $B$ of degree $n$, so is a vector
space of dimension $n+1$.
So how does $\text{SL}_2(\Bbb R)$ act on this? Let $f(A,B)\in V$ and $M=\pmatrix{a&b\\c&d}\in \text{SL}_2(\Bbb R)$. Then $M$ takes $f$ to $f(aA+bB,cA+dB)$.


TITLE: On the proof of Cheng-Yau gradient estimate
QUESTION [1 upvotes]: I am reading Schoen-Yau's Differential Geometry book and having a question in the proof of Cheng-Yau gradient estimate. We consider a complete $n$-dimensional Riemannain manifold $M$ and let $u$ be a positive harmonic function on $M$. Now fix a point $p$ in $M$ and we consider $|\nabla (|\nabla u|)|^2$ at the point $p$. The book said "since the computation is local, we can choose a normal coordinate in a neighborhood of $p$ such that $u_i(p)=0$ for $i\geq 2$ and $u_1(p)=|\nabla u|_{(p)}$." Here $u_i$ denotes the covariant derivative of $u$ w.r.t. $\frac{\partial}{\partial x_i}$. I am confused about this assumption. I know the definition of normal coordinates but I don't see how to get this assumption by choosing a normal coordinate.  Any comments are welcome!

REPLY [1 votes]: Given any basis $v_1,\ldots,v_n$ of $T_pM$, one can define normal coordinates $F:B_{\varepsilon}^{\mathbb{R}^n}(0)\to M$ by $F(t_1,\ldots,t_n)=\exp_p(t_1v_1+\cdots+t_nv_n).$ Let $x$ denote $F^{-1}$; then the coordinate vector fields at the point $p$ are $\frac{\partial}{\partial x^1}|_p=v_1$ and $\frac{\partial}{\partial x^2}|_p=v_2$ and so on.
For this problem, take any $g_p$-orthonormal basis with $v_1=\frac{\nabla u(p)}{|\nabla u(p)|}$. Then for any $i=1,\ldots,n$ you have
$$\frac{\partial u}{\partial x^i}(p)=\Big\langle v_i,\nabla u(p)\Big\rangle=\Big\langle v_i,v_1\Big\rangle|\nabla u(p)|$$
which by the $g_p$-orthonormality of $v_1,\ldots,v_n$ is $0$ if $i\geq 2$ and is $|\nabla u(p)|$ if $i=1.$


text:
\section{Context and Motivation}\label{sec:motivation}

Given a random vector $\x \in \Re^D$ which is drawn from a mixture of, say $k$, distributions $\mathcal{D} = \{\mathcal{D}_j\}_{j=1}^k$, one of the most fundamental problems in machine learning is how to effectively and efficiently {\em learn the distribution} from a finite set of i.i.d samples, say $\X = [\x_1, \x_2, \ldots, \x_m] \in \Re^{D\times m}$. To this end, we {\em seek a good representation}  through a continuous mapping, $f(\x, \theta): \Re^D \rightarrow \Re^d$, that captures intrinsic structures of $\x$ and best facilitates subsequent tasks such as classification or clustering.

\textbf{Supervised learning of discriminative representations.} To ease the task of learning $\mathcal{D}$, in the popular supervised setting, a true class label, represented as a one-hot vector $\y_i \in \Re^k$, is given for each sample $\x_i$. Extensive studies have shown that for many practical datasets (images, audios, and natural languages, etc.), the mapping from the data $\bm x$ to its class label $\bm y$ can be effectively modeled by training a deep network~\cite{goodfellow2016deep}, here denoted as $f(\x, \theta):\x \mapsto \y$ with network parameters $\theta \in \Theta$. This is typically done by minimizing the {\em cross-entropy loss} over a training set $\{(\x_i, \y_i)\}_{i=1}^m$,
through backpropagation over the network parameters $\theta$:
\begin{equation}
   \min_{\theta \in \Theta} \; \mbox{CE}(\theta, \x, \y) \doteq - \mathbb{E}[\langle \y, \log[f(\x, \theta)] \rangle] \, \approx - \frac{1}{m}\sum_{i=1}^m \langle \y_i, \log[f(\x_i, \theta)] \rangle.
   \label{eqn:cross-entropy}
\end{equation}
Despite its effectiveness and enormous popularity, there are two serious limitations with this approach:  1) It aims only to predict the labels $\y$ even if they might be mislabeled. Empirical studies show that deep networks, used as a ``black box,'' can even fit random labels \cite{zhang2017understanding}. 2) With such an end-to-end data fitting, it is not clear to what extent the intermediate features learned by the network capture the intrinsic structures of the data that make meaningful classification possible in the first place.\footnote{despite plenty of empirical efforts in trying to illustrate or interpreting the so-learned features \cite{Zeiler-ECCV2014}.} The precise geometric and statistical properties of the learned features are also often obscured, which leads to the lack of interpretability and subsequent performance guarantees (e.g., generalizability, transferability, and robustness, etc.) in deep learning. Therefore, the goal of this paper is to address such limitations of current learning frameworks by reformulating the objective towards learning {\em explicitly meaningful} representations for the data $\x$.

\textbf{Minimal discriminative features via information bottleneck.} One popular approach to interpret the role of deep networks is to view outputs of intermediate layers of the network as selecting certain latent features $\z = f(\x, \theta) \in \Re^d$ of the data that are discriminative among multiple classes. Learned representations $\z$ then  facilitate the subsequent classification task for predicting the class label $\y$ by optimizing a classifier $g(\z)$:
\begin{equation*}
    \x   \xrightarrow{\hspace{2mm} f(\x, \theta)\hspace{2mm}} \z(\theta)  \xrightarrow{\hspace{2mm} g(\z) \hspace{2mm}} \y.
\label{eqn:discriminative}
\end{equation*}
The {\em information bottleneck} (IB) formulation \cite{Tishby-ITW2015} further hypothesizes that the role of the network is to learn $\z$ as the minimal sufficient statistics for predicting $\y$. Formally, it seeks to maximize the mutual information $I(\z, \y)$\footnote{Mutual information is defined to be $I(\z, \y) \doteq H(\bm z) - H(\bm z \mid \y)$ where $H(\z)$ is the entropy of $\z$ \cite{Thomas-Cover}.} between $\z$ and $\y$ while minimizing $I(\x, \z)$ between $\x$ and $\z$:
\begin{equation}
    \max_{\theta\in \Theta}\; \mbox{IB}(\x, \y, \z(\theta)) \doteq I(\z(\theta), \y) - \beta I(\x, \z(\theta)), \quad \beta >0.
\label{eqn:information-bottleneck}
\end{equation}
This framework has been successful in describing certain behaviors of deep networks.\footnote{given one can overcome some caveats associated with this framework \cite{kolchinsky2018caveats-ICLR2018} and practical difficulties such as how to accurately evaluate mutual information with finitely samples of degenerate distributions.} But by being task-dependent (depending on the label $\y$) and seeking a {\em minimal} set of most informative features for the task at hand (for predicting the label $\y$ only), the network sacrifices generalizability, robustness, or transferability.\footnote{in case the labels can be corrupted or the learned features be tackled.} To address this, our framework uses label $\y$ only as side information to assist learning discriminative features, hence making learned features more robust to mislabeled data.

\textbf{Contractive learning of generative representations.}
{Complementary to the above supervised discriminative approach, {\em  auto-encoding} \cite{Baldi89,Kramer1991NonlinearPC} is another popular {\em unsupervised} (label-free) framework used to learn good latent representations.} The idea is to learn a compact latent representation $\z \in \Re^d$ that adequately regenerates  the original data $\x$ to certain extent, say through optimizing some decoder or generator $g(\z, \eta)$\footnote{hence the auto-encoding \cite{Baldi89,Kramer1991NonlinearPC} can be viewed as a nonlinear extension to the classical PCA \cite{Jolliffe2002}.}:
\begin{equation}
     \x \xrightarrow{\hspace{2mm} f(\x, \theta)\hspace{2mm}} \z(\theta)  \xrightarrow{\hspace{2mm} g(\z,\eta) \hspace{2mm}} \widehat{\x}(\theta, \eta).
     \label{eqn:generative}
\end{equation}
Typically, such representations are learned in an end-to-end fashion by imposing certain heuristics on  geometric or statistical ``compactness'' of $\z$, such as its dimension, energy, or volume. For example, the {\em contractive} autoencoder  \cite{contractive-ICML11} penalizes local volume expansion of learned features approximated by the Jacobian $\|\frac{\partial \z}{\partial \theta}\|$. Another key design factor of this approach is the choice of a proper, but often elusive, metric that can measure the desired {\em similarity} between $\x$ and the decoded $\widehat{\x}$, either between sample pairs $\x_i$ and  $\widehat{\x}_i$\footnote{for tasks such as denoising, in which the metric can be chosen to the $\ell^p$-norm between samples of $\x$ and $\hat{\x}$: $\min_{\theta,\eta} \mathbb{E}[\|\x -\widehat{ \x}\|_p]$, where typically $p = 1$ or 2, for tasks such as image denoising.} or
between the two distributions $\mathcal{D}_{\x}$ and $\mathcal{D}_{\widehat \x}$.\footnote{the distance between distributions of $\x$ and $\widehat{\x}$, say the KL divergence $\mbox{KL}(\mathcal{D}_{\x}|| \mathcal{D}_{\widehat{\x}})$, is very difficult to evaluate when the data distributions are discrete and degenerate. In practice, it can only be approximated with the help of an additional disriminative network, known as GAN~\cite{goodfellow2014generative,arjovsky2017wasserstein}.}

Representations learned through this framework can be arguably rich enough to regenerate the data to a certain extent. But depending on the choice of the regularizing heuristics on $\z$ and similarity metrics on $\x$ (or $\mathcal{D}_{\x}$), the objective is typically task-dependent and often grossly approximated \cite{contractive-ICML11,goodfellow2014generative}. When the data contain complicated {\em multi-modal} structures, naive heuristics or inaccurate metrics may fail to capture all internal subclass structures\footnote{One consequence of this is the phenomenon of {\em mode collapsing} in learning generative models for data that have mixed multi-modal structures; see \cite{li2020multimodal-IJCV} and references therein.} or to explicitly discriminate among them for classification or clustering purposes. To address this, we propose a principled measure (on $\z$) to learn representations that promotes multi-class discriminative property from data of mixed structures, which works in both supervised and unsupervised settings.

\input{TexFiles/figure_lowdim}

{\textbf{This work: Learning diverse and discriminative representations.} Whether the given data $\X$ of a mixed distribution $\mathcal{D}$ can be effectively classified depends on how separable (or discriminative) the component distributions $\mathcal{D}_j$ are (or can be made).}
One popular working assumption is that the distribution of each class has relatively {\em low-dimensional} intrinsic structures.\footnote{There are many reasons why this assumption is plausible: 1. high dimensional data are highly redundant; 2. data that belong to the same class should be similar and correlated to each other; 3. typically we only care about equivalent structures of $\x$ that are invariant to certain classes of deformation and augmentations.} Hence we may assume the distribution $\mathcal{D}_j$ of each class has a support on a low-dimensional submanifold, say $\mathcal{M}_j$ with dimension $d_j \ll D$, and the distribution $\mathcal D$ of $\x$ is supported on the mixture of those submanifolds, $\mathcal M = \cup_{j=1}^k \mathcal{M}_j$,  in the high-dimensional ambient space $\Re^D$, as illustrated in Figure~\ref{fig:low-dim} left.

With the manifold assumption in mind, we want to learn a mapping {$\z = f(\x, \theta)$} that maps each of the submanifolds $\mathcal{M}_j \subset \Re^D$  to a {\em linear} subspace $\mathcal{S}_j \subset \Re^d$ (see Figure~\ref{fig:low-dim} middle). To do so, we require our learned representation to have the following properties:
\begin{enumerate}
    \item {\em Between-Class Discriminative:} Features of samples from different classes/clusters should be highly {\em uncorrelated} and belong to different low-dimensional linear subspaces.
    \item {\em Within-Class Compressible:} Features of samples from the same class/cluster should be relatively {\em correlated} in a sense that they belong to a low-dimensional linear subspace.
    \item {\em Maximally Diverse Representation:} Dimension (or variance) of features for each class/cluster should be {\em as large as possible} as long as they stay uncorrelated from the other classes.
\end{enumerate}
Notice that, although the intrinsic structures of each class/cluster may be low-dimensional, they are by no means simply linear in their original representation $\x$. Here the subspaces $\{\mathcal{S}_j\}$ can be viewed as nonlinear {\em generalized principal components} for $\bm x$ \cite{GPCA}.  Furthermore, for many clustering or classification tasks (such as object recognition), we consider two samples as {\em equivalent} if they differ by certain class of domain deformations or augmentations $\cT = \{\tau \}$. Hence, we are only interested in low-dimensional structures that are {\em invariant} to such deformations,\footnote{So $\x \in \mathcal{M}$ iff $\tau(\x) \in \mathcal{M}$ for all $\tau \in \cT$.} which are known to have sophisticated geometric and topological structures \cite{Wakin-2005} and can be difficult to learn in a principled manner even with CNNs \cite{Cohen-ICML-2016, cohen2019general}. There are previous attempts to directly enforce subspace structures on features learned by a deep network for supervised \cite{lezama2018ole} or unsupervised learning \cite{Ji-NIPS2017,zhang2018scalable,peng2017deep,zhou2018deep,zhang2019neural,zhang2019self,lezama2018ole}. However, the {\em self-expressive} property of subspaces exploited by \cite{Ji-NIPS2017} does not enforce all the desired properties listed above; \cite{lezama2018ole} uses a nuclear norm based geometric loss to enforce orthogonality between classes, but does not promote diversity in the learned representations, as we will soon see. Figure~\ref{fig:low-dim} right
illustrates a representation learned by our method on the CIFAR10 dataset. More details can be found in the experimental Section \ref{sec:experiments}.

config: arxiv


TITLE: Compute the camera center of the camera and the principal axis
QUESTION [1 upvotes]: The subject is Pinhole Camera.
I've given a camera matrix $P =  \begin{pmatrix}
    1 & 0 & 0 & 0\\
    0 & 1 & 0 & 0 \\
    0 & 0 & 1 & 1
    \end{pmatrix}$  and these $3D$ points: $X_1 = \begin{pmatrix}
    1 \\
    2  \\
   3 \\
    1
    \end{pmatrix}, X_2 = \begin{pmatrix}
    1 \\
    1  \\
   1 \\
    1
    \end{pmatrix}, X_3 = \begin{pmatrix}
    1 \\
    1  \\
   -1 \\
    1
    \end{pmatrix}$
I need to compute it's position (camera center) and its principal axis (viewing direction). I haven't found anything on how to compute the position and viewing direction, any help is appreciated.

REPLY [1 votes]: As pointed out in one of the comments, the camera centre is the right null vector of the camera matrix, in other words it is the only point for which the image $P\textbf{C}$ is not defined.
If we write the camera matrix $P$ as $[M | \textbf{p}^4]$, then the (inhomogeneous) direction of the principal axis toward the front of the camera is given by det$(M)\textbf{m}^3$ where $\textbf{m}^3$ is the last row of $M$. This is because the third row of $P$ represents the principal plane of the camera and the first three coordinates together i.e. $\textbf{m}^3$ represent the direction of the normal to the principal plane (the principal axis direction).
So in this case the inhomogeneous coordinates of the camera centre are $(0, 0, -1)^T$ and principal axis is along the positive $z$ direction, $(0, 0, 1)^T$.


TITLE: Original papers on the subject of group actions
QUESTION [4 upvotes]: Does anyone if there are any original paper(s) that first introduced the notion of group action or permutation representation, and who the author(s) were? Any references I have found so far on e.g. Wikipedia are for current textbooks.
I am interested to find out the original motivation for introducing the concept. That is, whether group action was first introduced in its own right as a binary operation, which is a generalisation of the action of a permutation on a set, or whether the representation of an abstract group by a permutation group was considered first, and then the action defined later as an axiomatisation. (I realise these are equivalent concepts, but I am more interested in precisely which came first - of course this may be a 'chicken and egg' situation, but I just thought I would ask.) Many thanks

Just to clarify, the 'picture' I have in mind when explaining the possible origin of the notion is this: someone looked at the behaviour of a permutation $\sigma$ acting in the natural way on an element $x$ in a set $X$ to give $\sigma(x)$, with its properties such as $\sigma_1(\sigma_2(x))=(\sigma_1\sigma_2)(x)$, and then decided to generalise this to an arbitrary group by inventing a binary operation and forcing the group elements to obey the same sort of relations as the permutations, i.e. for the 'composition of maps' property $g.(h.x) = (gh).x $.

REPLY [0 votes]: I found a link to Arthur Cayley's first paper on group theory with comments, in an answer to a different question on MSE.