TITLE: A topological space is called a US-space provided that
QUESTION [1 upvotes]: A topological space is called a US-space provided that
each convergent sequence has a unique limit.

Each Fréchet $US$-space $X$ is a $KC$-space.

Proof. Suppose that $x \in K$ where $K$ is a compact subspace of $X$. Because
X is a Fréchet space, there is a sequence $(k_n)_{n∈N}$ of points of $K$ converging to
$x$. Since $K$ is compact, that sequence has a cluster point $c$ in $K$. There is a subsequence of $(k_n)_{n∈N}$ converging to $c$ . Hence $x = c \in K$, because $X$ is a $US$-space. so $K$ is closed and then $X$ is a $KC$-space.

Why "there is a subsequence of $(k_n)_{n∈N}$ converging to $c$"? How does it build?

REPLY [1 votes]: A US space is also $T_1$ since if $x$ is a point with another point $y$ in its closure then the constant sequence at $x$ converges to both $x$ and $y$, so by uniqueness of limits $x$ must equal $y.$ So points are closed.
Now, in a Fréchet-Urysohn space with the $T_1$-property a cluster point of a sequence is also a limit point of some subsequence (see my answer to your previous question).


TITLE: In how many ways can 5 identical balls be placed in a (3x3) grid such that each row contains atleast one ball?
QUESTION [1 upvotes]: I tried two methods here. First, the usual one of checking different arrangements with a 3,1,1 distribution and the other with a 2,2,1 distribution and calculating their respective combinations. Works out to a total of 108. Again a long method. So wanted a shorter one.
Then i tried the "Stars & Bars" approach. I first gave each row a ball so am left with 2 balls and 6 spaces to fill. But somehow I am unable to arrive at the right answer. Could someone help point out the flaw in reasoning? Thanks.

REPLY [0 votes]: First assign $1$ ball to each row and each of these three balls can take $3$ positions in each respective row.
So no of ways of assigning $1$ ball in each row is $3\cdot 3\cdot 3.$
 Now we are left with $2$ balls (as $3$ have been assigned) and $6$ places (as $3$ places already filled) so the number of ways in which $2$ balls arranged in $6$ places is $6P2$
Answer= $3^3 \cdot 6P2$


TITLE: conditions for long geodesics without self-intersections
QUESTION [10 upvotes]: Consider a Riemannian manifold $\Sigma$  of dimension two homeomorphic to a torus. When is there a non-closed geodesic on $\Sigma$ which does not intersect itself — are there reasonable necessary or sufficient conditions for this?
(This is a reference request. My guess is that this question does not have an amazing answer, but I am curious if it was considered in the literature. By the way, a flat torus obviously has this property, but I am interested in a more general setting.)

REPLY [8 votes]: This is more of a comment intended to provide examples.  One can construct a reasonably large class of examples of metrics on $T^2$ satisfying your property. I believe moreover that this class of metrics (described below) is stable with respect to small $C^{\infty}$ perturbation.
Infinite geodesics on surfaces without self intersections are closely connected to geodesic lamintations. Now, hyperbolic surfaces have plenty of geodesic laminations. So we can do the following. First take a hyperbolic metric on $T^2$ with a cusp and then smoothen the cusp  close to infinity. Then there still will be a geodesic lamination in the hyperbolic part of $T^2$. For a concrete example, look at Fig 11 here: (Geodesic lamination on surfaces by Bonahon)
https://www-bcf.usc.edu/~fbonahon/Research/Preprints/StonyBrookProc.ps

Fig.11. A geodesic lamination on the punctured torus.

Or one can use a different construction, identifying a part of $T^2$ with a part of hyperbolic surface of genus $2$ as on Fig 5 in the same text. In this case one will get in infinite geodesic in $T^2$ that accumulates to two closed geodesics on both ends.

I believe that both examples are stable under small perturbation of the metric.


text:
\section{Objective convergence rates}\label{sec:DRSconvergence}
In this section we will prove ergodic and nonergodic convergence rates of relaxed PRS when $f$ and $g$ are closed, proper, and convex functions that are possibly nonsmooth.

To ease notational memory, we note that the reader may assume that $\lambda_k = (1/2)$ for all $k \geq 0$.  This simplification implies that  $\Lambda_k = (1/2) (k+1)$, and $\tau_k = \lambda_k(1-\lambda_k)= (1/4)$ for all $k \geq 0$.

Throughout this section the point $z^\ast$  denotes an arbitrary fixed point of $\TPRS$, and we  define a minimizer of $f+g$ by the formula (Lemma~\ref{lem:PRSoptimality}):
\begin{align*}
x^\ast = \prox_{\gamma g}(z^\ast).
\end{align*}

The constant $(1/\gamma)\|z^\ast - x^\ast\|$ appears in the bounds of this section. This term is independent of $\gamma$: For any fixed point $z^\ast$ of $\TPRS$, the point $x^\ast = \prox_{\gamma g}(z^\ast)$ is a minimizer and $z^\ast - \prox_{\gamma g}(z^\ast) = \gamma \tnabla g(x^\ast) \in \gamma \partial g(x^\ast)$. Conversely, if $x^\ast \in \zer(\partial f + \partial g)$ and $\tnabla g(x^\ast) \in (-\partial f(x^\ast)) \cap \partial g(x^\ast)$, then $z^\ast = x^\ast + \gamma \tnabla g(x^\ast)$ is a fixed point. {Note that in all of our bounds, we can always replace $(1/\gamma)\|z^\ast - x^\ast\| = \|\tnabla g(x^\ast)\|$ by the infimum $\inf_{z^\ast \in \Fix(\TPRS)} (1/\gamma)\|z^\ast - x^\ast\|$ (although the infimum might not be attained).}

\subsection{Ergodic convergence rates}\label{sec:ergodic}

In this section, we analyze the ergodic convergence of relaxed PRS. The proof follows the telescoping property of the upper and lower fundamental inequalities and an application of Jensen's inequality.

\begin{theorem}[Ergodic convergence of relaxed PRS]\label{thm:drsergodic}
For all $k \geq 0$, let $\lambda_k \in (0, 1]$. Then we have the following convergence rate
\begin{align*}
- \frac{2\|z^0 - z^\ast\|\|z^\ast - x^\ast\|}{\gamma\Lambda_k} \leq f(\overline{x}_f^k) + g(\overline{x}_g^k) - f(x^\ast) - g(x^\ast) \leq \frac{1}{4\gamma\Lambda_k}\|z^0 - x^\ast\|^2.
\end{align*}
In addition, the following feasibility bound holds:
\begin{align}\label{thm:drsergodic:eq:feasibility}
\|\overline{x}_g^k - \overline{x}_f^k\|& \leq \frac{2\|z^0 - z^\ast\|}{\Lambda_k}.
\end{align}
\end{theorem}
\begin{proof}

Equation~\eqref{thm:drsergodic:eq:feasibility} follows directly from Theorem~\ref{eq:ergodicFPR} because $(z^j)_{j \geq 0}$ is Fej{\'e}r monotone with respect to $\Fix(T)$ and for all $k \geq 0$, we have $z^{k+1} - z^k = \lambda_k(x_f^k - x_g^k)$.

Recall the upper fundamental inequality from Proposition~\ref{prop:DRSupper} :
\begin{align}\label{thm:drsergodic:eq:upper}
4\gamma\lambda_k(f(x_f^k) + g(x_g^k) - f(x^\ast) - g(x^\ast) ) &\leq \|z^{k} - x^\ast\|^2 - \|z^{k+1} - x^\ast\|^2 + \left(1 - \frac{1}{\lambda_k}\right)\|z^{k+1} - z^k\|^2.
\end{align}
Because $\lambda_k \leq 1$, it follows that $(1-(1/\lambda_k)) \leq 0$.  Thus, we sum Equation~(\ref{thm:drsergodic:eq:upper}) from $i = 0$ to $k$, divide by $\Lambda_k$, and apply Jensen's inequality to get
\begin{align*}
\frac{1}{4\gamma\Lambda_k}(\|z^0 - x^\ast\|^2 - \|z^{k+1} - x^\ast\|^2) &\geq \frac{1}{\Lambda_k}\sum_{i=0}^k\lambda_i(f(x_f^i) + g(x_g^i) - f(x^\ast) - g(x^\ast)) \\
&\geq f(\overline{x}_f^k) + g(\overline{x}_g^k) - f(x^\ast) - g(x^\ast) .
\end{align*}

The lower bound is a consequence of the fundamental lower inequality and Equation~\eqref{thm:drsergodic:eq:feasibility}
\begin{align*}
f(\overline{x}_f^k) + g(\overline{x}_g^k) - f(x^\ast) - g(x^\ast) &\stackrel{\eqref{prop:DRSlower:eq:main}}{\geq} \frac{1}{\gamma}\dotp{\overline{x}_g^k - \overline{x}_f^k, z^\ast - x^\ast}  \numberthis \label{thm:drsergodic:eq:upper3} \stackrel{\eqref{thm:drsergodic:eq:feasibility}}{\geq} - \frac{2\|z^0 - z^\ast\|\|z^\ast - x^\ast\|}{\gamma\Lambda_k}. \qquad \qed
\end{align*}
\end{proof}

In general, $x_f^k \notin\dom(g)$ and $x_g^k \notin \dom(f)$, so we cannot evaluate $g$ at $x_f^k$ or $f$ at $x_g^k$. However, the conclusion of Theorem~\ref{thm:drsergodic} can be improved if $f$ or $g$ is Lipschitz continuous. The following proposition gives a sufficient condition for Lipschitz continuity on a ball:

\begin{proposition}[Lipschitz continuity on a ball]\label{prop:lipschitzCont}
Suppose that $f : \cH \rightarrow (-\infty, \infty]$ is proper and convex.  Let $\rho > 0$ and let $x_0 \in \cH$.  If $\delta = \sup_{x, y \in B(x_0, 2\rho)} |f(x) - f(y)| < \infty$, then $f$ is $({\delta}/{\rho})$-Lipschitz on $B(x_0, \rho)$.
\end{proposition}
\begin{proof}
See \cite[Proposition 8.28]{bauschke2011convex}.
\qed\end{proof}

To use this fact, we need to show that the sequences $(x_f^j)_{j \geq 0}$, and $(x_g^j)_{j \geq 0}$ are bounded.  Recall that $x_g^s = \prox_{\gamma g}(z^s)$ and $x_f^s = \prox_{\gamma f}(\refl_{\gamma g}(z^s))$, for $s \in \{\ast, k\}$. Proximal and reflection maps are nonexpansive, so we have the following simple bound:
\begin{align*}
\max\{ \|x_f^k - x^\ast\|, \|x_g^k - x^\ast\|\} &\leq \|z^k - z^\ast\| \leq \|z^0 - z^\ast\|.
\end{align*}
Thus, $(x_f^j)_{j \geq 0}, (x_g^j)_{j \geq 0} \subseteq \overline{B(x^\ast, \|z^0 - z^\ast\|)}.$ {By the convexity of the closed ball, we also have $(\overline{x}_f^j)_{j \geq 0}, (\overline{x}_g^j)_{j \geq 0} \subseteq \overline{B(x^\ast, \|z^0 - z^\ast\|)}.$}

\begin{corollary}[Ergodic convergence with single Lipschitz function]\label{cor:drsergodiclipschitz}
Let the notation be as in Theorem~\ref{thm:drsergodic}. Suppose that $f$ (respectively $g$) is $L$-Lipschitz continuous on $\overline{B(x^\ast, \|z^0 - z^\ast\|)}$, and let $x^k = x_g^k$ (respectively $x^k = x_f^k$). Then the following convergence rate holds
\begin{align*}
0 &\leq f(\overline{x}^k) + g(\overline{x}^k) - f(x^\ast) - g(x^\ast) \leq  \frac{1}{4\gamma\Lambda_k}\|z^0 - x^\ast\|^2 + \frac{2L\|z^0 - z^\ast\|}{\Lambda_k}.
\end{align*}
\end{corollary}
\begin{proof}
From Equation~(\ref{thm:drsergodic:eq:feasibility}),  we have  $\|\overline{x}_g^k - \overline{x}_f^k\| \leq (2/\Lambda_k)\|z^0 - z^\ast\|$. In addition, $(x_f^j)_{j \geq 0}, (x_g^j)_{j \geq 0} \subseteq \overline{B(x^\ast, \|z^0 - z^\ast\|)}$. Thus, it follows that
\begin{align*}
0 \leq f(\overline{x}^k) + g(\overline{x}^k) - f(x^\ast) - g(x^\ast) &\leq f(\overline{x}_f^k) + g(\overline{x}_g^k) - f(x^\ast) - g(x^\ast)  + L\|\overline{x}_f^k - \overline{x}_g^k\|\\
&\stackrel{\eqref{thm:drsergodic:eq:feasibility}}{\leq} f(\overline{x}_f^k) + g(\overline{x}_g^k) - f(x^\ast) - g(x^\ast)  + \frac{2L \|z^0 - z^\ast\|}{\Lambda_k}.
\end{align*}
The upper bound follows from this equation and Theorem~\ref{thm:drsergodic}.
\qed\end{proof}

\subsection{Nonergodic convergence rates}\label{sec:DRSnonergodic}

In this section, we prove the nonergodic convergence rate of the Algorithm~\ref{alg:DRS} whenever $\underline{\tau}  := \inf_{j \geq 0} \tau_j > 0$.  The proof uses Theorem~\ref{prop:averagedconvergence} to bound the fundamental inequalities in Propositions~\ref{prop:DRSupper} and~\ref{prop:DRSlower}.

\begin{theorem}[Nonergodic convergence of relaxed PRS]\label{thm:drsnonergodic}
 For all $k \geq 0$, let $\lambda_k \in (0, 1)$. Suppose that $\underline{\tau} := \inf_{j \geq 0} \lambda_k(1-\lambda_k) > 0$. Then we have the convergence rates:
 \begin{enumerate}
\item \label{thm:drsnonergodic:part:general} In general, we have the bounds:
\begin{align*}
-\frac{\|z^0 - z^\ast\|\|z^\ast - x^\ast\|}{2\gamma \sqrt{\underline{\tau}(k+1)}} \leq f(x_f^k) + g(x_g^k) - f(x^\ast) - g(x^\ast) \leq \frac{(\|z^0 - z^\ast\| + \|z^\ast - x^\ast\|)\|z^0 - z^\ast\|}{2\gamma \sqrt{\underline{\tau}(k+1)}}
\end{align*}
and $|f(x_f^k) + g(x_g^k) - f(x^\ast) - g(x^\ast) | = o\left({1}/{\sqrt{k+1}}\right).$
\item \label{thm:drsnonergodic:part:1d} If $\cH = \vR$  and $\lambda_k \equiv {1}/{2}$, then for all $k \geq 0$,
\begin{align*}
\frac{\|z^0 - z^\ast\|\|z^\ast - x^\ast\|}{\sqrt{2}\gamma (k+1)} \leq f(x_f^{k+1}) + g(x_g^{k+1}) - f(x^\ast) - g(x^\ast) \leq \frac{(\|z^0 - z^\ast\| + \|z^\ast - x^\ast\|)\|z^0 - z^\ast\|}{\sqrt{2}\gamma (k+1)}
\end{align*}
and $|f(x_f^{k+1}) + g(x_g^{k+1}) - f(x^\ast) - g(x^\ast) | = o\left({1}/({k+1})\right).$
\end{enumerate}
\end{theorem}
\begin{proof}
We prove Part~\ref{thm:drsnonergodic:part:general} first. For all $\lambda \in [0, 1]$, let $z_{\lambda} = (\TPRS)_{\lambda}(z^k)$. Evaluate the upper inequality in Equation~(\ref{prop:DRSupper:eq:main}) at $x = x^\ast$ to get
\begin{align*}
4\gamma\lambda(f(x_f^k) + g(x_g^k) &- f(x^\ast) - g(x^\ast) ) \leq \|z^{k} - x^\ast\|^2 - \|z_\lambda - x^\ast\|^2 + \left(1 - \frac{1}{\lambda}\right)\|z_\lambda - z^k\|^2.
\end{align*}
Recall the following identity:
\begin{align*}
\|z^{k} - x^\ast\|^2 - \|z_\lambda - x^\ast\|^2 - \|z_\lambda - z^k\|^2 &= 2\dotp{z_\lambda - x^\ast, z^k - z_\lambda}.
\end{align*}
By the triangle inequality, because $\|z_\lambda- z^\ast\| \leq \|z^k - z^\ast\|$,  and because $(\|z^j - z^\ast\|)_{j \geq 0}$ is monotonically nonincreasing (Corollary~\ref{cor:DRSaveragedconvergence}), it follows that
\begin{align*}
\|z_\lambda - x^\ast\| &\leq \|z_\lambda - z^\ast\| + \|z^\ast - x^\ast\| \leq \|z^0 - z^\ast\| + \|z^\ast - x^\ast\|. \numberthis \label{thm:drsnonergodic:eq:triangle}
\end{align*}
Thus, we have the bound:
\begin{align*}
f(x_f^k) + g(x_g^k) - f(x^\ast) - g(x^\ast) &\leq \inf_{\lambda \in [0, 1]}\frac{1}{4\gamma \lambda}\left(2\dotp{z_\lambda - x^\ast, z^k - z_\lambda} +2\left(1 - \frac{1}{2\lambda}\right) \|z_\lambda - z^k\|^2\right) \\
&\leq \frac{1}{\gamma }\|z_{1/2} - x^\ast\|\|z^k - z_{1/2}\|\\
&\stackrel{\eqref{thm:drsnonergodic:eq:triangle}}{\leq}  \frac{1}{\gamma }\left(\|z^0 - z^\ast\|+ \|z^\ast - x^\ast\|\right)\|z^k - z_{1/2}\|  \numberthis\label{eq:drsnonergodic:eq:oupper} \\
&\stackrel{\eqref{cor:DRSaveragedconvergence:eq:main}}{\leq}\frac{(\|z^0 - z^\ast\| + \|z^\ast - x^\ast\|)\|z^0 - z^\ast\|}{2\gamma \sqrt{\underline{\tau}(k+1)}}.
\end{align*}

The lower bound follows from the identity $x_g^k - x_f^k = ({1}/{2\lambda_k})(z^k - z^{k+1})$ and the fundamental lower inequality in Equation~(\ref{prop:DRSlower:eq:main}):
\begin{align*}
f(x_f^k) + g(x_g^k) - f(x^\ast) - g(x^\ast) \geq \frac{1}{2\gamma \lambda_k}\dotp{z^{k} - z^{k+1}, z^\ast - x^\ast}  &\geq -\frac{\|z^{k+1} - z^k\|\|z^\ast - x^\ast\|}{2\gamma\lambda_k}  \numberthis\label{eq:drsnonergodic:eq:olower} \\
&\stackrel{\eqref{cor:DRSaveragedconvergence:eq:main}}{\geq} -\frac{\|z^0 - z^\ast\|\|z^\ast - x^\ast\|}{2\gamma \sqrt{\underline{\tau}(k+1)}}.
\end{align*}

Finally, the $o(1/\sqrt{k+1})$ convergence rate follows from Equations~(\ref{eq:drsnonergodic:eq:oupper}) and~(\ref{eq:drsnonergodic:eq:olower}) combined with Corollary~\ref{cor:DRSaveragedconvergence} {because each upper bound is of the form (bounded quantity)$\times\sqrt{\text{FPR}}$, and $\sqrt{\text{FPR}}$ has rate $o(1/\sqrt{k+1})$.}

Part~\ref{thm:drsnonergodic:part:1d} follows by the same analysis but uses Theorem~\ref{thm:PPAconvergence} to estimate the FPR convergence rate.
\qed\end{proof}

Whenever $f$ or $g$ is Lipschitz, we can compute the convergence rate of $f + g$ evaluated at the same point. The following theorem is analogous to Corollary~\ref{cor:drsergodiclipschitz} in the ergodic case.  The proof essentially follows by combining the nonergodic convergence rate in Theorem~\ref{thm:drsnonergodic} with the convergence rate of $\|x_f^k - x_g^k\| = (1/\lambda_k)\|z^{k+1} - z^k\|$ deduced in Corollary~\ref{cor:DRSaveragedconvergence}.

\begin{corollary}[Nonergodic convergence with Lipschitz assumption]\label{cor:drsnonergodiclipschitz}
Let the notation be as in Theorem~\ref{thm:drsnonergodic}. Suppose that $f$ (respectively $g$) is $L$-Lipschitz continuous on $\overline{B(x^\ast, \|z^0 - z^\ast\|)}$, and let $x^k = x_g^k$ (respectively $x^k = x_f^k$).  Then we have the convergence rates of the nonnegative term:
\begin{enumerate}
\item \label{cor:drsnonergodiclipschitz:part:general} In general, we have the bounds:
\begin{align*}
0 \leq f(x^k) + g(x^k) &- f(x^\ast) - g(x^\ast) \leq \frac{\left(\|z^0 - z^\ast\| + \|z^\ast - x^\ast\| + \gamma L\right)\|z^0 - z^\ast\|}{2\gamma \sqrt{\underline{\tau}(k+1)}}
\end{align*}
and $ f(x^k) + g(x^k) - f(x^\ast) - g(x^\ast) = o\left({1}/{\sqrt{k+1}}\right).$
\item \label{cor:drsnonergodiclipschitz:part:1d}If $\cH = \vR$  and $\lambda_k \equiv {1}/{2}$, then for all $k \geq 0$,
\begin{align*}
0 \leq f(x^{k+1}) + g(x^{k+1}) &- f(x^\ast) - g(x^\ast) \leq \frac{\left(\|z^0 - z^\ast\| + \|z^\ast - x^\ast\| + \gamma L\right)\|z^0 - z^\ast\|}{\sqrt{2}\gamma (k+1)} \end{align*}
and $f(x^{k+1}) + g(x^{k+1}) - f(x^\ast) - g(x^\ast) = o\left({1}/({k+1})\right).$
\end{enumerate}
\end{corollary}
\begin{proof}
We prove Part~\ref{cor:drsnonergodiclipschitz:part:general} first. First recall that $\|x_g^k - x_f^k\| = (1/(2\lambda_k))\|z^{k+1} - z^k\|$. In addition, $(x_f^j)_{j \geq 0}, (x_g^j)_{j \geq 0} \subseteq \overline{B(x^\ast, \|z^0 - z^\ast\|)}$ (See Section~\eqref{sec:ergodic}). Thus, it follows that
\begin{align*}
f(x^k) + g(x^k) - f(x^\ast) - g(x^\ast) &\leq f(x_f^k) + g(x_g^k) - f(x^\ast) - g(x^\ast)  + L\|x_f^k - x_g^k\|\\
&= f(x_f^k) + g(x_g^k) - f(x^\ast) - g(x^\ast)  + \frac{L\|z^{k+1} - z^k\| }{2\lambda_k} \numberthis \label{cor:drsnonergodiclipschitz:oterm}\\
&\stackrel{(\ref{cor:DRSaveragedconvergence:eq:main})}{\leq}f (x_f^k) + g(x_g^k) - f(x^\ast) - g(x^\ast)  + \frac{\gamma L\|z^0 - z^\ast\|}{2\gamma  \sqrt{\underline{\tau}(k+1)}}. \numberthis \label{cor:drsnonergodiclipschitz:Oterm}
\end{align*}
Therefore, the upper bound follows from Theorem~\ref{thm:drsnonergodic} and Equation~(\ref{cor:drsnonergodiclipschitz:Oterm}). In addition, the $o(1/\sqrt{k+1})$ bound follows from Theorem~\ref{thm:drsnonergodic} combined with Equation~(\ref{cor:drsnonergodiclipschitz:oterm}) and Corollary~\ref{cor:DRSaveragedconvergence} {because each upper bound is of the form (bounded quantity)$\times\sqrt{\text{FPR}}$, and $\sqrt{\text{FPR}}$ has rate $o(1/\sqrt{k+1})$.}

Part~\ref{cor:drsnonergodiclipschitz:part:1d} follows by the same analysis, but uses Theorem~\ref{thm:PPAconvergence} to estimate the FPR convergence rate.
\qed\end{proof}

config: arxiv


TITLE: A function having poles at a sequence that has a limit point
QUESTION [1 upvotes]: This question was asked in my complex analysis end term exam which I got from senior and I was unable to solve it.

If f(z) has poles at  sequence of points ${z_n}$ and ${z_n} \to {z_0}$ , then prove that f(z) doesn't have pole at $z =z_0$.

I don't have any intuition on which result should I use.
It's my humble request to you to kindly tell on how should I approach it's proof.
Thanks!!

REPLY [3 votes]: False statement in case $z_n=z_0$ for all $n$. True if this case is excluded.
A pole is an isolated singularity. This means if $f$ has pole at $z_0$ it must be analytic throughout some disk containing $z_0$ except for the point $z_0$. So the conclusion follows just by definition!.


TITLE: Rayleigh and Exponential distributions?
QUESTION [6 upvotes]: If $X$ is Rayleigh distributed random variable. What is the distribution of $|X|^2$?
If $X$ is Exponential distributed random variable. What is the distribution of $|X|^2$?

REPLY [3 votes]: Many thanks to André Nicolas. I know now that $|X|^2$ is Exponential when $X$ is Rayleigh. Based on André Nicolas's answer, I will deal with the Exponential case. Please correct me if I did something wrong.
Let $X$ be Exponential with parameter $\lambda$. Then $X$ has density function $\lambda e^{-\lambda x}$ for $x\ge 0$, and $0$ elsewhere. Let $Y=X^2$. We find the density function $f_Y(y)$ of $Y$.
Let us find the cdf $F_Y(y)$ of $Y$. For $y\gt 0$, we have
$$F_Y(y)=\Pr(Y\le y)=\Pr(X^2\le y)=\Pr(X\le \sqrt{y})=\int_0^{\sqrt{y}} \lambda e^{-\lambda x}\,dx.$$
Now integrate. We find that for $y\gt 0$,
$$F_Y(y)=1-e^{-\lambda\sqrt{y}}.$$
This is I do not know which distribution it is.