ConvergenceInProbabilityIsPreservedUnderContinuousTransformations 2006-09-16 10:10:13 2006-09-16 10:32:23 Theorem convergence in probability \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amsthm} \usepackage{enumerate} %\usepackage{graphicx} %\usepackage{psfrag} %\usepackage{xypic} % define commands here \newcommand{\complex}{\mathbb{C}} \newcommand{\real}{\mathbb{R}} \newcommand{\rat}{\mathbb{Q}} \newcommand{\nat}{\mathbb{N}} \newcommand{\PP}{\mathbb{P}} \providecommand{\abs}[1]{\lvert#1\rvert} \providecommand{\absW}[1]{\left\lvert#1\right\rvert} \providecommand{\absB}[1]{\Bigl\lvert#1\Bigr\rvert} \providecommand{\norm}[1]{\lVert#1\rVert} \providecommand{\normW}[1]{\left\lVert#1\right\rVert} \providecommand{\normB}[1]{\Bigl\lVert#1\Bigr\rVert} \providecommand{\defnterm}[1]{\emph{#1}} \DeclareMathOperator{\D}{D} \DeclareMathOperator{\linspan}{span} \newtheorem{thm}{Theorem} \begin{thm} Let $g\colon \real^k \to \real^l$ be a continuous function. If $\{ X_n \}$ are $\real^k$-valued random variables converging to $X$ in probability, then $\{ g(X_n) \}$ converge in probability to $g(X)$ also. \begin{proof} Suppose first that $g$ is uniformly continuous. Given $\epsilon > 0$, there is $\delta > 0$ such that $\norm{g(X_n) - g(X)} < \epsilon$ whenever $\norm{X_n - X} < \delta$. Therefore, \[ \PP \bigl( \norm{g(X_n) - g(X)} \geq \epsilon \bigr) \leq \PP \bigl( \norm{X_n - X} \geq \delta \bigr) \to 0 \] as $n \to \infty$. Now suppose $g$ is not necessarily uniformly continuous on $\real^k$. But it will be uniformly continuous on any compact set $\{ x \in \real^k \colon \norm{x} \leq m \}$ for $m \geq 0$. Consequently, if $X_n$ and $X$ are bounded (by $m$), then the proof just given is applicable. Thus we attempt to reduce the general case to the case that $X_n$ and $X$ are bounded. Let \[ f_m(x) = \begin{cases} x\,, & \norm{x} \leq m \\ mx/\norm{x} \,, & \norm{x} \geq m \end{cases} \] Clearly, $f_m\colon \real^k \to \real^k$ is continuous; in fact, it can be verified that $f_m$ is uniformly continuous on $\real^k$. (This is geometrically obvious in the one-dimensional case.) Set $X_n^m = f_m(X_n)$ and $X^m = f_m(X)$, so that $X_n^m$ converge to $X^m$ in probability for each $m \geq 0$. We now show that $g(X_n)$ converge to $g(X)$ in probability by a four-step estimate. Let $\epsilon > 0$ and $\delta > 0$ be given. For any $m \geq 0$ (which we will \PMlinkescapetext{fix} later), \[ \PP\bigl( \norm{g(X_n) - g(X)} \geq \delta \bigr) \leq \PP\bigl( \norm{g(X_n^m) - g(X^m) } \geq \delta \bigr) + \PP\bigl( \norm{X_n} \geq m \bigr) + \PP\bigl( \norm{X} \geq m \bigr)\,. \] Choose $M$ such that for $m \geq M$, \[ \PP\bigl( \norm{X} \geq m \bigr) \leq \PP \bigl( \norm{X} \geq M \bigr) < \frac{\epsilon}{4}\,. \] (This is possible since $\lim_{m \to \infty} \PP \bigl( \norm{X} \geq m \bigr) = \PP \bigl( \bigcap_{m=0}^\infty \{ \norm{X} \geq m \} \bigr) = \PP(\emptyset) = 0$.) In particular, let $m = M+1$. Since $X_n^m$ converge in probability to $X^m$ and $X_n^m$, $X^m$ are bounded, $g(X_n^m)$ converge in probability to $g(X^m)$. That means for $n$ large enough, \[ \PP\bigl( \norm{g(X_n^m) - g(X^m)} \geq \delta \bigr) < \frac{\epsilon}{4}\,. \] Finally, since $\norm{X_n} \leq \norm{X_n - X} + \norm{X}$, and $X_n$ converge to $X$ in probability, we have \[ \PP\bigl( \norm{X_n} \geq m = M+1 \bigr) \leq \PP \bigl( \norm{X_n - X} \geq 1 \bigr) + \PP\bigl( \norm{X} \geq M \bigr) < \frac{\epsilon}{4} + \frac{\epsilon}{4} \] for large enough $n$. Collecting the previous inequalities together, we have \[ \PP\bigl(\norm{g(X_n) - g(X)} \geq \delta \bigr) < \epsilon \] for large enough $n$. \end{proof} \end{thm}