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'convergence in probability is preserved under continuous transformations'
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| Title of object: |
convergence in probability is preserved under continuous transformations |
| Canonical Name: |
ConvergenceInProbabilityIsPreservedUnderContinuousTransformations |
| Type: |
Theorem |
| Created on: |
2006-09-16 10:10:13 |
| Modified on: |
2006-09-16 10:16:31 |
| Classification: |
msc:60A10 |
Preamble:
\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} |
Content:
\begin{thm}
Let $g\colon \real^k \to real^k$ be a continuous function.
If $\{ X_n \}$ are real-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, 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
\[
f_m(x) = \begin{cases}
x\,, & \norm{x} \leq m \\
\frac{mx}{\norm{x}} \,, & \norm{x} \geq m
\end{cases}
\]
Clearly, $f_m$ is continuous; in fact, it can be verified that
$f_m$ is uniformly continuous on $\real^k$.
(It 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 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}
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