| Version 5 |
Version 4 |
| \begin{thm} |
\begin{thm} |
| Let $g\colon \real^k \to \real^k$ be a continuous function. |
Let $g\colon \real^k \to \real^k$ be a continuous function. |
| If $\{ X_n \}$ are real-valued random variables converging to $X$ |
If $\{ X_n \}$ are real-valued random variables converging to $X$ |
| in probability, then $\{ g(X_n) \}$ converge in probability to |
in probability, then $\{ g(X_n) \}$ converge in probability to |
| $g(X)$ also. |
$g(X)$ also. |
|
|
| \begin{proof} |
\begin{proof} |
| Suppose first that $g$ is uniformly continuous. |
Suppose first that $g$ is uniformly continuous. |
| Given $\epsilon > 0$, there is $\delta > 0$ such that |
Given $\epsilon > 0$, there is $\delta > 0$ such that |
| $\norm{g(X_n) - g(X)} < \epsilon$ whenever |
$\norm{g(X_n) - g(X)} < \epsilon$ whenever |
| $\norm{X_n - X} < \delta$. |
$\norm{X_n - X} < \delta$. |
| Therefore, |
Therefore, |
| \[ |
\[ |
| \PP \bigl( \norm{g(X_n) - g(X)} \geq \epsilon \bigr) |
\PP \bigl( \norm{g(X_n) - g(X)} \geq \epsilon \bigr) |
| \leq \PP \bigl( \norm{X_n - X} \geq \delta \bigr) \to 0 |
\leq \PP \bigl( \norm{X_n - X} \geq \delta \bigr) \to 0 |
| \] |
\] |
| as $n \to \infty$. |
as $n \to \infty$. |
|
|
| Now suppose $g$ is not necessarily uniformly continuous on $\real^k$. |
Now suppose $g$ is not necessarily uniformly continuous on $\real^k$. |
| But it will be uniformly continuous on any compact set |
But it will be uniformly continuous on any compact set |
| $\{ x \in \real^k \colon \norm{x} \leq m \}$ for $m \geq 0$. |
$\{ x \in \real^k \colon \norm{x} \leq m \}$ for $m \geq 0$. |
| Consequently, if $X_n$ and $X$ are bounded, then the proof just |
Consequently, if $X_n$ and $X$ are bounded, then the proof just |
| given is applicable. Thus we attempt to reduce the general case |
given is applicable. Thus we attempt to reduce the general case |
| to the case that $X_n$ and $X$ are bounded. |
to the case that $X_n$ and $X$ are bounded. |
|
|
| Let |
Let |
| \[ |
\[ |
| f_m(x) = \begin{cases} |
f_m(x) = \begin{cases} |
| x\,, & \norm{x} \leq m \\ |
x\,, & \norm{x} \leq m \\ |
| \frac{mx}{\norm{x}} \,, & \norm{x} \geq m |
\frac{mx}{\norm{x}} \,, & \norm{x} \geq m |
| \end{cases} |
\end{cases} |
| \] |
\] |
| Clearly, $f_m$ is continuous; in fact, it can be verified that |
Clearly, $f_m$ is continuous; in fact, it can be verified that |
| $f_m$ is uniformly continuous on $\real^k$. |
$f_m$ is uniformly continuous on $\real^k$. |
| (It is geometrically obvious in the one-dimensional case.) |
(It is geometrically obvious in the one-dimensional case.) |
|
|
| Set $X_n^m = f_m(X_n)$ and $X^m = f_m(X)$, |
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 $m \geq 0$. |
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. |
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), |
For any $m \geq 0$ (which we will fix later), |
| \[ |
\[ |
| \PP\bigl( \norm{g(X_n) - g(X)} \geq \delta \bigr) |
\PP\bigl( \norm{g(X_n) - g(X)} \geq \delta \bigr) |
| \leq \PP\bigl( \norm{g(X_n^m) - g(X^m) } \geq \delta \bigr) |
\leq \PP\bigl( \norm{g(X_n^m) - g(X^m) } \geq \delta \bigr) |
| + \PP\bigl( \norm{X_n} \geq m \bigr) |
+ \PP\bigl( \norm{X_n} \geq m \bigr) |
| + \PP\bigl( \norm{X} \geq m \bigr)\,. |
+ \PP\bigl( \norm{X} \geq m \bigr)\,. |
| \] |
\] |
|
|
| Choose $M$ such that for $m \geq M$, |
Choose $M$ such that for $m \geq M$, |
| \[ |
\[ |
| \PP\bigl( \norm{X} \geq m \bigr) \leq \PP \bigl( \norm{X} \geq M \bigr) < \frac{\epsilon}{4}\,. |
\PP\bigl( \norm{X} \geq m \bigr) \leq \PP \bigl( \norm{X} \geq M \bigr) < \frac{\epsilon}{4}\,. |
| \] |
\] |
| (This is possible since $\lim_{m \to \infty} \PP \bigl( \norm{X} \geq m \bigr) |
(This is possible since $\lim_{m \to \infty} \PP \bigl( \norm{X} \geq m \bigr) |
| = \PP \bigl( \bigcap_{m=0}^\infty \{ \norm{X} \geq m \} \bigr) = \PP(\emptyset) = 0$.) |
= \PP \bigl( \bigcap_{m=0}^\infty \{ \norm{X} \geq m \} \bigr) = \PP(\emptyset) = 0$.) |
|
|
| In particular, let $m = M+1$. |
In particular, let $m = M+1$. |
| Since $X_n^m$ converge in probability to $X^m$ |
Since $X_n^m$ converge in probability to $X^m$ |
| and $X_n^m$, $X^m$ are bounded, |
and $X_n^m$, $X^m$ are bounded, |
| $g(X_n^m)$ converge in probability to $g(X^m)$. |
$g(X_n^m)$ converge in probability to $g(X^m)$. |
| That means for $n$ large enough, |
That means for $n$ large enough, |
| \[ |
\[ |
| \PP\bigl( \norm{g(X_n^m) - g(X^m)} \geq \delta \bigr) < \frac{\epsilon}{4}\,. |
\PP\bigl( \norm{g(X_n^m) - g(X^m)} \geq \delta \bigr) < \frac{\epsilon}{4}\,. |
| \] |
\] |
|
|
| Finally, since $\norm{X_n} \leq \norm{X_n - X} + \norm{X}$, |
Finally, since $\norm{X_n} \leq \norm{X_n - X} + \norm{X}$, |
| and $X_n$ converge to $X$ in probability, |
and $X_n$ converge to $X$ in probability, |
| we have |
we have |
| \[ |
\[ |
| \PP\bigl( \norm{X_n} \geq m = M+1 \bigr) |
\PP\bigl( \norm{X_n} \geq m = M+1 \bigr) |
| \leq \PP \bigl( \norm{X_n - X} \geq 1 \bigr) + \PP\bigl( \norm{X} \geq M \bigr) |
\leq \PP \bigl( \norm{X_n - X} \geq 1 \bigr) + \PP\bigl( \norm{X} \geq M \bigr) |
| < \frac{\epsilon}{4} + \frac{\epsilon}{4} |
< \frac{\epsilon}{4} + \frac{\epsilon}{4} |
| \] |
\] |
| for large enough $n$. |
for large enough $n$. |
|
|
| Collecting the previous inequalities together, |
Collecting the previous inequalities together, |
| we have |
we have |
| \[ |
\[ |
| \PP\bigl(\norm{g(X_n) - g(X)} \geq \delta \bigr) < \epsilon |
\PP\bigl(\norm{g(X_n) - g(X)} \geq \delta \bigr) < \epsilon |
| \] |
\] |
| for large enough $n$. |
for large enough $n$. |
| \end{proof} |
\end{proof} |
|
|
| \end{thm} |
\end{thm} |
