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Let $X_n$, $X$ random variables on a probability space $(\Omega,\mathcal{A},\mathbb{P})$ and $(c_n)_n \subseteq \mathbb{R}$, $c \in \mathbb{R}$ such that $c_n \to c$ and $X_n \stackrel{d}{\to} X$. Then $c_n \cdot X_n \stackrel{d}{\to} c \cdot X$.

Here is my proof: By Lévy's continuity theorem it suffices to show that $$\forall \xi: \quad \hat{\mu}_n(c_n \cdot \xi) \to \hat{\mu}(c \cdot \xi) \qquad (n \to \infty)$$ where $\hat{\mu}_n$ (resp. $\hat{\mu}$) denotes the characteristic function of $X_n$ (resp. $X$). Using the tightness of the distributions one can show that $\{\hat{\mu}_n; n \in \mathbb{N}\}$ is uniformly equicontinuous. We have $$|\hat{\mu}_n(c_n \cdot \xi) - \hat{\mu}(c \cdot \xi)| \leq |\hat{\mu}_n(c_n \cdot \xi)- \hat{\mu}_n(c \cdot \xi)| + |\hat{\mu}_n(c \cdot \xi)- \hat{\mu}(c \cdot \xi)|$$ The first addend converges to zero as $n \to \infty$ by the uniform equicontinuity and the second one converges to zero since $X_n \stackrel{d}{\to} X$.

I was wondering whether there is an easier proof - my own proof seems rather like overkill to me. Thanks!

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  • $\begingroup$ I think I know the source of the question... $\endgroup$ – Did Mar 1 '13 at 9:13
  • $\begingroup$ @Did Probably you are correct with your guess. :) $\endgroup$ – saz Mar 1 '13 at 9:17
  • $\begingroup$ Which definition of weak convergence do you want to use? $\endgroup$ – Davide Giraudo Mar 2 '13 at 11:16
  • $\begingroup$ @DavideGiraudo $X_n \stackrel{d}{\to X} :\Leftrightarrow \forall f \in C_b(\mathbb{R}): \int f(X_n) \, d\mathbb{P} \to \int f(X) \, d\mathbb{P}$ as $n \to \infty$. $\endgroup$ – saz Mar 2 '13 at 12:40
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A quick way to prove this is to use Skorokhod's representation theorem, which asserts that $X_n\to X$ in distribution if and only if there exists some random variables $X'$ and $X'_n$, possibly defined on another probability space, such that each $X'_n$ is distributed as $X_n$, $X'$ is distributed as $X$, and $X'_n\to X'$ almost surely.

Now, if $c_n\to c$, then $c_nX'_n\to cX'$ almost surely hence $c_nX'_n\to cX'$ in distribution. But each $c_nX'_n$ coincides with $c_nX_n$ in distribution and $cX'$ coincides with $cX$ in distribution, hence $c_nX_n\to cX$ in distribution.

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  • $\begingroup$ Thanks, did, nice proof. I didn't know this theorem so far. $\endgroup$ – saz Mar 3 '13 at 17:22
  • $\begingroup$ You are welcome. Yes, this is a nice representation result. $\endgroup$ – Did Mar 3 '13 at 17:27
  • $\begingroup$ Thanks @did, this is well beyond my ken! BTW, in your comment (on my deleted answer), re the first problem you mention: is it fine if I say all "continuity points" of $x$ instead of $\forall x$? Obviously I found no easy way out of the second problem. $\endgroup$ – Bravo Mar 3 '13 at 19:37
  • $\begingroup$ @Shyam For all continuity points of the CDF of the limit $X$, yes. $\endgroup$ – Did Mar 4 '13 at 7:08
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A proof that uses more basic techniques for the case $c\neq 0$ is this:

Suppose $X_n\rightarrow X$ in distribution, and suppose $c_n\rightarrow c$ (where $X_n, X$ are random variables and $c_n,c$ real numbers with $c\neq 0$). We prove that $c_nX_n\rightarrow cX$ in distribution.

Proof: Without loss of generality assume $c>0$. The CDFs of $X$ and $cX$ are nondecreasing and hence have at most a countably infinite number of discontinuities. Fix $x \in \mathbb{R}$ such that the CDF of $cX$ is continuous at $x$. We want to show $\lim_{n\rightarrow\infty} P[c_nX_n\leq x] = P[cX\leq x]$.

1) Fix $\epsilon>0$ such that the CDF of $X$ is continuous at $x/c + \epsilon$ (since there are at most a countably infinite number of discontinuities, we can find arbitrarily small values of $\epsilon>0$ for which this holds). There is an index $N$ such that $$ c_n>0, \frac{x}{c_n} \leq \frac{x}{c} + \epsilon \quad \forall n \geq N$$ So for all $n \geq N$ we have \begin{align} P[c_nX_n \leq x] &= P[X_n \leq x/c_n]\\ &\leq P[X_n \leq x/c + \epsilon] \end{align} Taking a $\limsup$ as $n\rightarrow\infty$ and using the fact that $X_n\rightarrow X$ in distribution and the CDF of $X$ is continuous at $x/c+\epsilon$ gives $$ \limsup_{n\rightarrow\infty} P[c_nX_n\leq x] \leq P[X \leq x/c + \epsilon]$$ This holds for arbitrarily small $\epsilon>0$ and since the CDF of $X$ is continuous from the right we get: $$ \boxed{\limsup_{n\rightarrow\infty} P[c_nX_n\leq x] \leq P[X \leq x/c] = P[cX \leq x]}$$

2) Fix $\epsilon>0$ such that the CDF of $X$ is continuous at $(x-\epsilon)/c$ (again, since discontinuities are countable, there are arbitrarily small $\epsilon>0$ for which this holds). There is an index $N$ such that $$ c_n >0 , \frac{x-\epsilon}{c} \leq \frac{x}{c_n} \quad, \forall n \geq N $$ So for $n \geq N$ we have \begin{align} P[cX \leq x - \epsilon] &= P[X \leq \frac{x-\epsilon}{c}]\\ &\overset{(a)}{=}\liminf_{n\rightarrow\infty} P[X_n \leq \frac{x-\epsilon}{c}]\\ &\leq \liminf_{n\rightarrow\infty} P[X_n \leq \frac{x}{c_n}]\\ &= \liminf_{n\rightarrow\infty} P[c_nX_n \leq x] \end{align} where (a) holds because $X_n\rightarrow X$ in distribution and the CDF of $X$ is continuous at $(x-\epsilon)/c$. This holds for arbitrarily small $\epsilon>0$ and since the CDF of $cX$ is continuous at $x$ we get $$ \boxed{P[cX\leq x] \leq \liminf_{n\rightarrow\infty} P[c_nX_n\leq x]} $$ The two boxed inequalities imply the result. $\Box$

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