# $X_n \stackrel{d}{\to} X$, $c_n \to c$ $\implies c_n \cdot X_n \stackrel{d}{\to} c \cdot X$

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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I think I know the source of the question... –  Did Mar 1 '13 at 9:13
@Did Probably you are correct with your guess. :) –  saz Mar 1 '13 at 9:17
Which definition of weak convergence do you want to use? –  Davide Giraudo Mar 2 '13 at 11:16
@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 $n \to \infty$. –  saz Mar 2 '13 at 12:40

The quickest way might be 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.
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. –  Bravo Mar 3 '13 at 19:37
@Shyam For all continuity points of the CDF of the limit $X$, yes. –  Did Mar 4 '13 at 7:08