TITLE: If $X_n$ converges in distribution to $X$, is it true that $\alpha_n X_n$ converges to $ \alpha X$ as well?
QUESTION [6 upvotes]: Suppose we have a sequence of non-negative random variables $(X_n)_n$ converging weakly to the random variable $X$. Let also $(\alpha_n)$ be a sequence of positive numbers converging to $\alpha>0$. I'm stuck in proving or disproving that $\alpha_k X_n$ converge in distribution to $\alpha X$. Any suggestion?

REPLY [4 votes]: A slick way to do it is to use the following fact: If $X_n$ converges in distribution to $X$ then there is a probability space with $Y_n\sim X_n$ (that is, $Y_n$ has the same distribution as $X_n$) and $Y\sim X$ such that $Y_n \to Y$ almost surely.  That is, by moving to a new probability space we can "upgrade" convergence in distribution to almost sure convergence.  (One proof is to use the Skorohod representation for the $Y_n$ and $Y$ applying the inverse CDF for these random variables for a commonly chosen uniform random variable on $[0,1]$). 
With this set up, of course $a_n Y_n \to Y$ almost surely as $n\to \infty$ and, hence, $a_n Y_n\to aY$ in distribution.  But convergence in distribution doesn't depend on the underlying probability space, so $a_n X_n \to aX$ in distribution.  
Note a version of this argument can also prove Slutsky's theorem: if $W_n\to W$ in distribution and $W$ is almost surely constant, then $W_n X_n \to W X$ in distribution.