If $X_n$ converges in distribution to $X$, is it true that $\alpha_n X_n$ converges to $ \alpha X$ as well? 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?
 A: 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.
A: You just need to check that the cumulative distribution functions converge on points of mass zero. That is, you need to prove that: $$\mathbb{P}(\alpha_n X_n \le t) \rightarrow \mathbb{P}(\alpha X \le t),$$ for any $t$ such that ${P}(\alpha X =t) = 0.$ Now you divide by $\alpha,$ and since you know that: $$\mathbb{P}( X_n \le \frac{t}{\alpha}) \rightarrow \mathbb{P}(X \le \frac{t}{\alpha})$$ the result follows by showing that $$|\mathbb{P}( X_n \le \frac{t}{\alpha_n})-\mathbb{P}( X_n \le \frac{t}{\alpha})| \rightarrow 0$$
To see this just observe that (by assuming $\alpha_n \le \alpha$, and $n$ sufficiently large) what is written on the LHS is equal to: $$\mathbb{P}( X_n \in [\frac{t}{\alpha_n},\frac{t}{\alpha} ]) \le \mathbb{P}( X_n \in [\frac{t}{\alpha} - \epsilon,\frac{t}{\alpha} ]) \rightarrow\mathbb{P}( X \in [\frac{t}{\alpha} - \epsilon,\frac{t}{\alpha} ])$$ And the last term converges to zero by passing on the limit over $\epsilon.$
Please note that these last inequalities are just a trick to dispose of the limit of $X_n$ and $\alpha_n$ separately.
