Convergence of distribution implies convergence of a real sequence I just came across the statement:
If $(a_{n}),(b_{n})$ are real sequences and $X_{n}, Y,Z$ are random variables such that
$ P(X_{n} \le a_{n}x) \rightarrow P(Y\le x)$ and $P(X_{n} \le b_{n}x) \rightarrow P(Z\le x)$ for every $x \in \mathbb{R}$, then $\frac{a_{n}}{b_{n}}$ converges.
This is not obvious to me. Perhaps someone can help.
 A: You need to assume that the cdfs of $Y$ and $Z$ are continuous. Take $X_n\equiv -1$ and let $a_n=n^2$ and $b_n=n$. Then $\mathsf{P}(X_n/n\le x)=\to 1_{[0,\infty)}(x)$ and $\mathsf{P}(X_n/n^2\le x)\to 1_{[0,\infty)}(x)$. However, $a_n/b_n$ doesn't converge in $\mathbb{R}$. Alternatively, take $a_n=n(1+(-1)^n\epsilon)$ and $b_n=n$ for some $\epsilon\in (0,1)$. In this case the sequence $a_n/b_n$ is bounded (and is non-convergent).

In the following I suppose that $a_n,b_n>0$. If we assume that $F_Y$ and $F_Z$ are continuous, then the cdfs $F_{1,n}(\,\cdot\,):=\mathsf{P}(a_n^{-1}X_n\le \cdot)$ and $F_{2,n}(\,\cdot\,):=\mathsf{P}(b_n^{-1}X_n\le \cdot)$ converge uniformly to $F_Y$ and $F_Z$, respectively (by Polya's theorem). This fact implies that
$$
|F_Y(xa_n/b_n)-F_Z(x)|\le |F_{2,n}(x)-F_Z(x)|+|F_{1,n}(xa_n/b_n)-F_Y(xa_n/b_n)|\to 0
$$
as $n\to\infty$ for every $x\in\mathbb{R}$. Now choose $x$ s.t. $x'=F_Y^{-1}(F_Z(x))$ is a point of increase of $F_Y$, i.e. for any $\epsilon>0$, $F_Y(x'+\epsilon)-F_Y(x')>0$ and $F_Y(x')-F_Y(x'-\epsilon)>0$. Then
$$
F_Y(xa_n/b_n)\to F_Y(x')
$$
and, hence, $a_n/b_n$ must converge.
