$\Pr(X_n\leq x) \to \Pr(X\leq x)$ for $x\geq 0$ and $\Pr(X_n< x) \to \Pr(X\leq x)$ for $x<0$ imply convergence in distribution?

Question

Let $X_n$ be random variables and let $X\sim \mathcal{N}(0,1)$ such that $$\forall x\geq 0, \quad \Pr(X_n\leq x) \to \Pr(X\leq x)$$ and $$\forall x< 0, \quad \Pr(X_n< x) \to \Pr(X\leq x).$$

Does this imply that $X_n$ converges in distribution to $X$ ?

The $X_n$ are not assumed to be continuous, hence $\Pr(X=x)$ may be nonzero and I can't use the usual convergence criterion with cdfs.

For $x<0$, $\Pr(X_n\leq x) = \Pr(-X_n\geq -x) = 1-\Pr(X_n< -x) $, however I don't know whether $P(X_n< -x)$ converges.

I'm thinking the statement is not true, but I can't come up with a counterexample either.

Answer 1

Fix an $x<0$.

$P(X_{n}\leq x)\leq P(X_{n}<x+\frac{1}{m})\xrightarrow{n\to\infty} P(X\leq x+\frac{1}{m})$ for each $m$

And hence you have $\lim\sup_{n\to\infty}P(X_{n}\leq x)\leq P(X\leq x+\frac{1}{m})\xrightarrow{m\to\infty}P(X\leq x)$

Thus $\lim\sup_{n\to\infty}P(X_{n}\leq x)\leq P(X\leq x) $

And $P(X_{n}<x)\leq P(X_{n}\leq x)$ and the LHS tends to $P(X<x)$

Thus $\lim\inf P(X_{n}\leq x)\geq P(X<x)=P(X\leq x)$

Thus the result is true for all $x<0$ also. So you have $X_{n}\xrightarrow{d} X$.

Answer 2

Take an $x<0$, then

Step I: $\Pr(X_n = x) \to 0$.

To show this note that for every $\epsilon>0$, $$ 0\leq \Pr(X_n = x) \leq \Pr(X_n <x+\epsilon) - \Pr(X_n <x-\epsilon). $$ Taking $\limsup$, $$ 0\leq\limsup \Pr(X_n = x) \leq \limsup\Pr(X_n <x+\epsilon) - \liminf\Pr(X_n <x-\epsilon) \\ = \Pr(X<x+\epsilon) - \Pr(X <x-\epsilon). $$ You already have $\lim\Pr(X_n <x)= \Pr(X\leq x)$ for all $x<0$, therefore $$ 0\leq\limsup \Pr(X_n = x) \leq \Pr(X\leq x+\epsilon) - \Pr(X \leq x-\epsilon). $$ Now letting $\epsilon \to 0$, you get $\Pr(X_n = x) \to 0$.

Step II: Now $\Pr(X_n \leq x) = \Pr(X_n = x)+ \Pr(X_n < x)$. Using Step I, this converges to $0+\Pr(X \leq x)$. This concludes that $X_n$ converges to $X$ in distribution.