A basic question concerning convergence in probability

Question

I have a rather basic question concerning convergence in probability. It might be silly, but I'm not sure I'm getting to the correct conclusion.

Let $X_n$ be a sequence of real valued random variables with absolutely continuous distribution and let $x_0$ be a constant. Assume that $X_n$ converges to $x_0$ in probability, i.e. $$ \lim_{n \to \infty}\mathbb{P}(|X_n-x_0|>\epsilon)=0, \quad \forall \epsilon>0. $$

Question: What can we say about $\lim_{n \to \infty}\mathbb{P}(X_n<x_0)$? Can we claim that $\lim_{n \to \infty}\mathbb{P}(X_n<x_0)=0$?

Of course, since $X_n$ also converges in distribution to a degenerate random variable whose distribution is a Dirac delta at $x_0$, for any $x_{-}<x_0<x_+$ we would have that $$ 0=\lim_{n\to \infty}\mathbb{P}(X_n \leq x_{-}) \leq \liminf_{n \to \infty}\mathbb{P}(X_n<x_0) \leq \limsup_{n \to \infty}\mathbb{P}(X_n<x_0) \leq \lim_{n\to \infty}\mathbb{P}(X_n \leq x_{+})=1. $$ But sice $x_0$ is not a continuity point of the limiting distribution, I'm not sure that one can conclude that the answer to the above question is "yes", just letting $x_{-}\uparrow x_0$. Any comments in this regard? Does the limit in my question even exist?

Answer 1

Let $X$ have standard normal distribution and $X_n=x_0-\frac {X^{2}} n$. Then $X_n \to x_0$ almost surely, hence also in probability. But $P(X_n<x_0)=1$ for all $n$.

By taking $x_0-\frac {X^{2}} n$ for $n$ even and $x_0+\frac {X^{2}} n$ for $n$ odd we get an example where the limit does not exist.

Answer 2

You can actually get $\lim_{n\to \infty}\mathbb{P}(X_n < x_0) = \alpha$ for any $\alpha \in [0,1]$. Indeed, let $x_0 = 0$ and $\alpha \in [0,1]$ and consider the random variables $(X_n)$ with cumulative distribution function $$F(x) = \begin{cases} 0 &, x < -1/n,\newline \alpha nx + \alpha &, - 1/ n \le x < 0, \newline (1-\alpha)nx + \alpha &, 0 \le x < 1 / n,\newline 1 &, \text{ else}. \end{cases}$$ In words, $X_n$ is uniformly distributed in $[-1/n,0]$ with probability $\alpha$ and in $[0,1/n]$ with probability $1-\alpha$. Then $(X_n)$ converges to $0$ in probability, but $$ \mathbb{P}(X_n < 0) = \mathbb{P}(- 1/ n \le X_n < 0) = \alpha $$ for all $n\in\mathbb{N}$. Of course, you get the general case by considering $Y_n = X_n + x_0$.