Limit of $P(X_n > a_n)$ where $X_n \xrightarrow[n \to \infty]{d} X \sim{N(\mu,\sigma^2)}$ and $a_n\xrightarrow[n \to \infty]{} \infty$

Question

I've been working on following problem and could need some help.

Let $X_n$ be a sequence of RV with $$X_n \xrightarrow[n \to \infty]{d} X \sim{N(\mu,\sigma^2)}$$ for some $\mu \in \mathbb{R}$ and $\sigma^2 >0.$ Let $a_n$ be a sequence of real numbers with $\lim_{n \to \infty} a_n = \infty.$ I'd like to show that $$P(X_n > a_n) \xrightarrow[n \to \infty]{} 0$$

Is that even true though? It holds for the limit RV $X$ of course but I'm not so sure if it does for $X_n$, too, without knowing more about the speed of convergence of $X_n$. My problem is that I need the statement for a sequence $a_n$ that tends to infinity arbitrarily slow.

Thanks in advance!

Answer 1

The convergence in distribution $X_n\overset{d}\longrightarrow X$ is defined as $$P(X_n \le x) \underset{n \to \infty}\longrightarrow P(X\le x)$$ for every continuity point of $F_X(x):=P(X\le x)$. Therefore $$P(X_n>a_n)=1-P(X_n\le a_n) \underset{n \to \infty}\longrightarrow 1- P(X\le \infty)=1-1=0$$

Answer 2

For every $x$, the limiting distribution is continuous at $x$ hence $P(X_n\gt x)\to1-\Phi(x)$. There exists some finite $N$ such that $a_n\geqslant x$ for every $n\geqslant N$ hence $$\limsup_{n\to\infty}P(X_n\gt a_n)\leqslant\lim_{n\to\infty}P(X_n\gt x)=1-\Phi(x).$$ This holds for every $x$ hence $$\limsup_{n\to\infty}P(X_n\gt a_n)\leqslant\inf\limits_x(1-\Phi(x))=0.$$ Every $P(X_n\gt a_n)$ is nonnegative hence the inequality above proves that the limit exists and is zero.