# Prove that $X_n \sim N(0,\frac{1}{n})$ converges in probability to 0

Problem

Prove that $$X_n \sim N(0,\frac{1}{n})$$ converges in probability to 0.

Discussion

There a problems similar to this question, but I want to ask specifically about the convergence of an integral that arises in the solution.

Attempt

Let $$\epsilon>0$$. To show $$X_n \overset{p}{\to} 0$$, we must show $$P(|X_n| > \epsilon) \to 0$$.

\begin{align} P(|X_n|>\epsilon) &= P(X_n>\epsilon) + P(X_n < -\epsilon)\\ &= P(Z>\epsilon\sqrt{n}) + P(X_n<-\epsilon\sqrt{n}) \end{align} where $$Z$$ is the standard normal distribution. Thus we want to show $$P(|X_n|>\epsilon)=\int_{\epsilon\sqrt{n}}^\infty\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\text{dx} + \int_{-\infty}^{-\epsilon\sqrt{n}}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\text{dx}\to 0$$

I understanding intuitively that these integrals should converge to zero, but how can I show rigorously that they go to zero?

You can use the tail bound for a standard normal random variable: $$\int_x^\infty \exp(-y^2/2)\,dy \le x^{-1}\exp(-x^2/2),$$ which can be proved changing variables $$y = x + z$$ and using $$\exp(-z^2/2) \le 1$$.

• I am selecting this as my answer because it directly shows how to work with the integral, which is what I was specifically asking about. The other answers also offer great solutions. – StrugglingStudent42 Dec 4 '20 at 17:25

Your integrals contain spurious instances of $$\sqrt n$$. You are evaluating $$P(|Z|>\epsilon\sqrt n)$$ where $$Z$$ is standard normal, so the integrand should not involve $$\sqrt n$$. Anyhow, you can prove $$P(|Z|>\epsilon\sqrt n)\to0$$ rigorously using Markov's inequality.

$$E[X_n]=0$$

$$\lim_{n \to \infty}\mathrm{Var}[X_n]=0$$

This is a necessary and sufficient condition for

$$X_n\xrightarrow{L^2}0$$

Convergence in $$L^2$$ implies convergence in probability