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

Question

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?

Answer 1

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 $\exp(-z^2/2) \le 1$.

Answer 2

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.

Answer 3

$$E[X_n]=0$$

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

This is a necessary and sufficient condition for

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

Convergence in $L^2$ implies convergence in probability