Prove the convergence of a random variable

Question

Given $x_n \sim N(0, \frac{1}{n})$, is $x_n$ almost sure convergence or convergence in probability or convergence in distribution? How to prove it

Answer 1

By the Borel–Cantelli lemma, the sequence $X_n\sim N(0,n^{-1})$ converges to $0$ almost surely as $n\to\infty$ if $$ \sum_{n=1}^{\infty}\Pr\{|X_n|>\varepsilon\}<\infty $$ for each $\varepsilon>0$. Using Markov's inequality, we can obtain the following bound $$ \Pr\{|X_n|>\varepsilon\}\le\frac{\operatorname E|X_n|^4}{\varepsilon^4}. $$ Since $X_n\sim N(0,n^{-1})$, we know that $\operatorname E|X_n|^4=3(\operatorname E|X_n|^2)^2=3n^{-2}$. We obtain that $$ \sum_{n=1}^{\infty}\Pr\{|X_n|>\varepsilon\}\le\frac3{\varepsilon^4}\sum_{n=1}^\infty n^{-2}<\infty. $$ Almost sure convergence implies convergence in probability and convergence in probability implies convergence in distribution.