Convergence of Normal distribution in probability

Question

Hi Please help in solving below question.

Let $X_n ~ N(0,\frac{1}n)$. Does $X_n $ converge in probability to a constant $c$? If yes, what does it converge to?

I came up with a proof but it converges to $0$.

e below is epsilon

for any $\epsilon\geq 0$ using Markov inequality

$P(|X_n| > \epsilon) = P(|X_n|^2 > \epsilon^2) \leq E[X_n^2]/\epsilon^2 = (\frac{1}n)/\epsilon^2$

so when $n \rightarrow \infty$ then $X_n\rightarrow 0$

am i doing it write ?

Answer 1

Great job. Also, you have just derived the Chebyshev's inequality.

$$P(|X_n-0|> \epsilon) \le \frac{Var(X_n)}{\epsilon^2}=\frac1{n\epsilon^2}$$

$$\lim_{n \to \infty}P(|X_n-0|> \epsilon) \le 0$$

Hence $X_n\xrightarrow{P} 0.$