Convergence in probability, convergence in quadratic mean

Question

In the textbook All of Statistics by Wasserman, there's an exercise that asks the following:

Let $X_1, X_2, \ldots$ be a sequence of random variables such that $$ \mathbb{P}\left(X_n=\frac{1}{n}\right)=1-\frac{1}{n^2} \quad \text { and } \quad \mathbb{P}\left(X_n=n\right)=\frac{1}{n^2} . $$ Does $X_n$ converge in probability? Does $X_n$ converge in quadratic mean?

This is my attempt for the first part of the question:

We denote $X_n$ as the random variable and $x_n$ as its corresponding realization. We can then write the probability of $X_n$ as $$ P(X_n = x_n) = \left(1-\frac{1}{n^2}\right) \mathbb{I}\left(x_n = \frac{1}{n}\right) + \frac{1}{n^2} \mathbb{I}(x_n = n) $$ where $\mathbb{I}$ is the indicator function. Since the probability of $X_n$ is always bounded, it follows that $X_n$ converges in probability to 0.

I'm unsure about the correctness and I'm also having trouble proceeding with the second part.

Answer 1

What should be the approach to assess the quadratic mean convergence?

$\bullet$ Without resorting to any fancy trick, apply the definition, that is, check whether $\mathbb E|X_n-X|^2$ converges to $0.$

$\bullet$ Try for $X=0.$

$\bullet$ If $X_n\overset{\text{qm}}{\to}X, $ we know $X_n\overset{\mathbb{P}}{\to}X.$ So, if the preceding steps are affirmative, then evidently $X_n\overset{\mathbb{P}}{\to}0.$

$\bullet$ Otherwise, one has to assess the convergence in probability separately. For $X=0, $ one can choose $\varepsilon> n^{-1}$ in $\mathbb P[|X_n-0|>\varepsilon]$ possibly to use the limiting nature of $1/n.$ See what happens.