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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.

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  • $\begingroup$ I have added the self-study tag. In future, please do this by yourself for problems like this. $\endgroup$
    – User1865345
    Jan 11 at 11:44
  • $\begingroup$ I would say the first part is very obvious, yes, $X_n$ converges to 0 in probability; but your proof leaves me very confused and I can't follow it. You have written rewritten $P(X_n = 1/n) = 1-1/n^2$ and $P(X_n = n) = 1/n^2$ in what seems to me a correct, but unnecessarily convoluted way, using the indicator function. And then you say "the probability of $X_n$ is always bounded", a statement which I don't really understand. Note that probabilities are always bounded, since they're always real numbers between 0 and 1; and I don't know what "the probability of $X_n$" means. $\endgroup$
    – Stef
    Jan 11 at 14:58

1 Answer 1

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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.

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  • $\begingroup$ Why are you setting X to 0? Also can't we evaluate the limit without replacing X? $\endgroup$
    – neo
    Jan 13 at 4:11
  • $\begingroup$ You need something for it to (not) converge. $0$ is the easy one to go but not necessarily would work always. $\endgroup$
    – User1865345
    Jan 13 at 4:19
  • $\begingroup$ In that case, would this make sense $$ \mathbb{P}\left(\left|X_n-0\right| \geq \epsilon\right)=\mathbb{P}\left(X_n \neq 0\right)=\mathbb{P}\left(X_n = n\right)=\frac{1}{n^2} $$ and $$ \lim_{n \to \infty} \frac{1}{n^2} = 0. $$ Therefore, we have $$ \lim_{n \to \infty} \mathbb{P}\left(\left|X_n-0\right| \geq \epsilon\right)=0 $$? $\endgroup$
    – neo
    Jan 13 at 4:30
  • $\begingroup$ That is correct. Just that for this, $\varepsilon$ needs to be greater than $1/n.$ $\endgroup$
    – User1865345
    Jan 13 at 4:34

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