# Convergence in probability, convergence in quadratic mean

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.

• I have added the self-study tag. In future, please do this by yourself for problems like this. Jan 11 at 11:44
• 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.
– Stef
Jan 11 at 14:58

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.

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