# Help with this question of convergence of a random variable

I'm solving this assignment from this course from 2016 in MIT OpenCourseWare.

I didn't understand the first question:

For $$n \in \mathbb N^*$$, let $$X_n$$ be a random variable such that $$\mathbb P[X_n=\frac{1}{n}]=1-\frac{1}{n^2}$$ and $$\mathbb P[X_n=n]=\frac{1}{n^2}$$ Does $$X_n$$ converge in probability? In $$L^2$$?

### I have the following questions:

1. Using the definition of this random variable, we have $$\mathbb P[X_n=1]=0$$ and $$P[X_n=1]=1$$ in the same time. What's wrong?

2. My solution of the first part of the exercise is correct?

The definition of convergence in probability says if $$X_n$$ converges in probability to $$X$$,then $$\mathbb P(|Xn-X|>\epsilon)\rightarrow 0$$ for any $$\epsilon>0$$. Note first that we can discard the larger $$X_n$$'s because $$\frac{1}{n^2}\rightarrow 0$$. Furthermore, the probability of $$X_n$$ being inside the interval $$(-\epsilon, \epsilon)$$ for any $$\epsilon>0$$ is closer to $$1$$ as $$n$$ becomes larger since $$1-1/n^2\rightarrow 1$$. Thus, $$\mathbb P(|Xn-X|<\epsilon)\rightarrow 1$$ which is equivalent to the definition above.

1. I need help in the second part of the question:

Since we have $$X_n\xrightarrow{\mathbb P} 0$$, if $$X_n$$ converges in $$L^2$$ it must converges to $$0$$, then we have to prove $$\mathbb E(X_n)^2\rightarrow 0$$. I'm thinking to use LOTUS, but I don't know if this random variable is discrete or continuous. If it were discrete, we had

$$\sum\frac{1}{n^2}(1-1/n^2)+n^2\frac{1}{n^2}\rightarrow1$$

But even though treating $$X_n$$ as discrete, it misses the probability of the other possible values of the $$X_n$$. I need help.

The random variables $$(X_n)$$ are defined for $$n=2,3...$$. $$X_1$$ is not defined. I am not sure if you read the definitions correctly. For any particular $$n$$ the random variable $$X_n$$ takes just two values $$n$$ and $$\frac 1 n$$.
$$P(|X_n | >\epsilon) =P(X_n=n)$$ if $$\frac 1 n <\epsilon$$. Hence $$P(|X_n | >\epsilon) =\frac 1 {n^{2}}\to 0$$ which answers the first part. Thus $$X_n \to 0$$ in probability. [You have to specify what $$X$$ is in this part. You did some calculation without saying what $$X$$ is].
If $$X_n$$ converges in $$L^{2}$$ it can only converge to $$0$$ because $$X_n \to 0$$ in probability. Hence we should have $$EX_n^{2} \to 0$$. But this is false by direct calculation of $$EX_n^{2}$$. I will leave this calculation to you.
• Thank you very much for your answer. 1. I've just realized that $X_n$ is defined only to these two numbers. 2. Formally shouldn't you prove when $1/n>\epsilon$? 3. So the calculations I did in part 3 solves the question? Commented Oct 22, 2020 at 12:23
• $P(|X|_n| >\epsilon)= \frac 1 {n^{2}}$ for $n >1 /\epsilon$. This proves that $X_n \to 0$ in probability. Your calculation of $EX_n^{2}$ is correct but there is an extra $\Sigma$ sign which confused me . @user42912 Commented Oct 22, 2020 at 12:27
• I made a typo I'm sorry, I meant the case $1/n>\epsilon$ Commented Oct 22, 2020 at 12:31
• Let $\epsilon >0$ and $\eta >0$. Then $P(|X-n| >\epsilon) <\eta$ if $n >\frac1 {\epsilon}$ and $n >\frac 1 {\sqrt \eta}$. By definition this shows that $X_n \to 0$ in probability. Commented Oct 22, 2020 at 12:37