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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.

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1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$
    – user42912
    Commented Oct 22, 2020 at 12:23
  • $\begingroup$ $P(|X|_n| >\epsilon)= \frac 1 {n^{2}}$ 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 $\endgroup$ Commented Oct 22, 2020 at 12:27
  • $\begingroup$ I made a typo I'm sorry, I meant the case $1/n>\epsilon$ $\endgroup$
    – user42912
    Commented Oct 22, 2020 at 12:31
  • $\begingroup$ 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. $\endgroup$ Commented Oct 22, 2020 at 12:37

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