I am trying to solve the problem from MIT Open Coursware "Statistics for Applications" problem set. Specifically the first one:

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

So as for convergence in probability (from definition) I need to show that $P[|X_n - X| \geq \varepsilon]$ converges to 0 as $n$ goes to infinity for every $\varepsilon \gt 0$.

I am not sure if my approach is correct:

Every random variable $X_1, X_2, \ldots$ can have two outcomes. Either $\frac{1}{n}$ with probability $1-\frac{1}{n^2}$ or $n$ with probability $\frac{1}{n^2}$.

So as I start to calculate those probabilities for different values of $n$ I get:

For $X_1$: $P[X_1 = 1] = 0$ and $P[X_1 = 1] = 1$ (don't know how to understand that)

For $X_2$: $P[X_2 = \frac{1}{2}] = \frac{3}{4}$ and $P[X_2=2] = \frac{1}{4}$

For $X_3$: $P[X_3 = \frac{1}{3}] = \frac{8}{9}$ and $P[X_3=3] = \frac{1}{9}$

and so on...

Now I notice that as n gets larger $X_n$ tends to favor $\frac{1}{n}$ more and more.

So I choose a random variable $X$ such that $P[X=\frac{1}{n}] = 1$ which basically means that $X=\frac{1}{n}$.

Now plugging everything to the definition of probability convergence:

$$\lim_{n\to\infty} P[|X_n - X| \geq \varepsilon] = \lim_{n\to\infty} P[|X_n - \frac{1}{n}| \geq \varepsilon]$$

Now I can discard the absolute value because for $n \in N^*$ $X_n$ is either $n$ or $\frac{1}{n}$ and in this case $n-\frac{1}{n}$ and $\frac{1}{n} - \frac{1}{n}$ are always greater or equal to 0.


$$\lim_{n\to\infty} P[X_n - \frac{1}{n} \geq \varepsilon]$$

And now this part is a bit tricky - I look what happens in the limit. So as $n \to \infty$:

$$P[X_n=\frac{1}{n}] = 1 \;and\; P[X_n=n] = 0$$

Which basically means that $X_n = \frac{1}{n}$ as $n \to \infty$. So now: $$\lim_{n\to\infty} P[X_n - \frac{1}{n} \geq \varepsilon] = P[0 \geq \varepsilon]$$

Which equals to 0 because $\varepsilon \gt 0$. So $X_n$ converges in probability.

So my questions are:

  1. Is this approach correct? I feel like the beginning is reasonable but at the end it is kind of shady, especially the limit calculation - could it be done in more steps to describe the reasoning better?
  2. What would be a formally correct way to prove that $X_n$ converges in probability?
  3. How to understand that for $X_1$ I get $P[X_1 = 1] = 0$ and $P[X_1 = 1] = 1$?
  4. If random variable $X$ is such that $P[X=\frac{1}{n}] = 1$ is it correct to write $P[X=\frac{1}{n}] = 1 \Leftrightarrow X=\frac{1}{n}$?

EDIT: Thanks for all the responses, they really cleared up things for me. I am wondering about convergence in $L^2$ - now that we have established that $X_n$ converges in probability to 0 can I use that result to prove that it also converges in $L^2$?

So now I need to show that $E[X_n^2]$ goes to 0 as $n$ goes to infinity. So now because I know that $X_n$ converges to 0 in probability could I write: $$ \lim_{n\to\infty} E[X_n^2] = E[0] = 0$$ ? Or is probability convergence to weak to infer that?

The other soulution would be: $$E[X_n^2] = E[(\frac{1}{n}(1-\frac{1}{n^2}) + n \frac{1}{n^2} )^2 ] = E[( \frac{1}{n} - \frac{1}{n^3} + \frac{1}{n})^2 ] = E[(\frac{2}{n} - \frac{1}{n^3})^2] \xrightarrow[n \to \infty]{} E[0] = 0$$

Is it correct?



$Pr[X_n^2 = (\frac{1}{n})^2] = 1-\frac{1}{n^2}$ and $Pr[X_n^2=n^2]= \frac{1}{n^2}$.

Then: $E[X_n^2]= (\frac{1}{n})^2 (1-\frac{1}{n^2}) + n^2 \frac{1}{n^2} = \frac{1}{n^2} - \frac{1}{n^4} + 1 \xrightarrow[n \to \infty]{} 1$

So now this means that $X_n$ does not converge to 0 in $L^2$. Which also means that the previous approach was wrong and probability convergence was to weak.

  • $\begingroup$ add the self-study tag $\endgroup$ Commented Dec 21, 2019 at 18:36
  • 1
    $\begingroup$ Note that $1/n$ is not a constant, so you aren't really "converging to" it. What does $1/n$ itself converge to? $\endgroup$
    – jbowman
    Commented Dec 22, 2019 at 1:23
  • $\begingroup$ Right, that's a good point, thanks. $\endgroup$
    – Franek
    Commented Dec 22, 2019 at 17:46

1 Answer 1


You should let the target distribution be $Pr[X=0]=1$.

Let me first work out the CDF of $X_n$,

$$Pr(X_n \le \epsilon ) = \begin{cases} 0 &, \epsilon < \frac1n \\ 1-\frac1{n^2} & ,\frac1n \le \epsilon < n \\ 1 &, \epsilon \ge n\end{cases} $$

\begin{align}Pr(|X_n-X|> \epsilon)&=Pr(X_n>\epsilon)\\&=1-Pr(X_n \le \epsilon) \\&= \mathbb{1}_{\epsilon < \frac1n} + \frac1{n^2}\cdot \mathbb{1}_{ \frac1n \le \epsilon < n}\end{align}

Given any $\epsilon>0$, for any $n> \max(\lceil\frac1\epsilon \rceil, \lceil \epsilon\rceil) $ , we have $\frac1n \le \epsilon < n$, and $Pr(|X_n-X| > \epsilon ) =\frac1{n^2}.$

Taking the limit of $n$ to $\infty$ would give you the answer of whether it converges in probability.

You are right regarding $X_1$, the intention of the question could be to discuss $X_n$ where $n > 1$.

We usually write $X=\frac1n a.s.$ to denote $Pr(X=\frac1n)=1$, a.s. stands for almost surely. That is the set of possible exceptions may be non-empty, but it has probability $0$. The concept is essentially analogous to the concept of "almost everywhere" in measure theory.


Try to evaluate $E[X_n^2]$:

$$E[X_n^2]=\left( \frac1n\right)^2P(X_n=\frac1n)+n^2 P(X_n=n)$$

  • $\begingroup$ Hi, thanks for the explanation, really cleared up a lot of things for me. I expanded the main question to include $L^2$ convergence as well - could you please take a look at it? $\endgroup$
    – Franek
    Commented Dec 23, 2019 at 10:41
  • $\begingroup$ hi, good attempt. So far, you are not evaluating the expectation correctly. can you try to evaluate $E[X_n^2]$ again? $\endgroup$ Commented Dec 23, 2019 at 12:09
  • $\begingroup$ Thanks, I think the latest solution (EDIT2) is correct now. Another follow up question - so $X_n$ converges in probability to 0. This means it also converges in distribution. So $E[f(X_n)] \xrightarrow[n \to \infty]{} E[f(X)]$ for all continuous and bounded f. So in this case $f(x)=x^2$ so this doesn't apply because $f$ is not bounded? $\endgroup$
    – Franek
    Commented Dec 23, 2019 at 14:00
  • $\begingroup$ yes, you are right, convergence in probability does imply convergence in distribution. $\endgroup$ Commented Dec 23, 2019 at 14:29

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