Let $Y_1$, $Y_2$, ... be a sequence of random variables such that $P(Y_n=\frac{1}{n})=1-\frac{1}{n^2}$ and $P(Y_n=n)=\frac{1}{n^2}$. Does $Y_n$ converge in probability?

I am stuck because I don't know how to deal with RV with two outcomes. And I am confused with the different probability in which $Y_n=1/n$ and $Y_n=n$. I know how to prove convergence with a single $Y_n$ to Y. But really stuck in this one.

so $$ Y_1=\begin{cases} 1 &\mbox{p } = 0 \\ 1 &\mbox{p } = 1 \end{cases} $$ $$ Y_2=\begin{cases} 2 &\mbox{p } = \frac{1}{4} \\ \frac{1}{2} &\mbox{p } = \frac{3}{4} \end{cases} $$

and then I don't know how to prove whether $Y_n$ converges or not in probability.

any help will be great. thanks

  • $\begingroup$ Use the definition of convergence in probability. $\endgroup$ Feb 2 '20 at 6:44
  • $\begingroup$ When $n=1$, your equations yield $P(Y_1=1)=1$ and $P(Y_1=1)=0$, which is a contradiction. What does this mean? $\endgroup$
    – gunes
    Feb 2 '20 at 15:52
  • $\begingroup$ Yes, this is exactly where I am confused with. But just like the two outcomes of each $Y_n$, the sum of the probability is 1. $\endgroup$ Feb 2 '20 at 17:47

If we redesign your sequence as $Y_1=1$, and everything else remains the same, the contradiction disappears. Then, if $Y_n$ converges, it should converge to $0$. Here, we'll use the definition of convergence in probability we need to show the following: $$\lim_{n\rightarrow\infty}P(|Y_n-0|>\epsilon)=\lim_{n\rightarrow\infty} P(Y_n>\epsilon)=0$$

$Y_n$ is either $n$ or $1/n$. Now, let's fix some $n$, and calculate the above probability:

$$\begin{align}P(Y_n>\epsilon)&=P(Y_n>\epsilon|Y_n=n)P(Y_n=n)+P(Y_n>\epsilon|Y_n=1/n)P(Y_n=1/n)\\&=P(n>\epsilon)\frac{1}{n^2}+P(1/n>\epsilon)\left(1-\frac{1}{n^2}\right)\\&=\mathbb{I}(n>\epsilon){1\over n^2}+\mathbb{I}(1/n>\epsilon)\left(1-\frac{1}{n^2}\right)\end{align}$$

Now, we take the limit: $$\begin{align}\lim_{n\rightarrow\infty} P(Y_n>\epsilon)&=\underbrace{\lim_{n\rightarrow\infty}\mathbb{I}(n>\epsilon)}_{1}\underbrace{\lim_{n\rightarrow\infty}\frac{1}{n^2}}_0+\underbrace{\lim_{n\rightarrow\infty}\mathbb{I}(1/n>\epsilon)}_0\underbrace{\lim_{n\rightarrow\infty}\left(1-\frac{1}{n^2}\right)}_1\\&=0\end{align}$$

which is what we needed to prove.

  • $\begingroup$ Thanks a lot! This looks really great. I totally missed the conditional probability part. $\endgroup$ Feb 2 '20 at 18:52
  • 2
    $\begingroup$ yes. sorry, I'm kinda new to this. $\endgroup$ Feb 2 '20 at 19:46

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