# How to prove convergence in probablity

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

• Use the definition of convergence in probability. – StubbornAtom Feb 2 '20 at 6:44
• 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? – gunes Feb 2 '20 at 15:52
• 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. – Iwishworldpeace 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}