# Convergence in Probability!

I'm working on a problem and after couple of hours, I'm not still sure how I should approach it. Could you give me some hints?

Consider $X_1, X_2, \ldots$ as independent random variable where: $\Pr(X_n = k) = (1-p_n)p_n^k$ for $k = 0, 1, 2, \ldots$ ($p_n > 0$)

The goal is to show $X_n \rightarrow 0 \text{ in probability}$ IF and Only IF $p_n \rightarrow 0$

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Well, firstly we have to split it up into two directions since it is iff. I'll push in you in the correct direction for both ways.

The first important thing to notice is that $$\sum_{k=0}^{\infty} (1 - p_n)p_n^k = 1$$ so we know with probability $1$ that $X_n = k$ for some $k$.

"$\Rightarrow$" This time we're assuming $X_n \rightarrow 0$ in probability. This means that for any $\epsilon$, $$Pr(X_n \geq \epsilon) \rightarrow 0$$ So consider the specific case take $\epsilon = \frac{1}{2}$ or anything less than one. Then we know that $$Pr(X_n \geq \frac{1}{2}) \rightarrow 0$$ So what can we say about $Pr(X_n = k)$ for $k\geq1$ as $n$ gets very large? and in turn what can we then say about $p_n$?

"$\Leftarrow$" For this one we're assuming $p_n \rightarrow 0$ and we have to show convergence in probability. So fix $\epsilon > 0$ and what we need to show is that $$Pr(X_n \geq \epsilon)\rightarrow 0.$$ Let $M$ be the first one that is greater than $\epsilon$ (note that $M > 0$) then the probability that $X_n \geq \epsilon$ is the same as the probability that $X_n = a$ for some integer $a \geq M$.

So since $p_n \rightarrow 0$ we can choose $N$ large so that for $n$ bigger than $N$, $p_n < \frac{1}{r}$. In this case for large $n$ we have $$Pr(X_n = k) \leq \frac{1}{r^k}.$$ So then what can you say about the probability that $X_n = a$ for some $a \geq M$? In particular if you choose appropriately large $r$ you should be able to conclude exactly what we want.

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Hi @Deven Ware, very nice answer...Thanks. –  user48405 Nov 27 '12 at 6:01
@user48405 glad I could help! –  Deven Ware Nov 27 '12 at 6:02
Quick question Deven: what do you mean by "M be the first one that is greater than $\epsilon$? –  user48405 Nov 27 '12 at 6:12
@user48405 I mean let $M$ be the first integer which is greater than $\epsilon$, you could if you want just assume $0< \epsilon < 1$ and use $M = 1$ –  Deven Ware Nov 27 '12 at 6:23
Thanks a lot... –  user48405 Nov 27 '12 at 6:25