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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 $p_n \rightarrow 0$

I appreciate your help.

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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.2012-11-27
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    @user48405 glad I could help!2012-11-27
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    Quick question Deven: what do you mean by "M be the first one that is greater than $\epsilon$?2012-11-27
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    @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$2012-11-27
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    *then the probability that Xn≥ϵ is the same as the probability that Xn=a for some integer a≥M*... Nope, for every $a$ the latter is strictly smaller than the former..2012-11-27
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    @did perhaps I have a misunderstand then. How is it that the latter is strictly smaller than the former? We know with probability 1 that $X_n$ is a natural so why isn't the probability that $X_n \geq \epsilon$ be the probability that $X_n$ is some integer at least $\epsilon$?2012-11-27
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    $P(X\geqslant1)\gt P(X=a)$ for every $a\geqslant1$, not to be confused with the obvious identity $P(X\geqslant1)=P(\exists a\geqslant1,X=a)$.2012-11-27
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    @did I see; when I say "probability that $X_n = a$ for some $a \geq M$" I mean the latter of the two things you mention, the probability that there is $a \geq M$ such that $X_n = a$2012-11-27
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    Then I fail to see how a quantity such as $P(X_n=k)$ enters the picture, later on in your answer.2012-11-27
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    @did the probability that $X_n = a$ for some $a \geq M$ is the same as the sum of the probabilities $P(X_n = k)$ for $k \geq M$2012-11-27
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    I know, thanks... :-) The problem is not whether the result holds or not (it does), it is that your solution is ambiguously formulated. One does not need to decompose $P(X\geqslant M)$ into the sum of a series of $P(X=a)$ over every $a\geqslant M$, hence the insistence of your solution to do so seems to send the reader on a false track. All in all, when I read your solution, I am unable to decide whether you understood what happens or not. But the solution is accepted hence I guess all this is moot...2012-11-27
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    @did I'm not sure how this would send the reader on the wrong track... using what I say it's easy to make the sum of the series very small when we choose large $n$, which is the same thing as making the probability that $X_n > \epsilon$ very small when we choose large $n$.2012-11-27