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I have the following problem set at hand:

The random variable $\xi$ has Poisson distribution with the parameter $\lambda$. If $\xi=k$ we perform $k$ Bernoulli trials with the probability of success $p$. Let us define the random variable $\eta$ as the number of successful outcomes of Bernoulli trials. Prove that $\eta$ has Poisson distribution with the parameter $p\lambda$.

I feel confused about what to do exactly with the $\xi=k$ part of the question? I was trying to do a $\lambda = np$ substituion and let n go to infinity, but i cannot reach at the desired prove. Could someone help to guide me through this?

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Here's a hint: $$ P(\eta = i) = \sum_{k \ge 0} P(\eta = i \mid \xi = k) P(\xi = k). $$ The two pmfs on the right are provided in the question.

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  • $\begingroup$ This gives me: $P(\eta=i) = \sum_{k\geq0} {n\choose i}p^i(1-p)^{(n-i)} \cdot \frac{\lambda^k}{k!}e^{-\lambda}$. Now i would transform the binomial part into Poisson format letting $n$ go to infinity and then hoping to being able to combine the exponentials into the desired output? But not sure what to do with the sum? $\endgroup$
    – Oliver
    Apr 26, 2018 at 14:42
  • $\begingroup$ @Oliver I thought you said $\xi$ had a Poisson($p\lambda$) distribution and $\eta \mid \xi$ had a Binomial($k$,$p$) distribution? $\endgroup$
    – Taylor
    Apr 26, 2018 at 14:44
  • $\begingroup$ In my excercise it says that $\xi$ is Poissonian with paramter $\lambda$. The $p\lambda$ is the parameter for the $\eta$ distribution (which I need to prove). $\endgroup$
    – Oliver
    Apr 26, 2018 at 14:47
  • $\begingroup$ @Oliver okay I see your edit, but you still have one more mistake: $\eta \mid \xi = k$ is a Binomial($k$,$p$) not a Binomial($n$,$p$) $\endgroup$
    – Taylor
    Apr 26, 2018 at 14:54
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    $\begingroup$ @Oliver No it is not sensible. Try writing the first term of the sum (the one with $k=0$) explicitly when $i=3$, say, to see what issues might arise. $\endgroup$
    – Dilip Sarwate
    Apr 26, 2018 at 15:35
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Hint: $\eta$ could not possibly have value $i$ when $\xi$ has value smaller than $i$. You cannot get the silk purse of $i$ successes from the sow's ear of fewer than $i$ trials. So, that sum that @Taylor told you about can be simplified to a sum on $k \geq i$ since $P(\eta = i \mid \xi = k)$ has value $0$ for $0 \leq k < i$. Now, do some cancellations, change of variables, and pulling common factors out of the sum, in the sum on $k \geq i$ and you should get the desired result.

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