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The following experiment is performed. An observation is made of a poisson random variable N with parameter $\lambda$. Then N independent Bernoulli trials are performed, each with probability p of success. Let Z be the total number of successes observed in the N trials. Find the distribution of Z

So I formulated Z as a random sum, Z=$\xi_1+\xi_2+...+\xi_n$ where $\xi\sim $Bernoulli(p) and N$\sim$Poisson($\lambda$). and I then conditioned on Z to find the distribution $P(Z=z)=\sum_{n=1}^{\infty}P(N=n)P(Z=z|N=n)$. I'm assuming $Z=z|N=n$ is a binominal random variable but I don't know how to deduce it. If I do assume its binomial the sum works out and I get a poisson distribution.

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  • $\begingroup$ If you perform $\ N\ $ independent Bernoulli trials, each with probability $\ p\ $, then the distribution of the number $\ Z\ $ of successes will be $\ B(N,p)\ $, which is therefore the conditional distribution of $\ Z\ $ given $\ N\ $, if $\ N\ $ is chosen randomly. You seem to have come to this conclusion yourself already, so what is it about this process of deduction that you find unsatisfactory? $\endgroup$ Feb 4 '20 at 2:47
  • $\begingroup$ I'm trying to show it explicitly $\endgroup$
    – Jaider
    Feb 4 '20 at 2:49
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I assume by "show it explicitly", you have something like the following in mind.

Given $\ N=n\ $, the random variables $\ \xi_1, \xi_2, \dots, \xi_n\ $ are independent with $\ P\left(\left.\xi_j=1\,\right|N=n\right)=p\ $ and $\ P\left(\left.\xi_j=0\,\right|N=n\right)=1-p\ $—that is, if $\ i_1,i_2, \dots, i_n\ $ is a sequence of ones and zeros containing $\ m\ $ ones and $\ n-m\ $ zeros, then \begin{align} P\left(\xi_1=i_1, \xi_2=i_2, \dots, \xi_n=i_n |\,N=n\right)&=\prod_{j=1}^nP\left(\xi_j=i_j|\,N=n\right)\\ &=p^m(1-p)^{n-m} \end{align} Therefore \begin{align} P \left(\xi_1+ \xi_2+ \dots +\xi_n=m \,|\,N=n\right)&=\sum_{i_1,i_2, \dots, i_n\in \{0,1\}^n:\\ i_1+i_2+ \dots+i_n=m} p^m(1-p)^{n-m}\\ &={n\choose m} p^m(1-p)^{n-m}\ , \end{align} because $\ \left|\left\{i_1,i_2, \dots, i_n\in \{0,1\}^n\,|\, i_1+i_2+ \dots+i_n=m\right\}\right|= {n\choose m}\ $.

Is that the sort of thing you're looking for?

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  • $\begingroup$ thank you so much, that explains everything. I can take it from here $\endgroup$
    – Jaider
    Feb 4 '20 at 3:41

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