Can someone tell me how I have to prove this: If X is a poisson distribution with parameter (u) and Y|X is binomial distribution with parameters (n,p), then Y is a poisson distribution with paramater (pu). I’m supposed to use something like the convolution formula, but I don’t see how
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Shouldn't parameters of $Y|X$ depend on those of $X$? I don't see any dependence between $X$ and $Y$ and this obtains that $Y$ has the same binomial distribution of $Y|X$ also or the question has been badly stated... – Mostafa Ayaz Jan 15 '18 at 12:10
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I see that I wrote a mistake, Y|X = x ~ bin (x,p) is the correct formulation. But I still don’t know how I could find Y? – Mike D Jan 15 '18 at 12:19
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Simple like a piece of a cake! We have $$p(Y=y|X=x)=\binom{x}{y}p^y(1-p)^{x-y}\\p(X=x)=u^x\frac{e^{-u}}{x!}$$ so we can easily derive $p(Y=y,X=x)$ as follows: $$p(Y=y,X=x)=p(Y=y|X=x)p(X=x)=\binom{x}{y}p^y(1-p)^{x-y}u^x\frac{e^{-u}}{x!}$$ where $0\le y<\infty$ and $y\le x <\infty $. Now we calculate $p(Y=y)$ by summation of $p(Y=y,X=x)$ over x. We obtain:$$p(Y=y)=\sum_{x=y}^{\infty}p(Y=y,X=x)$$$$=\sum_{x=y}^{\infty}\binom{x}{y}p^y(1-p)^{x-y}u^x\frac{e^{-u}}{x!}$$$$=\frac{e^{-u}p^y}{y!}\sum_{x=y}^{\infty}\frac{1}{(x-y)!}(1-p)^{x-y}u^x$$$$=\frac{e^{-u}(pu)^y}{y!}\sum_{x=y}^{\infty}\frac{1}{(x-y)!}(1-p)^{x-y}u^{x-y}$$$$=\frac{e^{-u}(pu)^y}{y!}\sum_{n=0}^{\infty}\frac{1}{n!}(1-p)^{n}u^{n}=\frac{e^{-u}(pu)^y}{y!}e^{u-up}$$$$=\frac{e^{-up}(pu)^y}{y!}$$
Mostafa Ayaz
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