# Combination of Poisson and binomial distribution

I'm working on the following problem

Each time you flip a certain coin, heads appears with probability $p$. Suppose that you flip the coin a random number of $N$ times, where $N$ has the Poisson distribution with parameter $\lambda$ and is independent of the outcomes of the flips. Find the distributions of the numbers $X$ and $Y$ of the resulting heads and tails, respectively, and show that $X$ and $Y$ are independent.

What I tried, is conditioning on the value of $N$: \begin{eqnarray} \mathbb{P}(X=x) & = & \sum_{k=0}^{\infty}\mathbb{P}(X=x \ | \ N=k)\mathbb{P}(N=k)\\ & = & \sum_{k=0}^{\infty}\binom{k}{x}p^x(1-p)^{k-x}\frac{\lambda^ke^{-\lambda}}{k!}\\ & = & \sum_{k=x}^{\infty}\binom{k}{x}p^x(1-p)^{k-x}\frac{\lambda^ke^{-\lambda}}{k!}.\\ \end{eqnarray} Similarly, for $Y$ i found $$\mathbb{P}(y=y)=\sum_{k=y}^{\infty}\binom{k}{y}p^{k-y}(1-p)^y\frac{\lambda^ke^{-\lambda}}{k!}.$$ I tried to work this out but I didn't seem to go anywhere. The answer should be that $X \sim Pois(\lambda p)$ and because of symmetry we would have $Y \sim Pois(\lambda (1-p))$.

Can anyone provide some help about how to from where I came to $X \sim Pois(\lambda p)$? Thanks in advance.

I would suggest to use moment-generating functions (MGF): simpler, faster proof. Namely, you have, for $$t\in\mathbb{R}$$, \begin{align} \mathbb{E} e^{tX} &= \mathbb{E}[ \mathbb{E}[ e^{tX} \mid N ] ] \stackrel{(\dagger)}{=} \mathbb{E}[ (1-p+pe^{t})^N ]\\ &= \mathbb{E}[ e^{N\ln(1-p+pe^{t})} ] \stackrel{(\ddagger)}{=} \exp(\lambda(e^{\ln(1-p+pe^{t})}-1))\\ &= \exp(\lambda((1-p+pe^{t})-1))\\ &= \exp(\lambda p(e^{t}-1)) \end{align} where $$(\dagger)$$ uses the expression of the MGF of a Binomial distribution with parameters $$N$$ and $$p$$, and $$(\ddagger)$$ that of the MGF of a Poisson distribution with parameter $$\lambda$$ (applied to the argument $$t'\stackrel{\rm def}{=}\ln(1-p+pe^{t})$$).
At the end, you get that, for every $$t\in\mathbb{R}$$, $$\mathbb{E} e^{tX} = \exp(\lambda p(e^{t}-1)) \tag{\ast}$$ which is the MGF of a Poisson distribution with parameter $$\lambda p$$. As the MGF characterizes the distribution (when it exists), we have the result.
However, if you want to finish your computation: here how it goes. I assume $$p\neq 1$$, otherwise the answer is trivial. \begin{align} \mathbb{P}\{X=n\} &= \sum_{k=n}^\infty \binom{k}{n}p^n(1-p)^{k-n} \frac{\lambda^k e^{-\lambda}}{k!}\\ &= e^{-\lambda}\frac{p^n}{(1-p)^n}\sum_{k=n}^\infty \binom{k}{n}(1-p)^{k} \frac{\lambda^k}{k!}\\ &= e^{-\lambda}\frac{p^n}{(1-p)^n}\sum_{k=n}^\infty \frac{k!}{n!(k-n)!}(1-p)^{k} \frac{\lambda^k}{k!}\\ &= e^{-\lambda}\frac{p^n}{n!(1-p)^n}\sum_{k=n}^\infty \frac{1}{(k-n)!}(1-p)^{k} \lambda^k\\ &= e^{-\lambda}\frac{p^n}{n!(1-p)^n}\sum_{\ell=0}^\infty \frac{1}{\ell!}(1-p)^{\ell+n} \lambda^{\ell+n}\\ &= e^{-\lambda}\frac{(\lambda p)^n}{n!}\sum_{\ell=0}^\infty \frac{(1-p)^{\ell} \lambda^{\ell}}{\ell!}\\ &= e^{-\lambda}\frac{(\lambda p)^n}{n!}e^{\lambda(1-p)} = \boxed{e^{-\lambda p}\frac{(\lambda p)^n}{n!}} \end{align} and you get the probability mass function of a Poisson r.v. with parameter $$\lambda p$$, as desired.
• What you call the MGF is usually referred to as the characteristic function being related to the MGF by $\phi_X(t)=M_X(it)$, where $\phi$ is the CF and $M$ the MGF (of some RV). Or perhaps it was a typo, as you start out with $E(e^{itX})$ but end with the expression for the MGF—without any $i$. – Nap D. Lover Nov 6 '17 at 19:54