Cfrgtkky

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 $$mathbb{P}(y=y)=sum_{k=y}^{infty}binom{k}{y}p^{k-y}(1-p)^yfrac{lambda^ke^{-lambda}}{k!}.$$
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 $tinmathbb{R}$,
$$begin{align}
mathbb{E} e^{tX}
&= mathbb{E}[ mathbb{E}[ e^{tX} mid N ] ]
stackrel{(dagger)}{=} mathbb{E}[ (1-p+pe^{t})^N ]\
&= mathbb{E}[ e^{Nln(1-p+pe^{t})} ]
stackrel{(ddagger)}{=} exp(lambda(e^{ln(1-p+pe^{t})}-1))\
&= exp(lambda((1-p+pe^{t})-1))\
&= exp(lambda p(e^{t}-1))
end{align}$$
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 $tinmathbb{R}$,
$$
mathbb{E} e^{tX} = exp(lambda p(e^{t}-1)) tag{$ast$}
$$
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 $pneq 1$, 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.