Binomial random variable with number of trials being a Poisson random variable

Let $Y$ be the number of heads in a an $X$ toss sequence of flipping a coin with probability $p$ of heads. Show that $Y \sim \mathrm{Pois}(p \lambda)$ if $X \sim \mathrm{Pois}(\lambda)$.

Solution 1:

Note that $Y\mid X=x\sim\mathrm{bi}(x,p)$ for $x=0,1,\ldots$ and so $$ P(Y=k\mid X=x)=\binom{x}{k}p^k(1-p)^{x-k},\quad k=0,\ldots,x. $$ Thus $$ P(Y=k)=\sum_{x=k}^\infty P(Y=k\mid X=x)P(X=x)=\sum_{x=k}^\infty\frac{x!}{k!(x-k)!}p^k(1-p)^{x-k}\frac{\lambda^x}{x!}e^{-\lambda}, $$ which by letting $x!$ cancel out we can write as $$ \frac{p^k}{k!}e^{-\lambda}\sum_{x=k}^\infty \frac{1}{(x-k)!}(1-p)^{x-k}\lambda^x. $$ Now write $\lambda^x$ as $\lambda^{x-k}\lambda^k$ and do a change of index so the sum starts at $x=0$. Then you should end up with $$ \frac{(p\lambda)^k}{k!}e^{- p\lambda}. $$

Solution 2:

It is easiest to use probability generating functions $\mathcal{P}_Z(t) = \sum_{m=0}^\infty t^m \mathbb{P}\left(Z=m\right) = \mathbb{E}\left(t^Z\right)$. The probability generating function of a binomial random variable equals $\left(1-p + p t \right)^n$. Thus: $$ \mathcal{P}_Y(t) = \mathbb{E}\left(t^Y\right) = \mathbb{E}\left(\mathbb{E}\left(t^Y|X\right)\right) = \mathbb{E}\left( (1-p+t p)^X\right) = \mathcal{P}_X(1-p+ t p) = \exp\left(\lambda \left( \left(1-p + t p \right) -1 \right) \right) = \exp\left(\lambda p (t-1) \right) $$ The latter is the probability generating function of $\operatorname{Poi}\left(\lambda p \right)$.