show that the number of $1$’s transmitted interval is Poisson

Question

A transmitter send out either a $1$ with probability $p$, or a $0$ with probability $1 − p$, independently of earlier transmissions. If the number of transmissions within a given time interval is Poisson with parameter $λ$, show that the number of $1$’s transmitter in that same time interval is also Poisson, and has parameter $pλ$.

Answer 1

An answer without using the generating functions, let $N$ be the total number of emitted particles, and $N_1$ be the number of emitted 1's, then by marginalising we have $$ \begin{align} \mathbb{P}\left\{ N_1 = k \right\} &= \sum_{m=k}^{\infty}\mathbb{P}\left\{ N_1 = k , N = m \right\} \\ &= \sum_{m=k}^{\infty} \mathbb{P}\left\{N_1 = k | N = m \right\} \mathbb{P}\left\{ N = m \right\}. \end{align} $$ Now the first probability is the probability of observing $k$ successes from $m$ trials, and the second is just our Poisson probability giving $$ \begin{align} \sum_{m=k}^{\infty} \frac{m!}{k!(m-k)!}p^{k}(1-p)^{m-k} \frac{\lambda^m e^{-\lambda}}{m!} &= \frac{p^k}{k!}e^{-\lambda}\lambda^{k}\sum_{m=k}^{\infty}\frac{\lambda^{m-k}(1-p)^{m-k}}{(m-k)!} \\ &=\frac{p^k}{k!}e^{-\lambda}\lambda^{k}e^{\lambda(1-p)} \\ &= \frac{(p\lambda)^{k}}{k!}e^{- p\lambda }, \end{align} $$ which we recognise as a poisson random variable with parameter $\lambda p$.

Answer 2

Here's an outline: Let $X_1,X_2,X_3,\ldots$ be i.i.d. Bernoulli random variables with $\mathbb{P}(X_1 = 1) = p$. If we set $Y$ to be Poisson of parameter $\lambda$ and $$Z = \sum\limits_{k = 1}^Y X_k = \sum\limits_{k = 1}^\infty X_k \mathbf{1}_{k \leq Y}$$ then $Z$ has the same distribution as the number of $1$'s that appear. Now calculate the probability generating function for $Z$, $\mathbb{E}(s^Z)$ using this expression and independence of $Y$ from the $X_k$'s. If you show that $\mathbb{E}(s^Z) = e^{p\lambda(s-1)}$ then you may conclude that $Z$ is Poisson of mean $p\lambda$.