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

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 $\lambda p$.