Joint probability distribution of Bernoulli & Poisson Processes "Let the random process K(t) depend on uniform Poisson process N(t), with the mean arrival rate λ>0, as follows: Starting at t = 0, both N(t) = 0, and K(t) = 0. When an arrival occurs in N(t), an independent Bernoulli trial takes place with probability of success p, where 0>p>1. On success, K(t) is incremented by 1, otherwise K(t) is left unchanged. Find the PMF of the discrete valued random process K(t) at time t, that is, Pk (k ; t),for t ≥ 0." * * * So far, I've assumed the following... The probability of k arrivals in interval t follows the distribution **P(k arrivals in time t)=((λt) k e-λt)/k!** And PMF of K(t) conditioned on a fixed number of k arrivals from the Poisson process **P(K(t) = i | K = k) = C k,i(p)i(1-p)k-i** However, I am stuck trying to get the PMF of K(t) unconditioned on K.

$$\begin{align} P(K(t) = i) &= \sum_{n=0}^\infty P(K(t) = i \mid N(t) = n) P(N(t) = n) \\\ &= \sum_{n=i}^\infty \binom{n}{i} p^i (1-p)^{n-i} \cdot e^{-\lambda t} \frac{(\lambda t)^n}{n!} \end{align}$$ Can you take it from here?

> Continuing from above, use $\binom{n}{i} = \frac{n!}{i! (n-i)!}$ and some rearranging to obtain $$P(K(t)=i) = e^{-p \lambda t} \frac{(p \lambda t)^i}{i!} \cdot \underbrace{e^{-(1-p) \lambda t}\sum_{n=i}^\infty \frac{((1-p) \lambda t)^{n-i}}{(n-i)!}}_{=1}.$$ Do you recognize this PMF?