# 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) = Ck,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?
• @julez1234 Yes, it is just a Poisson distribution with mean $p\lambda$. – angryavian Jul 11 '19 at 16:50