Joint probability distribution of Bernoulli & Poisson Processes

Question

"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, P_k (k ; t),for t ≥ 0."

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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.

Answer 1

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