0
$\begingroup$

"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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ This is great, thanks! I didn't to split out the summation as you've done. That PMF looks like a specialized version of the Poisson distribution which incorporates the Bernoulli trials, but I don't know what its name is. $\endgroup$ – julez1234 Jul 11 '19 at 13:12
  • $\begingroup$ @julez1234 Yes, it is just a Poisson distribution with mean $p\lambda$. $\endgroup$ – angryavian Jul 11 '19 at 16:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.