0

Edit: Sorry for the Latex, I'm new to it and trying to fix it right now

I am trying to find the probability function of a random sum of Bernoulli variables in this scenario:

Starting from 9AM, clients arrive (independently), before a store opens, and stand in line to enter the store when it opens. An employee of the store randomly checks the line $M$ minutes after 9AM, with $M$ a Poisson variable with parameter $v$.

The number of clients that arrive during any $1$ minute interval is either $1$ or $0$ with probabilities $p$ and $1 - p$.

Let $S_m$ be the total number of clients in line when the line is checked at $M = m$ minutes after 9AM.

I need to find

1) $\Pr(S_m = k)$ for $k = 0,1,2,\ldots$

2) $\operatorname{E}(M \mid S_m = k)$

For 1) I found that $$\Pr(S_m = k) = \sum_{M=1}^\infty {M \choose k} p^k (1-p)^{M-k}\frac{{e^{-v}v^M }}{M!}$$

For 2), I started by trying to find $\Pr(M = m \mid S_m = k)$ but I'm having a hard time finding it.

I started by doing $$\Pr(M = m \mid S_m = k) = \frac{P(M = m, S_m = k)}{P(S_m = k)}$$ and plugged in the denominator my answer to $1$, but I don't know how to find the numerator and am really unsure of what I've done so far.

Could anyone help me out?

Thank you!

Michael Hardy
  • 1
nx__
  • 341

1 Answers1

0

$P(M=m,S_m=k)=P(S_m=k| M=m) \cdot P(M=m)$. You know the second term. For the first term, it's just $\binom{m}{k}p^k(1-p)^{m-k}$.

Alex R.
  • 32,771
  • Thank you for your answer, I am a bit confused though, did you mean P(S_m = k | M = m) * P(M = m) and not divided? Also, shouldnt I be multiplying by P(M = m) for the numerator you gave? – nx__ Jan 21 '16 at 19:18
  • @mc__: yep, sorry for the typo. – Alex R. Jan 21 '16 at 19:30
  • Do you mind telling me if my result for 1) is correct? Because if I do the calculation, I get a result of 1 because both numerator and denominator are the same – nx__ Jan 21 '16 at 19:32