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Let $X(t)$ be a Poisson process with rate $\lambda = 6$ describing arrivals per hour of customers at a bank. Let the probability of a customer being male be $2/3$. Suppose 10 males has arrived during the two first business hours. How many females would you expect to have arrived?

Attempt: Let $M$ be male customers in the first two hours and $F$ be the female ones. Then $M$ should be $Po(8)$ and $F$ should be $Po(4)$, right?

What we're looking for is $E[F | M+F = 10]$.

Two problems: 1) I don't think you can just divide $E(F) = 4$ by $P(X+Y=10)$ since X and X+Y aren't independent (right...?).

2) I'm not sure how to calculate $P(X+Y=10)$ for two Poisson processes.

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    $\begingroup$ Aren't you looking for $E[F\mid M=10]$? (wich equals $E[F]=4$ since $M$ and $F$ are independent) $\endgroup$
    – drhab
    Commented Aug 9, 2014 at 12:35
  • $\begingroup$ Yep. Got tangled up in my own brain here. Thanks... $\endgroup$ Commented Aug 9, 2014 at 12:41
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    $\begingroup$ If the question had been $E[F|M+F=10]$ then the answer would have been $\frac{10}{3}$ as the expectation of a binomial distribution with $n=10$ and $p=\frac13$. $\endgroup$
    – Henry
    Commented Aug 9, 2014 at 12:50

1 Answer 1

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When each point of a homogeneous Poisson point process (HPPP) of intensity (rate) $\lambda$ is marked (selected) with a probability $p$, independently of the other points, the result is two independent HPPPs (consisting of the marked and non-marked points) with intensities $\lambda p$ and $\lambda (1-p)$, respectively.

In this case, we have a HPPP representing the customer arrivals and each customer is randomly "marked" as being male. Therefore, $\mathbb{E}(F|M) = \mathbb{E}(F)$.

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