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Suppose $\lambda > 0$ and $p \in (0, 1)$. The number of fish caught in one day follows a $\operatorname{Poi}(\lambda)$ distribution. At the end of the day, each fish is examined independently. With probability $p$, each fish passes inspection. Let $X$ be the number of fish that pass inspection. What is the moment-generating function of $X$?

The moment-generating function is given by $\operatorname E(e^{tX})$. I tried applying the law of total expectation knowing that the number of fish gives a partition of the space. This gives $$ \operatorname E (e^{tX}) = \sum_{i=0}^\infty \operatorname E (e^{tX} \mathbin{|} N = i) \operatorname P ( N = i ) $$ where $N \sim \operatorname{Poi} (\lambda) $ is the number of fish caught in the day. The expectation under the sum is then exactly the moment-generating function of a binomially distribution variable with number of trials $i$. So I get, $$ \sum_{i=0}^\infty \operatorname E (e^{tX} \mathbin{|} N = i) \operatorname P ( N = i ) = \sum_{i=0}^\infty (1 - p + pe^t)^i e^{-\lambda} \frac{\lambda^i}{i!} $$ but I'm not sure how to proceed in the evaluation of this series.

One idea I have is to combine the factors to the power $i$ under the sum to try turning the sum into a sum of the probability mass function of $\operatorname{Poi}(\lambda(1 - p + pe^t))$, which will go to one. \begin{align*} \sum_{i=0}^\infty (1 - p + pe^t)^i e^{-\lambda} \frac{\lambda^i}{i!} &= e^{-\lambda} \sum_{i=0}^\infty \frac{(\lambda(1-p+pe^t))^i}{i!} \cdot \frac{e^{-\lambda(1-p+pe^t)}}{e^{-\lambda(1-p+pe^t)}} \\ &= e^{-(1-p+pe^t)} \sum_{i=0}^\infty e^{-\lambda(1-p+pe^t)} \frac{(\lambda(1-p+pe^t))^i}{i!} \\ &= e^{-(1-p+pe^t)} \end{align*}

Is this procedure legitimate? It seems wrong to me as taking the first derivative of the moment-generating function and evaluating at $t=0$ gives $-p$, meaning that the expectation of $X$ would be negative!

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I think I see some mistake here, you have $$e^{-\lambda}\cdot e^{\lambda(1-p+pe^t)}=e^{-(1-p+pe^t)},$$ while it should be $$e^{-\lambda(p-pe^t)}.$$

Then the first moment will be $\lambda p.$

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  • $\begingroup$ You are absolutely right. It was merely a calculation error on my part. $\endgroup$ – Jacob Errington Apr 19 '17 at 21:08

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