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This is exercise 5.2.3 (b) from One Thousand Exercises in Probability by Grimmett and Stirzaker:

Let $X_1,X_2,\ldots$ be independent identically distributed random variables with the logarithmic mass function $$f(k) = \frac{(1-p)^k}{k\log(1/p)},\quad k\geqslant 1, $$ where $0<p<1$. If $N$ is independent of the $X_i$ and has the Poisson distribution with parameter $\mu$, show that $Y=\sum_{i=1}^N X_i$ has a negative binomial distribution.

I computed the probability generating function of $X_1$: $$ P_X(s) := \mathbb E[s^{X_1}] = \sum_{k=1}^\infty \frac{((1-p)s)^k}{k\log(1/p)} = \frac{\log(1-s(1-p)}{\log p}, $$ and it is known that the probability generating function of $N$ is $P_N(s)=e^{\mu(s-1)}$. So the probability generating function of $Y$ is given by the composition $P_N\circ P_X$: \begin{align} P_Y(s) &= P_N(P_X(s))\\ &= P_N\left(\frac{\log(1-s(1-p)}{\log p}\right)\\ &= e^{\mu\left(\left(\frac{\log(1-s(1-p)}{\log p}\right)-1\right)}.\tag1 \end{align} The solution provided writes this in the form $$ G_Y(s) = \left(\frac p{1-s(1-p)} \right)^{-\mu/\log p}.\tag2 $$ I don't see how $(1)$ is equivalent to $(2)$; any hints?

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$$ e^{\mu\left(\left(\frac{\log(1-s(1-p))}{\log p}\right)-1\right)} = e^{\frac{\mu}{\log p} \left( \log(1-s(1-p))-\log p\right)} = e^{\frac{\mu}{\log p} \left(\log\left(\frac{1-s(1-p)}{p}\right)\right)} = e^{\log\left(\left(\frac{1-s(1-p)}{p}\right)^{\frac{\mu}{\log p}}\right)} = \left(\frac{1-s(1-p)}{p}\right)^{\frac{\mu}{\log p}} = \left(\frac{p}{1-s(1-p)}\right)^{-\frac{\mu}{\log p}} $$

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  • $\begingroup$ Thanks, it makes sense now. $\endgroup$ – Math1000 Jan 19 at 12:12

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