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I've been facing this - i don't even know how to call it - problem for a few hours now and I have know idea how to "do" this. enter image description here

I mean... I feel like this has something to do with binomality of Poisson distribution and MGF of Poisson distribution, but that as far as I could get. My question being: how do I differentiate the first equality to get the later one?

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    $\begingroup$ You use the formula for $\mu_r$ to 1) write down a formula for $\mu_{r+1}$, 2) write down a formula for $\mu_{r-1}$, and 3) find $d\mu_r/d\lambda$. Then you plug those three results into the recursion you're given, and check that the equation holds. $\endgroup$ – Gerry Myerson Apr 15 at 7:21
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To address your specific question, "how do I differentiate the first equality to get the later one," the calculation of the derivative is straightforward via logarithmic differentiation. Let $$a_x(\lambda) = (x - \lambda)^r \frac{\lambda^x e^{-\lambda}}{x!}.$$ Then $$\log a_x(\lambda) = r \log(x - \lambda) + x \log \lambda - \lambda - \log x!,$$ from which it easily follows that $$\frac{1}{a_x(\lambda)} \frac{da_x}{d\lambda} = \frac{r}{\lambda - x} + \frac{x}{\lambda} - 1.$$ All that remains is to simplify and perform the rest of the computations to establish the identity.

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