Expected value of Borel distribution

Question

Recall that the Borel distribution is defined by the pmf:

$$f_X(n) = \frac{(\mu n)^{n-1}}{n!} e^{-\mu n}$$

I have been informed that the expected value for such a distribution is

$$\frac{1}{1-\mu}$$

I have seen arguments that prove that fact using the Galton-Watson branching process, but it lacked detail, and I am pretty ignorant about the conventions. So I am a bit over my head. Can the expectation be evaluated from first principles? I tried and did not get far:

$$ E(X) = e^{1-\mu} \left(\sum_{n=0}^\infty \frac{(\mu e)^n}{n!} (n+1)^n\right) $$

(by cancelling $n$, reindexing, factoring,).

Answer 1

Let $\nu = \mu e^{-\mu}$. Without loss of generality $0 < \mu < 1$. The fact that $f_X$ is a pmf means that $$ \sum_{n=1}^\infty \frac{n^{n-1}}{n!} \nu^n = \mu .$$ (It also follows by the power series for the Lambert $W$ function, but we won't use that fact.)

Differentiate both sides with respect to $\nu$, to get $$ \sum_{n=1}^\infty \frac{n^{n}}{n!} \nu^{n-1} = \frac{d\mu}{d\nu} ,$$ or $$ E(X) = \sum_{n=1}^\infty n \frac{(\mu n)^{n-1}}{n!} e^{-n\mu} = e^{-\mu} \frac{d\mu}{d\nu} \\ = e^{-\mu} \left(\frac{d\nu}{d\mu}\right)^{-1} = e^{-\mu} \frac1{e^{-\mu} - \mu e^{-\mu}} = \frac1{1-\mu} .$$ P.S. Differentiating both sides with respect to $\nu$ should get you the variance of $X$.