What is the expected value and variance of this distribution? I was solving a past exam on probability and I got to a point where I need to know the variance and expected value of the following random variable:
$P(X=k) = \frac{\beta^a (a)_k}{k! (1 + \beta)^{a+k}}, k =0,1,...$
Here $(a)_k$ denotes the Pochhammer symbol.
This is a distribution that we used as an example in class a few times, but we never calculated $E(X)$ and $\text{var}(X)$. I think it would be nice to have those values written down for the exam, but I am unable to compute them.
Therefore, my questions is what are $E(X)$ and $\text{var}(X)$ and perhaps a hint as to how I can compute them.
 A: We can write the distribution as
$$P(X=k) = C\,\frac{a^{\overline k}}{k!}\left(\frac{1}{1+\beta}\right)^k, \qquad k =0,1,...,$$
where $C = (\beta/(1+\beta))^a$, and $a^{\overline k} = a(a+1)\cdots (a+k-1)$ is the rising factorial. By Taylor's theorem,
$$ \sum_{k\ge 0}\frac{a^{\overline k}y^k}{k!} = \frac{1}{(1-y)^a},$$ which enables us to write the probability generating function, $G(z) = \sum_{k\ge 0} P(X = k)\,z^k$, as
$$G(z) = \frac{\beta^a}{(\beta + 1 - z)^a}.$$
We can now compute $EX = G'(1)$ and $E(X(X-1)) = G''(1)$. We have
\begin{align}
G'(z) &= \frac{a\beta^a}{(\beta + 1 - z)^{a+1}} \\ 
G''(z) &= \frac{a(a+1)\beta^a}{(\beta + 1 - z)^{a+2}},
\end{align}
which gives
\begin{align}
EX &= \frac{a}{\beta} \\
{\rm var}(X) & = E[X(X-1)] + EX - (EX)^2\\
& =\frac{a(1+\beta)}{\beta^2}.
\end{align}
A: The PMF in question may also be written
$$\Pr[X = k] = \binom{a+k-1}{k} \left(\frac{\beta}{1+\beta}\right)^a \left(\frac{1}{1+\beta}\right)^k, \quad k \in \{0, 1, 2, \ldots \}. \tag{1}$$  In this form, it becomes evident that $$X \sim \operatorname{NegativeBinomial}\left(r = a, p = \frac{\beta}{1+\beta}\right) \tag{2}$$ where the parametrization is $$\Pr[X = x] = \binom{r+x-1}{r-1} p^r (1-p)^x, \quad x \in \{0, 1, 2, \ldots \} \tag{3}$$ and has the interpretation that for positive integers $r$, $X$ counts the random number of failures before observing the $r^{\rm th}$ success in a sequence of iid Bernoulli trials with success probability $p$.
As such we can refer to this question and answer, although it uses a different parametrization, to compute the expectation and variance.  Knowing that $(3)$ and hence $(1)$ sum to $1$ over the support, it is a straightforward application of the identity
$$k \binom{a+k-1}{k} = a \binom{a+k-1}{k-1} \tag{4}$$ that lets us conclude for instance
$$\begin{align}
\operatorname{E}[X] &= \sum_{k=0}^\infty k \Pr[X = k] \\
&= a \sum_{k=1}^\infty \binom{a+k-1}{k-1} \left(\frac{\beta}{1+\beta}\right)^a \left(\frac{1}{1+\beta}\right)^k \\
&= \frac{a}{1+\beta} \cdot \frac{1+\beta}{\beta} \sum_{k=0}^\infty \binom{(a+1)+k-1}{k} \left(\frac{\beta}{1+\beta}\right)^{a+1} \left(\frac{1}{1+\beta}\right)^k \\
&= \frac{a}{\beta}. \tag{5}
\end{align}$$
The calculation of $\operatorname{E}[X(X-1)]$ follows similarly, from which the variance $$\operatorname{Var}[X] = \operatorname{E}[X^2]-\operatorname{E}[X]^2 = \operatorname{E}[X(X-1) + X] - \operatorname{E}[X]^2 = \operatorname{E}[X(X-1)] + \operatorname{E}[X](1-\operatorname{E}[X]) \tag{6}$$ also follows.
