How to derived the mean and variance of Yule-Simon distribution?

Question

Yule-simon distribution is defined as $$f(x)=\rho B(x, \rho), \quad \text{for} \quad \rho > 0.$$

I've already show that this distribution sums to 1.

To show the moment and variance, I was trying to use the raw momemt and factorial momemt, but I failed.

Here is more information about Yule-simon distribution:

https://en.wikipedia.org/wiki/Yule%E2%80%93Simon_distribution

Answer 1

We know $(*)$ that if $X|W=w \sim \textrm{Geom}(e^{-w})$ and $W\sim \textrm{Exp}(\rho)$, then $X \sim \textrm{Yule-Simon}(\rho)$. Then if $\rho>2$ $$E[X]=E[E[X|W]]=E[e^{W}]=\rho\int_0^\infty e^{-x(\rho-1)}dx=\frac{\rho}{\rho-1}$$ $$E[X^2]=2E[e^{2W}]-E[e^{W}]=\frac{2\rho}{\rho-2}-\frac{\rho}{\rho-1}$$ $$V[X]=\frac{2\rho(\rho-1)^2-\rho(\rho-1)(\rho-2)-\rho^2(\rho-2)}{(\rho-2)(\rho-1)^2}=\frac{\rho^2}{(\rho-2)(\rho-1)^2}$$

---

$(*)$: to see this:

$$P(X= x)=E[P(X= x|W)]=E[(1-e^{-W})^{x-1}e^{-W}]=\rho\int_0^\infty(1-e^{-w})^{x-1}e^{-w(\rho+1)}dw$$ $$e^{-w}=y\implies dy=-e^{-w}dw$$ $$P(X= x)=\rho \int_0^1(1-y)^{x-1}y^{\rho}dy=\rho B(x,\rho+1)$$