What is the variance of $\ln(x)$ if $x$ has a gamma distribution? $Var(\ln(x))$

Question

I know that if $X \sim Gamma(\alpha,\lambda) \implies E(\ln(X))=\psi(\alpha)-\ln(\lambda)$

If $X \sim Gamma(\alpha,1) \implies E(\ln(x))=\psi(\alpha)$

But what happens with $Var(\ln(x))$?

Answer 1

Using Mathematica, I get a rather curious result: $$\operatorname{Var}[\log X] = \psi'(\alpha),$$ which apparently does not depend on the rate parameter $\lambda$. I think this is because the logarithm transforms the rate parameter into a location.

The naive computation would be to write

$$\operatorname{E}[X^k] = \frac{\lambda^\alpha}{\Gamma(\alpha)} \int_{x = -\infty}^\infty x^k e^{-\lambda e^x + \alpha x} \, dx,$$ compute this for $k \in \{1, 2\}$, then simplify.

Another strategy might be to look at the MGF of $Y = \log X$: clearly $$M_Y(t) = \operatorname{E}[e^{t \log X}] = \operatorname{E}[X^t] = \frac{\Gamma(\alpha+t)}{\Gamma(\alpha) \lambda^t}.$$ Then computing the derivatives at $t = 0$ gives

$$\operatorname{E}[Y] = \left[\frac{dM_Y}{dt}\right]_{t=0} = \psi(\alpha) - \log \lambda, \\ \operatorname{E}[Y^2] = \left[\frac{d^2 M_Y}{dt^2}\right]_{t=0} = \psi'(\alpha) + (\log \lambda - \psi(\alpha))^2.$$