Consider gamma random variable X∼Γ(α,θ). There are neat formulas for the mean, variance, and skewness:

E[X]=αθVar[X]=αθ2=1/α⋅E[X]2Skewness[X]=2/√α

Consider now a log-transformed random variable Y=log(X). Wikipedia gives formulas for the mean and the variance:

E[Y]=ψ(α)+log(θ)Var[Y]=ψ1(α)

via digamma and trigamma functions which are defined as the first and the second derivatives of the logarithm of the gamma function.

What is the formula for the skewness?Will tetragamma function appear?

(What made me wonder about this is a choice between lognormal and gamma distributions, see Gamma vs. lognormal distributions. Among other things, they differ in their skewness properties. In particular, skewness of the log of lognormal is trivially equal to zero. Whereas skewness of the log of gamma is negative. But how negative?..)

**Answer**

The moment generating function M(t) of Y=lnX is helpful in this case, since it has a simple algebraic form. By the definition of m.g.f., we have M(t)=E[etlnX]=E[Xt]=1Γ(α)θα∫∞0xα+t−1e−x/θdx=θtΓ(α)∫∞0yα+t−1e−ydy=θtΓ(α+t)Γ(α).

Let’s verify the expectation and the variance you gave. Taking derivatives, we have M′(t)=Γ′(α+t)Γ(α)θt+Γ(α+t)Γ(α)θtln(θ) and M″ Hence, \operatorname{E}[Y]=\psi^{(0)}(\alpha)+\ln(\theta),\qquad\operatorname{E}[Y^2]=\frac{\Gamma”(\alpha)}{\Gamma(\alpha)}+2\psi^{(0)}(\alpha)\ln(\theta)+\ln^2(\theta). It follows then \operatorname{Var}(Y)=\operatorname{E}[Y^2]-\operatorname{E}[Y]^2=\frac{\Gamma”(\alpha)}{\Gamma(\alpha)}-\left(\frac{\Gamma'(\alpha)}{\Gamma(\alpha)}\right)^2=\psi^{(1)}(\alpha).

To find the skewness, note the cumulant generating function (thanks @probabilityislogic for the tip) is K(t)=\ln M(t)=t\ln\theta+\ln\Gamma(\alpha+t)-\ln\Gamma(\alpha). The first cumulant is thus simply K'(0)=\psi^{(0)}(\alpha)+\ln(\theta). Recall that \psi^{(n)}(x)=d^{n+1}\ln\Gamma(x)/dx^{n+1}, so the subsequent cumulants are K^{(n)}(0)=\psi^{(n-1)}(\alpha), n\geq2. The skewness is therefore \frac{\operatorname{E}[(Y-\operatorname{E}[Y])^3]}{\operatorname{Var}(Y)^{3/2}}=\frac{\psi^{(2)}(\alpha)}{[\psi^{(1)}(\alpha)]^{3/2}}.

As a side note, this particular distribution appeared to have been thoroughly studied by A. C. Olshen in his *Transformations of the Pearson Type III Distribution*, Johnson et al.’s *Continuous Univariate Distributions* also has a small piece about it. Check those out.

**Attribution***Source : Link , Question Author : amoeba , Answer Author : Francis*