What is the expected value of the logarithm of Gamma distribution?

If the expected value of Gamma(α,β) is αβ, what is the expected value of log(Gamma(α,β))? Can it be calculated analytically?

The parametrisation I am using is shape-rate.

Answer

This one (maybe surprisingly) can be done with easy elementary operations (employing Richard Feynman’s favorite trick of differentiating under the integral sign with respect to a parameter).


We are supposing X has a Γ(α,β) distribution and we wish to find the expectation of Y=log(X). First, because β is a scale parameter, its effect will be to shift the logarithm by logβ. (If you use β as a rate parameter, as in the question, it will shift the logarithm by logβ.) This permits us to work with the case β=1.

After this simplification, the probability element of X is

fX(x)=1Γ(α)xαexdxx

where Γ(α) is the normalizing constant

Γ(α)=0xαexdxx.

Substituting x=ey, which entails dx/x=dy, gives the probability element of Y,

fY(y)=1Γ(α)eαyeydy.

The possible values of Y now range over all the real numbers R.

Because fY must integrate to unity, we obtain (trivially)

Γ(α)=Reαyeydy.

Notice fY(y) is a differentiable function of α. An easy calculation gives

ddαeαyeydy=yeαyeydy=Γ(α)yfY(y).

The next step exploits the relation obtained by dividing both sides of this identity by Γ(α), thereby exposing the very object we need to integrate to find the expectation; namely, yfY(y):

E(Y)=RyfY(y)=1Γ(α)Rddαeαyeydy=1Γ(α)ddαReαyeydy=1Γ(α)ddαΓ(α)=ddαlogΓ(α)=ψ(α),

the logarithmic derivative of the gamma function (aka “polygamma“). The integral was computed using identity (1).

Re-introducing the factor β shows the general result is

E(log(X))=logβ+ψ(α)

for a scale parameterization (where the density function depends on x/β) or

E(log(X))=logβ+ψ(α)

for a rate parameterization (where the density function depends on xβ).

Attribution
Source : Link , Question Author : Stefano Vespucci , Answer Author : whuber

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