Note: This section contains notes on the derivations made while authoring my book. The derivation details originate from public Q&A platforms and will naturally differ from those in the book chapters.
The goal is to calculate the expected value of the logarithm of gamma distribution, which is \[
E[lnX]
\] Where \(X\sim Gamma(\alpha,\beta)\)
Part I
To deal with the problem, we need to first figure out how to calculate the density of the transformed variable, which is the density \(f_Y(y)\) when \(Y=g(X)\).
Suppose \(Y=g(X)\), \(X\) is a random variable on probability space \((\Omega,\mathcal{F},\mathcal{P})\), and has continuous density \(f\).
\(\Rightarrow Y=g(X)\) has the distribution function \(F(g^{-1}(y))\).
Taking derivatives via the chain rule, we have \[
\begin{gather*}
f_Y(y)=\frac{dF(g^{-1}(y))}{dy}\\=\frac{dF(g^{-1}(y))}{dx}\frac{dx}{dy}\\
=\frac{dF(g^{-1}(y))}{dx}\frac{d[g^{-1}(y)]}{dy}\\
=f(g^{-1}(y))\frac{d[g^{-1}(y)]}{dy}
\end{gather*}
\] Recall that the derivative of an inverse function is \[
\begin{gather*}
\frac{d}{dy}[g^{-1}(y)]\\
=\frac{dx}{d[g(g^{-1}(y))]}\\
=\frac{1}{g'(g^{-1}(y))}
\end{gather*}
\] The density function of \(g(X)\) is then expressed as \[
\begin{gather*}
f_Y(y)=f(g^{-1}(y))\frac{1}{g'(g^{-1}(y))}\\
=\frac{f(g^{-1}(y))}{g'(g^{-1}(y))}
\end{gather*}
\]
Part II
Now that we have figured out how to calculate the density of the transformed variable, we can proceed to obtain the density of \(Y=lnX\).
Since \(X\sim \Gamma(\alpha,\beta)\), \(Y=g(X)=ln(X)\), we have \(X=e^y\). Because \(\beta\) is a scale parameter, its effect will be to shift the logarithm by \(log\beta\). Specifically, it will shift the logarithm by \(-log\beta\). So we can readily start with the case when \(\beta=1\).
That gives us the distribution of \(X\) as \[
f_X(x)=\frac{1}{\Gamma(\alpha)}x^{\alpha}e^{-x}\frac{1}{x}
\] Where \(\Gamma(\alpha)=\int_0^{\infty}x^{\alpha}e^{-x}\frac{dx}{x}\) is the normalizing constant.
Substituting \(x=e^y\) entails \(dy=\frac{dx}{x}\) as \(\frac{de^y}{dy}=e^y=x\), thus we have \[
\begin{gather*}
f_Y(y)=\frac{f(g^{-1}(y))}{g'(g^{-1}(y))}\\
=\frac{1}{\Gamma(\alpha)}e^{\alpha y}e^{-e^y}\frac{1}{e^y}/(\frac{1}{e^y})\\
=\frac{1}{\Gamma(\alpha)}e^{\alpha y-e^y}
\end{gather*}
\] The possible values of \(Y\) now range over all the real numbers \(\mathbb{R}\).
Part III
As \(f_Y\) must integrate to unity, as \[
\begin{gather*}
\int_{\mathbb{R}}f_Y(y)dy=\int_{\mathbb{R}}\frac{1}{\Gamma(\alpha)}e^{\alpha y-e^y}dy\\
=\frac{1}{\Gamma(\alpha)}\int_{\mathbb{R}}e^{\alpha y-e^y}dy\\
=1
\end{gather*}
\] we can obtain \[
\Gamma(\alpha)=\int_{\mathbb{R}}e^{\alpha y-e^y}dy
\] To calculate \(E(Y)=\int_{\mathbb{R}}yf_Y(y)dy\), we need to replace \(y\) with \(\alpha\) in order to eliminate the uncertainty.
To do so, we notice that \(f_Y(y)\) is a differentiable function of \(\alpha\), as \[
\frac{d}{d\alpha}e^{\alpha y-e^y}=ye^{\alpha y-e^y}\\
=y\Gamma(\alpha)f_Y(y)\\
\] Since we can obtain \(e^{\alpha y-e^y}=\Gamma(\alpha)f_Y(y)\) from \[
f_Y(y)=\frac{1}{\Gamma(\alpha)}e^{\alpha y-e^y}
\] Now that we can calculate the \(E(Y)\), as \[
\begin{gather*}
E(Y)=\int_{\mathbb{R}}yf_Y(y)dy\\
=\frac{1}{\Gamma(\alpha)}\int_{\mathbb{R}}\Gamma(\alpha)yf_Y(y)dy\\
=\frac{1}{\Gamma(\alpha)}\int_{\mathbb{R}}\frac{d}{d\alpha}e^{\alpha y-e^y}dy\\
=\frac{1}{\Gamma(\alpha)}\frac{d}{d\alpha}\int_{\mathbb{R}}e^{\alpha y-e^y}dy\\
=\frac{1}{\Gamma(\alpha)}\frac{d}{d\alpha}\Gamma(\alpha)\\
=\frac{d}{d\alpha}ln\Gamma(\alpha)\\
=\Psi(\alpha)
\end{gather*}
\] \(\Psi(\cdot)\) is the digamma function, which is the logarithmic derivative of the gamma function.
By restoring the scale parameter \(\beta\) value, we can obtain the general result of \(E[lnX]\) as \[
E[lnX]=\Psi(\alpha)-ln\beta
\]
Reference
How to calculate the density function of \(g(X)\)?
What is the expected value of the logarithm of Gamma distribution?