A log-gamma random variable, $Y$, is defined as,
\[
Y=\exp(-X)
\] where $X \in [0, \infty)$ follows a gamma distribution, and $Y \in (0, 1]$.
The probability density function of the log-gamma distribution is
\[
f_Y(y)=\frac{b^a\left\{\log(1/y) \right\}^{a-1}y^{b-1}}{\Gamma(a)},
\] where $b > 0$, and $a > 0$.
The expected value of $Y$ is
\[
\mathrm{E}[Y]=\int_0^1 yf_Y(y) \mathrm{d}y=\left( \frac{b}{1+b} \right )^a.
\] The variance of $Y$ is
\[
\mathrm{Var}[Y]=\mathrm{E}[Y^2]-(\mathrm{E}[Y])^2= \left( \frac{b}{2+b} \right)^a - \left( \frac{b}{1+b} \right)^{2a}.
\]
R code:
# shape = a, rate = b a <- 4 b <- 7 r <- -rgamma(100000, shape = a, rate = b) mean(exp(r)) (b/(1 + b))^a mean(exp(r - log((b/(1 + b))^a))) var(exp(r)) (b/(2 + b))^a - (b/(1 + b))^(2*a)
Result:
> mean(exp(r)) [1] 0.5855565 > (b/(1 + b))^a [1] 0.5861816 > > mean(exp(r - log((b/(1 + b))^a))) [1] 0.9989336 > > var(exp(r)) [1] 0.02245031 > (b/(2 + b))^a - (b/(1 + b))^(2*a) [1] 0.0223414
Note
\[
\int_0^1 yf_Y(y) \mathrm{d}y= \frac{b^a}{\Gamma(a)} \int_0^1 \left\{\log(1/y) \right\}^{a-1}y^b \mathrm{d}y
\] \[
\begin{align*} z &= \log(1/y) \\
\mathrm{d}z &= -\frac{1}{y} \mathrm{d}y \\
\Re(z) &> 0, \log(1/0)=\infty, \log(1/1)=0.
\end{align*}
\] \[
\begin{align*}
\int_0^1 yf_Y(y) \mathrm{d}y &= \frac{b^a}{\Gamma(a)} \int_{\infty}^0 -z^{a-1} e^{-z(1+b)} \mathrm{d}z \\
&= \frac{b^a}{\Gamma(a)} \frac{\Gamma(a)}{(1+b)^a} \\
&= \left( \frac{b}{1+b} \right)^a \end{align*}.
\]
