# Distribution of $Y=e^{-X}$ for Gamma-distributed $X$

Suppose we have a random variable $$X \sim \text{Gamma}(\alpha, \,\beta)$$ with shape $$\alpha>0$$, rate $$\beta > 0$$, and pdf \begin{align} f_X(x) = \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha - 1} e^{-\beta x}, \quad \text{for } x \in (0,\,\infty). \end{align} I'm curious about the properties of $$Y=g(X)=e^{-X}$$ and if it is another well-known distribution. Since $$g(X)$$ is strictly decreasing in $$X$$, and thus has inverse $$g^{-1}(X) = -\log(X) = \log\left(\frac1X\right)$$, we can find \begin{align} f_Y(y) = f_X\left(g^{-1}(y)\right)\left|\frac{d}{dy} g^{-1}(y)\right| = \frac{\beta^\alpha}{\Gamma(\alpha)} \left[\log\left(\frac1y\right)\right]^{\alpha - 1} y^{\beta -1}, \quad \text{for } y \in (0,\,1). \end{align} However, "neatly" obtaining the cdf $$F_Y(y)$$, mean $$\text{E}[Y]$$, variance $$\text{Var}[Y]$$, MGF $$M_Y(t)$$ etc. from this pdf seem rather difficult to do, as trying to integrate $$f_Y(y)$$ likely involves the incomplete gamma function. Is there some insight I'm missing (perhaps a change of variable or something similar) that would allow one to compute these integrals by hand?

EDIT: I actually realized after posting this that the moments of $$Y$$ are not too difficult to compute if one makes the substitution $$\log\left(\frac1y\right) = u$$. A similar thing can be done for the higher-order moments.

Using the notation on the wiki page, for the CDF, you can proceed by noting that $$F_Y(y) = P(Y \le y) = P(e^{-X} \le y) = P (X> -\ln y) = 1-F_X(-\ln y),$$ and since $$F_X(x) = \frac{1}{\Gamma(\alpha)} \gamma(\alpha,\beta x)$$, we have $$F_Y(y) = 1- \frac{1}{\Gamma(\alpha)} \gamma(\alpha,-\beta \ln (y))$$ For the expectation, we can recall that the MGF of a gamma random variable is $$m_X(t) = E(e^{tX}) = \left ( 1- \frac{t}{\beta} \right)^{-\alpha}, \quad t < \beta,$$ Then note that $$E(Y) = E(e^{-X}) = m_X(-1) = \left ( 1+ \frac{1}{\beta} \right)^{-\alpha}$$ Similarly, $$E(Y^2) = E(e^{-2X}) = m_X(-2),$$ and so you can use this to compute $$V(Y) = E(X^2) - [E(X)]^2$$