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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.

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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 $V(Y) = E(X^2) - [E(X)]^2$

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