What is the distribution of $T=\sum_{i=1}^{n} e^{-X_{i}}$?

Question

Let $ \ x_{1},x_{2},...,x_{n}$ a random sample with density $$f(x;\theta) = e^{-(x-\theta)} e^{-e^{-(x-\theta)}}$$ where $\theta \in \mathbb R$

What is the distribution of $T=\sum_{i=1}^{n} e^{-x_{i} }$

I try this way:

$m_{T}(t)=E(e^{tT})=E(e^{t\sum_{i=1}^{n} e^{-x_{i} }})=E(\prod_{i=1}^{n} e^{te^{-x_{i}}})= \prod_{i=1}^{n} E(e^{te^{-x_{i}}})= \prod _{i=1}^{n} m_{e^{-x_{i}}}(t)$

And then $m_{e^{-x}}(t)=E(e^{te^{-x}})= \int e^{te^{-x}} \left( e^{-(x-\theta)} e^{-e^{-(x-\theta)}} \right) dx$

I don´t sure what is the support.

Answer 1

Suppose $X$ has density

$$f(x;\theta)=e^{-(x-\theta)}\exp(-e^{-(x-\theta)})\quad,x \in \mathbb R, \,\theta\in \mathbb R$$

By change of variables, density of $Y=e^{-X}$ is

\begin{align}f_Y(y)&=f(-\ln y\,;\theta)\left|\frac{\mathrm d(-\ln y)}{\mathrm dy}\right|\mathbf1_{y>0} \\&=\frac{1}{y}e^{\ln y+\theta}\exp\left(-e^{\ln y+\theta}\right)\mathbf1_{y>0} \\&=e^{\theta}\exp\left(-ye^{\theta}\right)\mathbf1_{y>0} \end{align}

This means $Y$ is exponential with mean $1/e^{\theta}$.

As $e^{-X_i}$'s are i.i.d exponential, $T=\sum\limits_{i=1}^n e^{-X_i}$ has a Gamma distribution.