Prove that $f(x)=exp(-x-e^{-x})$ for $x\in \mathbb{R}$ is a probability density function and find the cumulative density function.
I think that by proving that $f(x)$ is a pdf, it should be fairly straightforward to find the cdf (if this is not true, would like a hint). I want to focus on the first part.
I am using the following definition.
A continuous random variable X is a RV whose c.d.f. satisfied
$F_X(x)=\mathbb{P}(X\leq x)=\int_{-\infty}^x f_X(u) du$
where $f_X:\mathbb{R} \rightarrow \mathbb{R}$ is a fct s.t. (a)$f_X(u) \geq 0$, $ \forall u \in \mathbb{R}$ and (b) $\int_{-\infty}^{\infty}f_X(u)du=1$
$f_X$ is called the p.d.f. of X
So my approach is to basicaly prove that $f(x)$ satisfies conditions (a) and (b).
Condition (a) seems fairly easy.
$exp(-x-e^{-x})=e^{-x-e^{-x}}=(\frac{1}{e})^{x+\frac{1}{e^x}}=(\frac{1}{e})^{x}(\frac{1}{e})^{\frac{1}{e^x}}$
Since $e$ is a positive constant, the function is also positive for all $x\in \mathbb{R}$. My question is how to prove the second condition, i.e. prove that
$\int_{-\infty}^{\infty}(\frac{1}{e})^{u}(\frac{1}{e})^{\frac{1}{e^u}}du=1$
My attempt so far:
Let $x=e^{-u}$
$dx=-e^{-u} \rightarrow -dx=e^{-u}$
$\int_{-\infty}^{\infty}e^{-u-e^{-u}}du=\int_{-\infty}^{\infty}e^{-u}e^{-e^{-u}}du=\int_{-\infty}^{\infty}-e^{-x}dx=e^{-x}|^{\infty}_{-\infty}=e^{-e^{-u}}|^{\infty}_{-\infty}$
What should I do from here?
Best Answer
Hint:
$$ \frac{d}{dx} exp\left(-\mathrm{e}^{-x}\right) = exp\left(-x -\mathrm{e}^{-x}\right) $$