$X$ has distribution function $F(x) = e^{-e^{-x}}$. Justify that such a probability measure on $\mathbb{R}$ exists

Question

How can I prove a probability measure exists? If $F(x) \rightarrow 1$ as $n \rightarrow +\infty$, does that mean $F(x)$ does exist? And how should I calculate $\mathbb{E}(F(X))$ and $Var(F(X))$?

Answer 1

you need to show that

1. F(x) is non-decreasing, ie that $x\geq y$ implies $F(x)\geq F(y)$
2. The maximum value is $1$ (in limit $x\to\infty$)
3. The minimum value is $0$ (in limit $x\to -\infty$)

Answer 2

The proof you give is going to depend on how deep you need to go, but if you want to go deep into measure theory then this is essentially an application of the Carathéodory extension theorem. Specifically, you will need to apply the following theorem, which can be proved using the Carathéodory extension theorem. (I leave the proof as an exercise for you.)

Definition: A function $F: \mathbb{R} \rightarrow \mathbb{R}$ is a (scalar) cumulative distribution function if it is non-decreasing and right-continuous with limits $\lim_{x \rightarrow -\infty} F(x) = 0$ and $\lim_{x \rightarrow \infty} F(x) = 1$.

Theorem: For any (scalar) cumulative distribution function $F: \mathbb{R} \rightarrow \mathbb{R}$ there exists a unique probability measure $\mathbb{P}$ defined on the Borel sets of $\mathbb{R}$ with:$$\mathbb{P}((a,b]) = F(b)-F(a).$$

So, as you can see, you need to establish that $F$ is a valid (scalar) cumulative distribution function. To do this, you need to show that the function is non-decreasing, right-continuous, and has the two required limits. Once this is established, the above theorem guarantees that the distribution function corresponds to a unique probability measure on the Borel sets for the real number line.

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Proving the properties: The function $F$ is differentiable, with derivative:

$$f(x) \equiv \frac{dF}{dx}(x) = e^{-x} \cdot e^{-e^{-x}} = e^{-x-e^{-x}} > 0.$$

Since the function is differentiable over its entire domain it is also continuous, and hence right-continuous. Since $f(x) > 0$ over all $x \in \mathbb{R}$ it is also non-decreasing. Finally, we have the limits:

$$\begin{aligned} \lim_{x \rightarrow -\infty} F(x) &= \lim_{x \rightarrow -\infty} e^{-e^{-x}} \\[6pt] &= e^{-e^{\infty}} \\[6pt] &= e^{-\infty} \\[6pt] &= 0, \\[6pt] \lim_{x \rightarrow \infty} F(x) &= \lim_{x \rightarrow \infty} e^{-e^{-x}} \\[6pt] &= e^{-e^{-\infty}} \\[6pt] &= e^{0} \\[6pt] &= 1. \\[6pt] \end{aligned}$$

This establishes that $F$ is a valid (scalar) cumulative distribution function, which means that it corresponds to a unique probability measure over the Borel sets on the real number line.