I'm trying to solve the following problem (Exercise 11.13) from Probability Essentials by Jacod and Protter: Let $X$ be a random variable (on $\mathbb{R}$) with distribution $F$ that is continuous. Show that the random variable $Y = F \circ X$ is uniform.
Here is what I've tried.
We first compute that
$$
F(X(x)) = P\{\omega \in \mathbb{R} : X(\omega) \le X(x)\}.
$$
We then find that if $G$ is the distribution of $F \circ X$, then
$$
G(x) = P\{P\{\omega \in \mathbb{R} : X(\omega) \le X(x)\}\le x\}.
$$
However, nested probabilities does not seem like the best route.
I then thought that I could use the following result: Given a random variable $X$ with density $f_X$ and a continuously differentiable $g : \mathbb{R} \to \mathbb{R}$ with a non-vanishing derivative, then the function $Y = g \circ X$ has the density
$$
f_Y(y) = f_X(h(y))|h'(y)|
$$
where $h = g^{-1}$.
I'd like to set $g = F$ in this result, and then I think the claim may follow.
However, this result requires that (i) my random variable $X$ has a density $f_X$ (which we haven't assumed), (ii) that $F$ is differentiable, (iii) that $F$ has a non-vanishing derivative. I can't believe that all three of these facts could be shown/taken for granted, so now I'm stuck and can't seem to find a solution.
Best Answer
For $0 < y < 1$, define $F^{−1}(y) = \inf\{x| F(x) \geq y\}$. Then we have 1) $\{x| F(x)<y\}=(-\infty,F^{−1}(y))$ and 2) $F(F^{−1}(y))=y$. The first equality is by definition of$F^{−1}(y)$ and the second is by continuity of $F$.
Let $x\to F^{−1}(y)^{-}$, $F(x)<y$, hence $F(F^{−1}(y))\leq y$ by continuity of $F$. Similarly, let $x\to F^{−1}(y)^{+}$, $F(x)\geq y$, hence $F(F^{−1}(y))\geq y$
Then, for any $0 < y < 1$, we have $P\{F(X) < y\} = P\{X < F^{−1}(y)\} = P\{X \leq F^{−1}(y)\} = F(F^{−1}(y)) = y$. The first equality follows from 1), the second follows from continuity of $F$, the last follows from 2)
Then by monotonicity of $F_Y$, we have $F_Y(y)=P\{F(X) < y\}=y, \, \forall 0<y<1$. So $Y$ is uniformly distributed.