Let $X$ be a random variable such that it's distribution function $F(x):=P(X \le x)$ is continuous.
Then is it true that $F(X)$ follows a uniform distribution ?
I can show this if $F$ is differentiable, but otherwise I'm not sure.
Please help.
Let $X$ be a random variable such that it's distribution function $F(x):=P(X \le x)$ is continuous.
Then is it true that $F(X)$ follows a uniform distribution ?
I can show this if $F$ is differentiable, but otherwise I'm not sure.
Please help.
Yes, it follows a uniform distribution.
We know that $F$ is non-decreasing, and we can further note that
$$P(F(X) \leq x) = P(X \in \{y \mid F(y) \leq x\})$$
Now set $M = \sup\{y \mid F(y) \leq x\} $ so that by continuity, we can obtain $F(M) = x$. It follows that $P(F(X) \leq x) = x$ for $0\leq x \leq 1$. In other words, $F(X)$ is uniformly distributed.
For convenience, consider that $F$ is absolutely continuous and strictly monotone. Then the random variable $U = F_{X}(x)$ has distribution given by \begin{align*} G_{U}(u) = \textbf{P}(U\leq u) = \textbf{P}(F_{X}(x)\leq u) = \textbf{P}(x \leq F^{-1}_{X}(u)) = F_{X}(F^{-1}_{X}(u)) = u \end{align*}
from whence we conclude $U = F_{X}(x)$ has uniform distribution.