The random variable $ Z = 1-F(X)$

Question

I will formulate the theorem (with no proof)

if $X \in \mathbb{R}$ is a random variable with continuous distribution function $F$ then the random variable $Z = 1-F(X)$ has a uniform distribution on $[0,1]$

but $Z = 1- P(X > X)$ or since $X$ is continuous $Z =1- P(X \geq X) $ in the first case $Z = 1$ and in the second case $Z = 0 $ ,but how can that be. I don't have the proof for this theorem, but something is confusing about $1- F(X)$. Could someone explain what Iam missing

Answer 1

First, $F(x)=\mathbb{P}(X\leq x)$, not $P(X>x)$.

$F(X)$ is not $P(X\leq X)$, because $P(X\leq X)$ is not a random variable, it's a number. $F(X)$ is a random variable. You cannot just plug $X$ in the definition of the cdf.

For example, let's say that $X$ has a standard exponential distribution. You have $F(x)=1-e^{-x}$. In this case $F(X)$ is the random variable $Y=1-e^{-X}$.

Finally, $1-F(X)\sim U([0,1])$ because if $V\sim U([0,1])$ then $1-V$ too. This is easy to prove.

Answer 2

$F(x)$ is the cdf of random variable $X$. So

\begin{align*} P(Z\leq z) = P(1-F(X)\leq z)&=P(F(X)\geq 1-z)\\ &=P(X\geq F^{-1}(1-z))\\ &=1-F(F^{-1}(1-z))\\ &=1-(1-z)\\ &=z, \end{align*}

where we used the fact that $F$ is continuous.