Consider the universality theorem of the uniform distribution. One way to formulate it is the following: Let $F:\mathbb{R}\rightarrow [0,1]$ be a right continuous, increasing function. Then, if $X\sim F$ (ie. $X$ is a random variable that has CDF $F$) then $F(X)\sim Uniform(0,1)$.

While I can prove it, I am confused about the concept of plugging $X$ (a random variable) into its CDF.

$X$ is a random variable, that is, X is a mapping from the sample space $S$ to $\mathbb{R}$ (considering 1-dimensional scenario). $X\sim F$ means $F(t)=P(X\le t)$ where "$X\le t$" is an event, "$X\le t$"$=\{s\in S: X(s)\le t\}$. The domain of $F$ is $\mathbb{R}$. How could we even imagine the concept of $F(X)$? $X$ is a function, coming from the space of functions, not the domain of $F$. And what would $F(X)$ even be, $P(X\le X)$? What event is "$X\le X$"? This is so weird and confusing.


$X$ is a function from $S$ to $\mathbb R$, $F$ is a function from $\mathbb R$ to $\mathbb R$, so $F(X)$ is just a composition of functions $F(X(s))$.

For example, let $X$ be an exponential random variable with unit mean. Then $F(t)=1-e^{-t}$ for $t\geq 0$. And $F(X)=1-e^{-X}$.

More concrete, for each $s\in S$, $X(s)\in\mathbb R$ is a real number, so $$ F(X(s)) = \mathbb P\{s'\in S~:~X(s')\leq X(s)\} $$ The event $\{X\leq X\}=S$ is not appeared here.

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    $\begingroup$ In short: trying to mix the shorthand notation for the event and the random variable itself is confusing, so use the set constructor representation to make clear which outcomes are being measured by $X$ $\endgroup$ – Graham Kemp May 3 at 1:54

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