Distribution function weirdness The distribution function of a random variable $X$ is defined by $F(t)=P(X\leq t)$. If I have another random variable $Y$, then $F(Y)$ is also a random variable. But, using the definition, we have $F(Y)=P(X\leq Y)$, which is a number. A simple example I invented myself: $X\sim\text{Exp}(1)$, $F(t) = 1 - e^{-t}$, $Y\sim\text{Exp}(1)$ independent of $X$, then $F(Y)=1-e^{-Y}$ and $F(Y)=P(X<Y)=1/2$. What am I doing wrong here? Didn't Aristotle taught us that a thing is equal to itself? Please, be gentle. I'm a total prob noob. Thank you very much.

Just following the excellent comments, if you formalize everything carefully, the apparent contradiction disappears. We have an underlying probability space $(\Omega,\mathscr{F},P)$ and a measurable space $(\mathbb{R},\mathscr{B})$, in which $\mathscr{B}$ is the Borel sigma-field. I'll suppose all random objects are defined in the same probability space.
Despite the name, by a random "variable" $X$ we mean a function
\begin{align}
  X &: \Omega \to \mathbb{R} \\
  &: \omega \mapsto X(\omega)
\end{align}
which is measurable, in the sense that $X^{-1}(B)=\{\omega\in\Omega: X(\omega)\in B\} \in \mathscr{F}$, for every $B\in\mathscr{B}$.
The same holds for the random variable $Y$.
Now, the distribution function of $X$ is the mapping $F_X:\mathbb{R}\to[0,1]$ defined by
$$
  F_X(x) = P\{\omega\in\Omega:X(\omega)\leq x\}.
$$
Composing $Y$ and $F_X$ we get a random variable $F_X(Y)=F_X\circ Y$, which is just the measurable function
\begin{align}
   F_X(Y) &: \Omega \to [0,1] \\
  &: \omega \mapsto F_X(Y)(\omega) = P\{\omega'\in\Omega : X(\omega')\leq Y(\omega)\}. \qquad\qquad (*)
\end{align}
Now, compare this to the probability $P\{\omega\in\Omega:X(\omega)\leq Y(\omega)\}$. The innocent dash $'$ in $(*)$ makes a huge conceptual difference.
Hence, if you define things with care, you (and Aristotle) will be just fine. No contradictions here.
