The definition of $F_X(X)$

probabilityprobability theory

I have been reading the Wikipedia article on the probability integral transform. Given a continuous random variable $X$ with cumulative distribution function $F_X$, the article defines a random variable $Y = F_X(X)$ and it shows that $Y$ has a standard uniform distribution. But by definition of the cumulative distribution function, $F_X(x) = P(X \leq x),$ so shouldn't $$F_X(X) = P(X \leq X) = 1,$$ no matter what value $X$ takes? In that case, the probability integral transform theorem wouldn't hold, so I suspect that there's something I'm getting wrong about the definition of $F_X(X)$.

(Research I've done on StackExchange prior to asking this question: I found a similar question on What is the meaning of $F_X(X)$?. However, I could not find an answer to my question in the comments there.)

Best Answer

Indeed, the symbol $F_X(X)$ is confusing.

A more proper way of expressing it:

Say we have two random variables $X, X'$, which are independent and have the same distribution, i.e. $F_X = F_{X'}$. We then define $Y = F_X(X')$, namely the function $F_X$ applied to the random variable $X'$.

We may calculate the distribution function of $Y$: \begin{eqnarray} F_Y(y) &=& P(Y \leq y)\\ &=& P(F_X(X') \leq y)\\ &=& P(X' \leq F_X^{-1}(y))\\ &=& F_{X'}(F_X^{-1}(y))\\ &=& F_X(F_X^{-1}(y))\\ &=& y. \end{eqnarray}

You will notice that the random variable $X'$ is kind of a dummy, which isn't very useful. In fact it doesn't really matter whether $X$ and $X'$ are independent or not, as we are only using $X$ to define its distribution function $F_X$.

Thus people tend to just write $F_X(X)$.

Related Question