Why isn't $F(X) = 1$ when $F$ is the cdf of $X$?

Question

This is a really simple question. For a random variable $X$ and $F$ its cumulative distribution function. Since $F$ is defined as $F(t) = \mathbb{P}(X \leq t)$ for all $t$. I was wondering why we don't have $F(X) = \mathbb{P}(X \leq X) = 1$ because $\Omega = \{X \leq X\}$?

Answer 1

This is one downside of the shorthand notation in probability theory, like $\{X\leq x\}$ or $\mathbb{P}(X \leq x)$, that easily confuses new learners. To clear doubt, we must uncover the abbreviated part and see what $F(X)$ stands for.

Let us first write the event $\{X \leq x\}$ in full-form. This is the set of all $\omega \in \Omega$ for which $X(\omega) \leq x$ holds, i.e.,

$$ \{X \leq x\} = \{\omega \in \Omega : X(\omega) \leq x\}. $$

Then, the corresponding probability is

$$ F(x) = \mathbb{P}(X \leq x) = \mathbb{P}(\{\omega \in \Omega : X(\omega) \leq x\}). \tag{1} $$

Now, we would like to substitute $x = X$. Since $X$ is a random variable, which is a function of samples in $\Omega$, we may invoke the function notation and write $X(\bullet)$, where $\bullet$ is the placeholder for the argument of $X$. Hence,

$$ F(X(\bullet)) = \mathbb{P}(\{\omega \in \Omega : X(\omega) \leq X(\bullet) \}). $$

Now we see why $F(X)$ need not be (and in fact, is in general not) equal to $1$, because $X(\omega)$ and $X(\bullet)$ need not be equal.