Is the cumulative distribution function of a r.v. X strictly increasing (X -) almost everywhere?

Question

Let $X$ be a random variable and $F_X(x) = P(X \le x)$ its cumulative distribution function (cdf). $P_X$ is the probability measure induced by $X$, which is defined by $P_X((a,b)) = P(X^{-1}((a,b))$ for $a,b \in \mathbb{R}$ .

1. Is it correct to say that $F_X$ is strictly increasing $P_X$ almost everywhere?
2. If so, what would be the correct way to express that formally?
3. Does it imply $ F(x) = P(X \le x) = P(F(X) \le F(x))$ ?

If have seen that the equality in the 3. question is used to prove the probability integral transform. I think that it is not as straight forward to show 3. if $F_X$ is not strictly increasing.

I attempted to answer these questions, considering some very helpful comments from whuber and Zhanxiong (who provided detailed explanations of the probability transform here and here).

My attempt goes so far:

Is it correct to say that $F_X$ is strictly increasing $P_X$ almost everywhere?

Not really, since almost everywhere statements refer to properties of individual points. Strict monotonicity is not a property of a single point.

What would be the correct way to express that formally?

It holds that for any $a \le b$ with $F_X(a) = F_X(b)$ that $$P_X((a,b)) \le P_X((a,b])) = F_X(b) - F_X(a) = 0$$ So it is better to say that "Strict monotonicity fails to hold only on sets of measure 0" (by whuber)

Does this imply $F_X(x) = P(X\le x) = P(F(X) \le F(x))$ ?

Yes. See: Zhanxiong's answer for a shorter version. Edit: I have adapted this attempt of a proof a couple of times following recommendations from the comments.

Consider a fixed $x \in \mathbb{R}$. Because $F_X$ is monotonically increasing, $$ \{ X > x, F_X(X) \le F_X(x) \} = \{ X > x, F_X(X) = F_X(x) \}$$ It holds that $$P(\{ X> x, F_X(X) = F_X(x) \}) = 0 $$ To see this consider $x_0 $ s.t. $$x_0 = \text{sup}\{y\mid y> x, F_X(y) = F_X(x) \}$$ Then $F_X(y) = F_X(x)$ for all $x_0> y > x$ and thus $$P(\{X> x, F_X(X) = F_X(x) \}) = P_X((x, x_0)) =0$$

We have by monotonicity $$\{X \le x \} \subseteq \{F_X(X) \le F_X(x) \} $$ It follows that $$P(\{ F_X(X) \le F_X(x) \})= P(\{X\le x \})$$

Is there an easier way to show 3.? I expected it to straightforward.

Another question I have is,

4. Can we say that $F_X$ is strictly increasing on the image of $X$ ?

I am not sure about this, but I guess we can remove all values from the image of $X$ that occur with probability 0 and then it would be true.

Answer 1

To your specific question "Is there an easier way to show 3.?", the answer is "Yes". The problem of your "proof" is not only its verbosity (for example, I don't think it is necessary at all to introduce "$\omega$"s into the proof), but more importantly, it is probably incorrect (in my opinion many steps are hard to justify).

Here is my proof by discussing the range of $F(x)$.

Case 1: $F(x) = 1$. In this case $P[X \leq x] = F(x) = 1$. On the other hand, clearly we have $P[F(X) \leq 1] = 1$. Thus $P[X \leq x] = P[F(X) \leq F(x)]$.

Case 2: $0 \leq F(x) < 1$. For simplicity, denote by $C$ the set of all continuous points of $F$ and define $x^* := \sup\{y \geq x: F(y) = F(x)\}$. Because $F(x) < 1$, we have $x^* < \infty$. There are the following two sub-cases:

• $x^* \in C$. In this case \begin{align*} P[F(X) \leq F(x)] = P[X \leq x^*] = F(x^*) = F(x). \end{align*} To help you understand the geometry of this case, see the graph of $F(\cdot)$ below:

• $x^* \in C^c$. In this case \begin{align*} P[F(X) \leq F(x)] = P[X < x^*] = F(x^*-) = F(x). \end{align*} As in sub-case 1, below is the plot of illustrating the above identity.

This completes the proof.