# Clarify one step in the probability integral transform for $Y = h(X)$: $P(Y \leq y ) = P( h(X) \leq y) = P(X \leq h^{-1}(y))$

in the proof of, e.g the probability integral transform, Sklar's theorem etc, there is a step I don't understand. Suppose that $$h()$$ is a monotonic transformation, hence invertible. Assume, for simplicity (although this can be generalized with the general inverse), that $$h()$$ is also a bijection.

Now suppose we have a random variables $$X,Y$$ such that $$Y = h(X)$$. We can write the probability distribution function of $$Y$$ as:

$$P(Y \leq y) = P(h(X) \leq y)$$

In the proofs then, there is a step which writes:

$$P(h(X) \leq y) = P(h^{-1}(h(X)) \leq h^{-1}(y)) = P(X \leq h^{-1}(y))$$

My problem is that I don't understand why this relation holds. Suppose $$y = 0.65$$, I can understand that if $$h(X) \leq 0.65$$ then $$X \leq h^{-1} (0.65)$$ because the inverse is also monotonic, hence if we apply the inverse on both sides, the inequality still holds. However, I cannot understand why the probability mass is also the same.

In other words. I agree that:

$$h(X) \leq 0.65\\ X \leq h^{-1} (0.65)$$

However, I dont see how the following identity holds from the fact that $$h()$$ is monotonic:

$$P(h(X) \leq 0.65) = P(X \leq h^{-1} (0.65))$$

In other words, why the amount of probability mass follows that equality.

Thanks to everybody

• I think it has to do first with the fact that the function $h$ is monotonic so it is measurable, and secondly with the definition of push-forward measure. Commented May 11, 2021 at 9:14
• Thank you, I am still getting familar with measure theory. How does the push-forward measure relates to this? It is the fact that the pushforward measure assigns similar measure when the function being applied is measurable? If so, where do I can find the proof? Commented May 11, 2021 at 9:33

Don't forget that the notation $$P(h(X)≤y)$$ is really shorthand for $$P(\{\omega \, | \, h(X(\omega))≤y\})$$ where $$\{\omega \, | \, h(X(\omega))≤y\} \subseteq \Omega$$ and $$P$$ is a measure on the probability space $$(\Omega, F, P)$$.
So if we're comfortable that $$$$\{\omega \, | \, h(X(\omega))≤y\} = \{\omega \, | \, X(\omega)≤ h^{-1}(y)\}$$$$ then you'd better hope that the $$P$$-measure of those sets are the same! If not, we would be in serious trouble.
Please note that you do have to be careful with $$h$$, it must be non-decreasing for this to hold as stated.