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

  • $\begingroup$ 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. $\endgroup$
    – Fiodor1234
    Commented May 11, 2021 at 9:14
  • $\begingroup$ 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? $\endgroup$
    – jdeJuan
    Commented May 11, 2021 at 9:33

1 Answer 1


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 \begin{equation} \{\omega \, | \, h(X(\omega))≤y\} = \{\omega \, | \, X(\omega)≤ h^{-1}(y)\} \end{equation} 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.

  • $\begingroup$ Thanks for your comment, really helpfull. I have recently started studying measure theory so not really confortable with all its concepts. Could you please ellaborate a bit more on your claim: "then you'd better hope that the P-measure of those sets are the same! If not, we would be in serious trouble". I agree with your first point, but dont really understand this later one. I guess it is because measure has to be preserved, but dont really know why $\endgroup$
    – jdeJuan
    Commented May 11, 2021 at 9:32
  • $\begingroup$ There isn't really anything to it. What does it mean for two sets to be equal? They are the same - you're just giving it two different names. If x = y then obviously f(x) = f(y) right! $\endgroup$ Commented Jun 1, 2021 at 1:46

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