Let $X$ be a random variable and $f$ and invertible function. The cumulative distribution (CDF) of $X$ is defined as $$F_{X}(x) = \mathrm{P}(X\leq x).$$

The CDF of $Y=f(X)$ is then $$F_{Y}(y) = \mathrm{P}(Y\leq y) = \mathrm{P}(f(X)\leq y) = \mathrm{P}(X \leq f^{-1}(y)) = F_{X}(f^{-1}(y)).$$

It would appear that this is always true when $f$ is invertible. However, is it so? Is there anything I am missing?

How would I continue from here to get the probability density function of $Y$ as well?


1 Answer 1


Your argument is correct, though it would be worth specifying an increasing invertible function, otherwise the inequality would flip. To get the density function, you only need to differentiate the cdf.

One thing interesting to note, if you grant me the use of a little measure theory, the pdf exists whether or on not the function is invertible. All that is needed is that $f$ is a measurable function (a rather weak condition). Thinking in terms of events:

$$ P(Y \in E) = P(X \in f^{-1}(E)) $$

Here, $\ f^{-1}$ stands not for the inverse function, but the inverse image operator, which takes sets in the range of a function to sets in its domain. All that is needed now is that the probability on the right hand side exists, but the inverse image of a measurable set under a measurable function is always measurable (which is the very definition of a function being measurable).

  • $\begingroup$ What happens if $f$ is not increasing (or decreasing)? (I guess it can be still invertible if neither of those conditions is true) $\endgroup$
    – mmh
    Sep 25, 2015 at 15:40
  • $\begingroup$ In most cases that would be of interest, invertible implies strictly increasing or decreasing. More precisely, if $f$ is continuous (mapping real numbers to real numbers), and it's image is a connected subset, then it is either strictly increasing or strictly decreasing. You can convince yourself of this by drawing a few pictures. $\endgroup$ Sep 25, 2015 at 15:46

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