# Let $X$ be a random variable and $f$ an invertible function. Then the CDF of a random variable $Y=f(X)$ always exists?

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?

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).
• What happens if $f$ is not increasing (or decreasing)? (I guess it can be still invertible if neither of those conditions is true)
• 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. Sep 25, 2015 at 15:46