Given a random variable $X$ and its density function $f(x),$ what is the distribution of $f(X)$?

Question

Probability integral transform states that if a random variable $X$ has a continuous distribution for which the cumulative distribution function (CDF) is $F_X$, then $F_X(X)$ has a standard uniform distribution, that is, $F_X(X)\sim U(0,1).$

My question is about its pdf instead of cdf.

Question: Given a random variable $X$ and its density function $f(x),$ what is the distribution of $f(X)$?

I have a feeling that $f(X)$ does not have a uniform distribution as density is the derivative of CDF. But I do not know what is the derivative of a uniform distribution.

Answer 1

It is not universal, and will depend on $X$, so there is no general statement as was the case for the cdf transformation.

For instance:

• if $X$ is uniform on $[0,1]$, then its pdf $f$ equals $\mathbf{1}_{[0,1]}$ and $f(X) = 1$ is constant a.s.

• if $X$ is say a standard exponential distribution, then its pdf is $f(x) = e^{-x}\mathbf{1}_{x \geq 0}$ and $f(X)$ is clearly not a constant r.v.: $$\forall y\in[0,1],\qquad \mathbb{P}\{ f(X) \leq y \} = \mathbb{P}\{ X \geq -\log y \} = y$$