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

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, $$F_X(X)\sim U(0,1).$$

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.

• Why do you say 'if any'? Every random variable has a distribution function. There is not much you can say abut the distribution of $f(X)$. It does not have a standard distribution function in general. – Kavi Rama Murthy Dec 28 '19 at 12:21
• For example, if $X$ follows a standard normal distribution, what is $E[f(X)]?$ – Idonknow Dec 28 '19 at 12:22
• I reckon the "if any" referred to the fact that the distribution may not be universal (as it was in the case with the cdf). Also, minor quibble: not every $X$ has a density to begin with. – Clement C. Dec 28 '19 at 12:22
• @ClementC. In this question, we assume that density function exists. – Idonknow Dec 28 '19 at 12:23

It is not universal, and will depend on $$X$$, so there is no general statement as was the case for the cdf transformation.
• 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.: $$\forall y\in[0,1],\qquad \mathbb{P}\{ f(X) \leq y \} = \mathbb{P}\{ X \geq -\log y \} = y$$
• Though your standard exponential distribution has $f(X)$ neatly with a uniform distribution on $[0,1)$. Others will not be as simple – Henry Jan 16 '20 at 10:22