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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.

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  • $\begingroup$ 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. $\endgroup$ – Kavi Rama Murthy Dec 28 '19 at 12:21
  • $\begingroup$ For example, if $X$ follows a standard normal distribution, what is $E[f(X)]?$ $\endgroup$ – Idonknow Dec 28 '19 at 12:22
  • $\begingroup$ 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. $\endgroup$ – Clement C. Dec 28 '19 at 12:22
  • $\begingroup$ @ClementC. In this question, we assume that density function exists. $\endgroup$ – Idonknow Dec 28 '19 at 12:23
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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.: $$\forall y\in[0,1],\qquad \mathbb{P}\{ f(X) \leq y \} = \mathbb{P}\{ X \geq -\log y \} = y$$

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  • $\begingroup$ Though your standard exponential distribution has $f(X)$ neatly with a uniform distribution on $[0,1)$. Others will not be as simple $\endgroup$ – Henry Jan 16 '20 at 10:22

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