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I'm reading and article and I came across a confusing formula

Suppose a realised value from a random variable is drawn. The variable is given by the cumulative distribution function F(x) with the support $[0,\bar{x}]$. The function $F$ is assumed to have the density $f$. Then the probability of drawing value x is

1) $∫_{0}^\bar{x}F^{-1}(x)f(x)dx$

Is this even correct? Why would the probability be the integral of the product of the inverse CDF and the PDF?

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  • $\begingroup$ I suppose $F_X^{-1}$ is the quantile function with domain $[0, 1]$. So the first thing you may need to clarify and check is the limits of the integral. And we know that $F_X^{-1}(U)$ has the same distribution as $X$ (where $U$ is the standard uniform), so $E[F_X^{-1}(U)] = E[X]$ and that is all I know. $\endgroup$ – BGM Aug 2 '17 at 16:27
  • $\begingroup$ The limits (i.e. xbar) are unspecified. I think they might be arbitrary and so not be restricted to [0,1]. $\endgroup$ – pafnuti Aug 2 '17 at 16:42
  • $\begingroup$ The integral can be interpreted as $E[F_X^{-1}(X)]$. As said, I just doubt if the quantile function is undefined outside $[0, 1]$. And secondly, if $X$ has a pdf and thus is a continuous random variable, I just doubt why you will ask the probability of drawing a particular value $x$, which is obviously $0$ unless I completely misunderstood the whole set up. Please correct me. $\endgroup$ – BGM Aug 3 '17 at 6:36

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