I have a grid approximation of a cdf, $F_x$. The cdf has support for $x>=0$

From there, calculating the $E[x]$ is straight forward with some std numerical integration techniques.

In my case, however, I have a variable $y$, which is randomly distributed according to a pdf $f_y$.

$y$ defines the percentile in which $x$ is likely to be. The relationship is $y = F_x(x)$

My goal is to calculate the $E[x]$ by knowing that $y$ is distributed as $f_y$. Can I just simply calculate the $E[x]$ integrating with $f_y(x)$ mapping $y$ to $x$ using $x = F_x(y)^{-1}$ (the quantile function). Or is there something that gets lost there (which is my concern)?


I have been reflecting on the problem formulation.

In a Bayesian framework where $c$ represents client data and $s$ represents sales

we have:

$$p(s|c) = \frac{p(c|s) * p(s)}{p(c)}$$

In the current set up: $p(s)$ is an empirical $cdf$ - which looks very weird and I would not try to fit - while $c$ is a ranking on clients. I was wondering if in this scenario we get to a numerical estimate of $E[s|c]$ and what is the role played by the probability integral transform ?

  • $\begingroup$ That's fine. In general, if $y$ has some distribution and $x$ is some function of $y$, $h(y)=x$ then $E[x] = \int h(y) f_y dy$. This is the so called law of the unconscious statistician! $\endgroup$ – CloseToC Jun 10 '15 at 17:55
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    $\begingroup$ If $$Y = F_X(X)$$ then by the probability integral transform $y$ is necessarily distributed as a $U(0,1)$ Uniform, not according to some arbitrary distribution. $\endgroup$ – Alecos Papadopoulos Jun 10 '15 at 18:07
  • $\begingroup$ @Alecos: This is true. This is the missing piece. Tnx. But in my current setup I have, in some cases, information on the ranking, the distribution on $y$. How to incorporate that? $\endgroup$ – IcannotFixThis Jun 10 '15 at 20:06
  • $\begingroup$ You mean you have information on realizations of $Y$? $\endgroup$ – Alecos Papadopoulos Jun 10 '15 at 20:07
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    $\begingroup$ Unfortunately, you are, because the distribution of $Y$ if $Y$ is defined as above, is always $U(0,1)$, no matter what any estimation of it based on any finite sample may show. $\endgroup$ – Alecos Papadopoulos Jun 10 '15 at 22:01

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