# A form of cumulative distribution

Let $f(x)$ and $g(x)$ be two probability density functions. Does the expression:

$$C = 2\int _{-\infty}^{\infty}\left[f(x)\int _{-\infty}^{x}g(y)\,dy\right]\,dx$$

have any meaningful graphical representation when $E[Y]>E[X]$, where $X$ has $pdf$ $f(x)$ and $Y$ has $pdf$ $g(x)$? Or, can it be expressed in a simpler fashion?

From some numerical simulations, it seems to approximately describe the "overlap" area of the two $pdf$'s. And if we let $g(x)==f(x)$ then $C$ is close to 1... The overlap is defined as, $$\int_{-\infty}^{\infty} \min(f(x),g(x)) dx$$

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There is a problem in the choice of variables: what is $\int_{-\infty}^xg(x)dx$? – Davide Giraudo Jul 1 '12 at 12:15
Yes, Henry, sorry. Will fix now. – Omri Jul 1 '12 at 12:28
You are missing that as I noted below we assume that $E[Y]>E[X]$. However, as I note in my question above, I suspect that $C$ is only approximately the 'overlap', so $C$ may be even slightly above 1. – Omri Jul 1 '12 at 22:07

If $X$ and $Y$ are independent continuous random variables with densities $f(x)$ and $g(y)$ then the probability that $Y$ is less than or equal to $X$ is
$$\Pr (Y \le X) = \int _{x=-\infty}^{\infty}[f(x)\int _{y=-\infty}^{x}g(y)dy]dx$$ and graphically is the probability of being below the line $y=x$.
$C$ is twice this. If $g(x)=f(x)$ for all $x$ then it is the probability of being above or below the line and (ignoring the zero measure of being on the line) is $1$.
So, given that $E[Y]>E[X]$, it does seem to portray the 'overlap area' of the two pdf functions, correct? – Omri Jul 1 '12 at 12:55
From Alice Through The Looking Glass: "When I use a word, it means just what I choose it to mean — neither more nor less.” Personally, I would not call this an "overlap" but so long as you are clear in your definition then you can use whatever language you like. You should note that it is possible to have $E[Y] \gt E[X]$ and $\Pr (Y \le X) \gt \frac12$ at the same time. – Henry Jul 1 '12 at 13:19
Thanks @Henry and Dilip. However, I am not sure my $C$ strictly equals the overlap, which I now define in the edited question above. – Omri Jul 1 '12 at 13:48
OK, I now verified via numerical methods that $C$ does not approximate the overlap, as defined in the original question, even for $E[Y]>E[X]$ (where $X$ has pdf $f(x)$ and $Y$ has $g(x)$). – Omri Jul 2 '12 at 13:53