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A form of cumulative distribution

Let $X$ and $Y$ be two continuous independent RVs with $f(x)$ and $g(x)$ as probability density functions, respectively. Assume that $E[Y]>E[X]$. Now, I have found numerically that the expression: $D=\frac{1}{2}\int_{-\infty}^{\infty} \min(f(x),g(x)) dx $ which describes half the 'overlap area' of the two densities, is a rough approximation of: $ \Pr (Y \le X) = \int _{-\infty}^{\infty}\left[f(x)\int _{-\infty}^{x}g(y)\,dy\right]\,dx $

How can I formally show that the approximation holds for any two densities for which $E[Y]>E[X]$, and how can I quantify the strength of the approximation?

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    @Dilip Thanks for the information. Asking again and again small variations of the same question seems to be all the rage these days... (About the condition E(Y)>E(X), you are right but one can modify slightly $f$ to get some $g$ with E(Y)>E(X), $P(Y\ge X)$ close to $\frac12$ and the former $D$ close to $1$.)2012-07-05

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Call $Q=\mathrm P(Y\leqslant X)$. Note that $2Q-1=2\displaystyle\int_{\mathbb R} f(x)\int_{-\infty}^x (g-f)\cdot\mathrm dx$ and $2-4D=\displaystyle\int_{\mathbb R}|g-f|$.

Since $-(g-f)^-\leqslant g-f\leqslant(g-f)^+$ and $f\geqslant0$, $-2\displaystyle\int_{\mathbb R} (g-f)^-\leqslant 2Q-1\leqslant2\int_{\mathbb R} (g-f)^+$. Now, $\displaystyle2\int_{\mathbb R} (g-f)^+=2\int_{\mathbb R} (g-f)^-=\displaystyle\int_{\mathbb R}|g-f|$ , hence $ |2Q-1|\leqslant2(1-2D), $ in full generality (the hypothesis that $\mathrm E(Y)\gt\mathrm E(X)$ is not needed). Thus, $Q\to\frac12$ when $D\to\frac12$ (that is, in a sense, when $f$ and $g$ are close). By contrast, the inequality above is not informative when $D\leqslant\frac14$.

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    Yes, but I assume (not explicitly stated) that$f$and$g$are continuous symmetric uni-modal distributions. I think in that case$D$approximates$Q$to some degree.2012-07-06