Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Possible Duplicate:
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?

share|cite|improve this question

marked as duplicate by Dilip Sarwate, Alex Becker, t.b., William, Rudy the Reindeer Sep 4 '12 at 18:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Are you sure about your finding? For example, if $f=g$, then $D=1$ and $P(Y\leqslant X)=\frac12$. – Did Jul 5 '12 at 16:06
Where does the factor of $2$ come from in your expression for Pr($Y \le X$)? – mjqxxxx Jul 5 '12 at 16:12
Sorry, I now fixed the factor of 2. – Omri Jul 5 '12 at 16:16
It's not a duplicate since now I am asking about showing the approximation formally, and quantifying it... – Omri Jul 5 '12 at 16:20
@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 $1$.) – Did Jul 5 '12 at 16:20

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 $D\leqslant\frac14$.

share|cite|improve this answer
Thanks much, did. However, as you also note, your bound on |2Q-1| is really only meaningful when Q is large. In my application, I need to show that (and see how closely) D approximates Q when they are both very small (say, < 0.1)... – Omri Jul 6 '12 at 11:14
Unfortunately, when $D$ is small you really can't say anything about $Q$. In fact you could have $D = 0$ if $f$ and $g$ are supported on disjoint sets (e.g. $f=0$ on the intervals $[n,n+1]$ where $n$ is odd and $g=0$ when $n$ is even), and $Q$ could be anywhere in $[0,1)$. – Robert Israel Jul 6 '12 at 17:16
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. – Omri Jul 6 '12 at 18:02

Not the answer you're looking for? Browse other questions tagged or ask your own question.