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I have two independent, identically distributed normal random variables $X \sim N(0,\sigma^2)$, $Y \sim N(0,\sigma^2)$. I want to know

$$Pr[X-Y>4\sigma \text{ and } X<3\sigma \text { and } Y<3\sigma]$$

Of course, the condition $Y<3\sigma$ seems to be redundant, since we must have $Y<-\sigma$ in order to have $3\sigma>X>4\sigma+Y$. I also know the fact that $X-Y$ is itself a normal random variable with $X-Y \sim N(0,2\sigma^2)$. However, $X-Y$ and $X$ are not independent, so I don't know how to separate out the "and"s in the probability condition...

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The last condition seems indeed redundant. Are you sure you didn't forget some absolute values? –  Eckhard Jan 11 '13 at 23:58
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1 Answer

up vote 2 down vote accepted

Let $g(x)=\frac1{\sqrt{2\pi}}\mathrm e^{-x^2/2}$, then the probability $p$ you are interested in is $$ p=\int_{-\infty}^3g(x)\int_{-\infty}^{x-4}g(y)\mathrm dy\,\mathrm dx=\frac1{2\pi}\int_{-\infty}^3\int_{-\infty}^{x-4}\mathrm e^{-(x^2+y^2)/2}\mathrm dy\,\mathrm dx. $$ Obviously, $\Phi(1)\Phi(3)\leqslant p\leqslant\Phi(3)$ but I see no reason why $p$ should have an analytic expression.

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Thanks! Did you assume $\sigma=1$? Also, where can I read about the function $\Phi$ (which I assume bounds the probability $p$)? –  jamaicanworm Jan 12 '13 at 21:21
    
The result does not depend on the value of sigma. // About Phi: try WP. –  Did Jan 12 '13 at 22:03
    
Yes, as already mentioned. –  Did Jan 14 '13 at 9:31
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