1
$\begingroup$

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...

Improve this question
$\endgroup$
  • $\begingroup$ The last condition seems indeed redundant. Are you sure you didn't forget some absolute values? $\endgroup$ – Eckhard Jan 11 '13 at 23:58
2
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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