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Suppose we have two independent random variables $X$ and $Y$. I am interested in calculating $P(X\leq x \leq Y)$. Is this correct?

$$P(X\leq x \leq Y) = P(X\leq x)P(Y \geq x) = P(X\leq x)[1 - P(Y \leq x)]$$

If X and Y are Gaussian random variables, then the CDF of X and Y are both monotonically increasing functions on $[0,1]$ and so $P(X\leq x \leq Y)$ will have a maximum, right?

Michael Hardy
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Jeremy
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  • If $\phi(x) = P(X \le x \le Y)$ then $\phi(x) \to 0$ as $|x| \to \infty$. if $X,Y$ have continuous distributions (they do) then $\phi$ has a maximum (regardless of independence). – copper.hat May 28 '15 at 16:49
  • @Jeremy $\phi(0) = \mathbb P(X\le 0\le Y) = \mathbb P(X\le 0)\mathbb P(0\le Y) = \frac12\frac12=\frac14$ as $X,Y$ are independent (although I'm assuming that $\mathbb EX=0=\mathbb EY$) – Dan Robertson May 28 '15 at 16:51
  • @copper.hat that's what I think as well. I just wanted to be sure. Thanks. – Jeremy May 28 '15 at 16:51
  • @DanRobertson sorry I forgot to mention that $X, Y > 0$ and x is defined on $[0,\infty)$. With these conditions, we should have $\phi(0) = P(X \leq 0) P(0 \leq Y) = 0*1 = 0, right? – Jeremy May 28 '15 at 16:55
  • Yes but then the distributions aren't Gaussian. – Dan Robertson May 28 '15 at 16:56
  • Right. They're actually just unimodal which I approximate as Gaussian and cutoff at 0. With the mean and variances I use, the pdfs are very small at the origin. – Jeremy May 28 '15 at 16:58

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