I accidentally asked my students a question that reduced to the following: find $P(X>Y, X>Z)$ where $X, Y$, and $Z$ are independent, $X$ is standard normal, and $Y$ and $Z$ are both normal with mean $1$ and variance $1$.

Then I sat down to write a solution to the question. Of course this probability can be rewritten as $P(X>\max(Y,Z))$. The CDF of $\max(Y,Z)$ is $\Phi(y-1)^2$ where $\Phi$ is the standard normal cdf. The joint density of $X$ and $\max(Y,Z)$ is therefore $\phi(x) {d \over dx} \Phi(y-1)^2$ or $2 \phi(x) \phi(y-1) \Phi(y-1)$. (Here $\phi$ is the standard normal density.) Integrating this joint density over the half-plane $x > y$ gives the expression $$ \int_{-\infty}^\infty \phi(x) \Phi(x-1)^2 \: dx $$ which can be evaluated numerically -- it's about $0.113202$ - but I don't recognize this number.

Is there some way to write this number in terms of, say, values of $\phi$ and $\Phi$ (without integrating)?

Also, the context here was as follows: let $X_1, \ldots, X_n$ be standard normal and let $Y_1, \ldots, Y_n$ be normal(1,1). Find the variance of the number of pairs $(i, j)$ such that $X_i > Y_j$. My method is to write that variance as a sum of $n^2$ indicators and find covariances of those indicators, leading to the probability I opened the problem with. Is there a solution to this problem that doesn't go via $P(X>Y, X>Z)$, which seems annoyingly hard to find?

  • $\begingroup$ Y and Z have mean 1 and variance 1; that was a typo. I'll fix it. $\endgroup$ Jul 23 '11 at 0:11
  • $\begingroup$ Here's a first thought; I'm not sure it gets you anywhere: $$ \begin{align} u & = \Phi(x-1) \\ du & = \phi(x-1)\,dx \end{align} $$ Then $$ \phi(x-1) = \text{constant}\cdot \phi(x) e^x. $$ So the integral becomes $$ \text{constant}\cdot\int_0^1 e^x u^2\,du. $$ $\endgroup$ Jul 23 '11 at 2:51
  • $\begingroup$ Here's another P.O.V.: The events $Y>X$ and $Z>X$ are conditionally independent given $X$. Therefore the conditional probability of their joint occurrence given the value of $X$ is $(1-\Phi(X))^2$. Consequently the probability of their joint occurrence is the expected value of that quantity. This is $$ \int_{-\infty}^\infty (1-\Phi(X))^2 \phi(x)\,dx. $$ $\endgroup$ Jul 23 '11 at 2:58

I'm going to answer my own question. This adapts Michael Hardy's second comment.

To find $P(X>Y, X>Z)$, note that the two events are conditionally independent given $X$. So

$$ P(X>Y, X>Z | X=x) = P(X>Y | X=x) P(X>Z | X=x) = P(Y<x) P(Z<x) = \Phi(x-1)^2. $$


$$ P(X>Y, X>Z) = \int P(X>Y, X>Z | X=x) \phi(x) \: dx $$

which is the integral I originally asked about. I don't think that there's a simpler expression for this number.


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.