# A nice expression for $P(X>Y, X>Z)$ where $X$ is standard normal, $Y, Z$ are normal with mean 1 and variance 1?

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?

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Y and Z have mean 1 and variance 1; that was a typo. I'll fix it. –  Michael Lugo Jul 23 '11 at 0:11
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 $$\text{constant}\cdot\int_0^1 e^x u^2\,du.$$ –  Michael Hardy Jul 23 '11 at 2:51
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 $$\int_{-\infty}^\infty (1-\Phi(X))^2 \phi(x)\,dx.$$ –  Michael Hardy 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

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

Then

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

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

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