TITLE: What is the probability that a Gaussian random variable is the largest of three Gaussian random variables?
QUESTION [6 upvotes]: Consider three independent,  normally distributed RVs: $YA \sim N(a,\sigma ^{2}),$ $%
YB\sim N(b,\sigma ^{2})$ and $YC\sim N(c,\sigma ^{2})$.
What is the probability that $YA$ is the maximum?: $$\Pr (YA>YB \;\;\mbox{and} \;\;YA>YC)=IA=\int_{-\infty }^{\infty }\Phi (\frac{Y-b}{%
\sigma })\Phi (\frac{Y-c}{\sigma })\phi (\frac{Y-a}{\sigma })dY$$
My first thought was to differentiate $IA$ with respect to $b$  and $c$, complete the square and exploit that $$\int_{-\infty }^{\infty }\frac{\sqrt{3}}{\sigma}\phi \left(\frac{Y-\frac{a+b+c}{2}}{%
\sigma /\sqrt{3} }\right)dY=1$$ to remove the integral over $Y$. 
Therefore, $$\frac{d^2IA}{dbdc}\propto \exp \left(-\frac{a^2-2 a y+b^2-2 b y+c^2-2 cy+3 y^2}{2 \sigma ^2}\right)$$.
Unfortunately, I cannot figure how to integrate with respect to $b$ and $c$ to get back to $IA$. 
Any suggestions would be much appreciated.

REPLY [2 votes]: Since
$\begin{pmatrix}YA-YB\\YA-YC\end{pmatrix}\stackrel{\text{dist}}{=}
N\left(\begin{pmatrix}a-b\\a-c\end{pmatrix},\begin{pmatrix}2\sigma^2&\sigma^2\\ \sigma^2&2\sigma^2\end{pmatrix}\right)$, then
\begin{align} \mathsf{P}((YA>YB)\cap(YA>YC))&=\mathsf{P}(YA-YB>0,YA-YC>0)\\
&=L\Bigl(\frac{b-a}{\sqrt2\sigma},\frac{c-a}{\sqrt2\sigma},\frac12\Bigr),
\end{align}
where
$$L(h,k,\rho)=\frac{1}{2\pi\sqrt{1-\rho^2}}\int_h^\infty dx\int_k^\infty\exp\Bigl[
-\frac{x^2-2\rho xy+y^2}{2(1-\rho^2)}\Bigr]\,dy.
$$
In the following book: S. Kotz et al., Continuous Multivariate Distributions,  Vol. 1, 2nd Ed(2000), The Ch.46, Sec.4, p.264--. It gave a complete discussion about above bivariate normal integral--table and approximation.