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.

independentin the statement of your question. $\endgroup$ – Liviu Nicolaescu May 26 '15 at 17:24