X is greater than Y, Z...X_n when all normal distributions

Question

Using the normal distribution. Let $X \sim N(\mu_x, \sigma^2_x)$, $Y \sim N(\mu_y, \sigma^2_y)$ and $Z \sim N(\mu_z, \sigma^2_z)$ where $N(\mu, \sigma^2)$ denotes the normal distribution with mean $\mu$ and variance $\sigma^2$. $X$, $Y$, and $Z$ are independent.

I know $P(X > Y)=P(X-Y>0)=\int_{0}^{\infty} N(\mu_x-\mu_y, \sigma^2_x+\sigma^2_y)$ (This is the cdf>0).

I'm curious how this generalizes to $X_n$ events. What is $P(X > (Y \>and\> Z))$?

I recall there being a generalized but complicated integral that required numerical integration, but I can't find the formula. I'm interested in a closed solution for 3 variables.

Answer 1

I'll do the case for $X,Y,Z$ standard normal. Let $\Phi$ be the standard normal distribution function and $\phi(x)=\frac{d}{dx}\Phi(x)$ be the standard normal density. Condition on $X$: \begin{align*} \Pr(X> \max \{Y,Z\})&= \text{E}(\Pr(\max \{Y,Z\} <X|X))\\ &=\int \Pr(\{Y<x\},\{Z<x\})\phi(x)dx\\ &=\int \Phi(x)^2\phi(x)dx\\ &= \left[\frac{\Phi(x)^3}{3} \right]|^{+\infty}_{-\infty}\\ &=\frac{1}{3}. \end{align*} I think general $X,Y,Z$ can be taken care of similarly with substitutions in the integral. Edit: I guess not, by Dilip's comment.

If you have $X,Y_i$ independent standard normal, then by the exact same logic, \begin{align*} \Pr(X>\max_{1\leq i\leq n} \{Y_i\})&=\int \Phi(x)^n\phi(x)dx=\frac{1}{n+1}. \end{align*}