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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.

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  • $\begingroup$ Do you mean to ask for $P(X > \max\{Y, Z\})$? You cannot get the answer in closed form, but the integral formula for this probability is straightforward to set up from first principles of conditional distributions of random variables. We have that $$P(X > \max\{Y, Z\} \mid X = x) = \Phi\left(\frac{x-\mu_Y}{\sigma_Y}\right)\Phi\left(\frac{x-\mu_Z}{\sigma_Z}\right)$$ and all that is left is to multiply this conditional probability by the density of $X$ and integrate over the real line. As you say, the numerical integration is not easy. $\endgroup$ – Dilip Sarwate Oct 31 '16 at 2:14
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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*}

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  • $\begingroup$ You're right I meant $X-Y$. I made an edit to hopefully clarify. I'm not sure what operator I mean by $and$. The abstract value I'm after is the probability that $X$ is greater than $Y$ and $Z$. $\endgroup$ – Mark Johnson Oct 31 '16 at 1:45
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    $\begingroup$ Actually, your detailed calculation to arrive at the $\frac 13$ is unnecessary. The probability that $X$ is the largest among three independent identically distributed random variables (standard normals in this case) is $\frac 13$ y symmetry! See my answer to this question too. $\endgroup$ – Dilip Sarwate Oct 31 '16 at 3:13
  • $\begingroup$ @Dilip Oops. I guess the difficult part is when they're not identical. Thanks! $\endgroup$ – user135912 Oct 31 '16 at 3:23
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    $\begingroup$ Thanks. When they're not all standard normal, the integral is non-trivial. $\endgroup$ – Mark Johnson Oct 31 '16 at 3:58

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