# Probability $P(X>Y,X>Z)$ for independent normal random variables $X$, $Y$, $Z$

There are several answers already given for working out the probability of one random variable being greater than another, but I can't make the leap to working out the probability of one random variable being greater than several others. My random variables are independent and normally distributed. For example:

Let $$X$$, $$Y$$ and $$Z$$ be independent normal random variables. What is $$P(X>Y,X>Z)$$?

The obvious (to me) answer, being to just multiply the two probabilities $$P(X>Y)$$ and $$P(X>Z)$$ does not work because the difference random variables $$(X-Y)$$ and $$(X-Z)$$ are not independent.

Edit

For $$P(X>Y)$$ the answer is: $${\rm P}(X > Y ) = \Phi \left(\frac{\mu_X - \mu_Y }{\sqrt{\sigma_Y^2 + \sigma_Y^2}}\right).$$ I'm hoping for a way of adding a third normal random variable to the equation. If this is possible I presume the answer can be easily expanded to add further random variables.

• Are the variances of $X,Y,Z$ the same or different? – Greg Martin May 7 '16 at 21:20
• These are identical random variables? If so, the answer is $\frac 13$. One of them has to be greatest, nothing to break the symmetry. – lulu May 7 '16 at 21:36
• lulu's symmetry argument applies to any case of independent and identically distributed random variables (normal is not necessary), provided $P(X=a)=0$ for all $a$. – Hagen von Eitzen May 7 '16 at 21:46
• @GregMartin: The variances (and means) are different. – Captain Normal May 8 '16 at 9:17
• @Did: Thank you for editing the question to improve it. – Captain Normal May 8 '16 at 9:18

$$P(X>Z,X>Y)= P(X> \min(Y,Z)) \ ( P(Y>Z) \text{ or } P(Y
$$P(X> \min(Y,Z)) =P(X>Y) \ P(X>Z)$$