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.


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.

  • $\begingroup$ Are the variances of $X,Y,Z$ the same or different? $\endgroup$ – Greg Martin May 7 '16 at 21:20
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    $\begingroup$ These are identical random variables? If so, the answer is $\frac 13$. One of them has to be greatest, nothing to break the symmetry. $\endgroup$ – lulu May 7 '16 at 21:36
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    $\begingroup$ 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 $a$. $\endgroup$ – Hagen von Eitzen May 7 '16 at 21:46
  • $\begingroup$ @GregMartin: The variances (and means) are different. $\endgroup$ – Captain Normal May 8 '16 at 9:17
  • $\begingroup$ @Did: Thank you for editing the question to improve it. $\endgroup$ – Captain Normal May 8 '16 at 9:18

Can be written in the below form ,

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

$$ P(X> \min(Y,Z)) =P(X>Y) \ P(X>Z) $$

Hope it helps.


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