5
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Are the variances of $X,Y,Z$ the same or different? $\endgroup$ – Greg Martin May 7 '16 at 21:20
  • 4
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
-1
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.