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 $P(X&gt;Y,X&gt;Z)$?


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

Edit

For $P(X&gt;Y)$ the answer is:
$$
{\rm P}(X  &gt; Y )  = \Phi \left(\frac{\mu_X - \mu_Y }{\sqrt{\sigma_Y^2 + \sigma_Y^2}}\right).
$$
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.
Can be written in the below form ,

$$P(X&gt;Z,X&gt;Y)= P(X&gt; \min(Y,Z)) \ ( P(Y&gt;Z) \text{ or } P(Y&lt;Z))$$

$$ P(X&gt; \min(Y,Z)) =P(X&gt;Y) \ P(X&gt;Z) $$

Hope it helps.
Can be written in the below form ,

$$P(X&gt;Z,X&gt;Y)= P(X&gt; \min(Y,Z)) \ ( P(Y&gt;Z) \text{ or } P(Y&lt;Z))$$

$$ P(X&gt; \min(Y,Z)) =P(X&gt;Y) \ P(X&gt;Z) $$

Hope it helps.