Actex 2009 Edition Section 5 Example 5-14

[MHailJ]

"The continuous random variable X has pdf f(x) = 1 0<x<1. X1,X2,X3 are independent random variables all with the same distribution as X. Y = max{X1,X2,X3} and Z=min{X1,X2,X3}. Find E[Y-Z]."

I have a quick question regarding how to solve for E[Y] and E[Z]...

Solution was to solve the cdf in order to solve for E[Y] and use the survival function to solve for E[Z].

Is this always the case when you are dealing with max and min with respect to random variables such as above?

max -> use cdf
min -> use survival?

Best regards,
j

[RossMoney34]

Yes. Try to picture it... If the maximum value of the 3 RV's is less than y, then all 3 RV's must be less than y, since the other values are less. The same works for the minimum value of the 3 RV's. If the minimum value of the 3 RV's is greater than y, then all 3 must be greater than y.