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Actex 2009 Edition Section 5 Example 5-14

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  • Actex 2009 Edition Section 5 Example 5-14

    "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
    Last edited by MHailJ; May 6 2012, 06:16 PM.

  • #2
    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.

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