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  • Median Question

    Question:

    The random variables X, Y and Z are independent and have an identical
    distribution that is uniform over (0, 1). Calculate the median of the
    smallest of the three.


    Solution:

    Let W = min(X, Y,Z). Then
    Pr(W > w) = Pr(min(X, Y,Z) > w)
    = Pr(X > w, Y > w,Z > w)
    = Pr(X > w)Pr(Y > w)Pr(Z > w) (independence)
    = (1 − w)^3 = 1/2
    when w = 1 − (1/2)1/3 = 0.206.

    I have 3 questions about this solution:
    -Where is the uniform distribution over (0, 1) used?
    -Why is it P(W>w) as opposed to P(W<w)?
    -Why is P(X>w)=P(Y>w)=P(Z>w)=(1-w)?

    After being unable to solve the question and then re-reading the solution, I tried P(W<w) instead where I assumed P(X<w)=P(Y<w)=P(Z<w)=w which gave an incorrect answer, but I'm unsure as to why so I'd appreciate it if somebody could help me answer these 3 questions.

  • #2
    Originally posted by kryptic91 View Post
    -Where is the uniform distribution over (0, 1) used?
    It's used when the probability is 1-w
    -Why is it P(W>w) as opposed to P(W<w)?
    Because this is easier to compute when you're working with a minimum. For the minimum to be *greater* than some value w, it means each of the individual variables must all be greater than that. If they're independent, then this just means that you take the product of the probability of each one individually being greater.
    -Why is P(X>w)=P(Y>w)=P(Z>w)=(1-w)?
    Because they are identically distributed (so all the same), and uniform on (0,1) (so the probability is (1-w), which is just the length of the interval from w to 1).

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