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Minimum of two random variable

  1. Mar 31, 2015 #1
    Hello, want to know if it's correct
    1. The problem statement, all variables and given/known data

    X and Y two random variables iid of common density f and f(x)=x*exp(-x²/2) if x≥0 and f(x)=0 if x≤0
    and Z=min(X,Y)
    Find
    -The density of Z
    -The density of Z²
    - E[Z²]

    2. Relevant equations


    3. The attempt at a solution
    1.

    FZ(u) = P(min(X,Y )≤u)
    = 1 − P(min(X,Y )>u)
    = 1 − P(X>u)*P(Y>u)
    = 1 − (1 − F(u))²

    And F(u)=1-exp(-u²/2) if u>0 , 0 if u≤0
    So FZ(u)=1-exp(-u²/2) if u>0 , 0 if u≤0
    The density is then
    fZ(u) = u*exp(-u²/2) 0 if u>0 , 0 if u≤0

    2. I'm not sure here
    u≤0
    FZ²(u)=P(Z²≤u)=0
    u≥0
    Z²≤u , -√u≤Z≤√u
    From -√u to 0 , FZ²(u)=0
    From 0 to √u , FZ²(u)=FZ(u)=1-exp(-u²/2)

    So Z² has the same density as Z ?

    Thanks
     
  2. jcsd
  3. Mar 31, 2015 #2

    RUber

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    Homework Helper

    I am not following how you went from F(u) to FZ(u) since FZ(u) = 1-(1-F(u))^2. This would give you 1-exp(-u²).
    Then fZ(u) = d/du (FZ(u)) = 2u exp(-u^2).
    You are looking for the probability that Z < sqrt(u), so FZ^2*=(u) = FZ(sqrt(u)) .
     
  4. Mar 31, 2015 #3

    Ray Vickson

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    Science Advisor
    Homework Helper

    (i) Standard textbook expression: ##f_Z(z) = 2 f(z) P(X>z)##, and in your case, ##P(X > z) = e^{-z^2/2}##. Your final expression is incorrect.
    (ii) Since ##Z \geq 0##, ##P(Z^2 \leq w) = P(Z \leq \sqrt{w})## for all ##w \geq 0##. Get the density of ##W = Z^2## by differentiation of ##P(W \leq w)##.
     
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