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Math Help - Find c in E[c|X-Y|] = 1

  1. #1
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    Find c in E[c|X-Y|] = 1

    Let X and Y by two independent identically distributed normal random variables with mean 1 and variance 1. Find c so that E[c|X-Y|] = 1.
    *****
    I went with the 'simple' route
    E[c|X-Y|] = cE[X-Y] + cE[X+Y] = 0 + c(1+1) = 2c
    And 2c = 1 so c = 1/2
    But that was wrong...

    I know there's a 1/sqrt(2pi) in the normal distribution but I can't see how that pi can get to the numerator in those integrals...
    ***
    Answer is sqrt(pi)/2
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  2. #2
    Super Member girdav's Avatar
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    Re: Find c in E[c|X-Y|] = 1

    The problem in what you did is in the first equality about the expectations. How would you justify that?

    Hint: what is the distribution of X-Y?
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  3. #3
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    Re: Find c in E[c|X-Y|] = 1

    Ah ok I made a mistake with the abs val...
    E[c|X-Y|] = cE[X-Y] + cE[Y-X] = 0 + 0 = 0

    The distribution of X-Y is another normal distribution with mean 0 and variance 2.

    I found this page but it's not helping..
    Normal Difference Distribution -- from Wolfram MathWorld
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  4. #4
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    Re: Find c in E[c|X-Y|] = 1

    From that site in my last post:
    1/[sqrt(2pi(sigx^2 + sigy^2)] * exp(-[u-(muX-muY)]^2/[2(sigx^2 + sigy^2)]

    In my question, since sigx^2 = sigy^2 = 1 and muX = muY = 0 this simplifies to:
    1/[sqrt4pi] * exp(-[u]^2/[4])

    Going backwards like this, I can sort of see that taking the double integral from the original Px-y with these values for mu and sig^2 could have led to:
    4/[sqrt(4pi)] * exp(-[u]^2/[4]) where exp(-[u]^2/[4]) = 1
    so E[X] = 4c/[sqrt(4pi)] = 2c/[sqrt(pi)]
    Setting E[X] = 1 gives the book's answer.

    But...I don't see myself getting a question like this right next time and I also didn't even take into account the absolute value. Honestly I have no idea... :/
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