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Conditional expectation and variance

  1. Jan 25, 2008 #1
    Let X, Y be independent exponential random variables with means 1 and 2 respectively.

    Z = 1, if X < Y
    Z = 0, otherwise

    Find E(X|Z) and V(X|Z).

    We should first find E(X|Z=z)
    E(X|Z=z) = integral (from 0 to inf) of xf(x|z).
    However, how do we find f(x|z) ?
  2. jcsd
  3. Jan 26, 2008 #2


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    Z is discrete. Since E[.|Z] and V[.|Z] are functions of Z, they too are discrete.

    E[X|Z=1] = E[X|X<Y].
    E[X|Z=0] = E[X|X>Y].

    Similarly for V[X|Z].
  4. Jan 27, 2008 #3
    I have found that

    E[X|Z=1] = E[X|X<Y] = 1/9
    E[X|Z=0] = E[X|X>Y] = 8/9
    by integrating, and conditioning on the random variable Y.

    So E(X) = E(E(X|Z)) = (1/9)(1/3) + (8/9)(2/3) = 17/27,

    which contradicts the fact that E(X) = 1, for X exponential with mean 1.

    I am wondering where is the error.
  5. Jan 28, 2008 #4


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    How are you computing Pr{Z=0} and Pr{Z=1}?
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