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Math Help - conditional expectation

  1. #1
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    conditional expectation

    if Z1 and Z1 are r.v with distribution N(0,1) and X1,X2 are r.v

    and X1 = a + bZ1 and X2 = c + dZ1 + eZ2

    then to calculate E[X2|X1=x1] can we just rearange and substitute so

    E[X2|X1=x1] = E[c + d((x1 -a)/b) + eZ2] = c + (d/b)E[X1] - a/b
    = c+ a/b(d-1)
    thanks
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  2. #2
    MHF Contributor
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    I think you have the right idea but im not sure your algebra is quite right


    To do this problem evaluate teh expectation on the right hand side, one term at a time.

    E(X2 | X1) = E(C|X1) + dE(Z1|X1) + eE(Z2|Z1)
    edit: (ive conditioned on Z1 in the last term for convenience later. This is ok as Z1 is the only random part of X1)

    E(X2 | X1) = C + dE(\frac{X1-a}{b}|X1) + eE(Z2|Z1)
    E(X2 | X1) = C + d\frac{X1-a}{b} + eE(Z2|Z1)

    E(X2 | X1) = C + d\frac{X1-a}{b} + e*0
    To get the last line you must assume that Z2 and Z1 are independant.



    In your derivation, you used E(X1|X1=x) = E(X1)=0. however, E(X1|X1=x) \neq E(X1)
    E(X1|X1=x1) = x1
    Last edited by SpringFan25; June 7th 2010 at 03:35 PM.
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