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Thread: Probability and conditional expectation

  1. #1
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    Probability and conditional expectation

    Y and V are independant, $\displaystyle E[V]=0$, $\displaystyle E[Y^2]<\infty$

    $\displaystyle X$$\displaystyle =c+$$\displaystyle aY^2$$\displaystyle +bV$
    a,b,c are constant.

    Find $\displaystyle E[X/Y]$

    I get $\displaystyle E[X/Y]$$\displaystyle =c$$\displaystyle +$$\displaystyle aE[Y^2 /Y]$

    Not sure if this is right
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  2. #2
    Moo
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    Hello,

    If Y and V are independent, then E[V|Y]=E[V] and for a suitable function f, E[f(Y)|Y]=f(Y).

    So there's just 1 simplification left.
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  3. #3
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    why does E[f(y)/y]=f(y)?

    in this case i would have thought that since f(y) depends explicitly on the values of y, this cannot be the case
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  4. #4
    Moo
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    It IS the case... f(Y) is Y-measurable.
    Go back to your definitions...
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