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Thread: Conditional Variance

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    Conditional Variance




    Hi,

    I have difficulty understanding the definition of conditional variance as:

    var(X|Y) = E[(X-E[X|Y])**2 | Y].

    Given the definition of variance of a random variable X:

    var(X) = E[(X-E[X])**2],

    and substituting X with X|Y, you get:

    var(X|Y) = E[(X|Y-E[X|Y])**2]. How can one get from this equation to the definition of the conditional variance given above? I would appreciate any help on this. Thank you!
    Last edited by fifosm; 11-09-2014 at 07:01 PM.

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    Re: Conditional Variance

    It seems that just a notation issue.

    When you write the conditional expectation, we will put the conditional part at the end of the operator. The reason why we write something like X|Y is because we can just regard the conditional distribution as a new distribution. Maybe just one more property to emphasize here:

    E[E[X|Y]|Y] = E[X|Y]

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    Re: Conditional Variance

    Thanks BGM. But I don't seem to be able to get it. Assuming that this is a notation convention, why the convention doesn't apply to the second term E[X|Y]? I couldn't use the law of iterated expectation that you mentioned here either. Could you please elaborate? Sorry I am new to the subject matter and am studying Dimitri Bertsekas's book - introduction to probability. Thanks!

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    Re: Conditional Variance

    The reason why you define the conditional variance like this is to the same as we define the ordinary variance

    Var[Z] = E[(Z - E[Z])^2]

    So I will say it is a notation issue as the concept is like what you have said, a "substitution" (although not really a substitution).

    And, of course we know that in general E[X|Y] and E[X] are different. The variance is calculating the average squared difference with respect to its own mean, so that why we use E[X|Y].

    If you carefully expand out the squared term, we have

    E[(X - E[X|Y])^2]

    = E[X^2 - 2XE[X|Y] + E[X|Y]^2|Y]

    = E[X^2|Y] - 2E[X|Y]E[X|Y] + E[X|Y]^2

    = E[X^2|Y] - E[X|Y]^2

    whereas

    E[(X - E[X])^2]

    = E[X^2 - 2XE[X] + E[X]^2|Y]

    = E[X^2|Y] - 2E[X]E[X|Y] + E[X]^2

    so indeed they are different.

    You may need to elaborate more on which part you stuck in iterated expectation if you still have some questions here.

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    Re: Conditional Variance

    Thanks again! I understood your point about E[X|Y] in the definition. If you clarify one point to me, I will get the whole thing. In your first expansion, how did you calculate the second term? Meaning, how did you get from E[2XE[X|Y]|Y] to 2E[X|Y]E[X|Y]? Thanks!

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    Re: Conditional Variance

    It is similar to something like

    E[cX] = cE[X]

    where c is a constant. We pulls out the constant. This is actually can be also be interpreted more generally as the tower property of conditional expectation.

    E[X|Y] is \sigma(Y)-measurable, and in this sense you can just treat it as like a "constant" inside the conditional expectation E[\cdot|Y]. But you should note that E[X|Y] itself is a random variable but not a constant.

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    fifosm (11-10-2014)

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    Re: Conditional Variance

    Got it. Thank you so much! :-)

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    Re: Conditional Variance


    Doing a 4th post to see if this works.

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