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Thread: Proof Var[X | Y=2]

  1. #1
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    Proof Var[X | Y=2]

    Can you prove to me that Var[X | Y=2] = E[X^2 | Y=2] - (E[X | Y=2])^2

    I see the relation of just say, var[x], but why is Y=2 not squared in the first part?
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  2. #2
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    Re: Proof Var[X | Y=2]

    $P[X|Y=2]$ is just a distribution like any other.

    so the formulas for expectation and variance are applied as usual.

    The $Y=2$ is just a conditioning expression. It's not part of the random object which is $X$.
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  3. #3
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    Re: Proof Var[X | Y=2]

    I see thanks.

    Also, why when say random variables X and Y are independent, the covariance between X and Y is 0?


    Cov[X,Y]= E[XY] -E[X]E[Y]

    so it is just saying if they are independent there is no E[XY]
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  4. #4
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    Re: Proof Var[X | Y=2]

    Quote Originally Posted by math951 View Post
    I see thanks.

    Also, why when say random variables X and Y are independent, the covariance between X and Y is 0?


    Cov[X,Y]= E[XY] -E[X]E[Y]

    so it is just saying if they are independent there is no E[XY]
    no. It says that

    $E[XY]=E[X]E[Y]$

    Note the reverse is not always true. Zero covariance does not imply independence.
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