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Thread: ExpectedValue

  1. #1
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    ExpectedValue

    Can anyone tell me how to calculate the expected value of A^2 when A=x*y and of B^2 when B=2x+2y.
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  2. #2
    MHF Contributor matheagle's Avatar
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    I assume that X and Y are rvs?

    Then $\displaystyle E(A^2)=E(X^2Y^2)$

    and if they are continuous, then.....

    $\displaystyle E(A^2)=E(X^2Y^2)=\int\int x^2y^2f(x,y)dxdy$

    If they are indep....

    $\displaystyle E(A^2)=E(X^2)E(Y^2)=\int x^2f_X(x)dx\int y^2f_Y(y)dy$

    Likewise $\displaystyle E(B^2)=4E[(X+Y)^2]=4\int\int (x+y)^2f(x,y)dxdy$
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  3. #3
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    Quote Originally Posted by matheagle View Post
    I assume that X and Y are rvs?

    Then $\displaystyle E(A^2)=E(X^2Y^2)$

    and if they are continuous, then.....

    $\displaystyle E(A^2)=E(X^2Y^2)=\int\int x^2y^2f(x,y)dxdy$

    If they are indep....

    $\displaystyle E(A^2)=E(X^2)E(Y^2)=\int x^2f_X(x)dx\int y^2f_Y(y)dy$

    Likewise $\displaystyle E(B^2)=4E[(X+Y)^2]=4\int\int (x+y)^2f(x,y)dxdy$
    Yes, they are both random v., independent and uniform on [0, 1].
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  4. #4
    MHF Contributor matheagle's Avatar
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    It does make a slight difference knowing the distribution........


    $\displaystyle E(A^2)=E(X^2)E(Y^2)=\int_0^1 x^2dx\int_0^1 y^2dy={1\over 9}$

    and $\displaystyle E(B^2)=4E[(X+Y)^2]=4\int_0^1\int_0^1 (x+y)^2dxdy$

    or $\displaystyle 4E[(X+Y)^2]=4\left[E(X^2)+2E(X)E(Y)+E(Y^2)\right]$

    $\displaystyle =4\left[ {1\over 3} +2\left({1\over 2}\right)\left( {1\over 2}\right) +{1\over 3} \right]$
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  5. #5
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    Thumbs up

    matheagle, I appreciate your help.

    Thank you
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