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Math Help - help with expected value

  1. #1
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    Exclamation help with expected value

    1. Let X,Y and Z be 3 random variables, and X and Z are independent. Prove that Cov(X+Y,Z)=Cov(Y,Z).

    For this i expanded into
    E(Z(X+Y)) - E(X+Y)E(Z)=E(YZ)-E(Y)E(Z)
    now i am not sure what to do here.....

    2. Prove that the variance of the uniform[L,R] distribution is given by the expression ((R-L)^2) / 12.

    3. Let X and Y be independent, X~Bernoulli(1/2), and Y~N(0,1). Let Z=X+Y, W=X-Y. Compute Var(Z), Var(W), Cov(Z,W), Corr(Z,W).
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  2. #2
    MHF Contributor harish21's Avatar
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    q1.

    Cov(X+Y,Z) = E[(X+Y)Z]-E[(X+Y)]E(Z)

    = E(XZ)+E(YZ)-[E(X)+E(Y)]E(Z)

    = E(XZ)+E(YZ)-E(X)E(Z)-E(Y)E(Z)

    =E(YZ)-E(Y)E(Z) + (E(XZ)-E(X)(EZ))

    =Cov(Y,Z)+ (\mbox{use the property of covariance of 2 random vars that are independent})
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  3. #3
    MHF Contributor matheagle's Avatar
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    1) It is true that Cov(X+Y,Z)=Cov(X,Z)+Cov(Y,Z), but they may want you to prove that.

    2) That's easy, just compute the first and second moments.
    Clearly the mean is the midpoint, (L+R)/2.

    3) Most of this is straight forward.
    Use Cov(X+Y,X-Y)=Cov(X,X)+Cov(X,-Y)+Cov(Y,X)+Cov(Y,-Y)= V(X)-Cov(X,Y)+Cov(X,Y)-V(Y)=V(X)-V(Y).
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  4. #4
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    for 3
    for var(Z)
    i am getting
    Var(Z) = Var(X+Y) = Var(X)+Var(Y)+2Cov(X,Y)
    =Var(X)+Var(Y)+2(E(XY)-E(X)E(Y))
    =Var(X)+Var(Y)+2(E(X)E(Y)-E(X)E(Y))
    =Var(X)+Var(Y)
    =E((x-0.5)^2) + E(y^2)

    Is this the last step or can i go further?
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  5. #5
    MHF Contributor matheagle's Avatar
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    X and Y are indep so their covariance is ZERO.

    So V(X+Y)=V(X)+V(Y)=(1/2)(1/2)+1.

    Likewise V(X-Y)=V(X)+V(Y)=(1/2)(1/2)+1.
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  6. #6
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    i am not understanding how your calculating the variances...
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