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Math Help - Proof

  1. #1
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    Proof

    Let X and Y be discrete random variables and a; b are real numbers.
    Prove E(aX + bY ) = aE(X)+ bE(Y ) and Var(aX + bY ) = a^2Var(X)+
    b^2Var(Y )+ 2abCov(X; Y ).

    Both these seem easy but I can't proof why E(X+Y)=E(X)+E(Y)
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  2. #2
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    Quote Originally Posted by whatsanihar View Post
    Let X and Y be discrete random variables and a; b are real numbers.
    Prove E(aX + bY ) = aE(X)+ bE(Y ) and Var(aX + bY ) = a^2Var(X)+
    b^2Var(Y )+ 2abCov(X; Y ).

    Both these seem easy but I can't proof why E(X+Y)=E(X)+E(Y)
    Look at the definition of expectation for discrete RVs:

    E(f(U))=\sum f(u_i) p(u_i)

    So (assuming suitable conditions allowing us to change orders of summation etc hold):

    E(X+Y)=\sum_{i,j} (x_i+y_j) p(x_i,y_j) = \sum_i \sum_j x_i p(x_i,y_j) + \sum_j \sum_i y_j p(x_i,y_j) =\sum_i x_i p(x_i) +\sum_j y_j p(y_j)

    (here we are writing the marginal distribution p(x_i)=\sum_j p(x_i,y_j) and similar for the other case)

    etc

    CB
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    Look at the definition of expectation for discrete RVs:

    E(f(U))=\sum f(u_i) p(u_i)

    So (assuming suitable conditions allowing us to change orders of summation etc hold):

    E(X+Y)=\sum_{i,j} (x_i+y_j) p(x_i,y_j) = \sum_i \sum_j x_i p(x_i,y_j) + \sum_j \sum_i y_j p(x_i,y_j) =\sum_i x_i p(x_i) +\sum_j y_j p(y_j)

    (here we are writing the marginal distribution p(x_i)=\sum_j p(x_i,y_j) and similar for the other case)

    etc

    CB
    Ya thats what I did for the expected value and got that. But what should I do for Var(X+Y)? Var(x)=E[(x-u)^2] but what is Var(X+Y) equal to?
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    Look at the definition of expectation for discrete RVs:

    E(f(U))=\sum f(u_i) p(u_i)

    So (assuming suitable conditions allowing us to change orders of summation etc hold):

    E(X+Y)=\sum_{i,j} (x_i+y_j) p(x_i,y_j) = \sum_i \sum_j x_i p(x_i,y_j) + \sum_j \sum_i y_j p(x_i,y_j) =\sum_i x_i p(x_i) +\sum_j y_j p(y_j)

    (here we are writing the marginal distribution p(x_i)=\sum_j p(x_i,y_j) and similar for the other case)

    etc

    CB
    We can of course shorten this by observing that by definition:

    E(X)+E(Y)=\sum_i \sum_j x_i p(x_i,y_j) + \sum_j \sum_i y_j p(x_i,y_j)

    CB
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  5. #5
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    Quote Originally Posted by whatsanihar View Post
    Ya thats what I did for the expected value and got that. But what should I do for Var(X+Y)? Var(x)=E[(x-u)^2] but what is Var(X+Y) equal to?
    Var(X+Y)=E([(X+Y)-(\overline{X}+\overline{Y})]^2 =E([(X-\overline{X})+(Y-\overline{Y})]^2) =E((X-\overline{X})^2 + 2(X-\overline{X})(Y-\overline{Y})+(Y-\overline{Y})^2)


    Now use what has already been proven (that the expectation of a sum is the sum of the expectations) to get:

    Var(X+Y)= =Var(X)+Var(Y) + 2 CoVar(X,Y)

    CB
    Last edited by CaptainBlack; February 23rd 2010 at 10:47 PM.
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  6. #6
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    But isnt that saying that Var(X+Y)=Var(X)+Var(Y)?

    When Var(X+Y)=Var(X)+Var(Y)+2Cov(V,Y)
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  7. #7
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    Quote Originally Posted by whatsanihar View Post
    But isnt that saying that Var(X+Y)=Var(X)+Var(Y)?

    When Var(X+Y)=Var(X)+Var(Y)+2Cov(V,Y)
    Sorry you got to see a work in progress, go back and look at the final form.

    CB
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  8. #8
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    Thanks, Thanks. It makes sense now .
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