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Math Help - a die is rolled...compute cov(x,y)

  1. #1
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    Question a die is rolled...compute cov(x,y)

    a die is rolled twice, let X be the sum of the outcomes and let Y be the first outcome minus the second outcome. Compute Cov(X,Y)
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    MHF Contributor harish21's Avatar
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    If A is the outcome of your first roll and B is the outcome of the second roll, then according to the question X=A+B \; and\; Y=A-B

    Now use the properties of covariancesto find:

    Cov(X,Y) = Cov(A+B, A-B)

    = Cov(A,A)-Cov(A,B)+Cov(A,B)-Cov(B,B)=....
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  3. #3
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    i did: Cov(A+B,A-B)=E(A+B)(A-B) - E(A+B)E(A-B) which equals Var(A^2)+ Var(B^2) is that correct also ?
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    Or you can say \displaystyle cov(X,Y) = E(X,Y)-E(X)\times E(Y)

    Find the expected values from these tables


    Code:
    		First					
    X		1	2	3	4	5	6
    Second	1	2	3	4	5	6	7
    	2	3	4	5	6	7	8
    	3	4	5	6	7	8	9
    	4	5	6	7	8	9	10
    	5	6	7	8	9	10	11
    	6	7	8	9	10	11	12
    							
    		First					
    Y		1	2	3	4	5	6
    Second	1	0	1	2	3	4	5
    	2	-1	0	1	2	3	4
    	3	-2	-1	0	1	2	3
    	4	-3	-2	-1	0	1	2
    	5	-4	-3	-2	-1	0	1
    	6	-5	-4	-3	-2	-1	0
    							
    XY		First					
    Second		1	2	3	4	5	6	
                 1	0	3	8	15	24	35
    	2	-3	0	5	12	21	32
    	3	-8	-5	0	7	16	27
    	4	-15	-12	-7	0	9	20
    	5	-24	-21	-16	-9	0	11
    	6	-35	-32	-27	-20	-11	0
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  5. #5
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by nikie1o2 View Post
    i did: Cov(A+B,A-B)=E(A+B)(A-B) - E(A+B)E(A-B) which equals Var(A^2)+ Var(B^2) is that correct also ?
    Seems like you're messed up.

    look at the second post and go over the link

    Cov(X,Y) = Cov(A+B, A-B)

    =Cov(A,A)-Cov(A,B)+Cov(A,B)-Cov(B,B)

    =Cov(A,A)-Cov(B,B)

    =Var(A)-Var(B)

    =0 \;\;\; WHY??
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