Probability Question
If X~Bin(2,0.5) and Y~Bin(6,0.5)
What is P(3X=Y)?
P(Y|3X=Y)
and Cov(X-Y,X+Y)
I think I know what to do but I would like to check if my method is correct.
Thanks
What is P(3X=Y)?
P(Y|3X=Y)
and Cov(X-Y,X+Y)
I think I know what to do but I would like to check if my method is correct.
Thanks
Original post by Harriettttttt1
If X~Bin(2,0.5) and Y~Bin(6,0.5)
What is P(3X=Y)?
P(Y|3X=Y)
and Cov(X-Y,X+Y)
I think I know what to do but I would like to check if my method is correct.
Thanks
What is P(3X=Y)?
P(Y|3X=Y)
and Cov(X-Y,X+Y)
I think I know what to do but I would like to check if my method is correct.
Thanks
Why don't you post your method and let us verify it for you?
1) I considered p(0,0)+p(1,3)+p(2,6) and I got 1/128
2) I found the expectation of Y with only y0,3 or 6 used and I got 33/32
3)I used cov(x-y,x+y)=E((x-y)(x+y))-E(x+y)E(x-y) expanded out using linearity to get: E(x^2)-E(y^2)-E(x)^2+E(Y)^2 to get-1
2) I found the expectation of Y with only y0,3 or 6 used and I got 33/32
3)I used cov(x-y,x+y)=E((x-y)(x+y))-E(x+y)E(x-y) expanded out using linearity to get: E(x^2)-E(y^2)-E(x)^2+E(Y)^2 to get-1
Original post by Harriettttttt1
1) I considered p(0,0)+p(1,3)+p(2,6) and I got 1/128
2) I found the expectation of Y with only y0,3 or 6 used and I got 33/32
3)I used cov(x-y,x+y)=E((x-y)(x+y))-E(x+y)E(x-y) expanded out using linearity to get: E(x^2)-E(y^2)-E(x)^2+E(Y)^2 to get-1
2) I found the expectation of Y with only y0,3 or 6 used and I got 33/32
3)I used cov(x-y,x+y)=E((x-y)(x+y))-E(x+y)E(x-y) expanded out using linearity to get: E(x^2)-E(y^2)-E(x)^2+E(Y)^2 to get-1
1) I get 21/128. Typo I presume.
2) You could use a symmetry argument - I get 3. Post working if you wish.
3) Agreed. In this case it works out to Var(X) - Var(Y)
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