Thread: Joint Probability Function
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Old October 11th, 2008, 05:25 PM
brd_7 brd_7 is offline
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Thanks so much, Ok this is what i got...


E[XY] = 1/3

E[X]^2 = (1/3 + y/2)^2 = y^2/4 + 2y/6 + 1/9

E[X^2] = 1/4 + y/3

E[Y] = 1/3 + x/2

And as a final answer.. i gained..

y^4/16 - 2y^3/24 - 13y^2/144 + 10y/216 + 10/324
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