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How to find E(XY) when X and Y are NOT indepdant?

by laura_a
Tags: indepdant
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May25-08, 07:34 PM
P: 64
1. The problem statement, all variables and given/known data

I have a joint pdf f_{XY}(x,y) = (2+x+y)/8 for -1<x<1 and -1<y<1

2. Relevant equations

I have to work out E(XY) but I have previously worked out that X and Y are NOT independant (that is f_{XY}(0,1) doesn't equal f_X{0}*f_Y{1}). I am using maxima so I don't need help with any integration, I just need to know what formula because I've read that E(XY) = E(X)E(Y) only when they're indepdant... so what happens when they're not?
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May26-08, 07:03 AM
P: 491
You integrate xy against the pdf. Do you not have the textbook?
May26-08, 07:15 AM
P: 64
No, there is no text book for this, I have bought some books, but none of them are written for people who aren't the best at statistics. I have no idea what you mean, isn't there an easier way using E(X) and E(Y) which I already have?

May26-08, 07:21 AM
P: 491
How to find E(XY) when X and Y are NOT indepdant?

No, there's not. Is this for a class?
May26-08, 09:32 AM
P: 352
Oh goody double posting! Laura you now have two people telling you the same thing-- integrate. I don't know why you had to start two threads on the same topic instead of just being patient.
Jan9-11, 09:36 AM
P: 2
You can use the definition of an expectation.
E(XY) = [tex]\oint\oint[/tex]x*y*f(x,y) dy dx
Or you could argue that since the function is symmetric about 0 and the intervals [-1, 1] are centred about 0 that E(XY) = 0
Jan9-11, 10:49 AM
HW Helper
P: 1,371
The density isn't symmetric about zero.

Laura, for any joint continuous distribution, whether or not [tex] X, Y [/tex] are independent, you can find [tex] E[XY] [/tex] as

\iint xy f(x,y) \, dxdy

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