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- Thread starter simonkmtse
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statdad

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[tex]

E[X \cdot Y] = E[X] \cdot E[Y]

[/tex]

I sense from the tone of your question something more is involved?

- #3

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But I wanna work out a proof of Expectation that involves two dependent variables, i.e. X and Y, such that the final expression would involve the E(X), E(Y) and Cov(X,Y).

I suspect it has to do with the Joint Probability distribution function and somehow I need to separate this function into a composite one that invovles two single-variate Probability distribution and one that involves correlation coefficient.

I just can't get beyond that step.

- #4

statdad

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[tex]

Cov[X,Y] = E[(X-E[X])\cdot(Y-E[Y])] [/tex]

[tex]

= E[X \cdot Y] - E[X \cdot E[Y]] - E[E[X] \cdot Y] + E[E[X] \cdot E[Y]] [/tex]

[tex] = E[X \cdot Y] - E[X] \cdot E[Y] [/tex]

= E[X \cdot Y] - E[X \cdot E[Y]] - E[E[X] \cdot Y] + E[E[X] \cdot E[Y]] [/tex]

[tex] = E[X \cdot Y] - E[X] \cdot E[Y] [/tex]

Thus,

[tex]E[X \cdot Y] = Cov[X,Y] + E[X] \cdot E[Y] [/tex]

Cheers,

-Emanuel

- #6

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E[XYZ] = E[X] * E[Y] * E[Z] + [term involving covariances]

Thanks for your help!

- #7

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E[XYZ] = E[X] * E[Y] * E[Z] + [term involving covariances]

Thanks for your help!

There is no such formula involving just covariances, you have to include higher order moments such as [itex]E[(X-E[X]) \cdot (Y-E[Y]) \cdot (Z-E[Z])] [/itex] for a 3-variable case.

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