Expectations on the product of two dependent random variables

  • Thread starter simonkmtse
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I am studying for the FRM and there is a question concerning the captioned. I try to start off by following the standard Expectation calculation and breakdown the pdf into Bayesian Conditional Probability function. Then i got stuck there. Anyone can help me to find a proof on it? Many thanks.
 

statdad

Homework Helper
1,493
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If the random variables [tex] X, Y [/tex] are independent then

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

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

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.
 

statdad

Homework Helper
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Sorry - I'm not sure how I did it, but when I first read your message I apparently saw, or thought I saw, a reference to independence.
 
71
0
No need to look at conditional PDFs. We have that:

[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]

Thus,

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

Cheers,

-Emanuel
 
3
0
Is anybody familiar with how this problem generalizes to multiple random variables? As a steppingstone, is there a formula for three random variables X, Y, and Z such that:

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

Thanks for your help!
 
71
0
Is anybody familiar with how this problem generalizes to multiple random variables? As a steppingstone, is there a formula for three random variables X, Y, and Z such that:

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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