Upper Bound Error for Expectation Approximation?

#1
Dear all,

I have a problem for the past few weeks and couldn't find the answer in the books or in the internet. I know that in general:
E(g(X1, X2, ...)) is not equal to g(E(x1), E(X2) ,...) where E() is the expectation operator and g() is a function. For example E(XY) is not equal to E(X)E(Y) and we are tolerating an error of Cov(XY) by this approximation. What is the error in general case I described above? I am forced to use such an approximation in a problem and I want to know how much error I am committing. I really appreciate your help and suggestions, I am really stuck!

Best Regards,
Mohammad
 

Dason

Ambassador to the humans
#2
Cov(X, Y) = E[XY] - E[X]E[Y]

so

E[XY] = E[X]E[Y] + Cov(X,Y)

there is no theoretical upper limit to the covariance without using additional information so I'm not sure where else there is to go from there.
 
#3
Thanks for the response. What I mean is that if approximate E[XY] with E[X]E[Y] I am tolerating an error of Cov(X,Y) (based on equation you provided). My question is about the general case of E[g(X1, X2, ...)]. What is the error of we approximate it with g(E(X1), E(X2), ...) ?


Cov(X, Y) = E[XY] - E[X]E[Y]

so

E[XY] = E[X]E[Y] + Cov(X,Y)

there is no theoretical upper limit to the covariance without using additional information so I'm not sure where else there is to go from there.
 
#5
It is interesting in general to know the answer but in my particular problem I have:
E[XY/Z] and I want to approximate it with E[X]E[Y]/E[Z] since there is no way to simplify XY/Z and make the expectation in a closed form. So in my problem g(X,Y,Z)=XY/Z.
 

Dason

Ambassador to the humans
#7
What do you know about X, Y, Z? Are there any distributional assumptions you make? Are you assuming any independence?
 

BGM

TS Contributor
#9
#10
Thank you so much. I'll try it.



Maybe you can try Taylor Expansion

http://en.wikipedia.org/wiki/Taylor_expansions_for_the_moments_of_functions_of_random_variables

I am very sure the exact assumption behind the method (e.g. does it require \( g \) to be an analytic function)

When it holds, you are just like doing the 0th order approximation for the expectation and you may try to give a bound for the remainder terms. I just doubt this method cannot be applied to very general distribution.