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QuoteOriginally posted by: DogonMatrix =?Whatizit ?

QuoteOriginally posted by: quantieQuoteOriginally posted by: DogonMatrix =?Whatizit ?yes, what is it ? or is there an expression in terms of cov(X,Y) , Var(X) and Var(Y)

Am I missing soemthing here? This thread doesn't make sense to me.

What is X... what is Y?how are they distributed?

Sorry guys may be indeed there is no simple solution to this. But what I was looking for is a way to express it as a function of cov(x,y) :something similar to this, cov( ax+b, cy+d) = a*c*cov(x,y).

For what it's worth, after a little algebra I get:

QuoteOriginally posted by: AthleticoFor what it's worth, after a little algebra I get:thanks. could you post the "little algebra" ?

You could go to the definition of covariance: Cov(X,Y) = [E(X*Y) - E(X)*E(Y)]/[SD(X)*SD(Y)], but you'll get a real mess.As a general matter, covariance is not sensitive to monotonic transformations of the data as long as the relation between X and Y is reasonably linear. You can think of the transformation as giving you an estimate of Cov(X,Y) that puts more weight on the points for which e^(a+b*X) is close to zero. As long as the correlation is close to the same for all values of X and Y, the difference in weighting usually won't make a difference.Where you get into trouble with this logic is if the correlation between X and Y changes. For example, suppose Y is N(0,1) and X = Y for -1<Y<1 and X = -Y otherwise. If a = 0 and b = 1, your expression will overweight the data between -1 and 1 and come up with a larger correlation, hence larger covariance, than Cov(X,Y).

QuoteOriginally posted by: AaronYou could go to the definition of covariance: Cov(X,Y) = [E(X*Y) - E(X)*E(Y)]/[SD(X)*SD(Y)], but you'll get a real mess.As a general matter, covariance is not sensitive to monotonic transformations of the data as long as the relation between X and Y is reasonably linear. You can think of the transformation as giving you an estimate of Cov(X,Y) that puts more weight on the points for which e^(a+b*X) is close to zero. As long as the correlation is close to the same for all values of X and Y, the difference in weighting usually won't make a difference.Where you get into trouble with this logic is if the correlation between X and Y changes. For example, suppose Y is N(0,1) and X = Y for -1<Y<1 and X = -Y otherwise. If a = 0 and b = 1, your expression will overweight the data between -1 and 1 and come up with a larger correlation, hence larger covariance, than Cov(X,Y).Thanks Aaron. I think I understood the general logic of your post but I am not sure how to apply it to get the answer. Is Athletico ' answer correct ?

I should have clarified my ans was a 2nd order approx:E[f(X)] = f(E[X]) + 1/2 Var(X) f''(E[X]) + ...Cov(f(X), Y) = f'(E[X]) Cov(X,Y) + ...Let Z = z(X) = exp(alpha + beta X), then we're looking for Cov(f(Z), Y) where f(z) = z/(1 + z)

I would do something simpler. I would assume the correlation is the same between the X and Y and the transformed X and Y (subject to the qualifications in my previous post). So I would just multiply by the standard deviation of the transformed X divided by the standard deviation of X.