Jon Cryer schrieb: > > This cannot be solved without some knowledge of the joint distribution > of X and Z > and of Y and Z. (At least the covariance in these distributions.)
Sorry, I forgot to write this in my first posting. Well, actually Z is independent of X and Y. In my scenario Z is white noise that is added to the signal Y. Consequently, I can use the posting of Hiu Chung - after reading the simple equation I really have to feel sorry for posting to the group. Nevertheless, your answers helped me a lot - thanks! In my case, since E[Z] = 0, so I get E[X*(Y+Z)] = E[ X*Y + X*Z ] = E[X*Y] + E[X*Z] = E[X*Y] + E[X] * E[Z] = E[X*Y] Consequently, cov(X,(Y+Z)) = cov(X,Y), since Z is independent of X and Y :-) Stefan . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
