Assume Z is independent of X. Then E[ZX] = E[Z] * E[X]. E[X*(Y+Z)] = E[ X*Y + X*Z ] = E[X*Y] + E[X*Z] = E[X*Y] + E[X] * E[Z]
In sci.stat.math Stefan Mangard <[EMAIL PROTECTED]> wrote: [snipped] > Since E(Y + Z) = E(Y), all I would need is some clou on how E(X*(Y+Z)) > looks like. But unfortunately, I don't know so far how X*(Y+Z) looks like. . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
