Patrick, If z = xy, then yes, E[z] = E[xy] = 0.
HOWEVER... E[xyz] = E[x^2*y^2]. -- T. Arthur Wheeler MathCraft Consulting Columbus, OH 43017 "Patrick Agin" <[EMAIL PROTECTED]> wrote in message W5aF7.1435$[EMAIL PROTECTED]">news:W5aF7.1435$[EMAIL PROTECTED]... > > Thank you very much Andrew for your reply, > I thought at this possibility before sending the post but my reasoning was: > > If cor(x,y)=0, it implies that cov(x,y)=0 => E[(x-mean(x))(y-mean(y))]=0 > but if mean(x)=mean(y)=0, then E[xy]=0. > So if z=x*y, E[z]=E[xy]=0, isn't it? Am I wrong? > > Patrick > > > "Andrew Schulman" <[EMAIL PROTECTED]> wrote in message > [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > > > I am interested in the following expression and conditions under which > it > > > equals 0: > > > E(x*y*z) where x,y and z are random variables and E(.) denotes > expectation. > > > > > > Here, x and y have mean 0 and the correlation between x and y is also > zero. > > > > > > Are these two conditions *sufficient* to ensure that E(x*y*z) = 0? > > *********************** > > > > No. Example: let Z=X*Y. > > > > -- > > To reply by e-mail, change "deadspam" to "home" > > ================================================================= Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =================================================================
