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
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
"And
> 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 ensur
Hi,
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(
