# Expectation of Independent Variables Equals Zero?

Given $n$ independent random variables, $X_1, X_2, ..., X_n$ , each having a normal distribution, why is it that the following expectation holds?

$$E[(X_i - \mu)(X_j - \mu)] = 0$$

where $i \neq j$

I saw this statement in a proof explaining why we divide by $n-1$ when computing the sample variance and of course there was no explanation. An intuitive explanation and/or a link to more detailed information about why this is true would be greatly appreciated

• The mean of the product of two independent random variables is the product of the means. Is that fairly intuitively clear to you? – André Nicolas Aug 15 '15 at 14:32
• Read your comment a few times and I haven't been able to understand yet. A little more information would probably help me out – Math_Illiterate Aug 15 '15 at 14:37
• An answer has been given by molarmass that completes the calculation, using mean of independent product is product of the means. – André Nicolas Aug 15 '15 at 14:39
• I see now, his explanation makes a lot of sense. Thanks – Math_Illiterate Aug 15 '15 at 14:40
• If $Y$ and $Z$ are independent then $\mathbb E(Y.Z)=\mathbb EY.\mathbb EZ$ (mean of product $=$ product of mean). Apply this on $Y=X_i-\mu$ and $Z=X_j-\mu$. – drhab Aug 15 '15 at 14:42