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
Best Answer
Since the random variables are independent, \begin{align}\operatorname{E}[(X_i-\mu)(X_j-\mu)]&=\operatorname{E}[X_i-\mu] \cdot \operatorname{E}[X_j-\mu]\\ &= (\operatorname{E}[X_i] - \mu)(\operatorname{E}[X_j]-\mu) \\ &= (\mu-\mu)(\mu-\mu)\\ &=0 \cdot 0\\ & = 0.\end{align}