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

  • 2
    $\begingroup$ The mean of the product of two independent random variables is the product of the means. Is that fairly intuitively clear to you? $\endgroup$ – André Nicolas Aug 15 '15 at 14:32
  • $\begingroup$ Read your comment a few times and I haven't been able to understand yet. A little more information would probably help me out $\endgroup$ – Math_Illiterate Aug 15 '15 at 14:37
  • $\begingroup$ An answer has been given by molarmass that completes the calculation, using mean of independent product is product of the means. $\endgroup$ – André Nicolas Aug 15 '15 at 14:39
  • $\begingroup$ I see now, his explanation makes a lot of sense. Thanks $\endgroup$ – Math_Illiterate Aug 15 '15 at 14:40
  • $\begingroup$ 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 $Z=X_j-\mu$. $\endgroup$ – drhab Aug 15 '15 at 14:42

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}

  • $\begingroup$ This makes sense, thanks! I'll accept your answer when the time limit lets me $\endgroup$ – Math_Illiterate Aug 15 '15 at 14:41

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