I'm trying to prove a theorem, where it is given that each $X_i$ is independent and identically distributed with mean $\mu$ and variance $\sigma^2$. Within this theorem, I have multiple sub-results to prove. One of the results is $${\rm Cov}(X_i - \overline{X},\ \overline{X})=0$$

Here's the work and where I get stuck:

In general, have that $${\rm Cov}(X,Y)=E\big[(X-E[X])(Y-E[Y])\big]=E[XY]-E[X]E[Y]$$ We know that $$\overline{X}=\frac{1}{n}\sum_{i=1}^n X_i$$ So it follows that \begin{align} {\rm Cov}(X_i - \overline{X},\overline{X})&={\rm Cov}(X_i,\overline{X})-{\rm Cov}(\overline{X},\overline{X}) \\ &=\bigg(\frac{1}{n}\sum_{j=1}^n{\rm Cov}(X_i,X_j)\bigg)-{\rm Var}(\overline{X}) \end{align}

It seems simple enough, but I have been unable to show that $$\frac{1}{n}\sum_{j=1}^n{\rm Cov}(X_i,X_j)={\rm Var}(\overline{X})$$

I end up at a result, for $\frac{1}{n}\sum_{j=1}^n{\rm Cov}(X_i,X_j)$, that is as such: $$\frac{1}{n}\sum_{j=1}^n{\rm Cov}(X_i,X_j)=\bigg(\frac{1}{n}\sum_{j=1}^nE[X_iX_j]\bigg)-\mu^2$$

Can anyone help me proceed from this step? Or maybe lead me in a alternate direction?


Because the iid assumption implies

$$\text{Cov}(X_i - \overline{X},\overline{X})=\text{Cov}(X_j - \overline{X},\overline{X})$$

for all $i$ and $j$, take the average of $\text{Cov}(X_i - \overline{X},\overline{X})$ (which the bilinearity of $\text{Cov}$ allows you to do by averaging the first argument $X_i - \overline{X}$) to obtain

$$\text{Cov}(X_i - \overline{X},\overline{X}) = \text{Cov}(\overline{X} - \overline{X},\overline{X}) = \text{Cov}(0, \overline{X}) = 0.$$

  • $\begingroup$ Am I missing something? $\bar X$ is a constant (the sample mean) isn't it? Doesn't it follow directly from the rules for covariances that ${\rm Cov}(X, {\rm constant})=0$? $\endgroup$ – gung - Reinstate Monica Jan 11 '15 at 22:23
  • 1
    $\begingroup$ @gung Since all the $X_i$ are random variables, so is $\overline{X}$. There are no additional assumptions that would make it a constant. In fact, you know that for sufficiently large $n$ it has an approximately Normal distribution. $\endgroup$ – whuber Jan 11 '15 at 22:25
  • $\begingroup$ Thank you, whuber! Much cleaner presentation then what I was going for $\endgroup$ – Savage Henry Jan 11 '15 at 22:40

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