1
$\begingroup$

Apologies if this may look like a silly question.

Consider a sample of size $N$. Let $y_i$ be the $i$-th realization of a random variable that is independently and identically distributed with mean $\mu$ and variance $\sigma^2$.

Let $\bar{y}$ be the mean of $y_i$ across all $i$. We know that $Var(\bar{y})=\sigma^2/N$.

How about $Cov(y_i,\bar{y})$? I would say that

$Cov(y_i,\bar{y})=Cov\left(y_i,\frac{\sum_i y_i}{N}\right)=\frac{1}{N}\left[Var(y_i)+(N-1)Cov(y_i,y_j)\right]=\sigma^2/N$

Where the latter follows because of the independence assumption. Is that right? Many thanks.

$\endgroup$
1
$\begingroup$

Overall this looks good. I have a few minor concerns and comments:

  • It would help to specify the definition $\overline{Y}= \frac{1}{N}\sum_{i=1}^N Y_i$. (At first I was confused since I thought you meant $\overline{Y}=E[Y_i]$).

  • I would personally prefer to write your equality as $$ Cov(Y_i, \overline{Y}) = \frac{1}{N}\left[Var(Y_i) + \sum_{j \neq i} Cov(Y_i, Y_j)\right] $$ This avoids the "dangling and undefined index $j$" in
    $$ Cov(Y_i, \overline{Y}) = \frac{1}{N}\left[Var(Y_i) + (N-1)\underbrace{Cov(Y_i,Y_j)}_{\mbox{undefined $j$}}\right]$$

  • The basic principles you are applying are \begin{align} Cov(Z, aW) &= a Cov(Z, W)\\ Cov\left(Z, \sum_{i=1}^N R_i\right) &= \sum_{i=1}^N Cov(Z, R_i) \end{align} for any random variables $Z, W, R_1, ..., R_N$ with finite means and variances, and any scalar $a$ and positive integer $N$.

$\endgroup$
  • $\begingroup$ Thanks, and apologies for being sloppy in my original formulation. $\endgroup$ – FscoA Dec 9 '17 at 19:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.