I was reading simple random sample without replacement.

Now, they are computing the $Var(\bar{X})$, where $\bar{X}=1/n \sum(X_i)$ and $X_i$ is the sample.

Going back to $Var(\bar{X})$.

They formulate as

$Var(\bar{X}) = 1/n^2 \sum _{i,j}Cov(X_i, X_j)= 1/n^2 (\sum_i Var(X_i) +\sum_{i \neq j}Cov(X_i,X_j)=\sigma^2/n +(n-1)/n*Cov(X_1,X_2)$.

I dont understand the last step, why $Cov(X_i,X_j) $ is indepedent of $i$ and $j$?

How come $Cov(X_i,X_j) =Cov(X_1, X_2)$? This doesn't make sense.

Reference of detail step: bottom of page 3 from http://dept.stat.lsa.umich.edu/~moulib/sampling.pdf

  • $\begingroup$ Haven't checked the reference. But $X_i$'s must be identically distributed. $\endgroup$ – Dhruv Kohli - expiTTp1z0 Jan 24 '17 at 18:25
  • $\begingroup$ Yes, even that is true, how come $Cov(X_1, X_2)$ they are independent of $i, j$? $\endgroup$ – wrek Jan 24 '17 at 18:40

Since $X_i$'s are identically distributed $Cov(X_i,X_j), i \neq j$ will be same as $Cov(X_{i'},X_{j'}), i' \neq j'$.

Using $i'=1$ and $j'=2$, $\sum\limits_{i\neq j} Cov(X_i,X_j) = n(n-1)Cov(X_1,X_2)$ and the result follows.


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