Variance of Sample mean (not independent)

Hi I have this question.

Suppose that the random variables $\displaystyle X_1, X_2, ... X_n$ are identically distributed, with mean $\displaystyle \mu$ and variance $\displaystyle \sigma^2$, but not independent. Assume the correlation between any pair is equal to $\displaystyle \rho$. i.e. $\displaystyle corr(X_i,X_j)=\rho$ for $\displaystyle i \neq j$.

1) Derive $\displaystyle Var(\bar{X})$ for this situation.

2) What is $\displaystyle Var(\bar{X})$ when \rho=0? Explain.

3) What is $\displaystyle Var(\bar{X})$ when \rho=1? Explain.

4) Use the rsult you have derived to comment on how small $\displaystyle \rho$ can be in this situation. Explain.

To me, this looks hard. I can do it for the case if the random variables are independent and identically distributed (because it is such a well-known result).

But how do you do it if the random variables are not independent? I'm a bit confused. Can someone lend me a hand and direct me to a suitable source for this question?

Thanks!!

Regards,

Lpd

Re: Variance of Sample mean (not independent)

Assuming Xbar is the sum of all Xi I think u'll need these relations. var(Xi) = sig^2:

p = corr(Xi,Xj) = cov(Xi,Xj)/[sqrt(var(Xi)var(Xj)]

cov(Xi,Xj) = E(XiXj) - E(Xi)E(Xj)

Var(Xi+Xj) = Var(Xi) + Var(Xj) + Cov(Xi,Xj)

Re: Variance of Sample mean (not independent)

For correlated variables, you can use the fact that the variance of the sum of correlated variables is equal to the sum of the covariances.

$\displaystyle Var(\bar{X_n})= Var(\dfrac{1}{n}\sum_{i=1}^n X_i)=\dfrac{1}{n^2}Var(\sum_{i=1}^n X_i)=\dfrac{1}{n^2} \sum_{i=1}^n \sum_{j=1}^n Cov(X_i,X_j)$

reading this link http://sites.stat.psu.edu/~dhunter/a...ures/asymp.pdf section 2.2.3 on page 54 might help

Variance - Wikipedia, the free encyclopedia

$\displaystyle \rho = 0 $ means the variables are independent and the variance will be given by $\displaystyle \dfrac{\sigma^2}{n}$