# Variance of Sample mean (not independent)

• Sep 15th 2012, 03:48 AM
lpd
Variance of Sample mean (not independent)
Hi I have this question.

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

1) Derive $Var(\bar{X})$ for this situation.
2) What is $Var(\bar{X})$ when \rho=0? Explain.
3) What is $Var(\bar{X})$ when \rho=1? Explain.
4) Use the rsult you have derived to comment on how small $\rho$ 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
• Sep 16th 2012, 08:22 PM
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)
• Sep 17th 2012, 09:15 AM
harish21
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.

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

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