# Covariance of two sample means

• Jun 19th 2009, 04:39 PM
B_Miner
Covariance of two sample means
Hello all!

I am looking to calculate VAR(xbar-ybar) where X and Y are not independent.

Want to try VAR(xbar) +VAR(ybar)+2COV(xbar,ybar).

I know VAR(xbar) = sigma^2 of X / n1

I know VAR(ybar) = sigma^2 of Y / n2

But the COV(xbar,ybar) has me stumped. I wanted to write
(1/n1)* (1/n2) * COV (SUM(Xi) 1 to n1, SUM(Yi) 1 to n2).

I have researched the additive rule of COV(X+Y,Z) = COV(X,Y) + COV(Y,Z) but not sure how to apply with a sum in both 'places' for COV(X,Y).

Anyone help?

Thanks!
• Jun 19th 2009, 08:19 PM
matheagle
You do need to know the covariance between $\displaystyle X_i$ and $\displaystyle Y_j$.

$\displaystyle Cov(\bar X,\bar Y)= Cov\biggl({\sum_{i=1}^n X_i\over n},{\sum_{j=1}^m Y_j\over m}\biggr)$

$\displaystyle ={1\over nm} Cov\biggl(\sum_{i=1}^n X_i,\sum_{j=1}^m Y_j\biggr)$

$\displaystyle ={1\over nm} \sum_{i=1}^n\sum_{j=1}^m Cov(X_i,Y_j)$.

NOW, if they are iid then...

$\displaystyle =Cov(X_1,Y_1)$.

AND VAR(xbar) +VAR(ybar)+2COV(xbar,ybar) is WRONG.
It's VAR(xbar) +VAR(ybar)-2COV(xbar,ybar).
• Jun 20th 2009, 05:04 PM
B_Miner
Thank you Matheagle!

One point of clairification, when you say if the variables are iid...does this mean the x's and y's are both iid? Could you explain how that then results in

COV(xbar,ybar) = Cov(X1,Y1)?

Thanks!~
• Jun 20th 2009, 05:29 PM
matheagle
iid means independent (i) and identically distributed (id)
Statisticians call that a random sample.
NOW the sample $\displaystyle X_1,\ldots ,X_n$ are iid
same for $\displaystyle Y_1,\ldots ,Y_m$ are iid.
BUT I'm assuming there is a common covariance between any
$\displaystyle X_i$ and $\displaystyle Y_j$ just like we assume there
is a common variance of the $\displaystyle X_i$'s and $\displaystyle Y_j$'s.
• Jun 20th 2009, 06:42 PM
B_Miner
Oh I see now. Say the covariance between X and Y is 10 (taken from a variance-covariance matrix) and there are 3 X's and 3 Y's:

(1/9) * (10+10+10+10+10+10+10+10+10) = 10

Thanks again!
• Jun 20th 2009, 07:03 PM
matheagle
taken from a variance-covariance matrix
isn't really right.
There is a variance-covariance matrix for the X's
and a different one for the Y's.
The variance-covariance matrix for the X's is $\displaystyle \sigma_X^2$ times the identity matrix of order n.
The variance of each $\displaystyle X_i$ is $\displaystyle \sigma_X^2$
while all the covariances between the X's are 0.