Covariance of two sample means

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June 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!
June 19th 2009, 08:19 PM
matheagle

You do need to know the covariance between $X_i$ and $Y_j$ .

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

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

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

NOW, if they are iid then...

$=Cov(X_1,Y_1)$ .

AND VAR(xbar) +VAR(ybar)+2COV(xbar,ybar) is WRONG.
It's VAR(xbar) +VAR(ybar)-2COV(xbar,ybar).
June 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!~
June 20th 2009, 05:29 PM
matheagle

iid means independent (i) and identically distributed (id)
Statisticians call that a random sample.
NOW the sample $X_1,\ldots ,X_n$ are iid
same for $Y_1,\ldots ,Y_m$ are iid.
BUT I'm assuming there is a common covariance between any
$X_i$ and $Y_j$ just like we assume there
is a common variance of the $X_i$ 's and $Y_j$ 's.
June 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!
June 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 $\sigma_X^2$ times the identity matrix of order n.
The variance of each $X_i$ is $\sigma_X^2$
while all the covariances between the X's are 0.

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