You do need to know the covariance between and .
.
NOW, if they are iid then...
.
AND VAR(xbar) +VAR(ybar)+2COV(xbar,ybar) is WRONG.
It's VAR(xbar) +VAR(ybar)-2COV(xbar,ybar).
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!
iid means independent (i) and identically distributed (id)
Statisticians call that a random sample.
NOW the sample are iid
same for are iid.
BUT I'm assuming there is a common covariance between any
and just like we assume there
is a common variance of the 's and 's.
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 times the identity matrix of order n.
The variance of each is
while all the covariances between the X's are 0.