computing the variance

Just redefine the var(x) as sum((x-mean(x))^2)/length(x)?
Or straightforward just use var(x)*(1-1/length(x))

As you already mentioned var(x) is now defined by 
sum((x-mean(x))^2)/(length(x)-1) which is an *unbaised* estimtor of COV.
While sum((x-mean(x))^2)/length(x) is a *biased* estimator with
Bias = -1/N COV

Denote population mean by  M
Proof: E[sum (Xj-mean(X))^2] = E[sum Xj^2 - n mean(X)^2]
                       = sum E[Xj^2] - n E[mean(X)^2]
                       = sum (COV + M^2) - n (1/n COV + M^2)
                       = (n-1) COV

Best regards,
Kristel

hi,
when i was computing the variance of a simple vector, i found unexpect 
result. not sure whether it is a bug.

 > var(c(1,2,3))

[1] 1  #which should be 2/3.

 > var(c(1,2,3,4,5))

[1] 2.5 #which should be 10/5=2

it seems to me that the program uses (sample size -1) instead of sample 
size at the denominator. how can i rectify this?

regards,
tianhua

______________________________________________
R-help at stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html

__________________________________________
Kristel Joossens        Ph.D. Student
Research Center ORSTAT  K.U. Leuven
Naamsestraat 69         Tel: +32 16 326929
3000 Leuven, Belgium    Fax: +32 16 326732
E-mail:  Kristel.Joossens at econ.kuleuven.be
http://www.econ.kuleuven.be/public/ndbae49

Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm