# [R] computing the variance

Kristel Joossens kristel.joossens at econ.kuleuven.be
Mon Dec 5 10:32:20 CET 2005

```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

Wang Tian Hua wrote:
> 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  #which should be 2/3.
>  > var(c(1,2,3,4,5))
>  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
>
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--
__________________________________________
Kristel Joossens        Ph.D. Student
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E-mail:  Kristel.Joossens at econ.kuleuven.be
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