Re: The sample s.d. always underestimates


Paul Sanchez wrote:
"Luis A. Afonso" wrote:
Jack wrote

*** Here is a counter-example. Suppose the population variance,
sigma^2 = 1. Then for a sample of size N=2, let x(1) = 0 and
x(2) = 10. The sample variance s^2 = sqrt(50) = 7.07... > 1.
I found one.***

My response

This is idiotic. Without knowing what are

The Distribution probability function (or density if it is continuous) in order to evaluate the
_____Population Mean
_____Population Variance

I have no way to compare your sample variance with the Population Variance!


What are the probabilities of to have x(1)=0 and x(2)=10? What are?

Your *soi-disant * counter-example is TRASH. You had found NOTHING.


______licas (Luis A. Afonso)

No, Jack was right. He told you that the population variance was 1, as
a given. There are an infinite number of distributions which have
population variance 1 which could yield 0 and 10 as outcomes. For
example, the simple distribution p(0) = (1 + sqrt(0.96)) / 2, p(10) = 1
- p(0) does the job nicely. If you do the math, you'll find that the
distribution I've specified has a variance of exactly 1 and both 0 and
10 occur with positive probabilities. QED, Jack's sample containing
{0,10} may be rare, but it's perfectly realizable. And it yields a
sample variance > 1, exactly as claimed.

Here is another counter-example. In R I did this:

set.seed(31233)
var(rnorm(100))
[1] 1.211056

The rnorm function generates normal variates, in this case
standard normal (var = 1). The sample variance in this
sample of 100 is clearly bigger than 1.

--
Kevin E. Thorpe
Assistant Professor, Department of Public Health Sciences
Faculty of Medicine, University of Toronto

.

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