To obtain an estimate for the variance of the population mean

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April 19th 2010, 07:25 AM
yus2k

To obtain an estimate for the variance of the population mean

To obtain an estimate for the variance of the population mean,
sigma^2, a natural idea would be to calculate

1/n(sum i=1 to n)(xi - (x bar))^2
.
It turns
out that this estimate has a mean of (n-1/n) sigma^2. This can be corrected
by multiplying 1/n(sum i=1 to n)(xi - (x bar))^2

by a factor so that the estimate is
unbiased.
What is the factor?
April 19th 2010, 08:28 PM
cpbrunner

I'm not sure what "factor" you are looking for. The estimator that you have for the sample variance is indeed unbiased, demonstrated by the following:

Let $V$ be a chi-square random variable. It can easily be shown that $V=\frac{(n-1)}{\sigma^2}S^2\implies S^2=\frac{\sigma^2}{(n-1)}V$ .

Then it follows that:
$E(S^2)=E\left(\frac{\sigma^2}{(n-1)}V\right)=\frac{\sigma^2}{(n-1)}E(V)=\frac{\sigma^2}{(n-1)}(n-1)=\sigma^2$ .

Since $E(S^2)=\sigma^2$ , it is an unbiased estimator of $\sigma^2$ .
April 19th 2010, 10:32 PM
CaptainBlack

Quote:

Originally Posted by yus2k

To obtain an estimate for the variance of the population mean,
sigma^2, a natural idea would be to calculate

1/n(sum i=1 to n)(xi - (x bar))^2
.
It turns
out that this estimate has a mean of (n-1/n) sigma^2. This can be corrected
by multiplying 1/n(sum i=1 to n)(xi - (x bar))^2

by a factor so that the estimate is
unbiased.
What is the factor?

You seem to be talking about the population variance not the sample mean's variance.

In which case the question is asking you:

What do you need to multiply:

$\left(n+\frac{1}{n}\right) \sigma^2$

by to obtain $\sigma^2$ ? An unbiased estimator has expectation equal to what you are trying to estimate.

CB
April 20th 2010, 06:13 AM
yus2k

Quote:

Originally Posted by CaptainBlack

You seem to be talking about the population variance not the sample mean's variance.

In which case the question is asking you:

What do you need to multiply:

$\left(n+\frac{1}{n}\right) \sigma^2$

by to obtain $\sigma^2$ ? An unbiased estimator has expectation equal to what you are trying to estimate.

CB

yer i got the factor as sumthing similar
but i had n/n-1