To obtain an estimate for the variance of the population mean

• Apr 19th 2010, 08:25 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?
• Apr 19th 2010, 09:28 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 $\sigma^2$.
• Apr 19th 2010, 11:32 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
• Apr 20th 2010, 07:13 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