estimators

**calliope** · April 29th 2008, 11:05 AM

A random sample $x_{1}...x_{n}$ , n=200, is taken from a distribution with mean $\mu$ and variance $\sigma^2$ , both are unknown. Find unbiased estimators of the mean and variance, given
$\sum_{i=1}^{200}x_{i}=308$ and $\sum_{i=1}^{200}x_{i}^2=3784$

I don't even know where to start. Can anyone give me any help please?

**CaptainBlack** · April 29th 2008, 08:17 PM

Originally Posted by calliope

A random sample $x_{1}...x_{n}$ , n=200, is taken from a distribution with mean $\mu$ and variance $\sigma^2$ , both are unknown. Find unbiased estimators of the mean and variance, given
$\sum_{i=1}^{200}x_{i}=308$ and $\sum_{i=1}^{200}x_{i}^2=3784$

I don't even know where to start. Can anyone give me any help please?

The sample mean is an unbiased estimator of the mean and the sample variance with (n-1) rather than n in the formulation is an unbiased estimator or the variance.

RonL

**calliope** · April 30th 2008, 02:20 AM

Thanks a lot =).

I got 1.54 as an estimate of the mean and 11.86.. for the variance. I'm hoping that sounds about right.

**CaptainBlack** · April 30th 2008, 04:02 AM

Originally Posted by calliope

Thanks a lot =).

I got 1.54 as an estimate of the mean and 11.86.. for the variance. I'm hoping that sounds about right.

The mean is correct, but you should have:

$\sigma_{n-1}^2=\frac{n}{n-1}\left( \overline{x^2}-\overline{x}^2 \right) =\frac{200}{199}\left( \frac{3784}{200}-1.54^2 \right)$

for the unbiased estimator of the variance.

RonL

**calliope** · April 30th 2008, 06:30 AM

Yeah, I realised since posting that last message that I made a stupid calculation error. Thanks a lot for your help.

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