1. ## estimators

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

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

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

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

3. 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.

4. 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:

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

for the unbiased estimator of the variance.

RonL

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