1
$\begingroup$

Let $X_1, \ldots, X_n \sim \mathcal{N}(\mu, \sigma^2)$ be the sample, when $\mu$, $\sigma$ are unknown.

We suggest assessment for $\sigma^2$: $$S^2 = \frac{\displaystyle\sum_{i=1}^n (X_i - \bar{X})^2}{n-1}$$

Now, I know this assesment is unbiased, so for calculating the MSE I have to calculate the expectation.

Any suggestions for this? I'm pretty stuck in this part.

$\endgroup$
0
$\begingroup$

I assume that MSE means mean-squared error, so I interpret this question as a request to show how to calculate the variance of the estimator $S^2$. I'd organise the algebra for this using the Kronecker delta $\delta _{ij}$, so that $$ \begin{eqnarray*} X_{i}-\bar{X} &=&\sum_{j=1}^{n}X_{j}\left( \delta _{ij}-\frac{1}{n}\right) \text{ .} \\ E\left( X_{i}X_{j}\right) &=&\sigma ^{2}\delta _{ij} \\ E\left( X_{i}X_{j}X_{k}X_{l}\right) &=&\sigma ^{4}\left( \delta _{ij}\delta _{kl}+\delta _{ik}\delta _{jl}+\delta _{il}\delta _{jk}+3\delta _{ij}\delta _{ik}\delta _{il}\right) \text{ .} \end{eqnarray*} $$ Then the variance of the estimator $S^2$ is

$$ \begin{eqnarray*} E\left( \left( S^{2}-\sigma ^{2}\right) ^{2}\right) &=&E\left( \left( \frac{% \sum_{i=1}^{n}\left( \sum_{j=1}^{n}X_{j}\left( \delta _{ij}-\frac{1}{n}% \right) \right) ^{2}}{n-1}-\sigma ^{2}\right) ^{2}\right) \\ &=&E\left( \left( \frac{\sum_{i,j,k=1}^{n}X_{j}X_{k}\left( \delta _{ij}-% \frac{1}{n}\right) \left( \delta _{ik}-\frac{1}{n}\right) }{n-1}-\sigma ^{2}\right) ^{2}\right) \end{eqnarray*} $$

The equations are a bit to long to write down, but the biggest term is

$$ E\left( \frac{% \sum_{i_{1},j_{1},k_{1},i_{2},j_{2},k_{2}=1}^{n}X_{j_{1}}X_{k_{1}}X_{j_{2}}X_{k_{2}}\left( \delta _{i_{1}j_{1}}-% \frac{1}{n}\right) \left( \delta _{i_{1}k_{1}}-\frac{1}{n}\right) \left( \delta _{i_{2}j_{2}}-\frac{1}{n}\right) \left( \delta _{i_{2}k_{2}}-\frac{1}{% n}\right) }{\left( n-1\right) ^{2}}\right) $$

and it's just a question of applying the equation for $E\left( X_{i}X_{j}X_{k}X_{l}\right)$ and working through the algebra.

$\endgroup$
0
$\begingroup$

Here is the solution using the mathStatica add-on to Mathematica:

$S^2$ is also known as the $2^{\text{nd}}$ h-statistic (unbiased estimator of the $2^{\text{nd}}$ central moment). Since $S^2$ is an unbiased estimator of population variance, $\text{MSE}[S^2]$ = $\text{Var}(S^2)$.

So we seek the $2^{\text{nd}}$ Central Moment (i.e. Variance) of the $2^{\text{nd}}$ h-statistic, so the answer is a one-liner:

MSE = CentralMomentToCentral[2, HStatistic[2][[2]] ] 

enter image description here

where $\mu_2$ and $\mu_4$ denote the $2^{\text{nd}}$ and $4^{\text{th}}$ central moments of the population. In the special case of the Normal distribution, just substitute in for $\mu_2$ and $\mu_4$:

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.