# Applying Lehmann-Scheffe Theorem to an example

Let me state the theorem first:

Let $T$ be a sufficient and complete statistic for the statistical model $\mathcal{P}$ and let $\tilde{\gamma}_1$ be an unbiased estimator for the parameter $\gamma = g(\theta) \in \mathbb{R}^k$ then the estimator \begin{equation}\hat{\gamma} = \hat{\gamma}(T)=E.(\tilde{\gamma}_1 \mid T) \end{equation}

has the smallest covariance matrix among all unbiased estimators for the parameter $\gamma = g(\theta)$

Here is the example: let $\boldsymbol{X}$ be a sample of independent $N(\mu,\sigma^2)$ distributed r.v's with parameter if interest $\theta = (\mu,\sigma^2)$

The arithmetic mean $\bar{X} = \dfrac{1}{n}\sum X_i$ and sample variance $S^2 = \dfrac{1}{n-1} \sum (X_i - \bar{X})^2$ are unbiased estimators.

We also know that ($\sum X_i, \sum X^2_i$) is sufficient and complete for the class of all normal distributions.

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Iam trying to apply the theorem above therefore i first set $\tilde{\gamma}_1 = (\bar{X},S^2)^T$ and $T = (\sum X_i, \sum X^2_i)^T$

but does $\hat{\gamma}(T)=E.(\tilde{\gamma}_1 \mid T)$ mean: $E\begin{pmatrix} \bar{X} \mid \sum X_i \\S^2 \mid \sum X^2_i \end{pmatrix}$ ? this cannot be the right , can someone tell me what Iam doing wrong?

Let me take another example: Let $(X_1,...,X_n)$ be iid acording to the uniform distribution over the interval $[0,\theta]$ then $\hat{\theta}_{MLE}= X_{max}$ is an unbiased estimator and $E_{\theta}X_{max} = \dfrac{n}{n+1}\theta$.

We know that $X_{max}$ is complete and sufficient , hence according to the theorem

$E[\frac{n+1}{n}X_{max} \mid X_{max} ] = \theta$ but that is just the parameter i should get an estimator, i think I am missing something fundamental

• I believe the Lehmann-Scheffe theorem requires completeness too and not just sufficiency. – dsaxton Jul 30 '15 at 20:44

In the context of the normal distribution both $\bar{X}$ and $S^{2}$ are already function of the UMVUE parameter (this is where your logic was wrong. You don't condition each separately but rather on the joint statistic $\vec{T(X)}=(\sum_{i}X_{i},\sum_{i}X_{i}^{2}$ ) and so they are already the UMVUE.
As far as the uniform family goes, you're computing the expectation wrong. $E[Y|Y]=Y$
• ok thanks @Nick !!, but just for the sake of it $\hat{\gamma}(T)=E.(\tilde{\gamma}_1 \mid T)$ means , in the case with the normal distribution example: $E[\bar{X} \mid \sum X_i , \sum X^2_i]$ and $E[S^2 \mid \sum X_i , \sum X^2_i]$ , – Danny Jul 30 '15 at 14:37