📂Mathematical Statistics

Minimum Variance Unbiased Estimator Uniqueness

Theorem ¹

If $W$ is the Best Unbiased Estimator for $\tau (\theta)$, then $W$ is unique.

Proof

Cauchy-Schwarz Inequality: For the Random Variable $X, Y$, the following holds: $$ \text{Cov} (X,Y) \le \text{Var} X \text{Var} Y $$ The necessary and sufficient condition for equality to hold is as follows: $$ \exist a \ne 0 , b \in \mathbb{R} : a X + b = Y $$

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Assuming $w '$ is another Best Unbiased Estimator for $W$, and considering $W^{\ast} := \left( W + W’ \right) / 2$, its expectation is $$ E_{\theta} W^{\ast} = \left( \tau (\theta) + \tau (\theta) \right) / 2 = \tau (\theta) $$ and its variance is $$ \begin{align*} \text{Var}_{\theta} W^{\ast} =& \text{Var}_{\theta} \left( {{ 1 } \over { 2 }} W + {{ 1 } \over { 2 }} W’ \right) \\ =& {{ 1 } \over { 4 }} \text{Var}_{\theta} W + {{ 1 } \over { 4 }} \text{Var}_{\theta} W’ + {{ 1 } \over { 2 }} \text{Cov}_{\theta} \left( W, W’ \right) \\ \le& {{ 1 } \over { 4 }} \text{Var}_{\theta} W + {{ 1 } \over { 4 }} \text{Var}_{\theta} W’ + {{ 1 } \over { 2 }} \sqrt{\text{Var}_{\theta} W \cdot \text{Var}_{\theta} W’} \\ =& \text{Var}_{\theta} W \end{align*} $$

If the inequality $<$ holds, this contradicts the premise that $W$ is the Best Unbiased Estimator, hence it suffices to show that the equality $=$ holds for all $\theta$. The necessary and sufficient condition for only the equality to hold is for some $a (\theta) \ne 0$ and $b(\theta) \in \mathbb{R}$, $a (\theta) W + b(\theta) = w '$ holds, and upon direct calculation, according to the Properties of Covariance, $$ \begin{align*} \text{Cov}_{\theta} \left( W, W’ \right) =& \text{Cov}_{\theta} \left( W, W’ \right) \\ =& \text{Cov}_{\theta} \left( W, a (\theta) W + b(\theta) \right) \\ =& \text{Cov}_{\theta} \left( W, a (\theta) W \right) \\ =& a (\theta) \text{Var}_{\theta} W \end{align*} $$ Having already established that $\text{Cov}_{\theta} \left( W, W’ \right) = \text{Var}_{\theta} W$, it follows that $a(\theta) = 1$, and since $E_{\theta} \tau (\theta)$ it implies $b(\theta) = 0$, which proves $W = w '$.

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