2

Consider the following statement:

Let $T_1,T_2$ be two unbiased estimators with common variance $\alpha \sigma^2$, where $\sigma^2$ is the variance of the UMVUE. Show that that $\rho(T_1,T_2) \geq \frac{2-\alpha}{\alpha}$, where $\rho$ is the correlation function.

Now, clearly $\alpha >1,$ since $\operatorname{Var} (T_1) = \alpha \sigma^2$ must be larger than the UMVUE's variance, $\sigma^2.$ Now, I'm kind of lost on how to show that the statement is true. We can write

\begin{align*} \rho(T_1,T_2) &= \frac{\operatorname{Cov}(T_1,T_2)}{\sqrt{\operatorname{Var(T_1)}} \sqrt{\operatorname{Var(T_2)}}}\\[0.3em] &= \frac{\operatorname{Cov}(T_1,T_2)}{\alpha \sigma^2}\\ \end{align*} Here we could write $\operatorname{Cov}(T_1,T_2) =\mathbb{E}(T_1 T_2) - \mathbb{E}(T_1) \mathbb{E}(T_2) $ where we know that $\mathbb{E}(T_1) =\mathbb{E}(T_2)= \theta$, if $T_1$ and $T_2$ are estimators for the parameter $\theta$ (since they are unbiased estimators). However I don't see how that helps. I've tried to reduce algebraically to something simpler, but don't really get anywhere.

If anyone could provide some guidance one how to proceed I would appreciate it very much.

Bergson
  • 1,647

2 Answers2

2

Let $T_3=\frac{T_1+T_2}{2}$ be an unbiased estimator for $\theta$. (This can easily be proved by taking expectations from both sides from: $T_3=\frac{T_1+T_2}{2}$

Then, use the fact that: V(U)

So: $V(U)<V(T_3)= \dfrac{V(T_1)+V(T_2)+2\operatorname{cov}(T_1,T_2)}{4} = \dfrac{\alpha\sigma^2+\operatorname{cov}(T_1,T_2)}{2}$

Now we have: $$V(U)=\sigma^2 \leqslant \dfrac{2\alpha\sigma^2+\operatorname{cov}(T_1,T_2)}{2}$$

Multiply both sides by 2 and with some basic algebra we get: $$\sigma^2 (2-\alpha)\leqslant \operatorname{cov}(T_1,T_2)$$

Now divide both sides by $\alpha \sigma^2$

Note: The inequality holds because: $\alpha>1$ and $0 \leqslant \sigma^2$

So: $$\dfrac{(2-\alpha)}{\alpha} \leqslant \dfrac{\operatorname{cov}(T_1,T_2)}{\alpha\sigma^2}=\rho(T_1,T_2)$$

As desired.

Michael Hardy
  • 1
PLanderos33
  • 174
1

Suppose you are estimating a parametric function $g(\theta)$. Let $T_0$ be the UMVUE of $g(\theta)$.

Define the estimator $$T'=\frac{1}{2}(T_1+T_2)\,,$$

so that $$\operatorname E_{\theta}(T')=g(\theta)\quad,\forall\,\theta$$

And \begin{align} \operatorname{Var}_{\theta}(T')&=\frac{1}{4}\left(\operatorname{Var}_{\theta}(T_1)+\operatorname{Var}_{\theta}(T_2)+\operatorname{Cov}_{\theta}(T_1,T_2)\right) \\&=\frac{\alpha\sigma^2}{2}\left(1+\rho_{\theta}(T_1,T_2)\right)\quad,\forall\,\theta \end{align}

Since $T_0$ has the minimum variance among all unbiased estimators of $g(\theta)$, we must have $$\operatorname{Var}_{\theta}(T')\ge \operatorname{Var}_{\theta}(T_0)\quad,\forall\,\theta$$

This leads to the answer.

StubbornAtom
  • 17,052