# Unbiased estimators - how to show unbiasedness?

Let $\hat{\theta}_i$ $(i = 1, 2)$ be two unbiased estimators of $\theta \in \mathbb{R}$, which are uncorrelated, with $V(\hat{\theta}_i) = \sigma^2_i > 0$ $(i = 1, 2)$. For $\alpha \in [0, 1]$, define $\hat{\theta} = \hat{\theta}(\alpha) = \alpha\hat{\theta}_1 + (1 − \alpha)\hat{\theta}_2$.

What is the quickest way to find a value of $\alpha$ which minimizes $V[\hat{\theta}(\alpha)]$ and to show that we have an unbiased estimator?

• All the techniques you need are exhibited in the closely related question at stats.stackexchange.com/questions/5392. – whuber Nov 8 '16 at 16:41
• But how do you develop a clean proof of that? – Bonsaibubble Nov 8 '16 at 17:05
• One way uses Lagrange Multipliers, as illustrated in a closely related situation. Another uses ordinary Calculus techniques--after all, $V(\hat\theta(\alpha))$ is a differentiable function of $\alpha$. The most elementary method observes that $$V(\hat\theta(\alpha))=(\sigma_1^2 + \sigma_2^2)\left(\alpha - \frac{\sigma_2^2}{\sigma_1^2+\sigma_2^2}\right)^2 +\frac{\sigma_1^2\sigma_2^2}{\sigma_1^2+\sigma_2^2},$$ which has an obvious unique minimum. – whuber Nov 8 '16 at 17:14

$$\text{Var} [\hat \theta (\alpha))] = \alpha^2 \sigma^2_1 + (1-\alpha)^2 \sigma^2_2$$
Taking the derivative with respect to $\alpha$ and setting it equal to zero we have the condition
$$2\alpha\sigma^2_1 -2(1-\alpha) \sigma^2_2 = 0 \implies \alpha^* = \frac {\sigma^2_2}{\sigma^2_1 + \sigma^2_2}$$
As regards unbiasedness, it comes immediately from the linearity property of the expected value, irrespective of whether we use the optimal $\alpha^*$ or not. Proving the linearity property in turn is proving the additive property in integration.