Given two unbiased estimators, show that a third value is also an unbiased estimator Consider $\hat\theta_a$ and $\hat\theta_b$ are uncorrelated, unbiased estimators of $\theta$. In addition, the variance of $\hat\theta_a$  is twice that of $\hat\theta_b$. Show that for any constant $k$, the weighted average $\hat\theta_a(k)+\hat\theta_b(1-k)=\hat\theta_c$ is also an unbiased estimator of $\theta$.

How should I approach this question? Should I start by figuring the variances into the equation somehow, or can it be shown just by manipulating the given equation? Thank you.
 A: $$\hat\theta_c = k\hat\theta_a + (1-k)\hat\theta_b$$
Taking expectation on both sides 
\begin{align*}
E(\hat\theta_c) = kE(\hat\theta_a) + (1-k)E(\hat\theta_b) && (1) 
\end{align*}
Since $\theta_a$ and $\hat\theta_b$ are unbiased estimator of $\theta$, hence $E(\hat\theta_a) = \theta$ and $E(\hat\theta_b) = \theta$. 
Substituting the same in (1) we get 
$$E(\hat\theta_c) = kE(\hat\theta_a) + (1-k)E(\hat\theta_b) = k\theta + (1-k)\theta = k\theta + \theta - k\theta = \theta$$
Since $E(\hat\theta_c) = \theta$, $\hat\theta_c$ is also an unbiased estimator for $\theta$. 
A: Start with the definition of unbiased estimator from, say, here. Then, for our model $\text{Pr}_{\theta}(x) = \text{Pr}(x|\theta)$, the bias for a given estimator $\hat{\theta}$, is: $$\text{Bias}_{\theta}(\hat{\theta}) = \mathbb{E}_{x|\theta}\left[\hat{\theta}\right] - \theta$$ in which $\mathbb{E}_{x|\theta}$ is the expected value over the distribution $\text{Pr}(x|\theta)$. 
The problem statement gives us that $\text{Bias}_{\theta}(\hat{\theta}_a) = \text{Bias}_{\theta}(\hat{\theta}_b) = 0$ and that, for $k\in\mathbb{R}$, $\hat{\theta}_c \equiv k\hat{\theta}_a + (1-k)\hat{\theta}_b$. That $\hat{\theta}_c$ be unbiased for $\theta$ means that $\text{Bias}_{\theta}(\hat{\theta}_c) = 0 \iff \mathbb{E}_{x|\theta}\left[\hat{\theta}_c\right]=\theta$. To that end, 
$$\begin{align}\mathbb{E}_{x|\theta}\left[\hat{\theta}_c\right] &= \mathbb{E}_{x|\theta}\left[k\hat{\theta}_a + (1-k)\hat{\theta}_b\right] \\ &= \mathbb{E}_{x|\theta}\left[k\hat{\theta}_a\right] + \mathbb{E}_{x|\theta}\left[(1-k)\hat{\theta}_b\right] \\ &= k\mathbb{E}_{x|\theta}\left[\hat{\theta}_a\right] + (1-k)\mathbb{E}_{x|\theta}\left[\hat{\theta}_b\right] \\ &= k\theta + (1-k)\theta \\ &= k\theta + \theta - k\theta \\ &= \theta \ \ \ \square \end{align}$$
by the linearity of expectation.
