A random sample consists of 3 independent observations $𝑋_1, 𝑋_2 and 𝑋_3$ follow Normal distribution 𝑁(πœ‡, $𝜎^2$) consider four estimators for population mean πœ‡ as follow:
$πœ‡Μ‚_1$ = $\frac{X_1 + 3𝑋_2 βˆ’ 2𝑋_3} {2}$ , πœ‡Μ‚2 = $\frac{5𝑋_1 βˆ’ 2𝑋_2} {3}$ , πœ‡Μ‚3 = $\frac{1}{2} 𝑋_1 + \frac{1}{2} 𝑋̅$, πœ‡Μ‚4 = $\frac{2𝑋_1 + 3𝑋_3 βˆ’ 2𝑋̅}{3}$
Where 𝑋̅ is the sample mean of $𝑋_1, 𝑋_2, 𝑋_3$.
(a) Which of the above is/are the unbiased estimator(s) for πœ‡ ?
(b) Which of the above is the best unbiased estimator for πœ‡ ?
(c) Provide an estimator for πœ‡ which is better than aforementioned πœ‡Μ‚1, πœ‡Μ‚2, πœ‡Μ‚3 and πœ‡Μ‚4. Justify your answer.

For part a, I found all are unbiased and for part b πœ‡Μ‚3 is the best. Are they correct?
But if they are correct, I think πœ‡Μ‚3 is already the best one and I can't think of an estimator better than aforementioned. How can I provide a better estimator?


$$\mathbb{E}[\hat{\mu}_1]=\frac{1\mu+3\mu-2\mu}{2}=\mu, \ \ \mathbb{V}[\hat{\mu}_1]=\frac{\sigma^2+9\sigma^2+4\sigma^2}{4}=\frac{7}{2}\sigma^2,$$

$$\mathbb{E}[\hat{\mu}_2]=\frac{5\mu-2\mu}{3}=\mu, \ \ \mathbb{V}[\hat{\mu}_2]=\frac{25\sigma^2+4\sigma^2}{9}=\frac{29}{9}\sigma^2,$$

$$\mathbb{E}[\hat{\mu}_3]=\frac{1}{2}\mu+\frac{1}{2}\frac{3\mu}{3}=\mu, \ \ \mathbb{V}[\hat{\mu}_3]=\frac{1}{4}\sigma^2+\frac{1}{4}\frac{3\sigma^2}{9}=\frac{12}{4\cdot 9}\sigma^2=\frac{1}{3}\sigma^2$$

$$\mathbb{E}[\hat{\mu}_4]=\frac{2\mu+3\mu}{3}-\frac{2}{3}\frac{3\mu}{3}=\mu, \ \ \mathbb{V}[\hat{\mu}_4]=\frac{4\sigma^2+9\sigma^2}{9}+\frac{4}{9}\frac{3\sigma^2}{9}=\frac{13\sigma^2+\frac{4}{3}\sigma^2}{9}$$

Yes $\hat{\mu}_3$ is best among them and unbiased. However, $\hat{\mu}_3$ is such that $$\lim_{N\to \infty}\mathbb{V}[\hat{\mu}_3]=\frac{1}{4}\sigma^2+\frac{1}{4}\lim_{N \to \infty}\mathbb{V}[\bar{X}]=\frac{1}{4}\sigma^2+\frac{1}{4}\lim_{N \to \infty}\frac{\sigma^2}{N}=\frac{1}{4}\sigma^2$$ So it is not consistent. So a better estimator would be for example $\bar{X}$.


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