A random sample consists of 3 independent observations $_1, _2
and _3$
follow Normal distribution (, $^2$) consider four estimators for population mean
as follow:
$̂_1$ =
$\frac{X_1 + 3_2 − 2_3}
{2}$
, ̂2 =
$\frac{5_1 − 2_2}
{3}$
, ̂3 =
$\frac{1}{2}
_1 +
\frac{1}{2}
̅$, ̂4 =
$\frac{2_1 + 3_3 − 2̅}{3}$
Where ̅ is the sample mean of $_1, _2, _3$.
(a) Which of the above is/are the unbiased estimator(s) for ?
(b) Which of the above is the best unbiased estimator for ?
(c) Provide an estimator for which is better than aforementioned ̂1, ̂2, ̂3
and ̂4. Justify your answer.
For part a, I found all are unbiased and for part b ̂3 is the best. Are they correct?
But if they are correct, I think ̂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 $\bar{X}$.