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Suppose we have a random sample of size $$2n$$ from a
population denoted by $$X,$$ and $$E(X)=\mu$$ and $$V(X)=\sigma^{2} .$$ Let

$$\overline{X}_{1}=\frac{1}{2 n} \sum_{i=1}^{2 n} X_{i}$$ and $$\overline{X}_{2}=\frac{1}{n} \sum_{i=1}^{n} X_{i}$$

be two estimators of $$\mu .$$ Which is the better estimator of $$\mu ?$$
Explain your choice.

$$f\left(\overline{x}_{1}\right)=E\left(\frac{1}{2n} \sum_{i=1}^{2n} x_{i}\right)=\frac{1}{2 n} E\left(\sum_{i=1}^{2 n} x_{i}\right)$$

$$=\frac{1}{2 n}(2 n \mu)=\mu \rightarrow$$ unbiased

$$E\left(\overline{x}_{2}\right)=E\left(\frac{1}{n} \sum_{i=1}^{n} x_{i}\right)=\frac{1}{n} E\left(\sum_{i=1}^{n} x_{i}\right)$$

$$=\frac{1}{n}(n \mu)=\mu \rightarrow$$ unbiased

$$V\left(\overline{x}_{1}\right)=\frac{\sigma}{2 n}$$

$$V\left(\overline{x}_{2}\right)=\frac{\sigma^{2}}{n}$$

$$v(\overline{x}_{1})<v\left(\overline{x}_{2}\right)$$

$$\rightarrow \overline{x}_{1} \rightarrow$$ is abetter estimator

Suppose that $$\hat{\Theta}_{1}$$ and $$\hat{\Theta}_{2}$$ are estimators of the parame-
ter $$\theta .$$ We know that $$E\left(\hat{\Theta}_{1}\right)=\theta, E\left(\hat{\Theta}_{2}\right)=\theta / 2, V\left(\hat{\Theta}_{1}\right)=10$$
$$V\left(\hat{\Theta}_{2}\right)=4 .$$ Which estimator is best? In what sense is it best?

$$E\left(\hat{\theta}_{1}\right)=\theta \rightarrow \hat{\theta}_{1}$$ unbiased

$$E\left(\hat{\theta}_{2}\right)=\frac{\theta}{2} \rightarrow \text {bias}=E(\hat{\theta}_{2})-\theta$$

$$=\frac{\theta}{2}-\theta=\frac{-\theta}{2}$$

for unbiasedness $$\rightarrow \theta_{1}$$ is better

Relative efficieny $$=\frac{M S E(\hat{\theta}_{1})}{M S E\left(\hat{\theta}_{2} \right)}$$

$$=\frac{V\left(\hat{\theta}_{1}\right)+b i as^{2}}{V\left(\hat{\theta}_{2}\right)+b i a s_{2}^{2}}$$

$$=\frac{10}{4+\left(-\frac{\theta}{2}\right)^{2}}=\frac{40}{16+\theta^{2}}$$

$$\frac{40}{16+\theta^{2}} \leq 1 \rightarrow \theta^{2} \geq 24$$

$$\theta \leq-4.899, \theta \geq 4.899$$

if $$-4.899 < \theta <4.899$$

$$\rightarrow$$ use $$\hat {\theta}_{2}$$

$$\text { otherwise } \rightarrow \hat{\theta}_{1}$$

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