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• Notes

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 $\mu ?$

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

$=\frac{1}{2 n}(2 n \mu)=\mu \rightarrow$ 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 $\hat {\theta}_{2}$

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