Point estimation

Activity 6.1   Prove the two expressions are equal.
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Answer 6.1  

$$(T) =E[(T-\theta)^2]$$

$$=E[\{(T-E(T))-(E(T)-\theta)\}^2]$$

$$=E[(T-E(T))^2-2(T-E(T)(E(T)-\theta)+(E(T)-\theta)^2]$$

$$=E[(T-E(T))^2]-2E[T-E(T)](E(T)-\theta) +(E(T)-\theta)^2]$$

$$=\operatorname{var}T- 0 + [ (T)]^2.$$

This derivation works by using the fact that $(E(T)-\theta)$ is a constant, and so can be taken outside all expected value operations.
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Activity 6.2   Work out why the above statement is true.
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Answer 6.2   The sample mean has MSE$(T)=\sigma^2/n$. The estimator $T=1.5$ has

$$(T) =\operatorname{var}T +[ (T)]^2$$

$$=0+(1.5-\mu)^2$$

$$=(1.5-\mu)^2.$$

$T$ has smaller MSE than the sample mean if and only if

$$(1.5-\mu)^2<\sigma^2/n.$$

If we know for sure that $\mu$ is between 1 and 2, then $(1.5-\mu)^2\le 1/4$, so if $\sigma > 1/2$, the MSE of $T$ is less than that of the sample mean.

But most people would not use $T$ all the same. The natural thing to use would be the sample mean, but replaced with 1 if it became less than 1, and with 2 if it became larger than 2.
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Activity 6.3   Find the mean squared error for $T$ and $\bar{X}$ for a pay structure in which the boss is paid £ 100, and the three other workers £20.
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Answer 6.3   The population mean pay is still 40. A sample $(100, 20, 20)$ gives mean 46.33 and estimate 40, and a sample $(20,20,20)$ gives both mean and estimate of 20. The mean squared error of the sample mean is

$$[3(46.33-40)^2+(20-40)^2]/4 =400/3.$$

The mean squared error of the estimator is

$$[3(40-40)^2+(20-40)^2]/4 =100.$$

So the estimator is still better than the sample mean for this pay structure.

However, if the pay structure is £80 for the boss, £ 50, £20 and £10 for the other workers, then the sample mean has a smaller mean squared error than the estimator. If EXCEL is available to you, you can try the EXCEL spreadsheet mse.xls for these calculations. Just replace the first four numbers in the first column by population values of your choice, and see the MSE come out.
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