**10.1** When three observations are taken without replacement, there are only four possible results. They are 1,3,5; 1,3,9; 1,5,9; and 3,5,9. The four sample means are 9/3, 13/3, 15/3, and 17/3. The expected value (each has probability 1/4) is 54/12 or 4.5, which equals the population mean. The four sample medians are 3, 3, 5, and 5. The expected value is 4, and so the median is biased.

**10.2**

**10.3** For a sample of size 3 from a continuous distribution, the density function of the median is 6ƒ(*x*)*F*(*x*)[1 − *F*(*x*)]. For this exercise, *F*(*x*) = (*x* − *θ* + 2)/4, *θ* − 2 < *x* < *θ* + 2. The density function for the median is

and the expected value is

where the first line used the substitution *y* = *x* − *θ* + 2.

**10.4** Because the mean of a Pareto distribution does not always exist, it is not reasonable to discuss unbiasedness or consistency. Had the problem been restricted to Pareto distributions with *α* > 1, then consistency can be established. It turns out that for the sample mean to be consistent, only the first moment needs to exist (the variance having a limit of zero is a sufficient, but not a necessary, condition for consistency).

**10.5** The mean ...

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