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# CHAPTER 10 SOLUTIONS

### 10.1   SECTION 10.2

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) = (xθ + 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 = xθ + 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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