When we derive the distribution of a single order statistic, we divide the underlying distribution into 3.
The observed value of the th order statistic is . ie . This is a random variable ( will have a different value from a new sample) with density . We have
Ordering and classifying by has produced a form similar to a multinomial distribution with 3 categories , ,,, and associated with these categories we have the entities ,,.
The multinomial density function for 3 categories is
The density of an order statistic has a similar form,
For 2 order statistics, , there are 5 categories,
From the same analogy to multinomials used for a single order statistic, there are 5 components to the density,
Since we know the pdf of
by (5.1), the marginal pdf of the th smallest
component, , can be found by integrating over the remaining
Notice the order of integration used is to first integrate over ,
and then (this is the part of (5.2)
by the parentheses). This is followed by integration over , then , and finally over . The limits of integration are otained
from the inequalities.
Hence using (5.3) and (5.3) in (5.2), we obtain
The probability density functions of both the minimum observation and the maximum observation are special cases of (5.5).
The integration technique can be applied to find the joint pdf of two (or more) order statistics, and this is done in 5.4. Before examining that, we will give an alternative (much briefer) derivation of (5.7).
Let the cdf of be denoted by . For any value in the range space of , the cdf of is
Of course in the above is just a dummy, and could be replaced by to give (5.7).
Exercise Use this technique to prove (5.6).
Let be a sample from the uniform distribution . Find a CI for using the largest order statistic, .