Order Statistic

In statistics, the nth order statistic of a sample is equal to the nth-smallest sample value. For example, if the sample values are:
6, 9, 3, 8
then the second order statistic is:
x(2) = 6.
The first order statistic (or smallest order statistic) is always the minimum of the sample. For a sample of size n, the nth order statistic (or largest order statistic) is the maximum. In conventional notation, one writes:
x1 = 6
x2 = 9
x3 = 3
x4 = 8
with no round brackets () in the subscripts, referring to the data in the order in which they appear, and:
x(1) = 3
x(2) = 6
x(3) = 8
x(4) = 9
with round brackets () in the subscripts, referring to the order statistics, i.e., the data sorted into increasing order.

Probability distribution

The kth order statistic

Let X_{1}, X_{2}, \ldots, X_{n} be iid continuously distributed random variables, and X_{(1)}, X_{(2)}, \ldots, X_{(n)} be the corresponding order statistics. Let f(x) be the probability density function and F(x) be the cumulative distribution function of Xi. Then the probability density of the kth statistic can be found as follows.
f_{X_{(k)}}(x)={d \over dx} F_{X_{(k)}}(x)
={d \over dx}P\left(X_{(k)}\leq x\right)={d \over dx}P(\mathrm{at}\ \mathrm{least}\ k\ \mathrm{of}\ \mathrm{the}\ n\ X\mathrm{s}\ \mathrm{are}\leq x)
={d \over dx}P(\geq k\ \mathrm{successes}\ \mathrm{in}\ n\ \mathrm{trials})={d \over dx}\sum_{j=k}^n{n \choose j}P(X_1\leq x)^j(1-P(X_1\leq x))^{n-j}
={d \over dx}\sum_{j=k}^n{n \choose j} F(x)^j (1-F(x))^{n-j}
=\sum_{j=k}^n{n \choose j}
\left(jF(x)^{j-1}f(x)(1-F(x))^{n-j} +F(x)^j (n-j)(1-F(x))^{n-j-1}(-f(x))\right)
=\sum_{j=k}^n\left(n{n-1 \choose j-1}F(x)^{j-1}(1-F(x))^{n-j} - n{n-1 \choose j} F(x)^j(1-F(x))^{n-j-1} \right)f(x)
=nf(x)\left(\sum_{j=k-1}^{n-1} {n-1 \choose j}
F(x)^j (1-F(x))^{(n-1)-j} - \sum_{j=k}^n {n-1 \choose j} F(x)^j (1-F(x))^{(n-1)-j}\right) and the sum above telescopes, so that all terms cancel except the first and the last:
=nf(x)\left({n-1 \choose k-1} F(x)^{k-1} (1-F(x))^{(n-1)-(k-1)}
- \underbrace\right) and the term over the underbrace is zero, so:
=nf(x){n-1 \choose k-1} F(x)^{k-1} (1-F(x))^{(n-1)-(k-1)}
={n! \over (k-1)!(n-k)!} F(x)^{k-1} (1-F(x))^{n-k} f(x).

Joint densities

Similarly, for i < j, the joint probability density function of two order statistics Xi and Xj can be shown to be
f_{X_{(i)},X_{(j)}}(u,v) = \frac{n!}{(i-1)!\,(j-i-1)!\,(n-j)!}F(u)^{i-1}(F(v)-F(u))^{j-i-1}(1-F(v))^{n-j}f(u)f(v)
The joint distribution of vector of order statistics is given by
f_{X_{(1)},X_{(2)},\ldots,X_{(n)}}(z_{1},z_{2},\ldots,z_{n}) = n! \prod_{i=1}^{n} f(z_{i}).\,

See also

External links

* Dr. Susan Holmes Order Statistics Retrieved Feb 02,2005

 

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