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Inside Collection (Textbook):

Textbook by: Paul E Pfeiffer. E-mail the author

# Order Statistics

Module by: Paul E Pfeiffer. E-mail the author

Summary: A basic treatment of order statistics.

In sampling statistics (see Sec 18.7), we deal with an iid class {Xi:1in}{Xi:1in} of random variables, where n is a prescribed positive integer known as the sample size. An observation of this class gives an n-tuple of numbers (t1,t2,,tn)(t1,t2,,tn). As an extension of the extreme values in the case of two variables (Ex 2.11), it is often useful to define a random variable Y1 whose value for any ω is the smallest of the Xi(ω)Xi(ω); a second random variable Y2 whose value at ω is the next smallest of the Xi(ω)Xi(ω), and so on through Yn whose value at ω is the largest of the Xi(ω)Xi(ω). We would like to be able to obtain the distributions for these new random variables in terms of the common distribution for the Xi. We formulate the problem as follows.

## Example 1: Order Statistics

Suppose {Xi:1in}{Xi:1in} is iid, with common distribution function F. Let

• Y1=Y1= smallest of X1,X2,...,XnX1,X2,...,Xn
• Y2=Y2= next larger of X1,X2,...,XnX1,X2,...,Xn
• . . .
• Yn=Yn= largest of X1,X2,...,XnX1,X2,...,Xn

Then Yk is called the kth order statistic for the class {Xi:1in}{Xi:1in}. We wish to determine the distribution functions Fk(t)=P(Ykt)1knFk(t)=P(Ykt)1kn. Now, YktYkt iff k or more of the Xi have values no greater than t. We may view the process as a Bernoulli sequence of n trials. There is a success on the ith trial iff XitXit. The probability p of a success is p=P(Xt)=F(t)p=P(Xt)=F(t). Hence

F k ( t ) = P ( Y k t ) = P ( k or more of the X i lie in ( - inf , t ] ) = j = k n C ( n , j ) F j ( t ) [ 1 - F ( t ) ] n - j F k ( t ) = P ( Y k t ) = P ( k or more of the X i lie in ( - inf , t ] ) = j = k n C ( n , j ) F j ( t ) [ 1 - F ( t ) ] n - j
(1)

Remark. Once the common distribution function F for the Xi is known, then the Fk are calculated in a straightforward manner. For that purpose we may use the MATLAB function cbinom.

## Example 2

Suppose the Xi are exponential (2). Then FX(t)=1-e-2tFX(t)=1-e-2t for positive t. Suppose n=5n=5. We calculate Fk(t)Fk(t) for t=0.1,0.3,0.5,0.7,0.9t=0.1,0.3,0.5,0.7,0.9.


n = 5;
t = 0.1:0.2:0.9;
m = length(t);
F = 1 - exp(-2*t);
for i = 1:m
FK(i,:) = cbinom(n,F(i),1:n);
end
disp([t' F' FK])    % k = 1    k = 2     k = 3     k = 4     k = 5
0.1000    0.1813    0.6321    0.2249    0.0445    0.0046    0.0002
0.3000    0.4512    0.9502    0.7456    0.4091    0.1324    0.0187
0.5000    0.6321    0.9933    0.9354    0.7364    0.3946    0.1009
0.7000    0.7534    0.9991    0.9852    0.9000    0.6400    0.2427
0.9000    0.8347    0.9999    0.9968    0.9653    0.8064    0.4052


The following special case is important in characterizing the Poisson process (see Sec 21.1).

## Example 3

Order statistics for uniformly distributed random variables

Suppose {Ui:1in}{Ui:1in} is iid, uniform on (0,T](0,T]. Determine the distribution functions for the order statistics.

SOLUTION

The common distribution function for the Ui is given by F(t)=t/T,0tTF(t)=t/T,0tT. According to the result in Ex 2.16, the kth order statistic Yk has the distribution function

F k ( t ) = P ( Y k t ) = j = k n C ( n , j ) t T j T - t T n - j 0 < t < T F k ( t ) = P ( Y k t ) = j = k n C ( n , j ) t T j T - t T n - j 0 < t < T
(2)

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