Download Order Statistics: Probability Distributions of Minimum and Maximum in a Random Sample and more Exams Statistics in PDF only on Docsity!
Order Statistics
1 Introduction and Notation
Let X 1 , X 2 ,... , X 10 be a random sample of size 15 from the uniform distribution over the interval (0, 1). Here are three different realizations realization of such samples.
Because these samples come from a uniform distribution, we expect them to be spread out “ran- domly” and “evenly” across the interval (0, 1). (You might think that you are seeing some sort of clustering but keep in mind that you are looking at a measly selection of only three samples. After collecting more samples I’m sure your view would change!)
Consider the single smallest value from each of these three samples, highlighted here.
Collect the minimums onto a single graph.
Not surprisingly, they are down towards zero! It would be pretty difficult to get a sample of 15 uniforms on (0, 1) that has a minimum up by the right endpoint of 1. In fact, we will show that if we kept collecting minimums of samples of size 15, they would have a probability density function that looks like this.
Notation: Let X 1 , X 2 ,... , Xn be a random sample of size n from some distribution. We denote the order statistics by
X(1) = min(X 1 , X 2 ,... , Xn) X(2) = the 2nd smallest of X 1 , X 2 ,... , Xn ..
. =
X(n) = max(X 1 , X 2 ,... , Xn)
(Another commonly used notation is X1:n, X2:n,... , Xn:n for the min through the max, respec- tively.)
In what follows, we will derive the distributions and joint distributions for each of these statistics and groups of these statistics. We will consider continuous random variables only. Imagine taking a random sample of size 15 from the geometric distribution with some fixed parameter p. The chances are very high that you will have some repeated values and not see 15 distinct values. For example, suppose we observe 7 distinct values. While it would make sense to talk about the minimum or maximum value here, it would not make sense to talk about the 12th largest value in this case. To further confuse the matter, the next sample might have a different number of distinct values! Any analysis of the order statistics for this discrete distribution would have to be well- defined in what would likely be an ad hoc way. (For example, one might define them conditional on the number of distinct values observed.)
2 The Distribution of the Minimum
Suppose that X 1 , X 2 ,... , Xn is a random sample from a continuous distribution with pdf f and cdf F. We will now derive the pdf for X(1), the minimum value of the sample. For order statistics, it is usually easier to begin by considering the cdf. The game plan will be to relate the cdf of the minimum to the behavior of the individual sampled values X 1 , X 2 ,... , Xn for which we know the pdf and cdf.
So, we have that the pdf for the minimum is
fX(1) (x) = (^) dxd FX(1) (x) = (^) dxd { 1 − [1 − F (x)]n}
= n[1 − F (x)]n−^1 f (x)
Going back to the uniform example of Section 1, we had f (x) = I(0,1)(x) and
F (x) =
0 , x < 0 x , 0 ≤ x < 1 1 , x ≥ 1.
The pdf for the minimum in this case is
fX(1) (x) = n[1 − x]n−^1 I(0,∞)(x).
This is the pdf for the Beta distribution with parameters 1 and n. Thus, we can write
X(1) ∼ Beta(1, n).
3 The Distribution of the Maximum
Again consider our random sample X 1 , X 2 ,... , Xn from a continuous distribution with pdf f and cdf F. We will now derive the pdf for X(n), the maximum value of the sample. As with the minimum, we will consider the cdf and try to relate it to the behavior of the individual sampled values X 1 , X 2 ,... , Xn.
The cdf for the minimum is FX(1) (x) = P (X(1) ≤ x).
Imagine a random sample falling in such a way that the maximum is below a fixed value x. This will happen if and only if all of the Xi are below x.
Thus, we have
FX(n) (x) = P (X(n) ≤ x)
= P ( X 1 ≤ x, X 2 ≤ x,... , Xn ≤ x )
= P (X 1 ≤ x)P (X 2 ≤ x) · · · P (Xn ≤ x) by independence
= [P (X 1 ≤ x)]n^ because the Xi are identically distributed
= [F (x)]n.
Take the derivative, we get the pdf for the maximum to be
fX(n) (x) = (^) dxd FX(1) (x) = (^) dxd [F (x)]n
= n[F (x)]n−^1 f (x)
In the case of the random sample of size 15 from the uniform distribution on (0, 1), the pdf is
fX(n)(x) = nxn−^1 I(0,1)(x)
which is the pdf of the Beta(n, 1) distribution.
Not surprisingly, all most of the probability or “mass” for the maximum is piled up near the right endpoint of 1.
4 The Joint Distribution of the Minimum and Maximum
Let’s go for the joint cdf of the minimum and the maximum
FX(1),X(n) (x, y) = P (X(1) ≤ x, X(n) ≤ y).
It is not clear how to write this in terms of the individual Xi. Consider instead the relationship
P (X(n) ≤ y) = P (X(1) ≤ x, X(n) ≤ y) + P (X(1) > x, X(n) ≤ y). (1)
We know how to write out the term on the left-hand side. The first term on the right-hand side is what we want to compute. As for the final term,
P (X(1) > x, X(n) ≤ y),
note that this is zero if x ≥ y. (In this case, P (X(1) ≤ x, X(n) ≤ y) = P (X(n) ≤ y) and (1) gives us only P (X(n) ≤ y) = P (X(n) ≤ y) which is both true and uninteresting! So, we consider the case that x < y. Note then that
P (X(1) > x, X(n) ≤ y) = P (x < X 1 ≤ y, x < X 2 ≤ y,... , x < Xn ≤ y)
iid = [P (x < X 1 ≤^ y)]n
= [F (y) − F (x)]n.
Another possibility is that we get X 5 = x and X 2 = y and the remaining X 1 , X 3 , X 4 in between x and y.
This would also happen with “probability”
f (x)[F (y) − F (x)]^3 f (y).
We have to add this “probability” up as many times as there are scenarios. So, let’s count them. There are 5! different ways to lay down the Xi. For each one, there are 3! different ways to lay down the remaining values in between that will result in the same min and max. So, we need to divide these redundancies out for a total of 5!/3! = (5)(4) ways to get that min at x and max at y.
In general, for a sample of size n, there are n! different ways to lay down the Xi. For each one, there are (n − 2)! different ways that result in the same min and max. So, there are a total of n!/(n − 2)! = n(n − 1) ways to get that
Thus, the “probability” of getting a minimum of x and a maximum of y is
n(n − 1)f (x)[F (y) − F (x)]n−^2 f (y),
which looks an awful lot like the formula we derived above!
5 The Joint Distribution for All of the Order Statistics
We wish now to find the pdf fX(1),X(2),...,X(n) (x 1 , x 2 ,... , xn).
This time, we will start with the heuristic aid.
Suppose that n = 3 and we want to find
fX(1),X(2),X(3) (x 1 , x 2 , x 3 ) “=” P (X(1) = x 1 , X(2) = x 2 , X(3) = x 3 ).
The first thing to notice is that this probability will be 0 if we don’t have x 1 < x 2 < x 3. (Note that we need strict inequalities here. For a continuous distribution, we will never see repeated values so the minimum and second smallest, for example, could not take on the same value.)
Fix values x 1 < x 2 < x 3. How could a sample of size 3 fall so that the minimum is x 1 , the next smallest is x 2 , and the largest is x 3? We could observe
X 1 = x 1 , X 2 = x 2 , X 3 = x 3 ,
or X 1 = x 2 , X 2 = x 1 , X 3 = x 3 ,
or X 2 = x 2 , X 2 = x 3 , X 3 = x 1 ,
or...
There are 3! possibilities to list. The “probability” for each is f (x 1 )f (x 2 )f (x 3 ). Thus,
fX(1),X(2),X(3) (x 1 , x 2 , x 3 ) “=” P (X(1) = x 1 , X(2) = x 2 , X(3) = x 3 ) = 3!f (x 1 )f (x 2 )f (x 3 ).
For general n, we have
fX(1),X(2),...,X(n) (x 1 , x 2 ,... , xn) “=” P (X(1) = x 1 , X(2) = x 2 ,... X(n) = xn)
= n!f (x 1 )f (x 2 ) · · · f (xn)
which holds for x 1 < x 2 < · · · < xn with all xi in the support for the original distribution. The joint pdf is zero otherwise.
The Formalities:
The joint cdf, P (X(1) ≤ x 1 , X(2) ≤ x 2 ,... , X(n) ≤ xn),
is a little hard to work with. Instead, we consider something similar:
P (y 1 < X(1) ≤ x 1 , y 2 < X(2) ≤ x 2 ,... , yn < X(n) < xn)
for values y 1 < x 1 ≤ y 2 < x 2 ≤ y 3 < x 3 ≤ · · · ≤ yn < xn.
This can happen if y 1 < X 1 ≤ x 1 , y 2 < X 2 ≤ x 2 ,... , yn < Xn < xn,
or if y 1 < X 5 ≤ x 1 , y 2 < X 3 ≤ x 2 ,... , yn < Xn− 2 < xn,
or...
Because of the constraints on the xi and yi, these are disjoint events. So, we can add these n! probabilities, which will all be the same, together to get
P (y 1 < X(1) ≤ x 1 ,... , yn < X(n) < xn) = n! P (y 1 < X 1 ≤ x 1 ,... , yn < Xn < xn).
Note that
P (y 1 < X 1 ≤ x 1 ,... , yn < Xn < xn) indep =
∏^ n
i=
P (yi < Xi ≤ xi) =
∏^ n
i=
[F (xi) − F (yi)].
So,
P (y 1 < X(1) ≤ x 1 ,... , yn < X(n) < xn) = n!
∏^ n
i=
[F (xi) − F (yi)] (2)
The left-hand side is ∫ (^) xn
yn
∫ (^) xn− 1
yn− 1
∫ (^) x 1
y 1
fX(1),X(2),...,X(n) (u 1 , u 2 ,... , un) du 1 du 2... , dun.
Taking derivatives (^) dxd 1 dx^ d 2 · · · (^) dxdn gives
fX(1),X(2),...,X(n) (x 1 , x 2 ,... , xn)
Differentiating both sides of (2) with respect to x 1 , x 2 ,... , xn gives us
fX(1),X(2),...,X(n) (x 1 , x 2 ,... , xn) = n!f (x 1 )f (x 2 ) · · · f (xn)
which holds for x 1 < x 2 < · · · , xn and all xi in the support of the original distribution. The pdf is zero otherwise.
for xi < xi+1 < · · · , xn− 1.
fX(i),...,X(n−2) (xi,... , xn− 2 ) =
∫ (^) ∞ xn− 2 fX(i),...,X(n−1) (xi,... , xn−^1 )^ dxn−^1
= (^) (i−n!1)! f (xi) · · · f (xn− 2 )[F (xi)]i−^1
∫ (^) ∞ xn− 2 f^ (xn−^1 )[1^ −^ F^ (xn−^1 )]^ dxn−^1
Letting u = 1 − F (xn− 1 ) and du = −f (xn− 1 ) dxn− 1 , we get
fX(i),...,X(n−2) (xi,... , xn− 2 ) = (^) (i−n!1)! f (xi) · · · f (xn− 2 )[F (xi)]i−^1
{ − 12 [1 − F (xn− 1 )]^2
}xn− 1 =∞ xn− 1 =xn− 2
= (^) 2(in−!1)! f (xi) · · · f (xn− 2 )[F (xi)]i−^1 [1 − F (xn− 2 )]^2
for xi < xi+1, · · · < xn− 2.
The next time through we will integrate out xn− 2 from xn− 3 to ∞. Note that
∫ (^) ∞ xn− 3 f^ (xn−^2 )[1 ︸ −^ F︷︷^ (x n−^2 )︸ u
]^2 dxn− 2 = − 13 [1 − F (xn− 2 )]^3
∣∣ ∣
xn− 2 =∞ xn− 2 =xn− 3
= 13 [1 − F (xn− 3 )]^3.
Thus,
fX(i),...,X(n−3) (xi,... , xn− 3 ) = n! (3)(2)(i − 1)!
f (xi) · · · f (xn− 3 )[F (xi)]i−^1 [1 − F (xn− 3 )]^3
for xi < xi+1 < · · · < xn− 3.
Continuing all the way down to the marginal pdf for X(i) alone, we get
fX(i) = (^) (n−i)!(n!i−1)! [F (xi)]i−^1 f (xi)[1 − F (xn− 3 )]n−i
for −∞ < xi < ∞. (← This may be further restricted by indicators in f (xi).)
The Heuristic:
We once again will think of the continuous random variables X 1 , X 2 ,... , Xn as discrete and fX(i) (xi) as the “probability” that the ith order statistic is at xi. First not that there are n! different ways to arrange the x′s. We need to put 1 at xi, which will happen with “probability” f (xi). We need to put i − 1 below xi, which will happen with probability [F (xi)]i−^1 and we need to put n − i above xi, which will happen with probability [1 − F (xi)]n−i. There are (i − 1)! different ways to arrange the x’s chosen to go below xi. These arrangements are redundant and need to be divided out. Hence, we have (i − 1)! in the denominator. There are (n − i)! different ways to arrange the x’s chosen to go above xi. These arrangements are also redundant and need to be divided out. Thus, we also have (n − i)! in the denominator.
7 The Joint Distribution of X(i) and X(j) for i < j
As in Section 6, one could start with the joint pdf for all of the order statistics and integrate out the unwanted ones. The result will be
fX(i),X(j) (xi, xj ) = n! (i − 1)!(j − i − 1)!(n − j)!
[F (xi)]i−^1 f (xi)[F (xj )−F (xi)]j−i−^1 f (xj )[1−F (xj )]n−j
for −∞ < xi < xj < ∞.
Can you convince yourself of this heuristically?
