1 Notes on Order Statistics

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1 1 Notes on Ode Statistics Fo X a andom vecto in R n with distibution F, and π S n, define X π by and F π by X π (X π(1),..., X π(n) ) F π (x 1,..., x n ) F (x π 1 (1),..., x π 1 (n)); then the distibution function of X π is F π, P (X π(1) x 1,..., X π(n) x n ) P (X 1 x π 1 (1),..., X n x π 1 (n)), so when densities exists, We say X is exchangeable if p π (x). X π d X fo all π S n, that is, if the distibution of the vaiables is pemutation invaiant, F π F. Independent and identically distibuted is a special case. If we take X with oint density p(x), each component of X is continuous and thee ae no ties, so thee exists a unique (andom) pemutation π such that X π(1) < < X π(n), and we let X (i) X π(i). To find the oint density of the vecto of ode statistics conside small nonovelapping and odeed intevals I 1,..., I n. Then {X (1) I 1,..., X (n) I n } {X π(1) I 1,..., X π(n) I n }, which ae disoint events, so that P (X (1) I 1,..., X (n) I n ) P (X π(1) I 1,..., X π(n) I n ) p π (x)dx π S I n I 1 n p π (x) dx I n I 1 dx; I n I 1 1

2 and so fo B {x : x 1 < x 2 < < x n }, we must have P ((X (1),..., X (n) ) B) dx B yielding the density of the ode statistics 1(x 1 < < x n ). Special cases, exchangeability is whee all the p π ae equal to, say, p and theefoe the density is fo iid we have and iid U(, 1), even simple, p(x)1(x 1 <... < x n ) n p(x )1(x 1 <... < x n ), 1 1(x 1 <... < x n ). Maginals, like p () that fo X (), can now be obtained by integating out uninteesting vaiables, as in (take the iid case fo instance) n p () (x) p(x )dx 1 d x 1 d x+1 dx n. x 1 < <x 1 <x<x < <x n 1 A pehaps easie way is to notice that n {X () x} { i : X i x }, so that P (X () x) P ( {i : X i x} ) P (exactly of X 1,..., X n ae x) P (Bin(n, F (x)) ) n F (x)(1 F (x)) n, 2

3 diffeentiate to get density n p () (x) F 1 (x)(1 F (x)) n p(x) n F 1 (x)(1 F (x)) n p(x) Second tem, eplacing by 1, gives +1 ( 1)!(n )! F 1 (x)(1 F (x)) n p(x) n 1 F (x)(n )(1 F (x)) n 1 p(x) +1 and cancellation of all but the th tem leaves p () (x) F (x) 1 (1 F (x)) n p(x).!(n 1)! F (x)(1 F (x)) n 1 p(x) F 1 (x)(1 F (x)) n p(x) We can now evisit the integal, and see how it is not as daunting as it looked at fist. Integating out the vaiables geate than x gives the n factos of 1 F (x), and the smalle ones fom the bottom the 1 factos of F (x); the x stays fixed giving the p(x), so to eally compute this integal we ust need to keep tack of constants, essentially. Special cases, natually, ae the maximum and minimum, fo which we have p (n) (x) nf (x) n 1 p(x) and p (1) (x) n(1 F (x)) n 1 p(x). Fo maximum and minimum, these ae the deivatives of P (X (n) x) F n (x) and P (X (1) x) 1 (1 F (x)) n espectively. Fo example, if floods have distibution F, then the expected size of the lagest flood in n yeas is EX (n) nxf (x) n 1 p(x). Looking at the density fo X () above, let s wite the coefficient as ( ) n!(n )! ( 1)!(n )! ( 1)!1!(n )! n, 1, 1, n 3

4 which we can intepet by saying that thee is exactly one vaiable at x, 1 less than x and n geate. Thee ae n 1,1,n ways to pick how that happens, and each way has pobability given by F (x) 1 fo those that go to the left of x, p(x) fo the one that hits x, and (1 F (x)) n fo those that go to the ight of x. We can apply this easoning to compute the oint distibution of two ode statistics (eithe this way o integate out the density fom the oint density above). Intepetation of density p (,s) (x, x s ) fo < s, we need 1 less than x, one exactly at x, and so on, to yield that p (,s) (x, x s ) equals ( ) n F (x ) 1 p(x )(F (x s ) F (x )) s 1 p(x s )(1 F (x s )) n s 1, 1, s 1, 1, n s Fom this expession, using a multidimensional tansfomation (to be discussed) we can find the distibution of e.g. the ange, X (n) X (1). Now we ae eady to talk about the Beta distibution. Since fo the unifom case we have F (x) x, the density of th smallest unifom ode statistics U () is x 1 (1 x) n, and since it is a density, we must have x 1 (1 x) n dx [ ] 1. And ( 1)!(n )! Γ(n + 1) Γ()Γ(n + 1), letting s n + 1, we have fo positive integes, s, x 1 (1 x) s 1 Γ()Γ(s) Γ( + s). Why not make a distibution fo all α, β positive by defining B(α, β) x α 1 (1 x) β 1 4

5 check existence fo positive paamete values, define density by b(x; α, β) xα 1 (1 x) β 1, x 1. B(α, β) We will lean late using a multivaiate tansfomation that B(α, β) Γ(α)Γ(β) Γ(α + β) fo all α, β positive. The unifom distibution U[, 1] is the special case B(1, 1) and the unifom ode statistics ae also, that is, U () B(, n + 1). 5

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