Distributions of Order Statistics


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1 Chapter 2 Distributios of Order Statistics We give some importat formulae for distributios of order statistics. For example, where F k: (x)=p{x k, x} = I F(x) (k, k + 1), I x (a,b)= 1 x t a 1 (1 t) b 1 dt B(a, b) 0 deotes the icomplete Beta fuctio. The probability desity fuctio of X k, is give as follows: f k: (x)=! (k 1)!( k)! (F(x))k 1 (1 F(x)) k f (x), where f is a populatio desity fuctio. The joit desity fuctio of order statistics X 1,, X 2,,...,X, has the form! f (x k ), < x 1 < x 2 < < x <, ad f 1,2,...,: (x 1,x 2,...,x )= 0, otherwise. There are some simple formulae for distributio of maxima ad miima. Example 2.1. Let radom variables X 1, X 2,...,X have a joit d.f. H(x 1,x 2,...,x )=P{X 1 x 1, X 2 x 2,..., X x }. The d.f. of M()=max{X 1, X 2,...,X } has the form P{M() x} = P{X 1 x, X 2 x,..., X x} = H(x,x,...,x). (2.1) Similarly we ca get the distributio of m()=mi{x 1, X 2,...,X }. Oe has P{m() x} = 1 P{m() > x} = 1 P{X 1 > x, X 2 > x,..., X > x}. (2.2) M. Ahsaullah et al., A Itroductio to Order Statistics, Atlatis Studies i Probability ad Statistics 3, DOI: / _2, Ó Atlatis Press
2 16 A Itroductio to Order Statistics Exercise 2.1. Fid the joit distributio fuctio of M( 1) ad M(). Exercise 2.2. Express d.f. of Y = mi{x 1,X 2 } i terms of joit d.f. H(x 1,x 2 )=P{X 1 x 1, X 2 x 2 }. Exercise 2.3. Let H(x 1,x 2 ) be the joit d.f. of X 1 ad X 2. Fid the joit d.f. of Y = mi{x 1,X 2 } ad Z = max{x 1,X 2 }. From (2.1) ad (2.2) oe obtais the followig elemetary expressios for the case, whe X 1, X 2,...,X preset a sample from a populatio d.f. F (ot ecessary cotiuous): ad P{M() x} = F (x) (2.3) P{m() x} = 1 (1 F(x)). (2.4) Exercise 2.4. Let X 1, X 2,...,X be a sample of size from a geometrically distributed radom variable X, such that P{X = m} =(1 p)p m, m = 0, 1, 2,... Fid P{Y r, Z < s}, r < s, where Y = mi{x 1, X 2,...,X } ad Z = max{x 1, X 2,...,X }. There is o difficulty to obtai d.f. s for sigle order statistics X k,.let F k, (x)=p{x k, x}. We see immediately from (2.3) ad (2.4) that F, (x)=f (x) ad F 1: (x)=1 (1 F(x)). The geeral formula for F k: (x) is ot much more complicated. I fact, F k: (x)=p{ at least k variables amog X 1, X 2,...,X are less or equal x} = m=k ( = m=k m Exercise 2.5. Prove that idetity P{ exactly m variables amog X 1, X 2,...,X are less or equal x} ) (F(x)) m (1 F(x)) m, 1 k. (2.5) m=k( m holds for ay 0 x 1, where ) x m (1 x) m = I x (k, k + 1) (2.6) I x (a,b)= 1 B(a, b) x 0 t a 1 (1 t) b 1 dt (2.7)
3 Distributios of order statistics 17 is the icomplete Beta fuctio with parameters a ad b, B(a,b) beig the classical Beta fuctio. By comparig (2.5) ad (2.6) oe obtais that F k: (x)=i F(x) (k, k + 1). (2.8) Remark 2.1. It follows from (2.8) that X k, has the beta distributio with parameters k ad k + 1, if X has the uiform o [0,1] distributio. Remark 2.2. Equality (2.8) is valid for ay distributio fuctio f. Remark 2.3. If we have tables of the fuctio I x (k, k + 1), it is possible to obtai d.f. F k: (x) for arbitrary d.f. F. Exercise 2.6. Fid the joit distributio of two order statistics X r, ad X s,. Example 2.2. Let us try to fid the joit distributio of all elemets of the variatioal series X 1,, X 2,,...,X,. It seems that the joit d.f. F 1,2,...,: (x 1,x 2,...,x )=P{X 1, x 1, X 2, x 2,..., X, x } promises to be very complicated. Hece, we cosider probabilities P(y 1,x 1,y 2,x 2,...,y,x )=P{y 1 < X 1, x 1, y 2 < X 2, x 2,..., y < X, x } for ay values y 1 < x 1 y 2 < x 2 y < x. It is evidet, that the evet A = {y 1 < X 1, x 1, y 2 < X 2, x 2,..., y < X, x } is a uio of! disjoit evets A(α(1),α(2),...,α()) = {y 1 < X α(1) x 1,y 2 < X α(2) x 2,...,y < X α() x },where the vector (α(1),α(2),...,α()) rus all permutatios of umbers 1, 2,...,. The symmetry argumet shows that all evets A(α(1), α(2),..., α()) have the same probability. Note that this probability is equal to Fially, we obtai that (F(x k ) F(y k )). P{y 1 < X 1, x 1, y 2 < X 2, x 2,..., y < X, x } =! (F(x k ) F(y k )). (2.9)
4 18 A Itroductio to Order Statistics Let ow cosider the case whe our populatio distributio has a desity fuctio f.it meas that for almost all X (i.e., except, possibly, a set of zero Lebesgue measure) F (x)= f (x). I this situatio (2.9) eables us to fid a expressio for the joit probability desity fuctio (p.d.f.) (deote it f 1,2,..., (x 1,x 2,...,x )) of order statistics X 1,, X 2,,...,X,.I fact, differetiatig both sides of (2.9) with respect to x 1, x 2,...,x, we get the importat equality f 1,2,...,: (x 1,x 2,...,x )=! f (x k ), < x 1 < x 2 < < x <. (2.10) Otherwise (if iequalities x 1 < x 2 < < x fail) we aturally put f 1,2,...,: (x 1,x 2,...,x )=0. Takig (2.10) as a startig poit oe ca obtai differet results for joit distributios of arbitrary sets of order statistics. Example 2.3. Let f m: deote the p.d.f. of X m,. We get from (2.10) that f m: (x)= f 1,2,...,: (x 1,...,x k 1,x,x k+1,...,x )dx 1 dx k 1 dx k+1 dx =! f (x) m 1 f (x k ) where the itegratio is over the domai k=m+1 f (x k )dx 1 dx m 1 dx m+1 dx, (2.11) < x 1 < < x m 1 < x < x m+1 < < x <. The symmetry of m 1 f (x k ) with respect to x 1,...,x m 1,aswellasthesymmetryof f (x k ) k=m+1 with respect to x m+1,...,x, helps us to evaluate the itegral o the RHS of (2.11) as follows: m 1 = f (x k ) 1 (m 1)! k=m+1 m 1 x = (F(x))m 1 (1 F(x)) m (m 1)!( m)! f (x k )dx 1 dx m 1 dx m+1 dx 1 f (x k )dx k ( m)! k=m+1 x f (x k )dx k. (2.12)
5 Distributios of order statistics 19 Combiig (2.11) ad (2.12), we get that! f m: (x)= (m 1)!( m)! (F(x))m 1 (1 F(x)) m f (x). (2.13) Ideed, equality (2.13) is immediately follows from the correspodig formula for d.f. s of sigle order statistics (see (2.8), for example), but the techique, which we used to prove (2.13), is applicable for more complicated situatios. The followig exercise ca illustrate this statemet. Exercise 2.7. Fid the joit p.d.f. f k(1),k(2),...,k(r): (x 1,x 2,...,x r ) of order statistics X k(1),, X k(2),,...,x k(r),,where1 k(1) < k(2) < < k(r). Remark 2.4. I the sequel we will ofte use the particular case of the joit probability desity fuctios from exercise 2.7, which correspods to the case r = 2. It turs out that! f i, j: (x 1,x 2 )= (i 1)!( j i 1)!( j)! (F(x 1 )) i 1 (F(x 2 ) F(x 1 )) j i 1 (1 F(x 2 )) j f (x 1 ) f (x 2 ), (2.14) if 1 i < j ad x 1 < x 2. Expressio (2.10) eables us also to get the joit d.f. F 1,2,...,: (x 1,x 2,...,x )=P{X 1, x 1, X 2, x 2,..., X, x }. Oe has where F 1,2,...,: (x 1,x 2,...,x )=! D f (u k )du 1 du, (2.15) D = {U 1,...,U : U 1 < U 2 < < U ; U 1 < x 1, U 2 < x 2,..., U < x }. Note that (2.15) is equivalet to the expressio F 1,2,...,: (x 1,x 2,...,x )=! du 1 du, D (2.16) where itegratio is over D = {U 1,...,U : U 1 < U 2 < < U ; U 1 < F(x 1 ), U 2 < F(x 2 ),..., U < F(x )}. Remark 2.5. It ca be proved that ulike (2.15), which eeds the existece of populatio desity fuctio f, expressio (2.16), as well as (2.9), is valid for arbitrary distributio fuctio F.
6 20 A Itroductio to Order Statistics Check your solutios Exercise 2.1 (solutio). If x y,the Otherwise, P{M( 1) x, M() y} = P{M() y} = H(y,y,...,y). P{M( 1) x, M() y} = P{M( 1) x, X y} = H(x,...,x,y}. Exercise 2.2 (aswer). where ad Exercise 2.3 (solutio). If x y,the If x < y,the P{Y x} = 1 H(x, ) H(,x)+H(x,x), H(x, )=P{X 1 x, X 2 < } = P{X 1 x} H(,x)=P{X 2 x}. P{Y x, Z y} = P{Z y} = H(y,y). P{Y x, Z y} = P{X 1 x, X 2 y} + P{X 1 y, X 2 x} P{X 1 x, X 2 x} Exercise 2.4 (solutio). We see that = H(x,y)+H(y,x) H(x,x). P{Y r, Z < s} = P{r X k < s, k = 1, 2,...,} =(P{r X < s}) =(P{X r} P{X s}) =(p r p s ). Exercise 2.5 (solutio). It is easy to see that (2.6) is valid for x = 0. Now it suffices to prove that both sides of (2.6) have equal derivatives. The derivative of the RHS is aturally equal to x k 1 (1 x) k B(k, k + 1) =!xk 1 (1 x) k (k 1)!( k)!, because B(a, b) = Γ(a)Γ(b)/Γ(a + b) ad the gamma fuctio satisfies equality Γ(k) = (k 1)!fork = 1, 2,... It turs out (after some simple calculatios) that the derivative of the LHS also equals!x k 1 (1 x) k (k 1)!( k)!.
7 Distributios of order statistics 21 Exercise 2.6 (solutio). Let r < s. Deote F r,s: (x 1,x 2 )=P{X r, x 1, X s, x 2 }. If x 2 x 1, the evidetly P{X r, x 1, X s, x 2 } = P{X s, x 2 } ad ) F r,s: (x 1,x 2 )= (F(x 2 )) m=s( m (1 F(x 2 )) m = 1 m F(x2 )(s, s + 1). Cosider ow the case x 2 > x 1. To fid F r,s: (x 1,x 2 ) let us metio that ay X from the sample X 1, X 2,...,X (idepedetly o other X s) with probabilities F(x 1 ), F(x 2 ) F(x 1 ) ad 1 F(x 2 ) ca fall ito itervals (,x 1 ], (x 1,x 2 ], (x 2, ) respectively. Oe sees that the evet A = {X r, x 1, X s, x 2 } is a uio of some disjoit evets a i, j, i j = {i elemets of the sample fall ito (,x 1 ], j elemets fall ito iterval (x 1,x 2 ],ad ( i j) elemets lie to the right of x 2 }. Recallig the polyomial distributio we obtai that! P{A i, j, i j } = i! j!( i j)! (F(x 1)) i (F(x 2 ) F(x 1 )) j (1 F(x 2 )) i j. To costruct A oe has to take all a i, j, i j such that r i, j 0ads i + j. Hece, F r,s: (x 1,x 2 )=P{A} = = i i=r j=max{0,s i} i i=r j=max{0,s i} P{A i, j, i j }! i! j!( i j)! (F(x 1)) i (F(x 2 ) F(x 1 )) j (1 F(x 2 )) i j. Exercise 2.7 (aswer). Deote for coveiece, k(0)=0, k(r + 1)= + 1, x 0 = ad x r+1 =. The f k(1),k(2),...,k(r): (x 1,x 2,...,x r )=! r+1 (F(x r+1 m ) F(x m 1 )) k(m) k(m 1) 1 r f (x m ), m=1 m=1 (k(m) k(m 1) 1)! m=1 if x 1 < x 2 < < x r,ad 0, otherwise. I particular, if r = 2, 1 i < j,adx 1 < x 2,the! f i, j: (x 1,x 2 )= (i 1)!( j i 1)!( j)! (F(x 1 )) i 1 (F(x 2 ) F(x 1 )) j i 1 (1 F(x 2 )) j f (x 1 ) f (x 2 ).
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