The distribution of the maximum of a first order moving average: the continuous case

No v em b er 2, 2 018 The distribu tion of the maxim um of a first order mo ving a v erage: the con tin uous case h ttp://arxiv.o rg/abs/0 8 02.052 3 b y Christopher S. Withers 1 Applied Mathematics Group Industr ial Researc h Limited Lo w er Hutt, NEW ZEALAND Saralees Nadara jah Sc ho ol of Mathematics Univ ersit y of Manc hester Manc hester M60 1QD, UK Abstract: W e giv e the distribution of M n , the maxim um of a sequence of n ob s erv at ions fr om a mo ving a verag e of order 1. Solutions are first giv en in terms of rep eated in tegrals and then for the case where the und erlying indep endent random v ariables ha ve an absolutely co ntin uous den s it y . When the correlation is p ositiv e , P ( M n ≤ x ) = ∞ X j =1 β j x ν n j x ≈ B x ν n 1 x where { ν j x } are the eigenv alues (singular v alues) of a F redh olm kernel and ν 1 x is th e eigen v alue of maximum magnitude. A similar result is giv en when the correlation is n egativ e. The result is analogous to large deviati ons expansions for estimates, since the maxim um need not be standardized to hav e a limit. F or the cont inuous case the integ ral e quations f or th e left and righ t eigenfun ctions are con v erted to fir st order linear differen tial equations. The eigen v alues satisfy an equation of the form ∞ X i =1 w i ( λ − θ i ) − 1 = λ − θ 0 for certain kno wn wei ghts { w i } and singular v alues { θ i } of a giv en m atrix. This can b e solv ed by truncating th e sum to an increasing n umb er of term s . 1 In tro duction and Su mmary Little is a v ailable in the literature on the b eha viour of extremes of correlated sequen ces apart fr om some sp ecial cases inv olving Gaussian pro cesses and some weak con vergence results. F or example 1 W ork b egun while visiting the Statistics Dept, UNC, Chap el Hill. 1 Leadb etter et al. (1983 ) p59 giv e a conv ergence in distribution for the scaled maxim um of a stationary sequence, and Resnick (1987) p 239 giv es a similar result for mo ving a verage s. This pap er giv es a p ow erfu l new method for giving the exact d istribution of extremes of n correlated obs erv ations as w eigh ted sums of n th p o w ers of asso ciated eigenv alues. The m etho d is illustrated h ere for a mo ving a verage of order 1. Let { e i } b e indep enden t and identica lly d istributed random v ariables from some distrib ution F on R . Consid er th e moving a v erage of ord er 1, X i = e i + ρe i − 1 where ρ 6 = 0 . I n Section 2 we giv e expr essions for the distribu tion of the maxim um M n = n max i =1 X i in terms of rep eated in tegrals. This is obtained via the r ecurrence relationship G n ( y ) = I ( ρ < 0) G n − 1 ( ∞ ) F ( y ) + K G n − 1 ( y ) (1.1) where G n ( y ) = P ( M n ≤ x, e n ≤ y ) , (1.2) I ( A ) = 1 or 0 for A true or false, and K is an inte gral op erator dep ending on x . (Dep endence on x is suppr essed.) F or this to work at n = 1 we defin e M 0 = −∞ so that G 0 ( y ) = F ( y ) . In Section 3 w e consider the case when F is absolutely contin uou s with densit y f ( x ) with resp ect to Leb esque measure. In th is case we s h o w that corresp onding to K is a F redholm kernel K ( y , z ). W e giv e a solution in terms of its eigen v alues and eigenfunctions. This leads easily to the asymptotic results stated in the abstract. Our expansions f or P ( M n ≤ x ) f or fixed x are large deviation results. If x is replaced by x n suc h that P ( M n ≤ x n ) tends to the GEV d istribution, then the expansion still holds, but not the asymptotic exp ansion in terms of a single eigenv alue, since this ma y approac h 1 as n → ∞ . Set R r = R r ( y ) dy . 2 Solutions using rep eated in tegrals. F or n ≥ 1, G n of (1.2) s atisfies G n ( y ) = P ( M n − 1 ≤ x, e n + ρe n − 1 ≤ x, e n ≤ y ) = P ( M n − 1 ≤ x, e n − 1 ≤ ( x − e n ) /ρ, e n ≤ y ) if ρ > 0 = Z y G n − 1 (( x − w ) /ρ ) dF ( w ) = P ( M n − 1 ≤ x, e n − 1 ≥ ( x − e n ) /ρ, e n ≤ y ) if ρ < 0 = Z y [ G n − 1 ( ∞ ) − G n − 1 (( x − w ) /ρ )] dF ( w ) = G n − 1 ( ∞ ) F ( y ) − Z y G n − 1 (( x − w ) /ρ ) dF ( w ) . (Thanks to the referee for noting a slip in the last line.) That is, f or n ≥ 1, (1.1 ) holds w ith K r ( y ) = sign( ρ ) Z y r (( x − w ) /ρ ) dF ( w ) . (2.1) 2 Our goal is to determine u n = P ( M n ≤ x ) = G n ( ∞ ) . In this section we give u n in terms of v n = [ K n F ( y )] y = ∞ . (2.2) F or examp le v 1 = − Z F ( z ) dF ( x − ρz ) = − I ( ρ < 0) + Z F ( x − ρz ) dF ( z ) . (2.3) The b eha viour of u n falls into t w o cases. The c ase ρ > 0 . F or n ≥ 1 , u n = v n (2.4) since G n ( y ) = K n F ( y ) . The marginal distribution of X 1 is u 1 = v 1 giv en b y (2.3). The c ase ρ < 0 . By (1.1), for n ≥ 0 , G n +1 ( y ) = u n F ( y ) + K G n ( y ) = a n ( y ) ⊗ u n + a n +1 ( y ) where a i ( y ) = K i F ( y ) , a n ⊗ b n = n X j =0 a j b n − j . (2.5) Putting y = ∞ giv es the recur rence equation for u n : u 0 = 1 , u n +1 = v n +1 + n X i =0 v i u n − i , n ≥ 0 . (2.6) The marginal distribution of X 1 is u 1 = 1 + v 1 of (2.3). An explicit solution for u n when ρ < 0 . Define th e generating functions U ( t ) = ∞ X n =0 u n t n , V ( t ) = ∞ X n =0 v n t n , Multiplying (2.6) b y t n and su mming from n = 0 giv es ( U ( t ) − V ( t )) /t = U ( t ) V ( t ), so that U ( t ) = (1 − W ( t )) − 1 V ( t ) where W ( t ) = tV ( t ) = ∞ X n =1 w n t n , w n = v n − 1 , tU ( t ) = (1 − W ( t )) − 1 W ( t ) = (1 − W ( t )) − 1 − 1 = ∞ X j =1 W ( t ) j . By d efinition, for j = 0 , 1 , · · · W ( t ) j = ∞ X n = j ˆ B nj ( w ) t n 3 where ˆ B nj ( w ) is th e p artial or dinary Bel l p olynomial in w = ( w 1 , w 2 , · · · ) tabled on p309 of Com tet (1974 ). F or example ˆ B n 0 ( w ) = δ n 0 , ˆ B n 1 ( w ) = w n , ˆ B n 1 ( w ) = w n 1 . So tU ( t ) = (1 − W ( t )) − 1 − 1 = ∞ X n =1 ˆ B n ( w ) t n where ˆ B n ( w ) = n X j =0 ˆ B nj ( w ) is the c omplete or dinary Bel l p olynomial . F or example ˆ B 0 ( w ) = 1 . T aking the co efficien t of t n giv es the exp licit solution u n − 1 = ˆ B n ( w ) , n ≥ 1 . (2.7) F or examp le u 0 = ˆ B 1 ( w ) = ˆ B 11 ( w ) = w 1 = v 0 = 1 , u 1 = ˆ B 2 ( w ) = ˆ B 21 ( w ) + ˆ B 22 ( w ) = w 2 + w 2 1 = v 1 + 1 . Similarly from Com tet’s table we can immediately read off u n , 1 ≤ n ≤ 9: u 0 = 1 , u 1 = v 1 + 1 u 2 = v 2 + 2 v 1 + 1 , u 3 = v 3 + (2 v 2 + v 2 1 ) + 3 v 1 + 1 , u 4 = v 4 + (2 v 3 + 2 v 1 v 2 ) + (3 v 2 + 3 v 2 1 ) + 4 v 1 + 1 , u 5 = v 5 + (2 v 4 + 2 v 1 v 3 + v 2 2 ) + (3 v 3 + 6 v 1 v 2 + v 3 1 ) + (4 v 2 + 6 v 2 1 ) + 5 v 1 + 1 , u 6 = v 6 + (2 v 5 + 2 v 1 v 4 + 2 v 2 v 3 ) + (3 v 4 + 6 v 1 v 3 + 3 v 2 2 + 3 v 2 1 v 2 ) + (4 v 3 + 12 v 1 v 2 + 4 v 3 1 ) +(5 v 2 + 10 v 2 1 ) + 6 v 1 + 1 , u 7 = v 7 + (2 v 6 + 2 v 1 v 5 + 2 v 2 v 4 + v 2 3 ) + (3 v 5 + 6 v 1 v 4 + 6 v 2 v 3 + 3 v 2 1 v 3 + 3 v 1 v 2 2 ) +(4 v 4 + 12 v 1 v 3 + 6 v 2 2 + 12 v 2 1 v 2 + v 4 1 ) + (5 v 3 + 20 v 1 v 2 + 10 v 3 1 ) + (6 v 2 + 15 v 2 1 ) + 7 v 1 + 1 , u 8 = v 8 + (2 v 7 + 2 v 1 v 6 + 2 v 2 v 5 + 2 v 3 v 4 ) + (3 v 6 + 6 v 1 v 5 + 6 v 2 v 4 + 3 v 2 3 + 3 v 2 1 v 4 + 6 v 1 v 2 v 3 + v 3 2 ) +(4 v 5 + 12 v 1 v 4 + 12 v 2 v 3 + 12 v 2 1 v 3 + 12 v 1 v 2 2 + 4 v 3 1 v 2 ) + (5 v 4 + 20 v 1 v 3 + 10 v 2 2 + 30 v 2 1 v 2 + 5 v 4 1 ) +(6 v 3 + 30 v 1 v 2 + 20 v 3 1 ) + (7 v 2 + 21 v 2 1 ) + 8 v 1 + 1 , u 9 = v 9 + (2 v 8 + 2 v 1 v 7 + 2 v 2 v 6 + 2 v 3 v 5 + v 2 4 ) + (3 v 7 + 6 v 1 v 6 + 6 v 2 v 5 + 6 v 3 v 4 + 3 v 2 1 v 5 + 6 v 1 v 2 v 4 +3 v 1 v 2 3 + 3 v 2 2 v 3 ) + (4 v 6 + 12 v 1 v 5 + 12 v 2 v 4 + 6 v 2 3 + 12 v 2 1 v 4 + 24 v 1 v 2 v 3 + 4 v 3 2 + 4 v 3 1 v 3 +6 v 2 1 v 2 2 ) + (5 v 5 + 20 v 1 v 4 + 20 v 2 v 3 + 30 v 2 1 v 3 + 30 v 1 v 2 2 + 20 v 3 1 v 2 + v 5 1 ) + (6 v 4 + 30 v 1 v 3 + 15 v 2 2 +60 v 2 1 v 2 + 15 v 4 1 ) + (7 v 3 + 42 v 1 v 2 + 35 v 3 1 ) + (8 v 2 + 28 v 2 1 ) + 9 v 1 + 1 . More generally any u n can b e obtained fr om (2.7) u sing the recurrence relation b n = w n ⊗ b n , n ≥ 1 , where w 0 = 0 , b n = ˆ B n ( w ) . (2.8) F or examp le since b 0 = 1, this giv es b 1 = w 1 , b 2 = w 2 1 + w 2 , b 3 = w 3 1 + 2 w 1 w 2 + w 3 . The recurr en ce relation (2.8) for the complete ordinary Bell p olynomials follo w s b y taking the co efficien t of t n in (1 − w ) − 1 − 1 = w (1 − w ) − 1 where w = W ( t ), and app ears to b e n ew. 4 3 The absolutely c on tin uous case. Our solutions (2.4), (2.6), (3.8) do not tell us h ow u n b eha ves for large n . Also they r equ ire rep eated in tegration. Here we giv e solutions that o v ercome these problems, u sing F r edholm in tegral theory giv en in App en d ix A. W rite (2.1) in the form K r ( y ) = Z K ( y , z ) r ( z ) dz where K ( y , z ) = ρI ( x ≤ y + ρz ) f ( x − ρz ) . (3.1) Since ||K|| 2 2 = Z Z K ( y , z ) K ( z , y ) dy dz = ρ 2 Z Z I ( x < y + ρz ) I ( x < z + ρy ) f ( x − ρz ) f ( x − ρy ) dy dz < ρ 2 Z Z f ( x − ρz ) f ( x − ρy ) dy dz = 1 , (3.2) K ( y , z ) is said to b e a F redholm k ernel w.r.t. Lebesgue measure, allo wing the F r edholm theory of the App endix to b e applied, in particular the f unctional forms of the Jordan form and singular v alue decomp osition. If sa y , 0 < ρ < 1, then one can show that ||K|| 2 2 = Z F ( x t ) dF ( t ) ↑ 1 as x ↑ ∞ wh ere x t = min( x − ρt, ( x − t ) /ρ ) . Let { λ j , r j , l j : j ≥ 1 } b e its eigen v alues (singular v alues) and asso ciated r igh t and left eigen- functions ord ered so that | λ j | ≤ | λ j +1 | . By App endix A th ese satisfy λ j K r j = r j , λ j l j K = l j , Z r j l k = δ j k , (3.3) where δ j k is the Kronec k er fu nction and we write R a ( y ) b ( y ) dy = R ab . S o { r j ( y ) , l k ( y ) } are biorthog- onal fu nctions with r esp ect to Leb esgue measur e. Set ν j = 1 /λ j . By (3.2) and (A.6), 1 > ||K|| 2 2 = ∞ X j =1 ν 2 j where ν j are the singular v alues, or if the J ordan form is diagonal, the eigen v alues. (W e sh all u se these terms in terc hangeably .) So | ν j | < 1 and 1 + ν j > 0. Consider the case where the Jordan f orm is d iagonal. Supp ose that the eigen v alue λ 1 of smallest magnitude has m ultiplicit y M (t yp ically 1). S et β j = r j ( ∞ ) Z F l j , B = M X j =1 β j . (3.4) Then by (A.8) for n ≥ 1, v n = ∞ X j =1 β j ν n j = B ν n 1 (1 + ǫ n ) (3.5) 5 where ǫ n → 0 exp onen tially as n → ∞ . (In fact b y (A.8) 1 = v 0 = P ∞ j =1 β j if this con verge s.) S o for n ≥ 1 , b y (2.4) for ρ > 0 , u n = ∞ X j =1 β j ν n j . (3.6) ν 1 is giv en by (A.10) w ith µ Leb esgue measure. (When ρ = 0 then (3.6) h olds with β j = δ j 1 , ν 1 = F ( x ). So we exp ect that ν 1 → F ( x ) as ρ ↓ 0 . ) No w supp ose that ρ < 0. By (3.5), for max ∞ j =1 | ν j t | < 1, V ( t ) = 1 + P ∞ j =1 β j ν j t/ (1 − ν j t ). So 1 − tV ( t ) = 1 − t − ∞ X j =1 β j ν j t 2 / (1 − ν j t ) = N ( t ) /D ( t ) where D ( t ) = Π ∞ j =1 (1 − ν j t ) , N ( t ) = Π ∞ j =1 (1 − w j t ) sa y . D ( t ) is the F redholm d eterminan t of K ( x, y ). (No w w j tak es on a different meaning than in Section 2.) So by the partial fr action expansion, assu m ing that { w j } are all different, N ( t ) − 1 = ∞ X j =1 c − 1 j (1 − w j t ) − 1 where c j = Π k 6 = j (1 − w k /w j ) , = ∞ X n =0 N n t n where N n = ∞ X j =1 c − 1 j w n j . (3.7) Also by F redholm’s first theorem - see for example, p 47 of P ogorzelski (1966), D ( t ) = 1 + ∞ X n =1 D n ( − t ) n /n ! , D n = Z · · · Z N  s 1 · · · s n s 1 · · · s n  ds 1 · · · ds n where N  s 1 ··· s n s 1 ··· s n  = det( N ( s j , s k ) , 1 ≤ j, k ≤ n ). Alternativ ely , a simple exp ansion giv es D n /n ! = X 1 ≤ j 1 < ··· 1, then B n ( d ) ≈ d n ≈ a 1 w n 1 as n → ∞ . An alternativ e appr oac h is to try a solution for u n of the form (3.5), sa y u n = ∞ X j =1 γ j δ n j (3.9) where δ j decrease in magnitude. Assuming that { δ j , ν j } are all d istinct, substitution into the recurrence relation (2.6) give s u s the follo wing elegan t relations. { δ j } are the r o ots of ∞ X k =1 β k / ( δ − ν k ) = 1 (3.10) and β k is given by (3.4). Ha ving found { δ j } , { γ j } are the ro ots of ∞ X j =1 γ j / ( δ j − ν k ) ≡ 1 . (3.11) The last equation can b e wr itten Aγ = 1 where A = ( A k j : k , j ≥ 1) , A k j = 1 / ( δ j − ν k ) . So a formal solution is γ = A − 1 1 . Numerical solutions can b e foun d by trun cating the infinite matrix A and infinite vecto rs 1 , γ to N × N matrix and N -v ectors, then increasing N un til the desired p recision is reac hed . Beha viour for large n and x indep endent of n . F or ρ > 0, (3.6) imp lies u n ≈ B ν n 1 and ν 1 > 0 , (3.12) where B is giv en by (3.4). Also ν 1 is giv en by (A.10 ) with µ Leb esgue measure. No w supp ose that ρ < 0. By (3.9), u n = γ 1 δ n 1 (1 + ǫ ′ n ) (3.13) 7 where ǫ ′ n → 0 exp onentia lly as n → ∞ and δ 1 has the largest magnitude among { δ n } . (The case where multiple δ n exist of magnitude | δ 1 | requires an obvious adap tation.) In tegral and differen tial equa t ions for the eigenfunctions and resolv ent. The right eigenfunctions satisfy ν j r j = K r j , that is ν j r j ( y ) = K r j ( y ) = sign( ρ ) Z y r j (( x − w ) /ρ ) dF ( w ) (3.14) F or examp le ν j r j ( ∞ ) = ρ Z r j ( z ) f ( x − ρz ) dz . Differen tiating giv es the non-standard linear first order differentia l equation ν j ˙ r j ( y ) = sign( ρ ) f ( y ) r j (( x − y ) /ρ ) , r j ( −∞ ) = 0 . (3.15) Similarly the left eigenfunctions satisfy ν j l j = l j K , that is ν j l j ( z ) = Z l j ( y ) K ( y , z ) dy = ρf ( x − ρz ) Z x − ρz l j ( y ) dy . (3.16) So l j ( −∞ ) = 0 if ρ > 0 , l j ( ∞ ) = 0 if ρ < 0 , (3.17) and by differen tiating, ν j ( d/dz )[ l j ( z ) /f ( x − ρz )] = ρ 2 l j ( x − ρz ) . (3.18) The resolve nt satisfies [ K ( y , z , λ ) − K ( y , z )] /λ = K K ( y , z , λ ) = K ( y , z , λ ) K . (3.19) So K ( − ∞ , z , λ ) = 0 , K ( y , ∞ , λ ) = 0 if ρ < 0 , K ( y , −∞ , λ ) = 0 if ρ > 0 and by differen tiation the r esolv en t satisfies th e first order partial differen tial equations ( ∂ /∂ y ) LHS(3.19) = sign( ρ ) f ( y ) K (( x − y ) /ρ, z , λ ) , ( ∂ /∂ z )[ LHS(3.19) /f ( x − ρz )] = ρ 2 K ( y , x − ρz , λ ) . These may in vo lve th e Dirac fu nction δ ( x ) since with x z = x − ρz , ( ∂ /∂ y ) K ( y , z ) = ρf ( x z ) δ ( y − x z ) , ( ∂ /∂ z ) K ( y , z ) = ρ 2 f ( x z ) δ ( y − x z ) − ρ 2 I ( x z < y ) ˙ f ( x z ) . F or s p ecial cases, it is p ossible to solv e (3.14) or (3.1 6 ) explicitly . 8 Example 3.1 Supp ose that F ( y ) = ae ay on ( −∞ , 0] wher e a > 0 , and that ν j < 0 , ρ < − 1 , y ≤ x. T aking r j (0) = 1 , a solution of (3.14 ) is r j ( y ) = e b j y wher e b j | ν j | /a = e b j x/ρ , b j > 0 . F ormal expressions for the eigenfunctions. W e no w give a formal solution of (3.15) for r j in terms of r j (0). (Th e v alue 0 is arbitrary: a similar solution can b e obtained in terms of r j ( y 0 ) for an y y 0 .) Set r ( y ) = r j ( y ) , c = λ j sign( ρ ) . Supp ose that f and r ha v e T aylor series expansions ab out 0. Denote the i th deriv ativ es of f ( y ) b y f .i ( y ) and set f i = f .i (0) , r i = r .i (0). Expanding ˙ r ( y ) = cf ( y ) r (( x − y ) /ρ ) ab out 0, for i ≥ 0 the co efficien t of y i /i ! is r i +1 = c X a + b = i  i a  f a r .b ( x/ρ )( − ρ ) − b = c ∞ X k =0 q ik r k = cf i r 0 + c ∞ X k =1 q ik r k where q ik = ρ − k min( i,k ) X b =0  i b  f i − b ( − 1) b x k − b / ( k − b )! . (3.20) F or l , k ≥ 1 set Q lk = q l − 1 ,k . Set Q = ( Q lk : l , k ≥ 1). Set f ′ = ( f 0 , f 1 , · · · ) , R ′ = ( r 1 , r 2 , · · · ) , Y ′ y = Y ′ = ( y / 1! , y 2 / 2! , · · · ) . (3.21) So R = f cr 0 + cQ R , R = ( I − cQ ) − 1 F cr 0 . But r ( y ) − r (0) = Y ′ R . So we obtain the j th right eigenfunction in terms of its v alue at 0: r ( y ) /r (0) = 1 + Y ′ ( c − 1 I − Q ) − 1 f , that is, r j ( y ) /r j (0) = 1 + Y ′ ( d j I − Q ) − 1 f where d j = ν j sign( ρ ) . (3.22) F or example for the extreme v alue distribu tion F ( x ) = e − e − x , f = e − 1 (1 , 0 , − 1 , − 1 , − 7 / 288 , − 31 / 4 , · · · ) ′ . Since r j is uniqu e only up to a constan t multiplier, we ma y tak e r j (0) ≡ 1 . The solution (3.22 ) can n o w b e implement ed by successive approximat ions. F or N ≥ 1 set r N j ( y ) /r j (0) = 1 + Y ′ N ( d j I N − Q N ) − 1 f N (3.23) where Y N , f N are the 1st N elemen ts of Y , f and Q N is the upp er left N × N elemen ts of Q . Then one exp ects that r N j ( y ) → r j ( y ) as N → ∞ , giving th e j th left eigenfunction. A similar treatmen t of (3.18) giv es an equ ation for the j th left eigenfunction in terms of its v alue at at an arbitrary p oin t, tak en h er e as x . Set c = ρ 2 λ j , l = l j , e ( y ) = f ( y ) − 1 . 9 By T a ylor exp ansions, l ( z ) e ( x − ρz ) = ∞ X i =0 ( z i /i !) X a + b = i  i a  l .a (0) e .b ( x )( − ρ ) b . By (3.18), l satisfies ( d/dz )LHS = cl ( x − ρz ) . T aking the co efficien t of z i /i !, f or i ≥ 0, X a + b = i +1  i + 1 a  l .a (0) e .b ( x )( − ρ ) b = cl .i ( x )( − ρ ) i . (3.24) By another T a ylor expansion, l .a (0) = ∞ X k =0 l .k + a ( x )( − x ) k /k ! . So LHS of (3.24) is P ∞ j =0 W ij l .j ( x ) = V i l ( x ) + ( W L ) i where we set W ij = min( j,i +1) X a =0  i + 1 a  e .i +1 − a ( x )( − ρ ) i +1 − a ( − x ) j − a / ( j − a )! , W = ( W ij : i, j ≥ 1) , U i = W i 0 = e .i +1 ( x )( − ρ ) i +1 , V j = W 0 j , L j = l .j ( x ) , L ′ = ( L 1 , L 2 , · · · ) , D r = diag ( r i : i ≥ 1) , r = − ρ. (3.25) So (3.24) for i ≥ 1 can b e written U l ( x ) + W L = cD r L so that L = ( cD r − W ) − 1 U l ( x ) giving in the n otatio n of (3.21), l ( z ) − l ( x ) = Y ′ z − x L = Y ′ z − x ( cD r − W ) − 1 U l ( x ) . That is, l j ( z ) = l ( z ) is giv en by l ( z ) /l ( x ) = 1 + Y ′ z − x ( cD r − W ) − 1 U . (3.26) Finally , the v alue of the multiplier l j ( x ) = l ( x ) is determined b y (A.3): 1 /l j ( x ) = Z r j ( z ) RH S (3.26) dz . An equation for the eigen v alues. Substituting in to (3.24) at i = 0, that is W 00 l ( x ) + V ′ L = cl ( x ) , w e obtain V ′ ( cD r − W ) − 1 U = c − W 00 . (3.27) The ro ots c of this equ ation are just { ρ 2 λ j } , so th is is the equation for the eigen v alues we hav e b een seeking. If { θ j , j ≥ 1 } are the singular v alues of D − 1 r W , an d if this has diagonal Jord an form D − 1 r W = R 0 Λ L ∗ 0 where Λ = diag( θ 1 , θ 2 , · · · ) , (see (A.1) b elo w), then (3.27) can b e written ∞ X i =1 w i ( c − θ i ) − 1 = c − θ 0 , where n ow the weigh ts { w i } are giv en b y w i = v i u i where v = ¯ R 0 D − 1 r V , u = L ∗ 0 U . 10 If { c N j : j =1 , ·· · ,N +1 } are th e ro ots of its N d imensional app ro ximation, say V ′ N ( cD ρN − W N ) − 1 U N = c − W 00 , then c N j → c j = ρ 2 λ j as N → ∞ . (This is essen tially a p olynomial in c of degree N + 1.) Ha ving obtained an eigenv alue, one can substitute it into (3.22) and (3.26) to obtain th e corresp onding eigenfunctions up to constan ts l ( x ) and r (0). As noted in the app end ix, either of these (but not b oth) can b e arbitrarily c hosen. The cond itions r ( −∞ ) = 0 an d (3.17) can b e verified numericall y . Example 3.2 Supp ose that f = φ , the density of a standar d normal N (0 , 1) r.v.. Then e .j ( x ) = φ ( x ) − 1 H ∗ j ( x ) wher e H ∗ j ( x ) = E ( x + N (0 , 1)) j = X k  j 2 k  x j − 2 k m 2 k is the mo difie d Hermite p olynomial and m 2 k = (2 k )! /k !2 k is the 2 k th moment of N (0 , 1) . Se e Withers and Mc Gavin (200 6). An alternativ e is to expand RHS(3.24) ab out x = 0, giving c ( − ρ ) i ( l (0) + U ′ L ) where w e s et L j = l .j (0) , L ′ = ( L 1 , L 2 , · · · ) , x k = x k /k ! , U ′ = ( x 1 , x 2 , · · · ) , A ia = l .a (0)[ e .b ( x )( − ρ ) b ] b = i +1 − a , A = ( A ia : i, a ≥ 1) , V i = A i 0 = [ e .b ( x )( − ρ ) b ] b = i +1 . Let 0 i denote the ro w i − v ector of zeros. F or i ≥ 1, (3.24) giv es V i l (0) + ( A i 1 , · · · , A i,i +1 , 0 , 0 , · · · ) L = c ( − ρ ) i (0 i − 1 , x 0 , x 1 , · · · ) L . That is V l (0) + A L = cD r X L where the i th row of the matrix X is (0 i − 1 , x 0 , x 1 , · · · ) . So L = B − 1 V l (0) where B = cD r X − A, l ( y ) /l (0) = 1 + Y ′ y B − 1 V . X is up p er triangular, while A is lo wer triangular except for the 1st sup er-diagonal. F or i = 0, (3.24) give s 1 X a =0 A 0 a l a = c ∞ X k =0 l k x k = cl (0) + c U ′ L . So we obtain as an alternative e quation for the eige nv alues A 00 + A 01 ( B − 1 V ) 1 = c + c U ′ B − 1 V w here A 00 = − ρe . 1 ( x ) , A 01 = e ( x ) . Unfortunately App endix A cann ot b e applied with µ = F since K G ( y ) = s ign( ρ ) R y G (( x − w ) /ρ ) dF ( w ) is not of the form R K ( y , z ) G ( z ) dF ( z ) . It w ould b e of great in terest, and in particular allo w a unified ap p roac h to this problem, if F redholm’s th eory can b e extended to the system KO r = ν r , K ∗ O ∗ l = ¯ ν l , l ∗ i O r j dµ = δ ij for K an q × q integ ral op erator with k ernel K ( y , z ) : R p × R p → C q × q with resp ect a measure µ , O a q × q op erator, wh ere ∗ is the transp ose of the complex conjugate, and ¯ ν is the complex conjugate of ν . F or ou r problem, one could then apply the theory with p = q = 1 , µ = F , K ( y , z ) = sign( ρ ) I ( z < y ) , O G ( w ) = G (( x − w ) /ρ ) . 11 APPENDIX A T o mak e the pap er self-con tained, we giv e h ere some theory for F redholm integ ral equatio ns with non-sym metric kernels. First consider the case w here K is any k × k complex matrix. Its singu lar v alue decomp osition is K = R Λ L ∗ where R R ∗ = I , LL ∗ = I , Λ = d iag( ν 1 , ν 2 , · · · ) , * den otes the complex conjugate tr an s p ose. Since K K ∗ R = R ΛΛ ∗ , K ∗ K L = L Λ ∗ Λ , the j th column of R is a r ight eigen vect or of K K ∗ with eigen v alue | λ j | 2 and the j th column of L is a righ t eigenv ector of K ∗ K w ith the same eigen v alue. If K is non-singular, its in verse is K − 1 = L Λ − 1 R ∗ . If it is singular, a pseudoinv ers e is giv en by K − = L Λ − R ∗ . Ho w ev er the sin gular v alue decomp osition do es n ot giv e a n ice form for p o wers of K . This dr a wbac k is o ve rcome by its Jordan decomp osition. Consider the case w here this is diagonal. Then K = R Λ L ∗ where R L ∗ = I , and Λ = diag( ν 1 , ν 2 , · · · , ) (A.1) is comp osed of the eigen v alues of K . Then for an y complex α , K α = R Λ α L ∗ . T aking α = n and α = − 1 giv es the n th p o w er an d in verse of K . No w let K ( y , z ) b e a real fu nction on Ω × Ω where Ω is a subset of R p . Su pp ose th at µ is a σ -finite measure on Ω and that 0 < ||K|| 2 2 = Z Z K ( y , z ) K ( z , y ) dµ ( y ) dµ ( z ) < ∞ . (This L 2 condition can b e c hanged to Z | K ( y , y ) | dµ ( y ) < ∞ at the exp ense of notational complexities that need not concern us h ere.) Th e corresp onding in tegral op erator K is defined by K φ ( y ) = Z K ( y , z ) φ ( z ) dµ ( z ) , φ ( z ) K = Z φ ( y ) K ( y , z ) dµ ( y ) . (A.2) The F red holm equations of the first kind, λ K r ( y ) = r ( y ) , λl ( z ) K = l ( z ) , 12 ha v e only a coun table num b er of solutions, sa y { λ j , r j ( y ) , l j ( z ) , j ≥ 1 } up to arbitrary constant m ultipliers for { r j ( y ) , j ≥ 1 } , and these satisfy Z r j l k dµ = δ j k where Z g dµ = Z R p g ( y ) dµ ( y ) . (A.3) These are called the singular values (or eig envalues) and right and left eigenfunctions of ( K, µ ) or K . Also K ( y , z ) = ∞ X j =1 r j ( y ) l j ( z ) /λ j (A.4) with conv ergence in L 2 ( µ × µ ), or more strongly u n der other conditions: see Withers (1974, 1975 , 1978) . This is the functional form of the Singular V alue Decomp osition for a s quare n on-symmetric matrix. F redholm equations of the second kin d, r ( y ) − λ K r ( y ) = f ( y ) , l ( z ) − λl ( z ) K = g ( z ) , can b e solv ed for λ not an eige nv alue using ( I − λ K ) − 1 = I + λ K λ where K λ f ( y ) = Z K ( y , z , λ ) f ( z ) dµ ( z ) , g ( z ) K λ = Z g ( y ) K ( y , z , λ ) dµ ( y ) , and the r esolvent K ( y , z , λ ) with op erator K λ is th e unique solution of ( I − λ K ) K λ = K = K λ ( I − λ K ) , that is, λ Z K ( y , u ) K ( u, z , λ ) dµ ( u ) = K ( y , z ) − K ( y , z , λ ) = λ Z K ( y , u, λ ) K ( u, z ) dµ ( u ) . If this ca n b e solv ed analytically or numericall y , then often one do es not need to compu te the eigen v alues and eige nf unctions. Alternativ ely , the resolve nt satisfies K ( y , z , λ ) = ∞ X j =1 r j ( y ) l j ( z ) / ( λ j − λ ) . (A.5) The F r e dholm determinant is D ( λ ) = Π ∞ j =1 (1 − λ/λ j ) = exp {− Z λ 0 dλ Z K ( y , y , λ ) dµ ( y ) } . Note that ||K|| 2 2 = ∞ X j =1 λ − 2 j . (A.6) 13 Also since (1 − λ/λ j ) − 1 − 1 = λ/ ( λ j − λ ), λK ( y , z , λ ) = ∞ X j =1 r j ( y ) l j ( z )[(1 − λ/λ j ) − 1 − 1] . If only a finite num b er of eigen v alues are n on-zero, the kernel K ( y , z ) is said to b e de gener ate . (F or example this holds if µ p uts weig ht only at n p oints.) If not, { l j } and { r j } typica lly b oth span L 2 ( µ ) = { f : R | f | 2 dµ < ∞} . F or f ∈ L 2 ( µ ), f ( y ) = ∞ X j =1 R j r j ( y ) = ∞ X j =1 L j l j ( y ) where R j = Z f l j dµ, L j = Z f r j dµ (A.7) with conv ergence in L 2 ( µ ). So K n f ( y ) = ∞ X j =1 R j r j ( y ) /λ n j , f ( y ) K n = ∞ X j =1 L j l j ( y ) /λ n j , n ≥ 0 . (A.8) So if | λ 1 | < | λ j | (A.9) for j > 1 then as n → ∞ , K n +1 f ( y ) / K n f ( y ) → λ − 1 1 , f ( y ) K n +1 /f ( y ) K n → λ − 1 1 . This is one wa y to obtain λ 1 arbitrarily closely . Another is to use λ − 1 1 = sup { Z g K hdµ : Z g hdµ = 1 } if λ 1 > 0 , (A.10) λ − 1 1 = inf { Z g K hdµ : Z g hdµ = 1 } if λ 1 < 0 . (A.11) The maximising/minimising fun ctions are the firs t eigenfunctions g = g 1 , h = h 1 . These are un ique up to a constan t m u ltiplier if (A.9) holds. If λ 1 is known, on e can use ( λ 1 K ) n f ( y ) → R 1 r 1 ( y ) , f ( y )( λ 1 K ) n → L 1 l 1 ( y ) , to appr o ximate R 1 r 1 ( y ) , L 1 l 1 ( y ). Also since l 1 ( y ) is only unique up to a m ultiplicativ e constan t, w e may c ho ose R 1 = 1 and so approximat e r 1 ( y ) , l 1 ( y ). One ma y no w rep eat the pro cedure on the op erator K 1 corresp onding to K 1 ( y , z ) = K ( y , z ) − r 1 ( y ) l 1 ( z ) to appro ximate λ 2 , r 2 ( y ) , l 2 ( z ), assum ing the next eigen v alue in magnitude, λ 2 , has multiplicit y 1. If say λ 1 has multiplicit y M > 1, then ( λ 1 K ) n f ( y ) → M X j =1 R j r j ( y ) , and one can adapt the metho d ab o v e. F or fu rther details see Withers and Nadara jah (In press). F or fu rther details on F red h olm theory f or symmetric kernels, see Withers (1974, 1975 , 1978). 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