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

Marc h 23, 2022 The distribu tion of the maxim um of a first order mo ving a v erage: the discre te case b y Christopher S. W ithers Applied Mathemat ics Group Industr ial Researc h Limited Lo w er Hutt, NEW ZEALAND Saralees Nadara jah Sc ho ol of Ma thematics 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 observ ations from a mo ving a v erage of order 1. Solutions are first giv en in terms of rep eated integ rals and then for the case where the underlying indep enden t r andom v ariables are discrete. A solution appropriate for large n tak es the form P r ob ( M n ≤ x ) = I X j =1 β j x ν n j x ≈ B x r n 1 x where { ν j x } are th e eigen v alues of a certain matrix, r 1 x is the maximum magnitude of the eigen- v alues, and I dep ends on the num b er of p ossible v alues of the underlyin g random v ariables. The eigen v al u es do not dep end on x only on its range. 1 In tro duction and summary W e give the d istribution of the maximum of a moving a v erage of order 1 f or discrete random v ariables. Section 2 summarises results for an y mo ving av erage of order 1 (d iscr ete or n ot) giv en in Withers and Nadara jah (2009 ). Tw o forms are giv en for the distrib u tion of the m axim um. Only one of these is appr opriate for large n . Th is form can b e view ed as a large d eviation expansion. It assumes that a related parameter v n can b e w ritten as a w eigh ted sum of n th p o w ers. Section 3 give s 3 sets of situ ation of increasing generalit y where this last assumption holds. Let { e i } b e indep endent and ident ically distributed random v ariables fr om some distribution F on R . Consid er the mo ving a v erage of order 1, X i = e i + ρe i − 1 where ρ 6 = 0 . So the observ ati ons tak e the v alues { x i + ρx j } . In Withers and Nadara jah (200 9) we ga v e expr essions f or th e distr ibution of the maxim um M n = n max i =1 X i 1 in terms of rep eated in tegrals. This was ob tained via the recurrence relationship G n ( y ) = I ( ρ < 0) G n − 1 ( ∞ ) F ( y ) + K G n − 1 ( y ) (1.1) where G n ( y ) = P r ob ( M n ≤ x, e n ≤ y ) , (1.2) I ( A ) = 1 or 0 for A tru e or false, and K is the in tegral op erator K r ( y ) = sign( ρ ) Z y r (( x − w ) /ρ ) dF ( w ) . (1.3) F or this to work at n = 1, define M 0 = −∞ so that G 0 ( y ) = F ( y ) . Note that dep endenc e on x is suppr esse d . Our purp ose is to find u n = P r ob ( M n ≤ x ) = G n ( ∞ ) , n ≥ 0 . Section 2 su mmarises and extends relev an t r esults in Withers and Nadara jah (2009). In Section 3 we consider the case where e 1 is discrete and deriv e a solution of th e form giv en in the abstract. Section 4 giv es a solution to G n of (1.2). F or any integrable function r , set R r = R r ( y ) dy = R ∞ −∞ r ( y ) dy , R x r = R x r ( y ) dy = R x −∞ r ( y ) dy . 2 Solutions using rep eated in tegrals and sums of p o w ers. Set v n = [ K n F ( y )] y = ∞ . (2.1) F or example v 0 = 1 , v 1 = − Z F ( z ) dF ( x − ρz ) = − I ( ρ < 0) + Z F ( x − ρz ) dF ( z ) . (2.2) The c ase ρ > 0 . (1.1) h as solution G n ( y ) = K n F ( y ) , n ≥ 0 , (2.3) so that u n = v n , n ≥ 0 . (2.4) F or example u 0 = 1. (The marginal distribution of X 1 is u 1 = v 1 giv en by (2.2).) 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 ) (2.5) where a i ( y ) = K i F ( y ) , a n ⊗ b n = n X j =0 a j b n − j . (2.6) 2 Putting y = ∞ , u n is giv en b y the r ecurrence equati on u 0 = v 0 = 1 , u n +1 = v n +1 + u n ⊗ v n , n ≥ 0 . (2.7) (The marginal distrib ution of X 1 is u 1 = 1 + v 1 of (2.2).) The solution of (2.7) is u n = ˆ B n +1 ( w ) , n ≥ 0 where w n = v n − 1 (2.8) and ˆ B n ( w ) is the c omplete or dinary Bel l p olynomial , a function of ( w 1 , · · · , w n ) generate d by (1 − ∞ X n =1 w n t n ) − 1 = ∞ X n =0 ˆ B n ( w ) t n . So { u n − 1 } ha v e generating function tU ( t ) = (1 − tV ( t )) − 1 − 1 where U ( t ) = ∞ X n =0 u n t n , V ( t ) = ∞ X n =0 v n t n . (2.9) F or example since v 0 = 1, reading from a table giv es 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 . (2.10) F or u n up to n = 9 and more details on compu ting ˆ B n ( w ), see Withers and Nadara j ah (2009). Note that 1 ≥ u n = P r ob ( M n ≤ x ) ≥ u n +1 ≥ 0 so that u n = 0 ⇒ u n +1 = 0 . (2.11) The solution (2.8) giv es n o indication of th e b eha viour of u n for large n . In Sectio n 3 we shall see that w e can usu ally write v n as a we ighte d sum of p o w ers, sa y v n = I X j =1 β j ν n − 1 j for n ≥ n 0 (2.12) where 1 ≤ I ≤ ∞ , n 0 ≥ 0. W e call this th e weig hte d-sum-of-p owers assumption. In this case u n generally has th e form u n = J X j =1 γ j δ n j for n ≥ max(0 , 2 n 0 − 1) w here J ≤ I ′ = I + n 0 , (2.13) and { δ j } are the ro ots of I X k =1 β k ν n 0 − 1 k / ( δ − ν k ) = p n 0 ( δ ) where p n +1 ( δ ) = δ n +1 − v n ⊗ δ n : (2.14) p 0 ( δ ) = 1 , p 1 ( δ ) = δ − 1 , p 2 ( δ ) = δ 2 − δ − v 1 , p 3 ( δ ) = δ 3 − δ 2 − δ v 1 − v 2 , · · · So (2 .14 ) can b e written as a p olynomial in δ of degree J wh ere J ≤ I ′ . Case 1: Assume that these J ro ots are all distinct. 3 Ha ving found { δ j } , { γ j } are the ro ots of J X j =1 A j n 0 ( ν ) γ j = q n 0 ( ν ) for ν = ν 1 , · · · , ν I (2.15) where A j n ( ν ) = δ n j / ( δ j − ν ) , q n +1 ( ν ) = ν n +1 + u n ⊗ δ n : q 0 ( ν ) = 1 , q 1 ( ν ) = ν + 1 , q 2 ( ν ) = ν 2 + ν + u 1 , q 3 ( ν ) = ν 3 + ν 2 + u 1 ν + u 2 , · · · (2.15) can b e wr itten A n 0 γ = Q n 0 where ( A n ) k j = A j n ( ν k ) , Q n = ( Q n 1 , · · · , Q nI ′ ) ′ , Q nk = q n ( ν k ) . (2.16) So if J = I , a solution is γ = A − 1 n 0 Q n 0 . (2.17) (If I = ∞ , numerical solutions can b e foun d by trun cating the infinite matrix A n and infi nite v ectors ( Q n , γ ) to N × N matrix and N -v ectors, then increasing N until the desired precision is reac hed.) The pro of, whic h is by su b stitution, assumes that { δ j , ν j } are all distinct. Th e p ro of r elies on the fact that if P J j =1 a j r n j = 0 for 1 ≤ n ≤ J and r 1 , · · · , r J are distinct, then a 1 = · · · = a J = 0, since det( r n j : 1 ≤ n, j ≤ J ) 6 = 0 . (2.13) extends a corresp ondin g resu lt in Wit her s and Nadara jah (2009 a). (If J < I , a s olution is giv en b y dropping I − J ro ws of (2.16). If J > I , th ere are n ot enough equatio ns for a solution, and the 2nd m etho d b elo w needs to b e used.) The v alues of u n for n < 2 n 0 − 1 ca n b e found from (2.7) or (2.8) or the extension of (2.10). Behaviour for lar ge n . If (2.12) holds then v n ≈ B r n 1 as n → ∞ wher e B = X j { β j ν − 1 j e iθ j n : | ν j | = r 1 } , r 1 = max I j =1 | ν j | , ν j = r j e iθ j . A similar result holds for u n of (2.13). Withers and Nadara jah (2009 b ) give a 2nd metho d of solution based on (2.9), that applies ev en wh en the w eigh ts β j in (2.12) are p olynomials in n . 3 The discrete case. Supp ose that e 1 is a discrete random v ariable, sa y e 1 = x i with probabilit y p i > 0 for i = 1 , 2 , · · · , P wh er e 1 ≤ P ≤ ∞ and P P i =1 p i = 1 . W e do not need to assu m e th at x 1 < x 2 < · · · . (The metho d extends in an ob vious w a y for i = · · · , − 1 , 0 , 1 , 2 , · · · w here P ∞ i = −∞ p i = 1 . ) So the observ ations X 1 , · · · , X n tak e their v alues fr om th e lattice { x i + ρx j , 1 ≤ i, j ≤ P } . Set I ( A ) = 1 or 0 for A tru e or false. T h e main task of this section is to giv e thr ee in creasingly general situations where the weig hted-sum-of-p ow er s assumption (2. 12 ) holds. F or an y function H , set A iH = sign( ρ ) p i H (( x − x i ) /ρ ) , A H = ( A 1 H , · · · , A P H ) ′ , q i ( y ) = I ( x i ≤ y ) , q ( y ) = ( q 1 ( y ) , · · · , q P ( y )) ′ , Q ij = sign( ρ ) q i (( x − x j ) /ρ ) p j , Q = ( Q ij : 1 ≤ i, j ≤ P ) . (3.1) 4 (Recall that dep endence on x is suppressed.) F or example Q 12 sign( ρ ) = I ( x 1 ≤ ( x − x 2 ) /ρ ) = 1 ⇐ ⇒ ρ > 0 , x ≥ x 2 + ρx 1 or ρ < 0 , x ≤ x 2 + ρx 1 . Note that K H ( y ) = A ′ H q ( y ) , K q ( y ) = Qq ( y ) , K n H ( y ) = A ′ H Q n − 1 q ( y ) for n ≥ 1 . So v 0 = 1 and for n ≥ 1, v n of (2.1 ) is giv en by v n = A ′ F Q n − 1 1 , wher e 1 ′ = (1 , · · · , 1) . (3.2) In terms of the b ackwar d op er ator B defin ed by B y n = y n − 1 , the recurrence r elation (2.7 ) can b e wr itten for n ≥ 0, u n +1 = v n +1 + C n u n where C n = I P + A ′ F B n 1 , B n = ( I P − Q n B n ) / ( I P − QB ) , (3.3) where I P is th e P × P ident ity m atrix. So B 0 = 0 . Alternativ ely , f rom (2.10) w e ha ve u 1 = 1 + A ′ F 1 , u 2 = 1 + A ′ F (2 I + Q ) 1 , u 3 = 1 + ( A ′ F 1 ) 2 + A ′ F (3 I + 2 Q + Q 2 ) 1 , u 4 = 1 + 3( A ′ F 1 ) 2 + 2( A ′ F 1 )( A ′ F Q 1 ) + A ′ F (4 I + 3 Q + 2 Q 2 + Q 3 ) 1 , and so on. Neither solution is satisfactory for large n . The idemp oten t case. W e shall see that Q frequen tly has the form Q = θ J w here θ is scala r and J 2 = J. (3.4) That is, J is idemp otent with eigen v alues 0 , 1. By (3.2) v 1 = A ′ F 1 , v n = θ n − 1 d for n ≥ 2 where d = A ′ F J 1 . (3.5) This is ju st (2.12) with I = 1 , n 0 = 2 , ν 1 = θ , β 1 = d/θ . By (2.13) a solutio n is u n = 3 X j =1 γ j δ n j , n ≥ 3 where { δ j , j = 1 , 2 , 3 } are the ro ots of δ 3 − ( θ + 1) δ 2 + ( θ − v 1 ) δ + ( v 1 − d ) θ = 0 and for A 2 , Q 2 of (2.16), ( γ 1 , γ 2 , γ 3 ) ′ = A − 1 2 Q 2 . An explicit s olution to a cubic is giv en in Section 3.8.2 p17 of Abr amowitz and Stegun (1964) . 5 Example 3.1 Supp ose that e 1 takes only two values, say 0 and 1. Then the observations take the values 0 , 1 , ρ, 1 + ρ and Q = sign ( ρ )  I (0 ≤ x/ρ ) p 1 , I (0 ≤ ( x − 1) /ρ ) p 2 I (1 ≤ x/ρ ) p 1 , I (1 ≤ ( x − 1) /ρ ) p 2  . The p ossible values of Q ar e ± Q i , 1 ≤ i ≤ 8 , wher e Q 1 =  p 1 p 2 p 1 p 2  , Q 2 =  p 1 p 2 0 p 2  , Q 3 =  p 1 p 2 00  , Q 4 =  0 p 2 00  , Q 5 =  0 p 2 0 p 2  , Q 6 =  p 1 0 00  , Q 7 =  p 1 0 p 1 0  , Q 8 =  p 1 p 2 p 1 0  . F or i = 1 , 3 , 5 , 6 , 7 and Q = Q i , (3.4) holds with θ = θ i , J i = Q i /θ i , θ 1 = 1 , θ 3 = p 1 , θ 5 = p 2 , θ 6 = θ 7 = p 1 . Also (3.4) holds with θ = 0 for Q 4 . Ther e ar e four c ases of ρ to c onsider. The case ρ ≤ − 1 : The observations take the values ρ < 1 + ρ < 0 < 1 . Q changes at these values of x . As x incr e ases thr ough x ≤ ρ, ρ < x ≤ 1 + ρ, 1 + ρ < x ≤ 0 , 0 < x ≤ 1 and 1 < x , Q changes fr om − Q 1 to − Q 2 to − Q 3 to − Q 4 to 0. The case − 1 < ρ < 0 : The the observations take the values ρ < 0 < 1 + ρ < 1 . As x incr e ases thr ough x ≤ ρ, ρ < x ≤ 0 , 0 < x ≤ 1 + ρ, 1 + ρ < x ≤ 1 and 1 < x , Q changes fr om − Q 1 to − Q 2 to − Q 5 to − Q 4 to 0. The case 0 < ρ < 1 : the observa tions take the v alues 0 < ρ < 1 < 1 + ρ. As x incr e ases thr ough x < 0 , 0 ≤ x < ρ, ρ ≤ x < 1 , 1 ≤ x < 1 + ρ, 1 + ρ ≤ x, Q changes fr om 0 to Q 6 to Q 7 to Q 8 to Q 1 . The case 1 ≤ ρ : The observations take the values 0 < 1 < ρ < 1 + ρ. As x incr e ases thr ough x ≤ 0 , 0 ≤ x < 1 , 1 ≤ x < ρ, ρ ≤ x < 1 + ρ, 1 + ρ ≤ x , Q changes fr om 0 to Q 6 to Q 3 to Q 8 to Q 1 . Consider the case 0 < ρ < 1 . So A iF /p i jumps fr om 0 to p 1 to 1 at x i + ρx 1 = x i and x i + ρx 2 = x i + ρ . Then ther e ar e 5 r anges of x to c onsider. x < 0 ⇒ Q = 0 , (3.4) holds with J = 0 , θ = 0 , A F = 0 , d = 0 , u n = v n = 0 for n ≥ 1 . 0 ≤ x < ρ ⇒ (3.4) holds with Q = Q 6 , θ = p 1 , J =  10 00  , A F = p 1 p, d = p 2 1 , u n = v n = p n +1 1 for n ≥ 1 . ρ ≤ x < 1 ⇒ (3.4) holds with Q = Q 7 , θ = p 1 , J =  10 10  , A F = p 1  1 0  , d = p 2 1 , u n = v n = p n 1 for n ≥ 1 . 1 ≤ x < 1 + ρ ⇒ Q = Q 8 , A F = p 1  1 p 2  , but (3.4 ) do es not hold. 1 + ρ ≤ x ⇒ (3.4) holds with θ = 1 , Q = J = Q 1 , A F = p, d = 1 , u n = v n = 1 for n ≥ 0 . Now supp ose that ρ < 0 . Then for i = 1 , 3 , 5 , 6 , 7 , Q = − Q i = θ i J i wher e θ 1 = − 1 , θ 3 = − p 1 , θ 5 = − p 2 , θ 6 = θ 7 = − p 1 . A gain, this de als with al l c ases exc ept for Q 2 , Q 4 , Q 8 . Also Q 2 4 = 0 so tha t for Q = ± Q 4 , v n = 0 for n ≥ 2 . F or 1 + ρ < x ≤ 1 , A F = (0 , − p 1 p 2 ) ′ , v 1 = − p 1 p 2 , V ( t ) = 1 − p 1 p 2 t. So by (2.9), tU ( t ) = D − 1 − 1 wher e D = 1 − tV ( t ) = (1 − p 1 t )(1 − p 2 t ) , giving u n = ( p n +2 1 − p n +2 2 ) / ( p 1 − p 2 ) for p 1 6 = 1 / 2 . So for p 1 = 1 / 2 , u n = ( n/ 2 + 1)2 − n . This il lu str ates our se c ond and most gener al metho d of solution, the use of (2.9). Final ly, the c ases Q 2 and Q 8 c an b e de alt with by the fol lowing metho d. 6 Our third s olution is in terms of the eigen v alues and left and r igh t eigen v ectors of Q , sa y { ν i , l i , r i : 1 ≤ i ≤ P } . The case of diagonal Jordan form ( for example distinct eigen v alues). In this case the P × P matrix Q has Jordan canonical form Q = R Λ R − 1 = R Λ L ′ = P X i =1 ν i r i l ′ i , R L ′ = I P , Λ = diag ( ν 1 , · · · , ν P ) , L = ( l 1 , · · · , l P ) , R = ( r 1 , · · · , r P ) . Then b y (3.2) Q n = L ′ Λ n R = P X i =1 ν n i r i l ′ i for n ≥ 0 , K n H ( y ) = P X i =1 b iH ( y ) ν n − 1 i for n ≥ 1 where b iH ( y ) = ( A ′ H r i ) ( l ′ i q ( y )) , (3.6) v n = P X i =1 β i ν n − 1 i for n ≥ 1 where β i = b iF ( ∞ ) = ( A ′ F r i ) ( l ′ i 1 ) . (3.7) So (2 .12 ) holds w ith I = P , n 0 = 1. So for n ≥ 1, if ρ > 0, then u n = v n of (3.7), and by (2.13), if ρ < 0, then u n = P +1 X j =1 γ j δ n j where { δ j } are the ro ots of P X k =1 β k ν k / ( δ − ν k ) = δ − 1 and γ is giv en by (2.17). So this metho d requ ires computing the left and right eigen v ectors of Q for its non -zero eigen v alues. In rare case s Q is symmetric so that L = R. One can show that this metho d agrees with the idemp oten t metho d w hen are b oth applicable. Example 3.2 L et us r e c onsider the pr evious example. Firstly, su pp ose that 0 < ρ < 1 . We c onsider 3 c ases. The case 0 ≤ x < ρ : Then Q = Q 6 has eige nvalues p 1 , 0 . F or ν = p 1 , l = r =  1 0  . So for n ≥ 1 , Q n = p n 1 ll ′ in agr e ement with the idemp otent metho d. The case ρ ≤ x < 1 : Then Q = Q 7 has eige nvalues 0 , p 1 . F or ν = p 1 , we c an take l =  1 0  , r =  1 1  . So for n ≥ 1 , Q n = lr ′ p n 1 in agr e ement with the idemp otent metho d. The case 1 ≤ x < 1 + ρ : Then Q = Q 8 has e i genvalues satisfying ν 2 − p 1 ν − p 1 p 2 = 0 so that 2 ν i = p 1 ± ( p 2 1 + 4 p 1 p 2 ) 1 / 2 . T ake r i =  ν i p 1  , l i =  ν i p 2  /c i wher e c i = p 1 ( ν i + 2 p 2 ) . Using ν 1 + ν 2 = p 1 , ν 1 ν 2 = − p 1 p 2 , we obtain Q n = 2 X i =1 ν n i  p 1 ( ν i + p 2 ) p 2 ν i p 1 ν i p 1 p 2  . 7 Also A ′ F = p 1 (1 , p 2 ) . So one obtains v n = 2 X i =1 ν n − 1 i β i for n ≥ 1 wher e β i = a i / ( ν i + 2 p 2 ) , a i = (1 + p 1 p 2 ) ν i + p 1 p 2 (1 + p 2 ) . Se c ond ly, supp ose that − 1 < ρ < x ≤ 0 . Then A ′ F = − ( p 2 1 , p 2 ) . Supp ose that p 1 6 = 1 / 2 . Then Q = − Q 2 , Q 2 = R Λ L ′ wher e Λ = diag ( p 1 , p 2 ) , R =  1 p 2 0 p 2 − p 1  , L ′ = R − 1 =  1 − p 2 ( p 2 − p 1 )) − 1 0 ( p 2 − p 1 ) − 1  , ( − Q ) n = R Λ n L ′ =  p n 1 p 2 a n 0 p n 2  wher e a n = ( p n 1 − p n 2 ) / ( p 1 − p 2 ) . So by (3.2 ), v n = ( − 1) n a n +2 . So 1 + tU ( t ) = (1 − tV ( t )) − 1 = (1 − ν 1 t )(1 − ν 2 t ) = 1 + t + p 1 p 2 t 2 giving u 1 = p 1 p 2 , u n = 0 for n ≥ 2 . If p 1 = 1 / 2 , then by a limiting ar gument, v n = ( − 2) − n ( n + 2) / 2 for n ≥ 0 , u 1 = 1 / 4 , u n = 0 for n ≥ 2 . Example 3.3 Supp ose tha t e 1 takes the 3 values 0, 1, 2. So the observations take the values 0 , 1 , 2 , ρ, 1 + ρ, 2 + ρ, 2 ρ, 1 + 2 ρ, 2 + 2 ρ and Q = sign ( ρ )   I (0 ≤ x/ρ ) p 1 , I (0 ≤ ( x − 1) /ρ ) p 2 I (0 ≤ ( x − 2) /ρ ) p 3 I (1 ≤ x/ρ ) p 1 , I (1 ≤ ( x − 1) /ρ ) p 2 I (1 ≤ ( x − 2) /ρ ) p 3 I (2 ≤ x/ρ ) p 1 , I (2 ≤ ( x − 1) /ρ ) p 2 I (2 ≤ ( x − 2) /ρ ) p 3   . Supp ose that 0 < ρ < 1 / 2 . Then Q changes e ach time x cr osses one of the nine values 0 < ρ < 2 ρ < 1 < 1 + ρ < 1 + 2 ρ < 2 < 2 + ρ < 2 + 2 ρ. So we ne e d to c onsider ten c ases, six of them idemp otent. The p ossible values of Q ar e ± Q i , 0 ≤ i ≤ 9 wher e Q 0 = 0 , Q 1 = p 1   1 0 0 0 0 0 0 0 0   , Q 2 = 2 p 1   1 0 0 1 0 0 0 0 0   / 2 , Q 3 = p 3   1 0 0 1 0 0 1 0 0   , Q 4 =   p 1 p 2 0 p 1 0 0 p 1 0 0   , Q 5 =   p 1 p 2 0 p 1 p 2 0 p 1 0 0   , Q 6 =   p 1 p 2 0 p 1 p 2 0 p 1 p 2 0   , Q 7 =   p 1 p 2 p 3 p 1 p 2 0 p 1 p 2 0   , Q 8 =   p 1 p 2 p 3 p 1 p 2 p 3 p 1 p 2 0   , Q 9 =   p 1 p 2 p 3 p 1 p 2 p 3 p 1 p 2 p 3   . Also Q 0 and J i = Q i /θ i ar e i demp otent for i = 1 , 2 , 3 , 6 , 9 wher e θ 1 = θ 3 = p 1 , θ 2 = 2 p 1 , θ 6 = p 1 + p 2 , θ 9 = 1 . Case 1: x < 0 ⇒ Q = 0 , A F = 0 , v n = 0 for n ≥ 1 . Case 2: 0 ≤ x < ρ ⇒ Q = Q 1 , A ′ F = ( p 2 1 , 0 , 0) , d = p , 1 v n = p n +1 1 for n ≥ 1 . Case 3: ρ ≤ x < 2 ρ ⇒ Q = Q 2 , A ′ F = ( p 1 ( p 1 + p 2 ) , 0 , 0) , d = p 1 ( p 1 + p 2 ) / 2 , v 1 = p 1 ( p 1 + p 2 ) , v n = (2 p 1 ) n − 1 d for n ≥ 2 . Case 4: 2 ρ ≤ x < 1 ⇒ Q = Q 3 , A ′ F = ( p 1 , 0 , 0) , v n = p n 1 for n ≥ 1 . 8 Case 5: 1 ≤ x < 1 + ρ ⇒ Q = Q 4 . Q 4 has distinct eigenvalues 0 , ν 2 = ( p 1 + ε ) / 2 , ν 3 = ( p 1 − ε ) / 2 wher e ε = ( p 2 1 + 4 p 1 p 2 ) 1 / 2 . Also A ′ F = ( p 1 , p 1 p 2 , 0) , 2 ε r ′ 2 = ( p 1 + ε, 2 p 1 , 2 p 1 ) = B ε say , 2 p 1 l ′ 2 = (2 p 1 , − p 1 + ε, 0) = C ε say , 2 ε r ′ 3 = − B − ε , 2 p 1 l ′ 3 = C − ε . So by (3.7), for n ≥ 1 , v n = P 3 i =2 β i ν n − 1 i wher e β 2 = p 1 b ε c ε / (4 ε ) , β 3 = − p 1 b − ε c − ε / (4 ε ) wher e b ε = p 1 + 2 p 1 p 2 + ε, c ε = 2 p 1 − p 1 p 2 + p 2 ε. Case 6: 1 + ρ ≤ x < 1 + 2 ρ ⇒ Q = Q 5 . This is the only example her e wher e the ge ne r al Jor dan form is ne e de d. Case 7: 1 + 2 ρ ≤ x < 2 ⇒ Q = Q 6 = 1 l ′ = θ 6 J 6 say, wher e l ′ = ( p 1 , p 2 , 0) , θ 6 = p 1 + p 2 , J 2 6 = J 6 . Also A ′ F = ( p 1 , p 2 , 0) . So by (3.5 ), v n = ( p 1 + p 2 ) n for n ≥ 1 . Case 8: 2 ≤ x < 2 + ρ ⇒ Q = Q 7 . Q 7 has distinct eigenvalues 0 , ν 2 = ( p 1 + p 2 + ε ) / 2 , ν 3 = ( p 1 + p 2 − ε ) / 2 wher e ε = (( p 1 + p 2 ) 2 + 4 p 1 p 3 ) 1 / 2 . Also A ′ F = ( p 1 , p 2 , p 1 p 3 ) , 2 p 1 ε r ′ 2 = ( a ε , b ε , b ε ) wher e a ε = p 1 p 2 + 2 p 1 p 3 + p 2 2 − p 2 ε, b ε = p 1 ( p 1 + p 2 − ε ) , 2 p 1 p 3 l ′ 2 = ( p 1 c ε , p 2 c ε , 2 p 1 p 3 ) wher e c ε = p 1 + p 2 + ε, 2 p 1 ε r ′ 3 = ( − a − ε , b − ε , b − ε ) , 2 p 1 p 3 l ′ 3 = ( p 1 c − ε , p 2 c − ε , 2 p 1 p 3 ) , so that (3.7) for n ≥ 1 , v n = P 3 i =2 β i ν n − 1 i wher e β 2 = B 2 ε C ε / (4 p 2 1 p 3 ε ) , β 3 = B 3 − ε C − ε / (4 p 2 1 p 3 ε ) , wher e B 2 ε = p 1 a ε + ( p 2 + p 1 p 3 ) b ε , B 3 − ε = − p 1 a − ε + ( p 2 + p 1 p 3 ) b − ε , C ε = ( p 1 + p 2 ) c ε + 2 p 1 p 3 . Case 9: 2 + ρ ≤ x < 2 + 2 ρ ⇒ Q = Q 8 . Q 8 has distinct eige nvalues 0 , ν 2 = ( p 1 + p 2 + ε ) / 2 , ν 3 = ( p 1 + p 2 − ε ) / 2 wher e ε = δ 1 / 2 , δ = ( p 1 + p 2 )( p 1 + p 2 + 4 p 3 = 1 + 3 p 3 ) . By (3.7) f or n ≥ 1 , v n = P 3 i =2 β i ν n − 1 i wher e A ′ F = ( p 1 , p 2 , p 3 ( p 1 + p 2 )) , 2( p 1 + p 2 ) εr ′ 2 = ( p 1 ( p 1 + p 2 + ε ) , p 1 ( p 1 + p 2 + ε ) , p 1 ) = B ε say , 2 p 1 l ′ 2 = (2 p 1 , 2 p 2 , − p 1 − p 2 + ε ) = C ε say , 2( p 1 + p 2 ) ε r ′ 3 = − B − ε , 2 p 1 l ′ 3 = C − ε . Case 10: 2 + 2 ρ ≤ x ⇒ Q = Q 9 . As note d, (3.4 hol ds with θ 9 = 1 . Also Q 9 1 = 1 , A ′ F = (0 , 0 , p 3 ) , d = p 3 . So by (3.5), v n = p 3 for n ≥ 1 . This le aves only Q 5 to de al. It wil l b e de alt with by the fol lowing metho d. 9 Our th ir d and general solution is in terms of the eigen v alues and left and righ t gener alize d eigen- v ectors of Q , say { ν i , l i , r i : 1 ≤ i ≤ P } . The general Jordan form. The general Jord an canonical form for a q × q m atrix Q is Q = R Λ R − 1 = R Λ L ′ = r X i =1 R i J m i ( ν i ) L ′ i , (3.8) where RL ′ = I , L = ( L 1 , · · · , L r ) , R = ( R 1 , · · · , R r ) , Λ = diag( J m 1 ( ν 1 ) , · · · , J m r ( ν r )) , and L i , R i are q × m i , J m ( ν ) = ν I m + U m , and U m is the m × m matrix with zeros ev erywher e except for ones on the diagonal ab ov e the leading diago n al: ( U m ) ij = δ i,j − 1 : J 1 ( ν ) = ν, J 2 ( ν ) =  ν 1 0 ν  , J 3 ( ν ) =   ν 1 0 0 ν 1 0 1 ν   , · · · J m ( ν ) has only one righ t eigen v ector. The i th blo c k in QR = R Λ is QR i = R i J m i ( ν i ). T aking its j th column gives Qr ij = ν i r ij + r i,j +1 , j = 1 , · · · , m i , wh ere r i,m i +1 = 0 and r ij is a q -v ector. S o one first computes th e right eigenv ector r i,m i and then the gener alize d eigen v ectors r i,m i − 1 , · · · , r i 1 recursiv ely , the Jordan c hain. F or n ≥ 0, the n th p o w er of Q is J m ( ν ) n = min( n,m − 1) X k =0  n k  ν n − k U k m , (3.9) Q n = R Λ n R − 1 = r X i =1 R i J m i ( ν i ) n L ′ i = r X i =1 min( n,m i − 1) X k =0  n k  ν n − k i W ik (3.10) where W ik = R i U k m i L ′ i . where U k m is the m × m matrix with zeros eve ryw here except f or ones on th e k th su p er-diagonal: ( U k m ) ij = δ i,j − k . S o U m m = 0. So Q n is a m atrix p olynomial in n of degree m − 1 where m = max r i =1 m i , and by (3.2), v n = r X i =1 min( n,m i ) − 1 X k =0  n − 1 k  ν n − 1 − k i w ik for n ≥ 1 where w ik = A ′ F W ik 1 . (3.11) F or diagonal Jord an form this reduces to (2.12). This lev el of generalit y is not needed if the eigen v alues of the non-diagonal Jordan blo c ks are zero, since we can rewrite (3.10) and (3.11) as Q n = X 1 ≤ i ≤ r, ν i 6 =0 min( n,m i − 1) X k =0  n k  ν n − k i W ik + X 1 ≤ i ≤ r, ν i =0 I ( n < m i ) W in , n ≥ 0 , (3.12) v n = X 1 ≤ i ≤ r, ν i 6 =0 min( n,m i ) − 1 X k =0  n − 1 k  ν n − 1 − k i w ik + X 1 ≤ i ≤ r, ν i =0 I ( n ≤ m i ) w i,n − 1 , n ≥ 1 . (3.13) 10 The solution for u n is a wei ghted su m of n th p o we rs where the weigh ts are constant s or p olynomials in n : see Withers and Nadara jah (200 9b) for details. Example 3.4 This c ontinues Case 6 of Example 3.3. Set q i = p i / ( p 1 + p 2 ) . By A pp endix A, (3.8) holds for Q = Q 5 with r = 2 , m 1 = 2 , m 2 = 1 , ν 1 = 0 , ν 2 = p 1 + p 2 , R 1 =   0 q 2 0 − q 1 q 2 − q 2 1   , R 2 = q 1   1 1 q 1   , L ′ 1 =  0 − p 1 /p 2 1 + p 1 /p 2 1 − 1 0  , L ′ 2 =  1 p 2 /p 1 0  . Also J 2 (0) n = 0 for n ≥ 2 . So Q n = ( p 1 + p 2 ) n R 2 L ′ 2 = ( p 1 + p 2 ) n − 1   p 1 p 2 0 p 1 p 2 0 p 1 q 1 p 2 q 1 0   for n ≥ 2 . Also A ′ F = ( p 1 , p 2 ( p 1 + p 2 ) , 0) . So by (3.2), v n = ν n − 1 v 1 for n ≥ 1 wher e ν = p 1 + p 2 , v 1 = p 1 + p 2 ( p 1 + p 2 ) . As note d, this is the only example her e wher e the gener al J or dan form is ne e de d. (It was implicit in the 2nd p art of Example 3.2 for the c ase p 1 = 1 / 2 , but was byp asse d by the limiting ar gument.) So V ( t ) = 1 + v 1 t/ (1 − ν t ) , 1 − tV ( t ) = L/ (1 − ν t ) , L = (1 − t )(1 − ν t ) − v 1 t 2 = (1 − tt 1 )(1 − t t 2 ) , t k = ( ν + 1 ± δ 1 / 2 ) / 2 , δ = ( ν + 1) 2 − 4( ν − v 1 ) = p 3 3 + 4 v 1 , 1 + t U ( t ) = (1 − tV ( t )) − 1 = (1 − ν t ) /L = P 2 k =1 c k / (1 − tt k ) , c k = (1 − ν /t k )(1 − t 3 − k /t k ) giving for ρ < 0 , u n − 1 = 2 X k =1 c k t n k , n ≥ 1 . So although m 1 = 2 , a sum of p owers solution stil l holds. As n oted f or the diagonal form, Q and its eigen v ectors only change v alue when x crosses one of the p ossible observ ation v alues, or equiv alentl y wh en ρ crosses one of the set { ( x − x j ) /x i , 1 ≤ i, j ≤ P } . If x is replaced b y x n , then aga in Q and its eig env alues do not dep end on n or x n except through its range. F or more on Jordan forms, see for example http: //en.wiki pedia.org /wiki/Jordan normal form Note 3.1 The singu lar values of J m ( ν ) ne e de d for its singular value de c omp osition, SVD, ar e quite differ ent fr om its eig e nvalues, which ar e al l ν . The SV D of Q give s an alternative form for its inverse, but is of no use in c omputing Q n . 4 A solution for G n ( y ) . A solution for G n ( y ) = P r ob ( M n ≤ x, e n ≤ y ) ma y b e of interest. If ρ > 0 then a solution is giv en b y (2.3): G n ( y ) = a n ( y ) where a n ( y ) = K n F ( y ) , n ≥ 1 , 11 Supp ose that Q of (3.1) has diago nal Jordan form. Th en by (3.6), a n ( y ) = A ′ F Q n − 1 q ( y ) = P X i =1 b iF ( y ) ν n − 1 i where b iF ( y ) = ( A ′ F r i ) ( l ′ i q ( y )) for n ≥ 1 . No w supp ose that ρ < 0. Then a solution is giv en by (2.5): G n +1 ( y ) = a n ( y ) ⊗ u n + a n +1 ( y ) , n ≥ 0 . Also b y Secti on 2, (2.12) holds w ith n 0 = 1: v n = I X j =1 β j ν n − 1 j for n ≥ 1 where β j = b j F ( ∞ ) = ( A ′ F r i ) ( l i 1 ) , so that by (2.13) u n = I ′ X j =1 γ j δ n j for n ≥ 1 . Substituting w e obtain u s ing (2.13), G n +1 ( y ) = 2 P X i =1 b iF ( y ) ν n − 1 i − I ′ X j =1 γ j c j ( y ) δ n +1 j for n ≥ 1 (4.1) where c j ( y ) = P X i =1 b iF ( y ) ν − 1 i ( ν i − δ j ) − 1 . (4.2) So its b eha viour for large n is d etermined b y the ν i or δ j of largest mo d ulus. A App endix: a MA PLE program to find the Jordan form. Here is the MAPLE pr ogram used to work out the J ordan form for Q = Q 5 /p 1 in Example 3.4. It can easily b e adapted to other examples. W e set c = p 2 /p 1 , M = L ′ . with(Lin earAlgebr a); Q:=Matri x(3,3,[[1 ,c,0],[1,c,0],[1,0,0]]); JordanFo rm(Q); R:= JordanFo rm(Q, outpu t=’Q’) ; M:=R^(-1 ); J:=simpl ify(M.Q.R ); quit; The last line is just to confirm that Q = RJ M . References [1] Abramowit z, M. and Stegun, I .A. (1964 ) Handb o ok of mathematic al functions. U.S. Depart- men t of Commerce, Nat ional Bureau of Standards, App lied Mathematics Series 55 . 12 [2] Withers, C. S. and Nadara jah, S. N. (2009a) Prepr in t. The distribution of the maxim u m of a first ord er mo ving a verag e: the con tinuous ca se. ht tp://arxiv.org/abs/0802 .0523 [3] Withers, C. S . and Nadara jah, S . N. (2 009b) Solutions to a conv olution p roblem. Subm itted. 13