A PROBABILITY PROBLEM

Transcription

1 A PROBABILITY PROBLEM A big superarket chai has the followig policy: For every Euros you sped per buy, you ear oe poit (suppose, e.g., that = 3; i this case, if you sped 8.45 Euros, you get two poits, if you sped 9. Euros, you get three poits, while if you sped 2.85 Euros, you get zero poits). After you collect a certai uber of poits, you ca redee the for a gift certificate. What is the actual price β of a poit? Geeral Discussio. The aout X spet per buy ca be cosidered a positive rado variable with a distributio fuctio F (x) ad fiite ea µ X = E[X] >. Recall the well-kow forula µ X = P {X x}dx = P {X > x}dx = which is valid for ay oegative rado variable X. [1 F (x)] dx, (1) If F ( ) := li x F (x) = P {X < } = 1, the it is ipossible to ear poits ad we ay say that i this case β =, which is totally urealistic ad uiterestig. Hece, i order for our proble to be eaigful we should at least have F ( ) < 1. Actually, fro ow o we will ake a slightly stroger assuptio, aely F () = P {X } < 1. (2) If oe visits a store ad speds X Euros, the uber of poits (s)he ears is where x deotes the greatest iteger x. Thus, Sice Y, forula (4) ca be slightly iproved: Y = X/, (3) X 1 < Y X. (4) ( ) + X 1 = (X )+ Y X, (5) where (x) + = ax{x, }. By takig expectatios (5) yields E [ (X ) +] 1 µ Y µ X, (6)

2 where µ Y = E[Y ]. Sice E [ (X ) +] = (x ) + df (x) = (x )df (x) = P {X > x}dx, (7) we have, i view of (2), E [ (X ) +] >. (8) Let us, also, otice that it follows easily fro (7) that E [ (X ) +] (µ X ) +. (9) Furtherore, equatio (3) iplies that µ Y = k P {Y = k} = k P {k X < (k + 1)} = P {X k}, (1) where the last equality follows by suatio by parts (otice, also, that P {X k} = 1 F (k )). Now, let X 1, X 2,..., X be the aouts spet per buy by a rado group of cosuers. We assue that these aouts are idepedet copies of X, aely that {X k } 1 k is a collectio of idepedet rado variables with coo distributio fuctio F (x). For these buys, the total aout spet is X 1 + X X (11) ad the total uber of poits eared is Y 1 + Y Y, where Y k = X k /, k = 1, 2,...,. (12) It is, the, reasoable to ifer that the actual value of a poit is β = li X 1 + X X Y 1 + Y Y. (13) Dividig the uerator ad the deoiator of the above fractio by ad ivokig the Law of Large Nubers, we obtai fro (13) that β = µ X µ Y. (14) Reark 1. The liit i (13) exists alost surely. Actually, for the a.s. existece of the liit it is eough to assue that {X k } is a sequece of pairwise idepedet rado variables with coo distributio fuctio F (x) (this is Eteadi s versio of the Strog Law of Large Nubers see, e.g., [1]). 2

3 Forulas (6) ad (9), applied to (14), give us soe bouds for β, aely β µ X E [ (X ) +] µ X (µ X ) + = ) +. (15) (1 µx Of course, the lower boud of β give i (15), aely β, is trivial. A equivalet way to write (15) is 1 λ µ X E [ (X ) +] 1 ) +, where we have set λ := (1 β. (16) µx Thus, if µ X is cosiderably larger that, the λ is close to 1, i.e. β is close to. Reark 2. (a) Suppose we have a whole faily (or a sequece) {F K (x)} of distributio fuctios satisfyig our basic assuptios (i) F K () =, (ii) F K () < 1, ad (iii) the expectatio associated to F K (x) is fiite, i.e. [1 F K (x)] dx <. Uder this setup, forula (16) iplies that if li µ X =, the li λ = 1. (17) K K (b) I the case where our faily {F K (x)} is such that li K F K ( ) = li K P {X < } = 1, oe ight expect that λ (i.e. β ). However, it is easy to fid exaples where λ stays fiite. Actually, we ca eve have li K µ X =, which by (17) will iply that λ 1. For exaple, for each K choose F K (x) so that P {X < } = 1 (2/K) ad P {X > K 2 } = 1/K. The, it is obvious that li K P {X < } = 1, while µ X K ad hece li K µ X = (thus, (17) iplies li K λ = 1). I fact, eve if, li K µ X = (which is, clearly, stroger tha li K P {X < } = 1), we have that the liit of λ, if it exists, ca take ay value 1 (icludig ). To see a exaple let us for coveiece take = 1: If F K (x) = (1 K 1 )1 [K 2, 1+K 2 )(x) + 1 [1+K 2, )(x), the li K λ = 1 (where 1 I (x) deotes the idicator fuctio of the iterval I), while if F K (x) = (1 K 2 )1 [K 1, 1+K 1 )(x) + 1 [1+K 1, )(x), the li K λ =. The Gaa Case. I order to get a ore precise estiate for β, we eed to have ore iforatio regardig the distributio of X. For istace, a plausible assuptio is that X follows a Gaa distributio with paraeters a > ad p >, aely that its probability desity fuctio is f(x) = 1 Γ(p) (ax)p 1 ae ax, x > (18) 3

4 (of course, f(x) = for x < ), where Γ( ) is the Gaa fuctio. Recall that, sice p >, we have Γ(p) = ξ p 1 e ξ dξ (19) ad that if X has the Gaa desity f(x) give by (18), the µ X = p a. (2) Therefore, i view of (14), i order to deterie the price β we have to calculate µ Y. Usig (18) i (1) yields or µ Y = 1 Γ(p) k (k+1) k (ax) p 1 ae ax dx = 1 Γ(p) µ Y = 1 Γ(p) By substitutig (2) ad (22) i (14) we obtai β = a ka which i ters of the ratio λ itroduce i (16) becoes Observe that (17) tells us that ad, i particular, i view of (2), k (k+1)a ka ξ p 1 e ξ dξ (21) ξ p 1 e ξ dξ. (22) pγ(p) ka ξp 1 e ξ dξ = Γ(p + 1) a ka ξp 1 e ξ dξ, (23) λ(a, p) := λ = β = Γ(p + 1) a ka ξp 1 e ξ dξ. (24) µ X iplies λ(a, p) 1 (25) li λ(a, p) = 1 ad li λ(a, p) = 1. (26) a + p It is rearkable that the above liits are ot so obvious fro (24). Lookig at forula (24) it sees that, uless we have specific uerical values for the paraeters, a, ad p, it is ot clear how to extract a precise estiate for the ratio λ(a, p). For this reaso, the qualitative behavior of λ(a, p) has soe iterest. Cojecture. The fuctio λ(a, p) of (24) is (i) strictly icreasig i a ad (ii) strictly decreasig i p. Furtherore, li a λ(a, p) = ad li λ(a, p) =. (27) p + 4

5 I a attept to obtai a forula for λ(a, p) which is ore explicit tha (24), let us suppose that X follows a Erlag distributio, aely The (24) becoes where we have set We clai that λ(a, ) = a F (x) := p = N := {1, 2,...}. (28)! ka ξ 1 e ξ dξ = 1 ( 1)! F (x) = S (x)e x, where S (x) := x a F (ka), (29) ξ 1 e ξ dξ,. (3) 1 l= x l l!. (31) For = 1 forula (31) is clearly true. To justify (31) for 2 we first observe that the defiitio (3) of F (x) iplies x 1 F (x) = ( 1)! e x ad F () = 1. (32) Now, fro the defiitio (31) of S (x) we have S (x) = S 1 (x) ad S () = 1. (33) Therefore, F (x) ad S (x)e x agree at x = ad by applyig (33) ad the defiitio of S (x) we get [ S (x)e x] = S (x)e x S (x)e x = S 1 (x)e x S (x)e x = x 1 ( 1)! e x. (34) By coparig (34) with (32) we get the validity of (31). Usig ow (31) i (29) yields λ(a, ) = a S (ka)e = ka a 1 l= (ka) l l! e ka (35) or, by iterchagig the order of suatio i the deoiator of the last expressio where we have set T l (x) := λ(a, ) = a 1 l= (a) l l! T l (a), (36) k l e kx, x >, l =, 1, 2,... (37) 5

6 (icidetally, T l (x) of (37) akes sese for ay coplex uber l, actually it is etire i l; also, for ay real l, T l (x) is strictly decreasig i x o (, ), with T l ( ) = ; fially, if l 1, the T l ( + ) =, while if l < 1, the T l () = ζ( l)). Fro (37) we have ad Therefore, T (x) = ad equatio (36) ca be writte as λ(a, ) = e kx = e x 1 e = 1 x e x 1 (38) T l+1 (x) = T l (x). (39) T l (x) = ( 1) l T (l) (x) (4) a 1 l= ( a) l T (l) l! (a). (41) However, fro (41), uless we specify it is still ot clear how λ varies with, a, ad. There is soethig peculiar i forula (41). If we cosider the Taylor polyoial Q (x; x ) := 1 l= (x x ) l l! T (l) (x ) (42) associated to T (x), the the su appearig i the deoiator of (41) is Q (; a), while T ( ± ) = ±. Fially, let us otice that by straightforward iductio we ca show where q (z) 1 ad T (l) (x) = ( 1) l q l (e x ), l =, 1,..., (43) (e x 1) l+1 q l+1 (z) = (l + 1)zq l (z) z(z 1)q l(z) l =, 1,.... (44) It is easy to see that q l (z) is a oic polyoial of degree l with q l () = ad q l () = 1 for all l 1. Furtherore, the coefficiets of q l (z)/z are strictly positive for all l 1 (also, q l (1) = l! ad q l (1) = (l + 1)!/2 for all l 1). I particular, we have q 1 (z) = z, q 2 (z) = z 2 + z, q 3 (z) = z 3 + 4z 2 + z, q 4 (z) = z z 3 + 9z 2 + z. (45) Exaple 1. Suppose = 1, so that X is expoetially distributed with paraeter a. The (41) yields λ(a, 1) = ea 1 a = 1 + a (a) ( + 2)!. (46) 6 =

7 Notice that λ(a, 1) is a icreasig fuctio of a. Furtherore, λ(a, 1) as a (equivaletly as µ X ), while λ(a, 1) 1 as a (equivaletly as µ X ), as expected. Exaple 2. Suppose = 2, so that X is Erlag with paraeters a ad 2. The (41) yields λ(a, 2) = a ea 1 a 1 e a = 2λ(a, 1) 1 + a 1 e a. (47) x Sice is strictly icreasig for x > (ad, hece, > 1), it follows fro (47) that 1 e x λ(a, 2) < λ(a, 1) ad, furtherore, that λ(a, 2) is strictly icreasig i a ad approaches as a (equivaletly as µ X ), while λ(a, 2) 1 as a (equivaletly as µ X ), as expected. Exaple 3. Suppose = 3, so that X is Erlag with paraeters a ad 3. The (41) yields λ(a, 3) = 1 + a 1 e a [ a 1+e a 1 e a 2 ] ea 1 a = 3λ(a, 1) [ 1 + a 1 e 1 + a a 1+e a 1 e a 2 (48) 1+e x is strictly icreasig for x > (ad, hece, > 1), it follows fro (47) 2 x Sice 1 e x ad (48) that λ(a, 3) < λ(a, 2) ad, furtherore, that λ(a, 3) is strictly icreasig i a ad approaches as a (equivaletly as µ X ), while λ(a, 3) 1 as a (equivaletly as µ X ), as expected. ]. Refereces [1] R. Durrett, Probability: Theory ad Exaples, Third Editio, Duxbury Advaced Series, Brooks/Cole Thoso Learig. Belot, CA, USA, 25. 7