UllHllUllll'lllllfllllllllllllllll i 1293 02058 6206 This is to certify that the dissertation entitled Super-Replication of EurOpean Exotic Options presented by Chanho Park has been accepted towards fulfillment of the requirements for Ph-D- degree in StatiStiCS film flaw Major professor Shlomo Levental [hue December 9, 1999 MS U is an Affirmative Action/Equal Opportunity Institution 0-12771 PLACE IN RETURN BOX to remove this checkout from your record. TO AVOID FINE return on or before date due. MAY BE RECALLED with earlier due date if requested. DATE DUE DATE DUE DATE DUE 11:00 mm.“ SUPER-REPLICATION OF EUROPEAN EXOTIC OPTIONS By Chanho Park A DISSERTATION Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Statistics and Probability 1999 ABSTRACT SUPER-REPLICATION OF EUROPEAN EXOTIC OPTIONS By Chanho Park “- We study the continuous time problem of hedging a European style Asian call option in the presence of transaction costs. Under the assumption that the price process of the relevant stock both fluctuates and does not fluctuate with positive probability, we find a portfolio that super-replicates the option. Most important, we prove that the portfolio that we found is optimal in the sense that it requires the smallest initial investment among all the super-replicating portfolios. ACKNOWLEDGMENTS I would like to extend gratitude and thanks to my adviser Professor Shlomo Levental for his guidance and his help with my English. I also thank my thesis committee members for careful reading of my thesis, which is hard to read. iii TABLE OF CONTENTS INTRODUCTION ........................................................... 1 Chapter 1 The model, basic definitions and main results ........................ 4 Chapter 2 Proof of the theorem about the option ................................ 10 APPENDIX 22 A.1 The tools of the proof .............................................. 22 A.2 Equivalent forms of Assumption 1.8 ............................. 55 A3 Two simple inequalities ............................................ 58 BIBLIOGRAPHY .......................................................... 59 INTRODUCTION In their fundamental paper, Black and Scholes (1973) discovered how to price options in continuous—time financial markets where the stock price follows a geometric Brownian motion and the market is free of transaction costs. By “option,” we mean here a contract between a buyer and a seller whose value at some future date, the “exercise time,” will be equal to a given function of the underlying stock. The value of the option when it will be exercised will be transferred from the seller to the buyer. For the right to receive that transfer of wealth in the future, the buyer pays the seller a certain amount of money which is the option price. The main idea in Black and Scholes (1973) is that the option price should be the exact difference between the value of the option at the exercise time and the “capital gain” achieved from some “replicating portfolio.” This replicating portfolio is based on the underlying stock and money market account. By using the replicating portfolio, the seller is able to “hedge” his or her liability; namely, the seller will not lose any money from the option contract. The main problem in the Black and Scholes theory is that the replicating portfolio demands continuous trading. This makes the theory not practical in the presence of transaction costs that are proportional to the monetary value of the trades. The replicating portfolio will create an infinite amount of trading and hence an infinite amount of transaction costs. The transaction costs are called “two-sided” when they are being charged in both buying and selling of shares. They are called “one-sided” when they are being charged only in buying shares (or only in selling shares). We only consider the two—sided transaction costs case in this thesis. It was discovered in Bensaid, Lense, Pages, and Scheinkman (1992), in the context of a discretetime model, that if the requirement of exact replication portfolio is relaxed, it is sometimes possible to lower the option price. That is why we will only require here that the hedging portfolio will dominate almost surely (“super-replicate”) the value of the option at payoff time. This is, of course, enough protection from the seller’s point of view so that is why we will not deal with exact replication in this thesis. Davis and Clark (1994) has formally conjectured the “conventional wisdom” concerning hedging of options in the presence of transaction costs. More precisely, they conjectured that the only possible way to hedge a European style call option is by a trivial hedging portfolio: buy one share and hold it till expiration day. Soner, Shreve, and Cvitanic (1995) have proved the conjecture in a setup where the stock price is modeled by a geometric Brownian motion. In their proof they have used some ideas from convex function theory. We like their proof and we believe that their methods can be applied to other problems as well. An example would be a problem where the super-replication requirement is relaxed. Levental and Skorohod (1997) (hereforth referred to as LS) deal with both generalized American and European style call options. They require only that the stock price will be modeled by a non-degenerate, continuous, positive semimartingale rather than by a geometric Brownian motion. LS only use only fundamental properties of stochastic integrals of continuous semimaningales. It is very exciting from the mathematical point of view, but their approach has its limitations. Their method works well only in super-replication context and it will be less useful when this requirement is relaxed. Finally Civtanic and Karatzas (1996) state a general result about the minimal price that is needed to super-replicate European style options. They found essentially that this price is the supremum of the expected discounted value of the claim with respect to all equivalent probability measures under which all portfolios are superrnartingales. In our model we use the same model as LS with one difference. We use a stronger assumption on the fluctuations of the stock price process than the one used by LS. The option that we use is “European style Asian call option” which has a payoff similar to the classical call option with one difference: The stock price at payoff time is being replaced by averaging the stock price throughout the option’s lifetime. Our achievement is that we find a portfolio that super-replicate the option and, most important, we prove that among all possible hedging portfolios it is the one that requires the smallest initial investment. Chapter 1 The model, basic definitions and main results We consider a financial market in which one stock is traded in the time interval 0 S t S 1. The price of this stock is represented by a stochastic process Z = {Z(t): O S t S 1}, which is defined on a complete probability space ((2, F, P). Assume that Z is a continuous semimartingale with respect to a filtration { F ,2 O _<_ t S 1} that is right continuous, and such that F, contains all P null sets, 0 S I S 1, and F0 is the trivial o- algebra. Since Z represents a price of stock, we will assume that Z is strictly positive process. For simplicity we will assume that Z(O) = 1. We will assume that the interest rate equals 0 in our model. Since one can always work with discounted price, rather than the actual ones, this entails no loss of generality. We want to assume that Z both fluctuates and does not fluctuates with positive probability. We will give a precise definition later in this chapter. Definition 1.1 A portfolio is an adapted stochastic process N = {N(t): 0 S t S 1}, which has almost surely (a.s.) left limit and right continuous sample paths, and satisfies (1.2) P(j|dN|(z) < oo) = 1. We denote the class of all portfolios by F V. For N e F V, We define two processes N + and N _, which are associated with N: 1 N(0) + N(t)+ j |dN|(s) N+(t)= 2 ° , N(O)—N(t)+ j |dN|(s) N‘(t)= 2 ° The process M and N - are nondecreasing as. and satisfy: (1.3) N =N+— N‘, lle =dN+ + dN', N +(O) = MO), and N '(O) = O. The process N +(t) (respectively N '(t)) represents the accumulated number of shares that the owner of the N portfolio has bought (sold) up to time t, and N(I) represents the number of shares in the account at time I. Let 0 < A < 1, O < u < 1. In what follows A, respectively p, represents the fractional transaction costs when one is buying, respectively selling, shares. Remark: It is assumed that no transaction costs are being paid due to holding of N(O) shares at time I = 0. The accumulated capital gain generated by a portfolio N is a stochastic process {S~(t): O <15 1} defined by, (1.4) SN (1) = :[N(s)dZ(s) — A ]Z(s)dN* (s) - u ]Z(s)dN' (s). The financial interpretation of (1 .4) is the following: N(t) dZ(t) represents instant gain (lost) of the portfolio due to the change of the share price dZ(t), while 7» Z(t) dM(t) (u Z(t) (IN—(0) represents the transaction cost paid at time t due to buying (selling) of (IMO) (respectively, dN "(t)) shares. We describe now the European type Asian call option that we deal with in this paper. This option is a contract between two persons: a seller and buyer. The option can be exercised only at time t = 1. At that time the seller has the obligation to pay the buyer g( [2mm where g(x) = ((1 — #) x — q): where q is a given positive number. The function g is called the payoff function of the option. From now on, when we mention option, we mean the option that we have just defined. Remark. We find it easier to work with the option above. However some readers will 1 think that the payoff function g( I Z (t)dt) with g(x) = (x — q)+ is a more natural choice. 0 Those readers are directed to the Corollary 1.9 at the end of this chapter. The question that we are asking here is, what price should the seller charge the potential buyer at t = O, for the right to own the option? The idea is that the seller will charge the minimal amount of money that will allow him or her to hedge their liability. This means that the seller will create a portfolio skillfully. This portfolio’s capital gain at 1= 1 plus the money received from the buyer at t = 0 will be at least as large as the payment that the seller has to transfer to the buyer at t = 1. In that way there is a certainty that seller will not lose any money. More precisely we define for each N eFV. l (1.5) xN = inf{x e R :x + S~(1) _>_g( jzmao} 0 If the set is empty, M will be taken to be 00. We define the selling price of the option to be (1.6). bg=inf{x~:Ne FV}. Observe that b5 5 1. To show this we take M!) = l — t (so N +(t) = 1 and N " (t) = t) and we get g( (:jzmdz) = {(1- p) (:jzuylz — W 30— MJZUW’ = JZ(t)dt — u JZUW = [(1-1)d2(z)+1— p jZ(t)dN’(t) = l + SN(1). So we conclude that b5 5 1. In this paper we will assume that the stock price Z(t), in addition to being a positive semimartingale, also satisfies some extra assumptions. To define them precisely, we need some notation. For every 0 < d S l and 8, y > O and stopping time T < d, we define the following stopping time. inf{T s: s d : 2(1) = e'5 Z(T) or 2(1) = e7 Z(T)‘, d if no such 1 exists. (1.7) T5, ={ We denote T: = T5“, ,TM = T5", and T8 = Ta'. The following basic assumption on Z will hold throughout this paper. Assumption 1.8 For every 5 > 0, d > O and stopping time 0 S T S 1, the following holds (1) On the event {T < d} we have 3.5. Pa," < (1,205"): 2(T)e‘5 /FT ) > 0, Pa," < d,2(Tg’) = 2(T)e6 /FT) > 0. (ii) P(T8 =1 /FT)> 0. It will be convenient for us to modify Assumption 1.8. We will state here an equivalent form of Assumption 1.8. In the proof of our main theorem we will use this equivalent form. We state it here as Lemma A.25. The proof of this lemma will appear in the Appendix. Lemma A.25 If Assumption 1.8 is satisfied, then for every 0 < a < 1, there exists 8°(e) > O that satisfies: e250 S (l — 28) / (l — 3c), 1 — 635° S u, e350 — l S 2. so that for every stopping times 0 S T S e, and 1 — e S r < l, we have (i) P(T5o = 1 /FT) > 0 as, (ii) Paw < 1, 2mg) = e‘B 2(1) / F.) > 0 as, where (3 = —[log(l — p) + 5°], and (iii) P(1 — e 3 T50 < 1, 2(T50) = 55° Z(T) / FT) > 0 as. Finally, we state the main result of this thesis. Theorem 2.1 If Assumption 1.8 is satisfied then b5 = 1. We already saw that b); S 1. We need then to show that b); 2 1. The idea behind the proof of that is simple but the details are complicated. First we create a discrete version of the problem. We are doing it in Lemma A.1. Then we need to know how to handle that discrete version. This is done with the help of Lemma A.6. The actual proof the theorem will appear in Chapter 2. The proofs of all the lemmas that we use during the proof of the theorem in Chapter 2 will appear in the Appendix. 1 Corollary 1.9 Let the payoff function be: ( IZ(t)dt— q)+. Under Assumption 1.8 we have 0 1 b5:— 1— u ' Proof. We have (IZ(t)dt— q)+= -1—]—— {(1 — u) IZ(t)dt— (l — u) q}+. o - [.1 0 Now observe that for every random variable H 2 O and constant a 2 0 we have bE(aH) = a bE(H), where bE(H), the selling price of a European option with payoff H, is l defined by (1.5) with H replacing g( [2(z)dz) and by (1.6). By taking 0 l H = {(1 — 11) IZ(t)dt— (l — u) q}+ , we see that it is enough to prove that bE(H) = l but 0 . this follows form Theorem 2.1 when we use (1 — p) q instead of q. Chapter 2 Proof of the theorem about the option In this chapter we will prove that b5 2 1. This will be achieved by proving that xN 2 l — e for any N e F V where O < e < 1 is arbitrary. This implies that xN 2 l and since N is arbitrary we conclude that b5 2 1. In Chapter 1 we have already proved that b5 S 1. Putting the two together gives b); = 1. We will quote here four lemmas (A. l , A3, A5 and A.6) that are essential to our proof. The proof of these lemmas will be given in the Appendix. We hope that this will make our proof easier to read. The idea behind the proof is simple. First we create a discrete version of the problem in Lemma A. 1 , then we handle that discrete version in Lemma A.6 and finally we convert the result on the discrete version back into the setup of our original continuous—time problem in Lemmas A3 and A.5. In order to state Lemma A.1 we need a new definition. Let N e F V, and let T S r be stopping times. We define: SN(T,r)= jN(s)d2(s)—x [Z(s)dN*(s)—p [2(s)dN‘(s). (Tm) (T31 (T31 SN (T,r) is the capital gain generated by the portfolio N between the T and 1. Lemma A.1 Let N e F V and O S T S r S 1 be two stopping time so that e‘8 Z(T) S Z(t) S e8 Z(T), for all te [T , t] as, where 5 > 0 satisfies 1 — e—zs S u and e26 — 1 S 71., then SN(T,r) _<_ N(T) (2(1) - Z(T)). 10 Furthermore, (i) If2(t) = e‘5 Z(T) then S~(T,T) S N(T) (Z(T) - Z(T)) - 7t Z(T) (N(T) - N(T))+ - Hts Z(T) (N(T) - N(T)) _ , where us = e25 (p. + e—28 — 1) 2 0. (ii) IfZ(‘t) = e25 Z(T) then SN(T,T) S N(T) (Z(T) - Z(T)) - ls Z(T) (N(T) - N(T))+ - H Z(T) (N(T) - N(T)) — , where x, = a” (x + 1 — e25) 2 0. Lemma A3 is an asymmetric extension of Lemma A. 1. Lemma A.3 Let N e FV and o s T s ‘C s 1 be two stopping time so that e‘” Z(T) s Z(t) s e7 Z(T), T _<_ z s 1' a.s, and Z(T) = a‘B Z(T), where B, y > 0 satisfy that 1 — e45”) s it, then SN(T,r) S N(T) (Z(r) — Z(T)). Lemma A.5 shows that after each stopping time there is a positive probability that our capital gain will be non-positive regardless of the trading strategy that we are using. Lemma A.5 Let N e F V and O S T S 1 be a stopping time. Then P(SN(T,1) S O / FT) > O. The last lemma that we quote is Lemma A.6. To state it we need the following notations. Let 0 < e < 1 and let 6 > 0 so that 1 — 628 S )1, t228 - l S A as in Lemma A.l. Letb>l/(1—u)sothatczs l—EE— q 4 (1—u)b > 1 — e, where q is the strike price of the option. The quantities 11;, and 15 are as in Lemma A.1. Lemma A.6 There exists an integer M = M(e,5) 2 1 and a sequence of measurable functions Zk : R" —> R+, k 2 0, so that 20 = 1, and for every sequence of numbers M, k 2 O, we have 11 2 N,...,N_ . . k( 0 H) =2”, k215uchthat1f1nf Sn>g—1, Zk-I (Nos-"9 Nk_2) OSnSM then 3 0 S n S M for which (i). z,, = b, and (11). C2 (b — l)——p.b(N,, -C2)_ 2 Sn, n—l where S, = ZNk(Zk+l ‘21.) k=0 — {21m = 66 Zk}[7»5 (Nk+l — Nit)+ + H (Nk+l - Nit) _] Zk+1 — {Zk+1 = 6—5 ZkHA (NM - Nk)+ + l~lti(Nk+1 — Nit) _] Zk+1, and Zk = Zk(No,...,Nk_1), k 2 0. Finally we are able to prove our main theorem. Theorem 2.1 If Assumption 1.8 is satisfied then b); = 1. Proof. Let N e F V and O < e < 1. To apply Lemma A.6 we select some constants. Let 8 > 0 satisfies 1 — 2'25 S u and e225 — 1 S 7t. Choose b > 1 /(1 - u) so that C251-3—8— q 4 (l—u)b satisfies 02 > 1 — e . We denote d = e / 4 for notational simplicity. Next we define a sequence of stopping times To = 0, Tk+1 = (110:, k 2 0. By Lemma A.1, we have, for every n 2 O SN(O,Tn) S S" as. on {In < d}, where 12 n—l 5. = ZNtrnthna—Ztr.» — {Z(Tk+1) = 68 Z(Tk)} [la (N(Tk+1) - N(Tk» + + H (N(tw) - N(Tk» -] Z(Tk+1) - {Z(Tm) = 6‘5 Z(Ik)} [7L (N(Tkfl) - N(Tt» + + H8 (N(TkH) - N(Tt)) '] Z(Tt+t)- By Lemma A.6, there is integer M(e,6) Z l, and there are measurable functions 2;, : Rk—)R+,k20,sothatif inf Sn >e—1 thenElO S n S Mforwhich OSnSAI (2.2). 2,, = b, and cz(b — l)—p.b(Nn —cz)_ 2 S", where S" = SNUtt )(zM —zk) _ T2,.” = e5 2k} [la (N(rk+t) — N(Tk» + + it (N(TkH) - N(Tk» _] Zk+1 — {21m = a“ at [A (N(Tk+1)— Nun) * + H5 (Neat) — NW» '1 a... 2,. = 2,.(N(t0),. . .,N(‘Ck_1)), k 2 0, and 2,... = e:6 2., k 2 0. We will show that (2.3). P(Z(rk) = 2),, 1 S k S M, m < d) > 0. In principle, (2.3) follows because the price process Z fluctuates according to Assumption 1.8 (i). We will prove it formally by induction. To start the induction, we assume that P(Ak) > O, for some k between 1 and M — 1, where Ak = {2(9) = z,-, 1 S i S k, r). < d} 6 Since P0410 = P(Ak, Zk+1 = 56 Zk) + P(Ak, Zk+l = 6‘8 Zk), we will assume without loss of generality, that P(A/., 21m = e6 2),) > 0. l3 Since 21.4.] e F“ , it follows that (At, 21.4.1 = e5 2;.) e F“ . From Assumption 1.8 (i), we get P(A,., 2,... = e5 2,, 2m“) = e5 2(a), n+1 < d) > 0. But Ak+1 2 {Aka zit-+1 = e5 Zk, Z(TkH) = 38 Z(Tk), Tk+1 < d}, so P(Ak+.) > 0. By induction we get (2.3). From (2.2) and (2.3) we conclude that El 0 S n S M so that either P(S,,S 8—1)>Oor (2.4). P(Z(t,,) = b, 62 (b — 1) — u b (N(rn) — cz) _ 2 S") > 0. We claim that xN 2 1 — e. The proof will be divided to 3 cases. Remark. Throughout the proof all the inequalities will be understood to hold with positive probability and we will not repeat it. Case 1: HS 3 s —1)>0. By using Lemma A.5, we have SN(O,1) S SN(O,r,,). We have already seen that SN(0,t,,) S 5". So we conclude that Sit/(0,1) S e — 1. It follows from the positivity of the payoff function g that xN 2 1 — a under the assumption of Case 1. Next we assume that (2.4) holds and we split it into two cases based on the value of Mr"). Case 2: P(2(t,,) = b, c2 (b — 1) — a b (N(rn) — c2) “ 2 Sn, Me.) 3 0) > 0. We denote T = In, N‘ = Mr”) and z = 2(1). We need to estimate SN(O,1). l4 We claim that (2.5). SN(O,1) S c; (1 — u) b — Q. We choose 8°, [3 > 0 so that e25° s (1 -e/2)/(1 — 3e/4),1—e‘35°sn,e35° —1S}tand (““5”): 1—11. By using our assumption on the process Z in the form of Lemma A.25 (i) with e / 4 instead of e, we have P(T50 = l / F r) > 0. Now we use Lemma A.1 and (2.4) with (N. — cz) ’ = cz — N. as follows from N‘ S O and C; > 1 — e. We get mom) s 5,, 3 c2 (b — 1) — 11 (c2 —N‘) b =cz(b—1)—uczb+uN.b. We apply again Lemma A.1 and we get SN(T,1) SN‘ (2— b) SN. (55° — 1) b <— nN‘ b, because T5. = 1, N’ s 0, z 2 e'5° b, and 11 2 1 — 635° > 1 — 65°. So we have SN(O,1) = SN(0,T) + SN(T,1) IA c2(b—1)—j.tczb Cz(1—|.l)b—Cz. Thus we get (2.5). 1 Next we will estimate the payoff g( Z 2,) where ZS, == J'Z(t) dt. 0 We calculate 1423.120 —1t) [Z(tidt—q 15 (2.6). 2(1 —n)(1—e/2)e‘5°b—q, bye—SObSZ(t)S esob,TStSTao a.s,TSe/4and T50=1. Finally we claim that g( Zé) — SN(0,1) Z l — e. To see this, we use (2.5) and (2.6) g(Zti)-S~(0,1) 2(1—11m—8/2)e‘6°b~q—(02(1-11)b—C2) q (1_“)b(1—H)b—q+cz =(I-uitl—g—e5°(I—3f)>e‘5°b+ (use 62 =1—3—8— q 4 (l—p.)b ) 2 (1—p.)(1—§2——6280(1—§§-))e'80b+q—q+cz 2 (1— n)(1—e /2--(—1——HE2(1—3—8))e‘5°b+c2 3e (1"?) (usee25°s(1—s/2)/(1—3s/4)) =(1—n)(1—s/2—(1—e/2))e‘5°b+c2 =Cz>1—8. So we haveg(Z(',)—SN(O,1)2 1 —8. It follows now that xN 2 l — 8 under the assumption of Case 2. Case 3: P(2(t,,) = b, c2 (b — 1) — a b (N(Tn) — e2) ‘ 2 Sn, Nu.) > 0) > 0. Again we denote T = 13,, and M = N(rn). We introduce now some quantities to simplify the calculations. We choose 5°, [5 > O as in Case 2 and the following: c9= (1—p)cze£’°b—cz, 16 C3 =Cz(b—1)-H(N. —C2)— b, and c4 =N’ (55° — 1) b —— as» N’ e‘5° b. By Lemma A.1 and the assumption of Case 3, we get (2.7). c, 2 SN(O,T). The first calculation is to compare C9 with C3 and C4. We claim that under the assumption N. > O, we have (2.8). C9 2 C3 + C4. Proof of (2.8). We split the proof into two cases according to the relation between N‘ and C2, First we assume that N. 2 C2 > 0. Here (N. — C2) " = 0. So we have C3+C4=Cz(b—1)+N.(e—6° —- l)b *uaoN‘ e'sob s e2 (b — 1) + Q (55° — 1) b — usage—5° b (use N. 2 C2 and 65° — l < O) = C2 (8-50b — 1) — 115° 626—50 b = (l — 115°) C2 e‘éob — C2 = e25° (l — p.) C2 6—80 b — C2 (use 1 — as: =1 — e25° (a + {25° — 1) = 1 — 85°11 — 1 + (225° = 225° (1 — 11)) =(1— )4) C2 eéob —C2 = Cg. To finish the proof of (2.8) we assume that O < N. < C2. 17 Here (N‘-Cz)-=C2-N.. So we have C3+c4 = C2 (b—l) — “(c2—N‘)b +N'(e_5°—l)b —u5oN*e”5°b = (cz—N‘)b+N‘b+N’(e‘5°—1)b —tt(e2—N’)b —ttaoN’e“5°b— c2 = (1 ~11) (cz—N‘)b+N‘ 65°12 —ttsoN‘e‘5°b—c2 =(1—uitc2-N‘)b+(1—uso)N’e‘5°b—c2 = (1 —11) (ca-N‘)b+e25°(1 —u)N‘e“"°b—c2 (usel ~us°=825°(1 —11)) s (1 —n)(e2—N')e5°b+(1 —n)1v"e“°b—c2 (usel —u >OandC2>N‘) = (1 —n)(c2—N‘ +N")e‘5°b—c2 (1 —n)c2e5°b —c2 H = Cg. After establishing (2.8) we will estimate SN(0,1). We start by defining a stopping time t = Tao. Then we use Lemma A.25 (iii) with e / 4 instead of e, and we get that P(1—s/4Sr<1,Z(r)=e’5°b/FT)>O. Next we denote N = Mr) and define a stopping time L I . if 1150. {I n fiso, [3.8 We claim that on the event {N. > O} (2.9). P(SN(T,L) S C4 / FT) > 0. Proof of (2.9). 18 First we assume N S 0, so (N — N‘)+ = 0, (N — N‘)" = N. - N and L = I. To see (2.9), we use Lemma A.1 (i) with 8° SN(T,L) = smut) s N‘ (65° — 1) b — 1150 (N‘ — A7) 55° b = C4 + 145° Ne—6° b S C4, by N S 0. Now we assume N > 0 (namely L = 1923c). By using our assumption in the form of Lemma A.25 (ii) (with e / 4 instead of e) we get P(L < 1, Z(L) = 6‘3 2(1) / F t) > 0. Using Lemma A3 with [3, 8°, e’wW) = l — u, and Z(L) = e‘B Z(r), we get (2.10). SN(T,L) 3 A7 (e‘B — 1) e‘°° b. Using Lemma A.1 (i) with 8° and (2.10) we have (2.11). SN(T,L) = SN(T,T) + SN(r,L) s N‘ (65° — 1) b — A (N — N‘)+ 65° b — 1150(A7 — N‘)‘ e"5° b + A7(e'B — 1) e"5° b. We need to show that the RHS of (2.1 l) is less than C4. We do it first under the assumptionN < N'. Here (A7 — AI‘)+ = 0, (A7 — N‘)‘ = N‘ — A7, so (2.11) gives SN(T,L) s N‘ (155° — 1) b —— as (N‘ — A7) e‘5° b + Ame“3 — 1) e?“ b C4 + twelve—5°b+N(e—B —— l)e—5°b C4 + (H8°+e—B ~1)Ne—60b c. + (e“’ —- (1 — 1150))1‘7 e"‘° b 5 C4, by 1 — 115° = e250 (1 * H) = 8‘6“” > e'B and N> 0. To finish the proof of (2.9), we work with the assumption N > N. which gives 19 (A7 — N‘)+ = A7 — N‘, (N — N’)‘ = 0. So (2.11) gives SN(T,L) SN‘ (e‘°° — 1) b — A (N —N") 65° b + A7(e"3 —1)e‘5° b s N” (55° — 1) b + N“ (e‘B — 1) e"°° b (useA7 >A/">0,e‘B — 1 <0) =N‘ (e‘5° — 1) b — (1 —e"’)N‘ ea" b S C4, by 1 —e“B>naoandN‘>0. We have established (2.9). We will use (2.7) (2.8) and (2.9) to estimate SN(O,1). Under the assumption of Case 3 we have (2.12). SN(O,L) = SN(O,T) + SN(T,L) S C3 + C4 S Co. By using Lemma A5 and (2.12) we finally have (2.13). SN(O,1) = SN(O,L) + SN(L,1) S C9. Next we will estimate the payoff function. We observe that (2.14). g( ZS.) =(1— a) ths1ds — q 2(1—n)(1—e/2)e“"’°b—q, by e'5°2(T)s2(t)se5°2(T),rgist,Ts 8/4and1-e/4St. Finally we claim that g( 22,) — SN(0,1) 2 l — e. To see this, we use (2.13) and (2.14) 20 Z(l—u)(l—e/2)e_5°b—(1—u)C2e5°b—q+C2 =1_ -5° _3_ _3_3 25° q _ 5° _ ( we 110 2 (1 4)e )+(1_“)b(l Me b q+cz (useC2=l—3e/4——q/[(l—u)b]) 1—8/2 21- e'°b1—e/2—1—3e/4 ( 11) ( ( )1_38/4 )+qea° -q+cz (usee25°s(1—e/2)/(1—3s/4)) 2(1—u)e"°°b(1 —e/2—(1—e/2))+C2 = 62 >1 -8. So we get that g( 23,) — SN(O,1) 2 1 — s. It follows now that xy 2 1 —C under the assumption of Case 3. By combing the 3 cases we see that xN 2 1 — e. As we explained in the first paragraph of this chapter this leads to b); = 1. 21 APPENDIX A] The tools of the proof In this appendix, we will prove four lemmas that will be useful for us. The first lemma is Al It is a simple result of the integration by parts formula. It allows us to create a discrete—time version of the problem by looking at the hedging portfolio at the times where the price process Z is going up or down by a factor of as, where 8 > O is related to the order A and p. The second lemma is A.3. This lemma is an asymmetric extension of Lemma A.1. The third lemma is A.5. We call this lemma the “closing lemma,” since this lemma helps us to finish the proof after finding a stopping time in which our goal is achieved. The fourth lemma is a main lemma. This lemma (Lemma A.6) shows how to deal with that discrete time version of our problem. Recall that SN(T,1:)= [WA/(s)d2(s)—A LTJIZ(s)dN+(s)—u LTJIZ(s)dN‘(s), where N e F V, and T s. t are stopping times. SN(T,r) is the capital gain generated by the portfolio N between the T and 1. Lemma A.1 Let N e F V and O S T S r S 1 be two stopping time so that {5 Z(T) S Z(t) S e5 Z(T), T S t S r as, where 8 > 0 satisfies 1 — 6‘28 S u and e28 — l S A, then SN(T,t) s N(T) (Z(t) — Z(T)). 22 Furthermore, (i) 1r2(t) = e“5 Z(T) then S~(T.t) S N(T) (Z(r) — Z(T)) - 7» Z(I) (N(t) — N(T))+ - H8 2(1) (Mr) — N(T)) ‘ , where as = e28 (11 + a” — 1) 2 0. (ii) IfZ(t) = e5 Z(T) then S~(T,r) S N(T) (Z(r) - Z(T)) — As 2(1) (Mr) - N(T))+ - u Z(r) (N(r) — N(T)) ‘ , where A5 = 525 (A + 1 - e”) 2 0. Proof. We will use the following notations: Z: = min{Z(t): T S t S t}, Z. = max{Z(t): T S t S t}, h. = N”(r) — NWT). h2 = N "(1) - N“(1‘), h3 = ht — h2 = N(r) — N(T). By integration by part and the definitions of Z., Z. we have: S~(T,T) = N(T) (Z(T) - Z(T)) + [(1,1(2 (I ) — 2(5)) dN+(S) — LTJ](Z(‘C) — Z(s)) dN'(s) — A LTJ]2(s)dN*(s) — a LWZQ) dN’(s) = N(T) (Z(T) - Z(T)) + jm](Z(T) - (1+ 7») Z(S)) dN+(s) — (was) — (1 — u) 20)) dN‘ts) s N(T) (Z(r) - Z(T)) + (”ga ) - (1 + A) 2.) ems) 23 — (mtzc ) — 2’0 — 11)) dN'(s) = N(T) (Z(T) — Z(T)) + (Z(T) - (1 + 71) 2.) hi - (Z(T) - (1 - 11) Z.) 122 = N(T) (Z(r) — Z(T)) + Z(r) h3 — (1 + A) Z. h1+(1— p.) Z* h2 = N(T) (Z(T) - Z(T)) + Z(T)/13 -(1+ 1») Z11 h1+(1- 102.013 - ht) = N(T) (2(1) — Z(T))+(Z(1)+(1— 11) Z‘) ’13 — ((1 + A) 21 — (1 — 11) 2‘) hi. Since {5 Z(T) s 2. s Z(T) s 2“ s e8 Z(T), we have 2’/2. SeS/éa =628$(1+A)A1/(1-},l). This implies that (1 + A) Z- — (1 — p) Z‘ 2 0, so the last term is maximized when h] = h3+ (and then necessarily h2 = hf). We conclude that (A.2). SN(T,r) S N(T) (Z(r) — Z(T)) + (Z(t) — (l + A) Z.) h3+ — (Z(r) — (1 — p.) Z.) 113‘. The RHS of (A2) is an increasing in Z. (decreasing in Z), so we can use e° Z(T) and e“8 Z(T) instead of Z’ and 211 respectively for estimation. Since Z(r) — (1 + A) Z. S O and 2(1) — (1 — p.) Z‘ 2 O, we get immediately that SN(T,t) s N(T) (Z(T) — Z(T)). Proof of(i). We use (A2) and the assumptions 2. = 2(t) = e‘5 Z(T) and 2‘ s e5 Z(T), and we get S~(T.1-) s N(T) (2(1) — Z(T)) — A 2(1) ha“ — (Z(r) — (1 — 11) e8 Z(T)) I13“ = N(T) (Z(I) - Z(T)) - A 2(1) hi - (Z(t) - (1 - 11) 82° 2(1)) hs’ = N(T) (2(1) — Z(T)) — A 2(1) In" — (1 — e26 (1 — 11)) 2(1) h; = N(T) (Z(T) - Z(T)) - 7» Z(T) (N(T) - N(T))+ - 115 Z(T) (N(T) - N(T))_ - 24 Proof of(ii). We use (A2) and the assumptions 2“ = 2(1) = e8 Z(T) and 2. 2 e‘5 Z(T), and we get SN(T,‘t) s N(T) (Z(t) — Z(T)) + (2(1) — (1 + A) e“8 Z(T)) I13+ — 11 2mm: = N(T) (2(1) - Z(T)) + (2(1) — e‘” (1 + A) 2(1)) 113+ — 11 2(1»; = N(T) (2(1) — Z(T)) + (1 — 6-21 (1 + A))Z(1) hf— 11 Z(T)h:1’ = N(T) (2(1) — Z(T)) — A1 2(1) (N(1) — N(T))“ — 11 2(1) (N(1)— N(T))“ . Lemma A.3 Let N e F V and O S T S r S 1 be two stopping time so that e‘B Z(T) S Z(t) S e7 Z(T), T s 1 s ‘t as, and Z(t) = e‘B Z(T), where (3,31 > 0 satisfy that 1 — 12'1””) g u, then SN(T,r) S N(T) (Z(t) — Z(T)). Proof. We will use the notation from Lemma A. 1. By integration by part, the definitions of 21, Z. we have: SN(T,‘C) s N(T) (Z(t) — Z(T)) + Z(t)/13 — (1 + A) 2. h, + (1 — 11) 2" 112 = N(T) (Z(t) - Z(T)) + (2(1) + Z‘(1 — 11)) 1., — (21(1 + A) - Z‘(1 — 11)) hi. Since 1.1—1’ Z(T) s 2. s Z(T) s 2‘ s eY Z(T), we have 2‘/2. set/e—B =eB*‘Ys 1 /(1 -—u). This implies that (1 + A) 21 — (1 — p) Z. 2 0, so the last term is maximized when h] = h3+ (and then necessarily h2 = h3_). We conclude that (A.4). 5,4111) 5 N(T) (Z(t) - Z(T)) + (2(a) - (1 + A) 2.) It; — (2(1) — (1 — u) 2‘) 113‘ . 25 The RHS of (A.4) is an increasing in 2', so we can use el Z(T) instead of Z‘ for estimation. We use (A.4) and the assumptions Z = 2(1) = e"B Z(T) and Z' S e7 Z(T), and we get SN (T11) S N(T) (Z(T) - Z(T)) - A Z(t) 113+ - (Z(T) - (1 - 11) 6’ Z(T)) h3— = N(T) (Z(T) - Z(T)) - A Z(T) 113+ - (e—B - 6y (1 - 11)) Z(T) 123‘ S N(T) (Z(T) - Z(T)), bye—B—ey(l—u)20. I Next we state and prove the “closing lemma.” Lemma A.5 Let N e F V and 0 S T S 1 be a stopping time. Then P(SN(T,1) S O /F1~) > 0. Proof. Let a > 0 satisfies 1 — 63“ S p and e3“ — 1 S A, let 0 S T S 1 be a stopping time and let I = T,. Case 0: N(T) = 0. We may use Lemma A.1 with N(T) = O and we get SN(T,r) S N(T) (Z(T) — Z(T)) = O. The result now follows because P(r = l / F T) > 0 via Assumption 1.8 (ii). Case 1: N(T) > 0. We can assume that r < 1, 2(1) = e‘“ Z(T) and Ta = 1 with positive probability, by Assumption 1.8. We denote that p = Z(T), N ‘ = N(T), N = Mr) and z = 2(1) for notational simplicity. We calculate using Lemma A.l SN(T,1) SN‘(e‘°‘ —1)p—(A7 —N‘)‘u,e‘“p —- (A7 —N‘)*Ae’°‘p + N(z—e_ap). 26 Now we split case 1 into 3 sub cases: Case1(i):N 2N‘ >0. Here (N —N‘)—=0,and(N —N')+=N —N'. So SN(T,1) sN‘(e‘°‘ — l)p-(N -N‘)Ae‘“p +A7(z-e‘°‘p) sN"(e’°‘ —1)p—(A7 —N‘)Ae‘°‘p +A7(1—e‘°‘)p (usezSpandN >0) =07 —N‘)(1—e'°‘)p—(N —N’) Ae‘“p =(A7 —N’)(1-e‘°‘ —Ae’°‘)p s 0, byNZN.andAe‘“2(em—l)e_0L =62“ —e’Cl >1—e'a. Casc1(ii):N‘>A'I 20. Here (A7 —N‘)*=0, and (A7 —A/‘)‘=N‘ —A7. So 511(11) SN’W‘ —1)p—(N‘ —A7)11.e"°‘p+N(z—e‘“p) SN‘(e‘°‘ —1)p—(N‘ -A7)11.e‘°‘p+N(I—e‘“)p (usezSpandN’>N20) =(N‘ —A7)(e‘“ —1)p—(N‘ —A7)11.e‘°‘p =(N‘ 4111121 —1—11.e‘“)p<0. bye‘“ —1<0,A/‘ —A7 >0andu,>0. Case 1 (iii): N‘ > 0 >N. Here (A7 —N‘)*=0,(A7 —N‘)‘=N‘—A7. So S~(T.1)s1v‘ (e’°‘ —1)p-(N‘— A7111. e‘“p+A7(z—e‘°‘p) sN‘(e‘°‘ —1)p—(N‘ —N)u,e‘°‘p+iv(e‘“ —1)e"°‘p (use N< 0 and z 2 e’zap) 27 sN‘(e‘°‘ —1—u,e‘°‘)p +A7(e‘°‘ —1+u,)e‘°‘p<0, byN">O>N,e_‘ll —1<0and 11,: eza(u+e'2°‘—l) 2 e2“(1—e_3a+e"2a—l)=l—e_a >0. Case 2: N(T) > 0. We can assume that r < 1, 2(1) = e‘1 Z(T) and Ta = 1 with positive probability, by Assumption 1.8. We denote that p = Z(T), N. = N(T), N = Mr) and z = 2(1) for notational simplicity. We calculate using Lemma A.1 SN(T,1) sA/‘(ea — l)p—(N —N‘)‘ u Cap—(N —N‘)* 11.,er + N(z—eap). Now we split Case 2 into 3 sub cases: Case 2 (i): N> 0 > N’. Here (N -N‘)' = O, and (N -—N‘)+ =N —N.. So SN(T,1) SN'(e°‘ — l)p-—(N —N‘) A,e“p+A7(z—e°‘p) SN‘(e°l — l)p—(N —N‘)A,,e°‘p+N(eOI —1)e°‘p (useN >Oanszezap) =N.(e‘ll — l +A,,,e°‘)p+N(eCl —1—A,,)e°‘p<0, byA7 >0>N‘,e°‘ —1>0and A,=e'2°‘(A+ 1 —e2“) 2 e‘2“(e3°‘ —1+1—e2°‘)= e“ —1>0. Case 2 (ii): 0 2 A7 > N‘. Here (N —N.)' = O, and (N —N')+ =N —N’. So SN(T,1)SN'(e° —1)p—(N -N‘)A,e°‘p+A7(z — cap) 28 sN’(e“ —1)p—(A7—N‘)A,e“p+A7(1 —e°‘)p (useNSOandzZ p) =(A7 -N‘)(1—e°‘)p-(N—N‘)A.e“p =(A7 —N‘)(1—e°‘ —A,,ea)p<0, byA7>N‘,1—e“<0and).,>0. Case2(iii):0>N‘2A7. Here(A7—N‘)"=0,and(A7 —A/‘)‘=N"—A7. So S~(T,1)SN'(e°‘ —1)p—(N‘ —A7)11e°‘p +A7(z—e°‘p) sN‘(e°‘ —1)p—(N‘ —A7)ue°‘p +A7(1—e°‘)p (useN eat —1. We conclude that P(SN(T,1) S O / FT) > O. I We will start with the setup cf Lemma A.6. Let0<8 < l andlet8 >Osothat1—e_26 S u,e2°—1SAasinLemmaA.l. Letb>1/(1 —p.)sothatC221 —3e/4—q/[(1—u)b] >1 —ewhereqisthestrike price of the option. Finally recall the notation of Lemma A.1: 115 = e25 (u + 1.1-25 — 1), and A5 = e“28 (A +1 — a”). 29 Lemma A.6 There exists an integer M = M(e,8) 2 l and a sequence of measurable fimctions Zk : R" —) R+, k 2 0, so that 20 = 1, and for every sequence of numbers N1, k 2 O, we have 2,(N,,...,N,_,) =64 2,_,(N,,...,N,_,) , kZI such thatifoinf Sn >e—l, SnSAI then 3 0 S n S M for which (i). 2,, = b, and (II). Cz (b — 1)—pb(N,, —Cz)— 2 Sn, —‘ where S, = N1(Zt+t ”21) O a. ll — (21.1: e521,}[A5 (NM —N,.)+ + 11 (N111 —N1)‘] 2111 - {Zk+1 = 8‘8 Zk}[7t (IV/1+1 - Nk) + + 116 (IV/1+1 - N11) _] Zk+1, and 2,, = Zk(No,. . .,N1_1), k 2 0. Proof. Leta>OsothatC1 E 1 —e+a_ 1. lr=0 Let qk = max{l S i S k: L,-_1 = 11}, (= O ifthe set is empty). Let N‘k =max{N,-:q1 S iS k} ika= 1, =min{N,-:q1 S iS k} iflk= 1. Next, we define the following sequence: 520:0. n-l sf = ZNsz —z,),n 21. k=0 31 Claim 1 SI” 2 S", n 2 0. Proof of Claim 1. Ika = 1 (namely 2111 > 21) then N11, = max{N,: q), S i S k} 2 N1, and Iflk =1 (namely Zk+1 < 2011161 N1], = min{N,-: q], S i S k} S Nk. We can conclude that N‘k (2111 - 21) 2 N), (21.11 — 21,), k 2 0, and $2,, 2 S,,, n 2 0. Claim 2 The sequence {5%,} satisfies the following: (A.7) (i) If m > k and z,,, < 21,, then 32m _.. S’k Sflzm) (Zm — Zk) < CI (Zm — 211)- (ii) If m > k and 2", > 21,, then S’m — SI), S f(z,,,) (2m - 21,) < C2 (2,,, — 2),). (iii) Ifm 2 k and z,,, = 21,, then 5’," — 5'21, S 0. (iv) Assume (w.l.o.g.) 3 0 < r, 3 EN so that a = 6%, b = 55.1fa S 2,, S b, 0 S k S n, then 5“,, S C2(b—a) — (n—(s+r))0/2, where = JUE t(f(e‘**"“ ) — f(e’° ))(e‘**”‘"’ — e“ )1. Proof of Claim 2. We first observe that N‘k S flzk) if L), = 1. To see this, we calculate: If L1, = land q), S i S k then f(z,~) S f(z,,), and N11,: max{N,:q1, Si S k} Smax{/(z,-): qk S i S k} 32 =f(21,). We also observe that N11, 2 fizk) if I), = 1. Similarly, we calculate: Iflk = land q), S i S k thenflz,) Zflzk), and M1: min{N,: q), S i S k} 2 min{f(z,-): q), S i S k} = flzk)° So for every k 2 0 we have Slk+l —Szk = Mk (PI/1+1 - Zk) Sf(th) (2H1 - 211) (A8). S f (Z111) (21,11 — 21,) < C] (21.11 — 21,), if 21.1] < 2;, and 2,,, S 2),, or S f (2",) (21+, — 21,) < C2 (21H — 21,), if 21,11 > 21, and z,,, 2 21,. We also observe that Ifzi, = z,,,+1 and 21+] = 2”,, k at m, then (A9) (SI/1+1 _Szk) + (52ml 357111) S —[f(2k+l) — f(Zk)l (21m - 21) < 0- To see (A9), we calculate (521+, —SZ,,) + (Sim. —S?,,,) Sf (21,) (21,1, - 21) + f (211.) (2,1,1. —- Zm) = f (2,.) (21+, — 2,.) + f(zk+1)(z;,- 21.11) = —[f(Zk+l) - f (211)] (21., - 21)- 33 Next we define, for every integer v and O S k < m, m-l u(v’k’m) : Z{(Z" ’Zn+l) = (8‘6 £04115”, n=k m—l d(v,/em) =Z{(z..z...) =(e“'*"5 ,6... )1. n=k where we identify sets with their indication functions. In word, u(v, k, m) and d(v, k, m) are the number of changes e"8 T 604-013, em”8 11 e"°, respectively, of the sequence (21,. . .,z,,,). Now we verify (A.7) (i). We define for each v that satisfies 2;, > e"5 2 2m: n(v) = min {n 2 k: (2,,, 2,11) = (Cow’s, e"5)}. We get n+1 m-l 53—53:}:(52 —S:) n=k < 26 S130)“ '53”) z,>e‘ 22,, I = Z Nn(v)(zn(")+1 —Z"(")) 2, >e‘” 22,, S 2 f(zm )(Zn(v)+l — Zn(V)) z, >e‘” 22,, =flzm) (2m ‘ Zk) < C] (Zn, — Zk). The first inequality follows from (A9) and the fact that z), > e"° 2 2,, implies d(v, k, m) = u(v, k, m) + I, while that 6‘6 2 z), or z,,, > e"° implies d(v, k, m) = u(v, k, m). The second inequality follows from (A8). The proof of (A.7) (ii) is similar to the proof of (A.7) (i) and will be omitted. We prove (A.7) (iii). Since each v is either e"8 2 z), = 2", or 21, = z,,, > e"°, this implies d(v, k, m) = u(v, k, m), and S2”, — S‘k S 0. 34 Next we prove (A.7) (iv). We have 3—1 11 = £110,011) + u(t,0,n) 3-1 s 22[d(t,0,n)/\u(t,0,n)]+l. So (A.10) r=—r s—l _ 21109.11) 4 u(t,0,n) 2 $53 Next we define A, = {0 S k S n - l: (21,, 21,11) = (6’5, (3mm) or (cums, e’°)}, —r S t S s —1 By using (A8) and (A9) we get Z(SZ-H .- 53) S C2 (e‘m’s — 8’5 )— [d(t,0,n) /\ u(t,0,n)p. [1614, Using (A.10) we now have 5-] S: : 22(Sf+1_512) !=-rkeA, SC2 (e58 — e"5)—(n—(s+r))0/2 =C2(b — a)—(n—(s+r))0/2. This is the end of the proof of Claim 2. We will use now (A.7) to prove Lemma A.6. Wedefine:M=[(s+r)+2(C2(b — a)+l—e)/0] +1. First we need to divide Lemma A.6 into three cases. 35 Case 1: Ifa=e"° < zk(s+r)+2(C2(b — a)+1—e)/0. So inf S, Se —1 in this case. OSnSM Case 2: Ifthere is O S n S M such that 2,, = a, then by A.7 (i) and a < e / 4, we have S,, 3 5°, SC1(a—l) (l—8+a)(a -—l) a—ae+az—1+8—a 8—l—a(8—a) < e—l. So inf S, Se —1 in this case. OSnSM Case 3: Ifthere is 0 S n S M— 1 such that 2,,11 = b, then we claim that S,,+, S C2 (b — 1) — u b (NM — C2) ‘. We will call this inequality the “main inequality.” We need to go through some very long and painfill steps to achieve the main inequality. First we need some notations. Let H“, = (M11, — N',)+, 0 s k s 11—1. Let h‘, =(N‘111— N11,) ‘, 0 S k S n—l. Let Bk = N11,(Zk+1- Zk) - [L], 21.5 I‘ll/(1'11, [.15 Wk] Zk+1, 0 S k S n—l . 36 Next we compare [3,, with 011,. This is a basic element of the proof. Claim 3 Bi, 2 011, 0 SkS 11—]. Proof of Claim 3. We divide the proof into the two cases. Case (i): L), = 1. Here 2111 = e5 2,, and N, S N11, < f(z;,). We observe that N‘M 2 N11,, and M11, = N111 v N11, 0 S k S n—l. To see these, we calculate that IfN1+1 2 fl2111) then q1+t = k+l, N‘k+l =Nk+l Z lek-tl) >f(Zk) >Mk, and N'k+l =Nk+l V Mk, Ika+1 < flzm) then qm = q), and MM = max{N,: q), S i Sk+l} =N1+t v N11, 2N‘k. We also observe that H“), S H1, 0 S k S 11—]. To see this, we calculate H'k = (NI/(+1 - N'It)+ = {UV/1+1 v N's—N11)" = (IV/1+1 - Mk) + s (N111 — N1) + = H,, by N,, S N‘k. So we can easily get the claim: 13k = N‘k (Zk+l - 2k) - Ab H111 th+1 2 N1, (21,11 — 21,) — A5 H121,” — u h), 2111: 011,, 0 S k S n—l, by L], =1,Zk+] >Zk, Mk ZN], and If], S H],. 37 The second case is very similar to the first one. Case (ii): 11, = 1. Here 21,11 = 8—8 21,, N1, 2 NH, Zflzk). We can easily have that N‘1+1 S N11,, and MM = N‘k /\ N11,], 0 S k S n—l by a Similar calculation. We observe that 11‘), S h1,0 S k S 11-]. To see this, we calculate 171/1 = (NI/1+1 - N111)— = {(N'k A Nk+l) — N111}— = (NkH - N111)— S (IV/1+1 - Nk) — = hit, by M, 2 M1,. So we easily have the claim: [3,, = N1), (PI/1+1 — Zk) — 115 h‘ka+l 2 N1, (2111— 21,) — (.15 h), 2114— A H), 21,11 = 011,, 0 S k S n—l, by 1,, =1, 21,11 < 21,, N11, S N), and W, S h. Thus we conclude that [3,, 2 011,, O S k S 11—]. This is the end of the proof of Claim 3. The next claim is of the fundamental importance. Claim 4 S1n+1 2 SW”. n+1 n-l Here 11 = inf{k: 2,1, = b} and 5' =1, +2131 k=0 38 , where y, = N‘, (2,1, — 2,) — (A, (N,,+1 — N‘,)+ + u (N,., — N‘,)‘] 2,1,. We have: L, = 1, 2,11 > 2,, and N, S N‘, < j(z,,). Proof of Claim 4. We need to split the proof into three cases. The first case is a trivial one. Case (I): N,,+1 2 N‘, Here (NM — N‘,)" = 0. We observe y, 2 01,: Yr! Nln (Zn+l — Zn) _ )1-8 (Nn+l _ Mn)+ Zn+l IV Nn (Zn+l ‘Zn) " 7“-8 (Nn+l ’ Nn)+ Zn+l " “1 (Nn+l _ N110— : an, by L, =1, N1, 2 N, and z,,+, >z,,. We will use this to prove our claim: 1 n-1 n—1 n Sn+1 : 23k +Yn 2 Zak +0." : Zak = Sn+1‘ k=0 k=0 k=0 We get that SB,“ 2 S n+1 if N,“ 2 N‘, Before we discuss the other 2 cases, we need one more notation. Letp, = max{q, S kS n: N, = N‘,}. In words: p, is the last time when the number of shares in the portfolio {Nkz q, S k S n} is maximized. In particular: N ,', = N p" . Case (11): M1,, < N‘, and p, = 11. Here N‘, = N,,, (N,., —- 11/1,) * = (N,.. -— N,) + = 0, and L, = 1. It is similar to the previous case and it is also easy. 39 We observe that Yn = N1n (Zn+l " Zn) '_ 1'1 (Nn+l — Nln)— Zn+l : Nn (Zn+l ’Zn) — ”1 (Nn+l ‘ Nn)— Zn+l : an- So we have n—1 n—1 n l _ — — Sn+l - ZBk +a’n Z Zak +an — Zak _ Sn+l' k=0 k=0 k=0 We get that S1,,“ 2 S 11+1 in this case. Next we go to the hardest case. It has a long proof. Case (111): N,,+1 < N‘, and p,, < n. Here(N,,+1— N‘,)+ = 0, {k: p, S k S n —l} :t q), and when q, S k S n we have: L1, = 1, q), = q,,. First we observe the property NI 1, related with p,,. By definition of p,,, when p, S k S n we have: N‘1,= Np” andh‘, =H‘1, =0. So we have (A.ll). B), = Np" (2111—21,), p, S k Sn—l. Next, we need some notations. Let N1=max{N,: kSiSn}, p,S k S n, 11‘, =(N111— A71.) p, s k s 11 —1, a), = N1, (2111 ~21) — u h} 21,11, p, S k S n -—1. We first observe that when p, S k S n -l we have 40 Nk=Nk ka+t 2 N1“, 11‘, = A7,. — A7,..., 2 0, and (A12) ak = Nk(zk+1 - 2k) — 11 (NA - Nkfl) Zk+1~ Next we observe that: N‘kZNk, ankSn. To see this, we calculate N1), = max{N,: q, S i S k} 2 Np" = N‘, = max{N,-: q, S i S n} Zmax{N,-:k SiSn} = N1, bqu =q,, SP2 S k S n. Next we observe that h‘, s 11,, ankSn-l. Indeed, [1.11 =(A7/1+1 — Nit) _ = {Nkfl — (N/t V NH!»— = (A7/1+1 - Nit)— 5 (IV/1+1 * Nk)— = hit, by N111 .>_ N111. We also observe that 01,201,, ankSn—l. 41 Indeed, a, = N1 (Zk+l - Z11) - 11 h.k Zk+l Z Nk (Ii/1+1 - Zk) — 11 hit Zk+l — 715 Hit Zk+l = 0111 by L], =1, [)7], _>. Nk, Zk+1> Z], and ht], S [1],. Next we define a temporary summation that is useful for us. .2 [1,," "TI Let s“, = 213, + 211, +ot,,. k=0 k=p,, First, we observe that 5,11 2 5,11. To see this, we calculate V 1),-1 n-1 n-1 n-1 n S,+l = EB, + Ea, +01, 2201, + 201, +01, = 201, 2 SM. lr=0 k=p,, k=0 11:11, k=0 To finish Case (III), all we need is to Show that: S‘,+1 2 5,1,. We start with some basic simplifications. First we observe r1. = N‘11(z11+1 — 211) — 11 (Nn+l — N11.) ‘ 21.11 = N111 (Zn+l — Z11) - 11 (N‘, — N11+1) Z11+l, because in our case N,,+1 < N‘, We also observe 0111 = N1 (211+1 - 2,) - 715 (Mn - N11) + Z11+l - 14 (Mn — N11)— 211+] 5 N11(Z11+l -— Z11) - 11 (Nn+l — N11) ‘ Z11+11 because L, = 1. So we have Yn " an 2 (Mn — Nn) (Zn-H — Zn) " 111 [(Mn _ Nn+l) — (NIH-1’ Nn)_] 211+] 2 (Mn _ Nn) (Zn+l _Zn) _ 11 (Mn _Nn) Zn+l 42 (A.l3) = (an — I.) (2,.l — 2,) — 11 (Np, — N,) 2..., where the second inequality follows from (A26), and the equality follows from N, = N, N and from N,» gm: NP" 3 N s N1 ,SoN =N =N‘,. p" p" p" p" Next we observe 01-— a1 = 117,,(2111-21) — (N1 (211. —z1) — 11(N1 — Naomi) (A.l4) = (N,,—N1)(z1+1—21)+11(N1. — N111)zi+1, 111.: k s 11-1, by (A.1 1), (A.12) and A7,," = N p,' In particular B," -ap" =p.(an — N11,“) 2111+! ~ ~ k-l ~ ~ Now we use (A.l3), (A.l4) and N," —N, = Z(N, —N,.,,), p,+ 1 _<_ k s 11, i=p1 and we prove that S‘,+t 2 3,11. Indeed, Sn+l_ v n+l=(ZBk +Yn)- (ZBk+k§-:ak +0. n) n- : Z(Bk Tat)+(y11 -an) k=p1 n-l ~ ~ n-l ~ ~ 2 (an —N, )(2111 — z, ) + 11 Z(N, - N111 >21... k=p,,+l k=P11 +(N, —N )(z... —z .)— 11(N, —N ..)z... ==Z(Np" —N,,)(z,,+,- zk)+P:Z(N/1 TNk+l)th+lT 11:2:(Nk _Nk+l)zn+l _Z=(N,—N1)(z...—z.)- 11207 -N...)(z...— z...) T : Z(fil-fii+l)(zk+1uzk)flp:(fik ‘1V1.1)(Z,.. T2111) k=p,, +li=p,, i=1), =ZZ(Zk+I— Zk)(Ni TNiT+l) 142(1),: TNk+l)(z11+lT 21111) 43 (use index change) ~ 3 (211+: _Zi+l)(Ni TN1‘+1)Tl’l Z(ZnH -Zk+l)(Nk —Nk+l) k=pn u 'M .. fi II I E 3 ~ ~ "—1 ~ ~ (21m _zk+l)(Nk T Nk+|)Tl1 Z(znn -Zk+l)(Nk —Nk+l) k=p1 : (l—H):(zn+l _Zk+l)(fik _filwl) Z O k=pn k: 'fi : , as follows from z,,+1 >zk+1,1\71, 2 NM, p" S k S n—l. We get that S‘nfl _>. 5',“ 2 Sm if N,,+1 < 1V‘,, and p" < n. 80 we conclude that S1,.“ 2 S,.+1 in all the 3 cases. We call {N11,} the “dominant portfolio” of {N1}. This is the end of the proof of Claim 4. Finally, we can prove our main inequality. We will divide this proof into two cases according to the value of 2 q” . Case (I): zqn S 1. First we need to divide S‘,,+1 into two sub summations: S‘nH = W0 + (S‘mq — W0). There exists 0 so that 20 = 1, o 2 q" and we define W0 = 0:] [3 11 . k=0 We get o—l (A.lS) W0 sZN,1(z,,, —z,)=sj so, k=0 by Bk = N‘1,(z1,+1 — 21,) - [£11 la [1‘]: +111 #5 h‘kl 211+), and (A.7) (iii) with zm = 2,, = l, m = 0, and k = 0. 44 As for the other summation (the more interesting one) S1n+l" W0 — n— : Bk +Yn *. O 3 Nkl (zk+l _ Zk ) ‘7‘5 Hlizk+l =0 ‘- + N1" (ZnH — Zn) “ l8(1Vn+l ‘ 1V1”)r Zn+l — “(Nn+l_N1n)_zn+l SZN;(ZI(+I _Zk)-H(Nn+l _eri)_zn+| i=0 (use NB, 2 AN and zk+1 > zk when q,, _<_ o _<_ k S n—l) = N1n(zn+l — Zo) — l1 (Nn+l — N‘n ) _ Zn+1 (A.l6) =N‘n(b-1) -H(Nn+1—N‘n)‘ 19, by 2,.“ = b, 20 =1. From (A.lS) and (A.l6) we get (A.l7). S‘m 5N1" (b— 1) —u(N,,+1—N‘,,)_ b. We call this the “basic inequality.” We will use the basic inequality to prove our main inequality. Recallthatb>1/(1—u)>1,namely:b—1—ub>0. Since cz =f(oo) > f(z,,) > N‘n, we only need to consider three sub cases. Sub case (I.i): N,,+1 2 c; > M". Here (NM — N‘n) ‘ = O and (NM — cz)_ = 0. Using (A.l7) with (NM — N‘n) ’ = O, we have S‘nH S N‘n (b— 1) S 62(b—1)— “Mn(Nn+l—CZ)H’ 45 by b >1, c; > Mn and (NM — cz)‘ = 0. Sub case (I.ii): c; > NM 2 N1". Here (NM. — N‘,) = O and (NM — c2)‘ = cz — NM. Using (A.l7) with (NM — N‘n) “ = O, we get $1,,+l 3 Nu, (b — 1) |/\ N‘n(b-1)+(Cz-N‘n)(b-1-Hb) (usecz >N‘,,andb—1—ub>0) cz(b—l)—pb(cz—N‘,,) |/\ C2 (b— 1)—ub(cz—N,,+1) (use NW ZN‘n) = cz(b—1)—ub(N,,+l — cz)‘. Sub case (I.iii): c; > NE, > NM. Here (NW —N‘,,)’ =N‘,, — NM, and (NM —c2)_ =c2 —N,,+.. Using (A.l7) with (NM — N‘n)’ = N” — NM, we also get S‘n+1 S N‘n(b- 1) -H(N‘n—Nn+1)b S N‘n(b—1)-H(N‘n-Nn+n)b+(62—N‘n)(b-1-l~lb) (usecz >N‘,,andb—l—pb>0) =cz (b—l) —p.b(cz—N,,+1) =cz (b— 1) —ub(N,,+l — cz)". SOSInH S C2(b—l)—Hb(Nn+1—C2)_ iqu" $1. We conclude that SW S S‘N S C; (b — l) — u b (NM — cz)’ if zq" S 1. 46 Case (11): zq" > 1. We need another long and painful steps for this result. First we observe simple facts. Since z,,_1= e‘26 b, 2,, = e’5 b and 2,,“ = b, q" S n —1. By definition ofq,, and N‘k, Lk =1, 2m > 2;. and N‘m 2 N‘k when q,, S k S n —1. Next we need new notations. We denote Na =f( zq" ) for notational simplicity. Leth=N‘k vN., q,,S k S n. We first observe that N°k=N‘ka.Sf(zk)vflzqn)=f(zk), ankSn, by zk 2 zq". We also observe that N°k+|=le+1VNt ZMk VNt=N°k, an k S "-1, byN'kH ZMk. ThUS(N°k+|-N°k)+=N°k+1—N°k20, ankSn-l. LetH°k=N°k+1—N°k, ankSn—l. LetB‘:_I = N;,—I (zqu — zqn_l)—p.5(N1 —N.) z qn-l q. ’ Let Bok =N°k(Zk+1—Zk)—}»5H°Zk+1, an k S 11—]. We compare 13°]. with Bk, it is a basic element of this step. Claim5B°k 2 Bk,q,,-1S k S 72—]. Proof of Claim 5. We first consider q,, - 1 case: 47 [32-1 = N(i,.-l(zq,.— zqn-l) —p'5(N¢i..-l _N') 24» N;n‘l(zqn— zq"—l) —“6(N;n—l —N;") Zq" : Bqn—l, by NJ," zkand Mk 2 Hok. This is the end of the proof of Claim 5. Next we define new summation. qn—Z "’1 LetSLL = 213,. + ZBE’ +73, k=0 k=q,,—l WhCl’C 70,, = N°n(Zn+1-Z,,)— K5(Nn+1 — 1V0")+ Zn+1 - ll (Nn+1 - Non)‘ Zn+|. Claim 6 5°,“ 2 S‘m. Proof of Claim 6. We divide the claim proof into two cases. Case (i): ND, 2 N. or NM 2 Ni. We first observe that 7°" = y” if N‘,, 2 N.. Here N°,, = N‘n v N:- = N‘,,. So we have You :Non (Zn+l_zn)-}-8 (Nn+1 —NO,,)+Z,,+] _H(Nn+l _ Non)-Zn+l 48 = N3. (Zn+1— Zn) — k5 (Nn+1 — N‘») + Zn+1 - u (Nn+1 —- N‘n) ' Zn+1 = in. We also observe that y°,, 2 7,, if NM 2 N. > N‘,,. Here N°,, = N‘,, v N: = Nw, and (NM — N°,,)‘ = 0. So we also have 7°" = N°n (Zn+1 — Zn) — la (NM — N°n) +231+] 2 Mn (Zn+l — Zn) " 7&6 (MM - N‘n)+ Zn+1 2 7m by N°,, 2 Nb, and 2,,” > 2”. We get that 7°" 2 y", ifN‘n 2 N. or NM 2 M. We will use this fact to prove Claim 6: qn-Z n—1 n-1 S2” = 23!: + 2 Bl? +7: 22B]: +7" =Srll+l' k=0 k=0 k=qn—1 So we get S°,.+. Z 51,.“ ifN‘n 2 N. or NM 2 Na Case (ii): (N‘n v Nn+1)< Na HereN°k=N‘ka- = N5 q,,S kSn. It is a hard part of the proof of Claim 6. Sinceqn Sn— 1,n an+ 1. We denote that q = q,, and p = zq" for notational simplicity in this case. By the definition of q, zq_1 = zq+1 = e6 p, zq+2 = 828 p. Ifn = q + 1 then let N‘q+z = N‘q+1 v Nq+2. By the definition ofq and (ND, v NM) < M, NH > N: =f(zq) > MW 2 NE,“ 2 N‘q. First we need two constants. Let V” = Mq_1 (Zq — Zq_|) + Nt (Zq+2 - Zq) - 1.15 (N144 — Ni) Zq — ’1 (N. — Mq+2) zq+2 49 (A.18) = ~24 (1 - e5») + N. (e25 —1)p — 115(N‘q—1 — Nap - u (N- — N‘w) e281). Let U” = N‘q_1 (zq — zq_1) + N‘q (zq+1 — 2,) + N‘q+1 (zq+2 — zq+1) — 115(N1q4 — NE) 2,, — k5 (N‘q+1 — N‘q) zq+1 — A5 (N‘q+2 — N‘q+1)zq+2. We start with some basic simplifications. By MW 2 N‘q and qu > zq we get U” S N‘q_1 (zq — 204) + NE,“ (zq+2 — zq) — p5 (N‘q_. — N1'(,)zq (A.19) = NH (1— ebp + N2,“ (e26 — 1)p — us (NEH — N'q)p. Here we explain the meaning of these constants. Ifn = q + 1 then Vn = Boa-2 + Boa—1 + You + 11(Nn+1 —Nln)— Zn+l, and Un = Bn-z + Bn—i + Y» + p (NW - N‘n)‘ Zn+l. Ifn Z q + 2 then V,, = [5°44 + [3°q + B°q+1* 11 (N‘q+2 *N° q+1)~Zq+2, and U": Bq-1+ Bq + Ba“- We claim V,. 2 U". To see this, we use (A.18) and (A.19) V» - Un 2 MM (1 — e5)p + N. (e26 — 1)p — 116 (N‘q—n - N*)p - P (N. - N‘q+2) 825p — [Nari (1— e5») + N2,“ (e25 — 1)p — us N‘q+2 2 N2,“ 2 Mg) = (N:- — Mq+2>p (e25 — 1 + w. — u e”) = 0. bY(1-Hs)=1-625(11+€_25—1)=1—€28(14—1)-1=628(1—H)- We will use this to prove that S°n+i Z S‘m. We split this case into two sub cases. 50 Sub case (ii.a): n = q + 1. Here N,,+1 = Nq+2, N‘n = N‘q+1. We observe q-Z SSH :ZBk +Vn —“‘(Nn+l _eri)-zn+l k=O q—2 2 2151 +U. - MN.” - Nl)'z,... k=0 :51 n+l' We get that S°,,+1 Z S‘m in this sub case. Sub case (ii.b): n 2 q + 2. First we define the following temporary summation —2 n—l SIN-l = Bk+Vn+ 28k +Yn‘ k k=q+2 Q ll 0 Here we explain the meaning of SW. SM is a capital gain from portfolio {1%}, where A7,. = NH, k 2 0 except NC, = NW = N... When q S k S n — 1, L, = 1, and the dominant portfolio of {M} is {N°k} as follows from max {A7,-2 q S i S k} = N. v N‘k = N°k. By the same argument of Case (I) we conclude Son-+1 Z Sn+l. We also observe that 5,,” 2 S‘m. To see this, we calculate: 51 k=0 k=q+2 q-2 n-l ZZBI: +Un + ZBI: +Yn k=0 k=q+2 n-l l : Bk +Yn _ Sn+1 k=0 So we get S°,,+1 2 SW 2 S‘,,+l in this sub case. We conclude that S°,.+1 Z S ‘,.+1 2 S,“ in all cases. This is the end of the proof of Claim 6. We use q,, instead of q after this point. Finally, we can prove the main inequality in this case. Recall that we claim that SW S S°,,+1 S c; (b — 1) — p b (NM — c2) ". First we need to divide S°,,+1 into two sub summations: S°,.+1 = W," +(Son+]- Wq"). qu We define W," = Z [3,, +35", k=0 We get qn-l (A20) W, 5 EN: (2,... — 2.) =51; s fez, )(z, — 1), k=0 by [35,14 = N;"_, (2,," — zq"_l)—— p5 (Nip: —N.) z," and (A.7) (ii) withm =qn, k=0, 2,": z, andzk= 1. As for the other summation (the more interesting one) 52 : Nli)(zk+1—Zk)_}\'6HI?Zk+1 +Non (Zn+l_zn)_)‘v8 (Nn+1 _Non)+zn+l —H(Nn+l —Non)m Zn+1 $2N2(Zk+1 —Zk)_i'l(Nn+1 _Nr?)—Zn+l k=qn (use N°,, Z N”), and zk+| > 21,, when q,, S k S n) =N°,,(z,,+1— Zq") — “(Nn+1—Non)-Zn+l (A.21) =N°n(b— 2,) -u(Nn+1—N°n)“b, by zn+1 = b. From (A20) and (A.21) we get 5°... = (Sm - W, ) + W, SN°,, (b— zqn)—u(N,,+1—N°,,)"b +f(zqn)(zqn— 1) (A22). S N°,, (b — 1) — u (NM — N°,, ) ‘ b, byN°nZ N: =flzqn)andb> zqn> 1. We will use this basic inequality to prove our main inequality. Recallthatb>1/(1 —u)>1,namely:b—1—pb>0. Since cz =f(oo) > flzn) > N°,,, we only consider three sub cases. Sub case (II.i): NM 2 c; > N°,,. Using (A.22) with (N,,+1 — N°,,) ‘ = O, we get S°n+1$N°n(b—1) Scz(b—1)— ub(Nn+1-Cz)~, 53 by b >1, cz >N°,, and (NM — cz)_ = 0. Sub case (II.ii) cz > NM 2 N°,,. Here(N,,+1 — N°,,)— = O, and (NM — cz)‘ = cz - NM. Using (A.22) with (NW — N°,,) ’ = 0, we have S°,,+l SN°,, (b— 1) SN°,,(b—1)+(cz —N°,,) (b— l —p.b) (usecz >N°,,andb—1—pb>0) =62 (b— 1)—pb(cz —N°,,) Scz(b—1)—ub(cz —N,,+1) (use NM 2 N°,,) = c; (b— 1) — p. b (N,,+1 — 02)”. Sub case (II.iii) cz > N°,, > NM. Here (NM —N°,,)" =N°,, —N,,+1,(N,,+1 — c2)": cz — N,,+1. Using (A.22) with (NW — N°,,) “ = N°,, — NM, we also have S°n+1$N°n(b-1)-H(N°n—Nn+i)b SN°n(b-1)*H(N°n-Nn+1)b+(cz —N°n)(b— 1 “HM (usecz>N°,,andb-—1—ub>0) =cz (b— 1) — pb(cz— NM) :62 (b— 1) — 11 b(Nn+1_ Cz)_- So we have S°,,+1 S c; (b —- 1) — u b (NM — cz)‘ in all three case. We getthatSnH SS°,,+1 S cz(b—1)— ub(N,,+1 — cz)’ ifz >1. qn We conclude that S,“ S c; (b — l) — u b (NM — cz)‘ if 2,,“ = b. 54 A.2 Equivalent forms of Assumption 1.8 Finally we modify Assumption 1.8 to a more convenient form and we need three lemmas for that purpose. The form of Assumption 1.8 that is achieved in the third lemma is the one that we use in the proof of the main theorem. Lemma A.23 If Assumption 1.8 (i) is satisfied, then for every I, m 2 1, 8 > O, O < d S 1 and stopping time T, we have, on the event {T < d} as. (i) P( T51,5 < d, Z( Tg‘m, ) = {’5 Z(T) / FT) > 0 and (ii) P( Tgfl, < d, Z( T5,", ) = e'"5 Z(T) / FT) > 0. Proof of Lemma A.23 (i). First we define the followings. Let T0=T, T, =(T,_,)g,k21. Let A, = {T, < d, Z(T,) = e“8 Z(T,_1),1SiS k},1S k s 1. Then A,1 = {A,, T,+1 < d, Z(T,+1) = e“5 Z(T,)}, 1 s k s 1— 1. We claim that P(A, / FT) > O as. We will prove it formally by induction. By Assumption 1.8 (i), P(A. / FT) > O as. We assume that P(A, / FT) > 0 as. for some k between 1, I— 1. P(AM /FT) = 1P0)”, < d,Z(T,,+,) = 8‘5 2(1", )/Fr, )(x)P(dx/ FT) > O as, by AssumptionAkl .8 (i) with T, ( < d) and P(A, / FT) > 0 as. So P(A1/ FT) > 0 as, by induction. Let B = {13”, < d, Z( T3,, ) = e‘18 Z(T)}. We have 55 B 3 A], and P(B / FT) 2 P(A, / FT) > O as. So we have proved Lemma A.23 (i). The proof of Lemma A.23 (ii) is similar to the proof of Lemma A.23 (i) and will be omitted. Lemma A.24 If Assumption 1.8 is satisfied, then for every 5, 52, 53 > O, O < d S 1 and stopping time T, we have, on the event {T < d} as. (i) P(Tg’z, < d, Z(Tg’z, ) = e‘52 Z(T) / FT) > 0, (ii) P( T5153 < d, Z(Tg’z, ) = e53 Z(T) /FT) > 0, and (iii) P(T5 = 1 /FT) > 0. Proof of Lemma A.24 (i). Assume (w.l.o.g.) that 3 I, m 2 1 so that (l — l) 5 < 62 S 15 and 83 = m 5. By Lemma A.23 (i), P( Tgm < d, Z( Tgfl, ) = {’5 Z(T) / FT) > () 3.3. Since {T5233 < d, Z( T51,” ) = (2‘52 Z(T)} 3 {Tgm < d, Z(Tg‘m, ) = {’5 Z(T)}, P( T51, < d, Z( T52, ) = e‘52 Z(T) / FT) > 0 as. We prove Lemma A.24 (i). The proof of Lemma A.24 (ii) is similar to the proof of Lemma A.24 (i) and will be omitted. The proof of Lemma A.24 (iii) is trivial by Assumption 1.8 (ii). Lemma A.25 If Assumption 1.8 is satisfied, then for every 0 < e < 1, there exists 5°(e) > O that satisfies: e250 S (1 — 28) / (1 — 3e), 1 — 6-380 S u, e380 — 1 S A so that for every stopping times 0 S T S e , and 1— e S I <1, we have (i) P(Tao = 1 /FT) > 0 as, 56 (ii) P(ngo < l, Z(TB'50) = e'B Z(T) / F,) > 0 as, where [3 = —[log(1 — p.) + 23°], and (iii) P(l — s 3 T, < 1, Z(T5o) = e"°° Z(T) / FT) > 0 as. Proof. By Lemma A.24 (iii) with 5 = 6°, the proof of (i) is trivial. By using Lemma A.24 (i) with d = 1, 52 = B and 63 = 5°, the proof of (ii) is trivial. We need some calculation for (iii). We define stopping time L = (1 — e) /\ T50. By P(T5o =1 /FT)> Oa.s,P(L=1— e/F-r) >0 a.s. e‘(B+b°) = 1 35 -25°. — p S e‘ o implies that (e-B S e If L: 1 —e thenT5o21—8,Z(L)Se5°Z(T)and e‘“ Z(L) s e‘B e5° Z(T) 3 525° e5“ Z(T) s e’°° Z(T). So P(1 — e s T50 < 1; Z(Tgo) = 55° Z(T) / FT) 2 P(Lw < 1, Z(L,,,o) = e"B Z(L) / L = 1 — e) P(L = 1 -— e /F1~) > 0 as, byP(L=]—8/FT)>Oa.s,and(ii)withT=Lon{L=1-—e}. 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