inequality for geometric Brownian motion

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(3.27) y 0 2(x) 0 y 0 1(x) 0for all x > 0 . Thus the difference x 7! y 2 (x) 0 y 1 (x) is increasing on ]0;1[ . Therefore:(3.28) max8)9y 3(x 0 ) 0 y 1 (x 0 ) ; y 2 (x 0 ) 0 y3(x 0 y 2 (x 0 ) 0 y 1 (x 0 ) y 2 (x " ) 0 y 1 (x " ) " .Thus, either y 1 (x 0 ) or y 2 (x 0 ) is within " of y3(x 0 ) . This completes the claim.4. The payoff and the optimal stopping strategyIn this section we prove that the explicit formulas for the payoff and the optimal stoppingstrategy guessed in Section 2 are correct. This establishes the validity of the principle of smoothfit and the maximality principle for the optimal stopping problem under consideration. The resultsobtained here are further applied in the following section. The fundamental result of the paper isformulated in the theorem as follows.Theorem 4.1Let X = (X t ) t0 be geometric Brownian motion with drift < 0 and volatility > 0 asdefined in (2.1), and let S = (S t ) t0 be the maximum process associated with X as defined in(2.4). Consider the optimal stopping problem with the payoff given by:(4.1) V (x; s) := sup Ex;s0S 0c1for s x > 0 given and fixed ( under P x;sthe process (X; S) starts at (x; s) ), where thesupremum is taken over all stopping times for X .1. Then V (x; s) < 1 if and only if < 0 . The optimal stopping strategy in (4.1) (thestopping time at which the supremum is attained) is given by the following formula:(4.2) 3 = inf8t > 0 j X t g3(S t )9where s 7! g3(s) is the maximal solution of the differential equation:(4.3) g 0 (s) = K g(s)1+1s 1 (s > 0)0g(s) 1under the condition 0 < g(s) < s , with 1 = 1 0 2= 2given by the following formula:(4.4) V (x; s) = 2c1 2 2 xg3(s)!10 logxg3(s)= s , if 0 < x g3(s) .and K = 1 2 =2c . The payoff is!10 1!+ s , if g3(s) < x s2. The optimal stopping boundary s 7! g3(s) is strictly increasing on ]0;1[ and satisfies0 < g3(s) < s for all s > 0 , as well as the following limiting conditions:10
(4.5) g3(0+) = g 0 3(+1) = 0(4.6) g 0 3(0+) = 1 .Moreover, the following estimates are shown to be valid:(4.7)s1 +101 K11=1 g 1013(s) s K11=1s 101=1for all s > (101)=K1 (the left-hand inequality holds for all s>0 as well). Finally, the optimalstopping strategy 3 is exponentially integrable, and the following estimate holds true:(4.8) E x;s exp 2 1 283! !1=2xg3(s)whenever g3(s) < x s . ( The left-hand side equals 1 for 0 < x g3(s) .)Proof. Since the facts from the second part of the theorem will be needed in the proof of thefirst part, we shall begin with the second part.For the second part it should be noted that (4.5)-(4.7) follow directly from our construction ofthe maximal solution which is presented in Section 3. ( The estimates (4.7) are nothing but theestimates (3.4) just rewritten, while (4.5)+(4.6) is easily deduced by (4.7).)To show (4.8) assume that g3(s) < x s are given and fixed. The key point is to note that:(4.9) 3 g 3(s) := inf8t > 0 j X t = g3(s)9since s 7! g3(s) is increasing. Moreover, by (2.1) we find:(4.10) g3(s) = inf8t > 0 j B t = at0b9:= a;bwith a = ( 2 02)=2 and b = log(x=g3(s)) 1= , both strictly positive. Finally, it is well-known(see [8], p.70) that we have:(4.11) E exp a22 a;b!= exp0 ab1.Now, matching (4.9)-(4.11) together, we get (4.8). This completes the proof for the second part.For the first part note that V (x; s) < 1 for < 0 follows by (2.7). To prove the converse,it is enough to show that V (x; s) = +1 when = 0 . For this, denote the payoff associatedwith the given by V (x; s) , and the corresponding optimal stopping boundary by g3;(s) .Then clearly V 0 (x; s) V (x; s) , while from (4.7) we find that lim "0 g3;(s) = 0 . Thereforefrom (4.4) (being proved below) we get lim "0 V (x; s) = +1 , thus proving the claim.It remains to show the validity of (4.2) and (4.4). For this, we shall denote the function on theright-hand side in (4.4) by V3(x; s) . It should be recalled that this function satisfies (2.10)-(2.13).Since clearly V3(x; s) s , we may write:11
Page 1 and 2: J. Appl. Probab. Vol. 35, No. 4, 19
Page 3 and 4: (2.2) dX t = X t dt + X t dB t .The
Page 5 and 6: first derivative, due to the smooth
Page 7: for all x > 0 . Thus in our proof b
Page 12 and 13: (4.12) V (x; s) = sup Ex;s0S 0c1 su
Page 14 and 15: fixed vertical level s > 0 its rest
Page 16 and 17: Remark 4.4The referee kindly pointe
Page 18: linear operators on martingales: Ac

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