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1 Otimal exercise boundary for an American ut otion Rachel A. Kuske Deartment of Mathematics University of Minnesota-Minneaolis Minneaolis, MN Joseh B. Keller Deartments of Mathematics and Mechanical Engineering Stanford University Stanford, CA Phone:(45) Fax: (45) Running Title: Otimal exercise boundary for American ut otion. Keywords: ut otion, exercise boundary, American otion, free boundary Abstract The otimal exercise boundary near the exiration time is determined for an American ut otion. It is obtained by using Green's theorem to convert the boundary value roblem for the rice of the otion into an integral equation for the otimal exercise boundary. This integral equation is solved asymtotically for small values of the time to exiration. The leading term in the asymtotic solution is the result of Barles et al. []. An asymtotic solution for the otion rice is obtained also. Suorted in art by an NSF Mathematical Sciences Postdoctoral Research Fellowshi.

2 . Introduction and results We shall determine the otimal exercise boundary S = (T ) and the rice P (S; T ) for an American ut otion on a share of rice S at time T near the exiration time T F. Our result for the boundary is, in arametric form with arameter : (T F, T ) e, 6 = r 8 < (T ) K :, (T F, T ) ; (.) = 9 = ; ; (T F, T ) : (.) Here K is the strike rice, is the volatility, and r is the interest rate. To use these formulas we choose large values of to ensure that the inequality in (.) will hold, comute T from (.) and (T ) from (.). To obtain (T ) exlicitly, we can solve (.) for in terms of t = (T F, T )=. By taking logarithms in (.) and solving the resulting equation iteratively, we get after three iterations (t) ( log, log 6r t log, log log 6r t 6r t, log!) = : (.3) The use of (.3) in (.), with the nal log omitted, gives an exlicit exression for (T ). By using only the rst term in (.3), we get (T ) K 8 < :, (T F, T ) log 6 = r [ (T F, T ) =] = = 9 = ; : (.4) This is the dimensionless form of the result of Barles et al. [], with an imroved constant in the argument of the logarithm. Figure shows (t) obtained from (.) and Figure shows (T ) obtained from (.) and (.). Figure 3 shows P (S; T ) based uon (.3) with (x; t) given by (3.). To obtain (.) and (.) we derive an integral equation for the otimal exercise boundary (T ), using the Green's function of the artial dierential equation adjoint to that satised by P. Then we solve this equation asymtotically for T near T F. We use the asymtotic solution for (T ) in the integral reresentation (3.) of to get P. Our results can also be obtained directly without the use of the integral equation, as we show in section 5. Previously McKean [] and Van Moerbeke [3] used the integral equation method to analyze the rice of an American call otion. In that case the boundary is given by (.) with a constant, so the boundary is arabolic for T near T F. However for the ut, tends to innity as T tends to T F, as (.) shows, so the boundary

3 (t).5.4 = :4 r.3. = : r t Figure : (t) as a function of t obtained from (. ) for =r = : (lower curve) and =r = :4 (uer curve) = : r (T)=K.85.8 = :4 r (T F, T)= Figure : The otimal exercise boundary (T )=K as a func tion of t = (T F,T )= given by (.) with determined by (. ) for =r = : (uer curve) and =r = :4 (lower curve). 3

4 = :4 r...8 P=K S=K Figure 3: The rice P (S; T )=K as a function of S=K for (T F,T )= =: (uer curve) and (T F, T )= =: (lower curve ) for =r = :4. The curves are based uon (.3) with given by (3.). The solid straight line is the rice at T F, T = and is given by P=K =, S=K. 4

5 is not arabolic near T = T F. In fact we show that the equation analogous to that derived in [3] for, onthe assumtion that is constant, has no solution. There is a considerable literature on the otimal exercise boundary, both analytical and numerical. See for examle Kim [4], Barone-Adesi and Elliott [5], Barles et al. [] and Wilmott et al. [6]. A recent list of references, together with numerical aroximations, is given by Aitsahlia and Lai [7]. The integral equation (3.3) for the boundary can be solved numerically, starting at T F. Equations (.) and (.) rovide the behavior of the boundary near the starting oint. Once the boundary has been found, P (S; T ) can be obtained from (.3) and (3.).. Formulation The boundary value roblem for P (S; T ) and for the exercise boundary S = (T ) can be formulated as follows (Wilmott, Dewynne, and Howison [6]):,P T = S P SS + rsp S, rp ; <T <T F ; S >(T ) (T F ) = K P (S; T F ) = ; S K P [(T );T] = K, (T ); P S [(T );T]=,; T T F : (.) For S (T ); P (S; T )=K, S: This roblem (.) can be written in a simler form by introducing the arameter =r=, the new variables x and t, and the new deendent functions (x; t) and b(t). They are dened by S = Ke x ; T = T F, t (.) P (S; T ) = Ke,t [ + (x; t)], S (.3) b(t) = log[(t )=K] : (.4) Then (.) becomes the following roblem for the transformed rice (x; t) and transformed exercise boundary x = b(t): t = xx +(, ) x ; x>b(t); <t< T F = (.5) b() = (.6) (x; ) = e x, ; x (.7) [b(t);t] = e t, ; t T F = : (.8) x [b(t);t] = ; t T F = : (.9) 5

6 3. The integral equation for b(t) The roblem (.5){ (.9) can be converted into an integral equation for b(t) by making use of the causal fundamental solution or Green's function for (.5). This solution G(x; t; ; s) isgiven by G(x; t; ; s) = [x,+(,)(t,s)] q 4(t, s) e, 4(t,s) ; s<t: (3.) It has a singularity at x =, t = s. We aly Green's theorem to G[x; t; ; s] and (; s) in the region > b(s), < s < t. Then we use the values of and x on the boundary = b(s), which are given by (.8) and (.9), and the initial value of, given by (.7). We also use the fact that is bounded by its initial value, which follows from the maximum rincile for (.5). Then we obtain (x; t) =, + Z, x, b(s) (t, s) + (, ) + b (s) (e s, )G[x; t; b(s);s]ds (e, )G[x; t; ;]d (3.) This is a reresentation of in terms of the boundary function b(t). We use (3.) in the boundary condition (.8) to get e t, =, + Z, b(t), b(s) (t, s) + (, ) + b (s) (e s, )G[b(t);t; b(s);s]ds (e, )G[b(t);t; ;]d : (3.3) Equation (3.3) is an integral equation for b(t). We rewrite it by calling the rst integral I(t) and using (3.) for G, obtaining e t, +I(t) = Z e b(t)+(,(,))t Z e,z = dz,, b(t)+(3,)t t 4. Asymtotic solution of the integral equation, b(t),(,)t t e,z = dz : (3.4) We shall solve (3.4) asymtotically for t. In the aendix we show that I(t) t= for t. Then the left side of (3.4) is asymtotically t +t= = 3t=. Next we assume that,t,= b(t) for t, which we shall verify later. Then we use the asymtotic value of the error function, Z y e,z = dz = e,y = [y, + O(y,3 )] for y (4.) 6

7 to evaluate the right side of (3.4). Then (3.4) becomes s 3t t t b (t) b (t) e, 4t for t : (4.) It is convenient to rewrite this equation for b(t) in terms of (t), dened by b(t) =,t = (t) : (4.3) We use (4.3) for b(t) in (4.), set =r= and relace t by its denition in (.), which is t = (T F, T ) =. The result is (.). The rst term of the exression (.4) for (t) shows that,t,= b(t) = (t) for t. This veries the assumtion we made in using (4.). Next we use (.4) to exress (T ) in terms of b(t), and then use (4.3) to exress b(t) in terms of. This yields exactly (.). Thus we have obtained both (.) and (.). To get P (S; T )we use (3.) in (.3). 5. Another derivation Now we shall resent another derivation of (.) and a modied form of (.) which may hel us to understand why they arise. We begin by seeking a function ~(x; t) which satises the dierential equation (.5), the initial condition (.7) for x, and the initial condition ~(x; t) =forx : ~ t = ~ xx +(, )~ x ; <t< T F = (5.) ~(x; ) = e x, ; x ~(x; ) = ; x : The solution of this roblem is ~(x; t) = Z (e, )G(x; t; ;)d: (5.) We note that ~ is just the second term on the right side of (3.). Next we seek the curve ~ b(t) on which (5.) has the boundary value (.8): ~[ ~ b(t);t] = e t, : (5.3) When we use (5.) for ~ in (5.3) the resulting equation for ~ b(t) is e t, = Z (e, )G[ ~ b(t);t; ;]d: (5.4) This is just (3.3) with the rst integral, I(t), omitted. Proceeding as in Section 4, we evaluate the integral in (5.4) asymtotically. Then we write ~ b =,t = ~(t) (5.5) 7

8 and we nd from (5.4) that ~(t) satises e, ~ 4r t : (5.6) ~ This is just (.) with the correct constant 6 relaced by 4. Therefore (T ) comuted from (.4) with b(t) relaced by ~ b(t), is still given by (.){(.4) with 6 relaced by 4. As (.4) shows, the leading term in (T ) is unaected by this change. Furthermore (5.5) and (5.6) show that ~ b()=,so (.6) is satised. We must still verify (.9) so we evaluate (5.) using (3.) for G and we obtain for t ~(x; t) = Z e x+(,(,))t Z e,z = dz,, x+(3,)t t, x,(,)t t e,z = dz : (5.7) >From (5.7) we comute ~ x [ ~ b(t);t] to get ~ x ( ~ b(t);t) = Z e ~ b(t)+(,)t +, ~ b(t)+(3,) t t e,z = dz s e, (~ b(t)+[3,])t) + ~ b(t)+(,)t 4t, e, (~ b(t),[,]t) 4t 4t : (5.8) We use (4.) to evaluate the integral in (5.8) and then use (5.5) for ~ b(t) with ~ given by the rst term in (.3). In this way we get ~ x [ ~ b(t);t] e, ~ ~ + ~ e, ~ + o( t)=o( t~ 3 )=O[ tj log tj 3= ] (5.9) Thus ~ x [ ~ b(t);t] is asymtotically equal to zero for t, so (.9) holds asymtotically. Summarizing, we have constructed functions ~(x; t) and ~ b(t) which satisfy (.5)- (.8) exactly and which satisfy (.9) asymtotically. Thus ~; ~ b is an asymtotic solution of (.5)-(.9). Relacing (x; t) by ~(x; t) in the exression (.3) for P (S; T ) yields an exression similar to that for the value of the Euroean ut. In fact, if the exression for the Euroean ut, rather than ~(x; t) is used in (.8) to determine ~ b(t), the resulting equation yields a solution which is also asymtotic to b(t) for small t. This rocedure was used in [] to obtain a lower bound on the free boundary. The result (.4) shows that the boundary is not arabolic for t small. There has been some confusion about this oint. It arises because the assumtion that b(t) t for t, with constant, leads to an equation for for the American call in [] and [3]. Suose we assume that b(t) t in the integral equation 8

9 (3.3) and evaluate both sides asymtotically for t. The leading order terms yield Z t,t z, (, z) + ex (, z) [, (, z) =(4(, z))] q dz z 4(, z) + Z, +(,) t Z, e,u du + t,(3,) t, e,u du : (5.),(3,) t Here we have used the substitution s = tz to eliminate t from the rst integral on the right. The last term on the right in (5.) is O( t) while the left side is O(t). Since the left and right sides do not have the same asymtotic behavior for t, there is no solution. Acknowledgments We thank Professor Jonathan Goodman for bringing this roblem to our attention and for helful discussions of it, and Dr. F.A. Aitsahlia and Professor Tse L. Lai for a rerint and discussion of their work. References [] G. Barles, J. Burdeau, M. Romano, Nicholas Samsoen, Critical stock rice near exiration, Mathematical Finance , (995). [] H.P. McKean, Jr., Aendix A: A free boundary roblem for the heat equation arising from a roblem in mathematical economics. Industrial Management Review, 6,.3-39 (965). [3] P. Van Moerbeke, On Otimal Stoing and Free Boundary Problems, Arch. Ration. Mech. Anal. 6,. -48, (976). [4] I.J. Kim, The analytic valuation of the American otions, Rev. Fin. Stud (99). [5] G. Barone-Adesi and R. Elliott, Aroximations for the values of American otions, Stochastic Analysis and Alications 9,. 5-3, (99). [6] P. Wilmott,J. Dewynne, S. Howison, Otion Pricing: Mathematical Models and Comutation, Oxford Financial Press, Oxford, 993. [7] F. Aitsahlia and T. Lai, Aroximation for American otions, rerint, 996. [8] C.M. Bender and S.A. Orszag, Advanced mathematical methods for scientists and engineers, McGraw-Hill, NY

10 Aendix We shall now show that in (3.3), I(t) t= for t, as we stated in section 4. We divide the integral into two arts, I (t) and I (t): I = + Z,, b(t), b(s) (t, s) b(t), b(s) (t, s) First, we bound ji j: ji j = Z Z s + (, ) + b (s) + (, ) + b (s) b(t), b(s), + (t, s) (, ) + b (s) b(t), b(s) (t, s) t (t;) q ds + 4(t, s) (e s, )G[b(t);t; b(s);s]ds (e s, )G[b(t);t; b(s);s]ds I + I : (e s, )G[b(t);t; b(s);s]ds (t;) s (, ) t q 4(t, s) ds (A.) + t (t;) jb (s)j sq ds 4(t, s) (A.) where (t; ) =!, (, )t > : (A.3) b(t) Using the result that b(t), t log t, in (A.) yields ji j = o(t). Next we consider I. For <<twe write the integral as f(s; t)e,(s;t) ds (A.4) where = (t)=4 and (s; t) = h i b(t),b(s)+(,)(t,s). (t)(t,s) We evaluate this integral = asymtotically for, noting that the main contribution comes from the uer limit. (See e.g. Bender and Orszag [8].) Thus I 4 b (t)+ (, ) t 4 b (t)+ (, ) t t, s e, j (t;t)j(t,s) ds q j (t; t)j Z j (t;t)j(t,) e,z dz : (A.5) Since j (t; t)j = (b (t) +, ) =4, the uer limit of the last integral in (A.5) tends to innity ast tends to zero. Then the asymtotic behavior of I is I f(t; t) q j (t; t)j t for t : (A.6)

11 Thus I(t) =I + I = o(t)+t= for t, as we stated.