## Other American Options

The arguments we have just given for the American put apply, with appropriate modifications, to any vanilla option or combination of options with payoff A(£), or even A(S, t). Including a constant dividend yield, the option value V(S, t) satisfies the Black-Scholes inequality

Where exercise is optimal, V(S,t) = A(S) and the inequality is strict (< rather than <). Otherwise V(S,t) > A(S), and the inequality is an equality. At the free boundary (or boundaries, for there may be several of them if the payoff is complicated enough), both V and dV/dS must be continuous. The specification of the problem is completed by the terminal condition

V(S,T)=A(S), together with appropriate conditions at infinity. We see other examples of such partial differential inequalities in later chapters on exotic options.

Technical Point: Options with Discontinuous Payoffs.

The condition that the A for an American option must be continuous assumes that the payoff function itself has a continuous slope. That is, it is possible for the option value to meet the payoff tangentially only if the payoff has a well-defined tangent at the point of contact.

As an example, consider an American cash-or-nothing call option with payoff given by

The payoff is discontinuous. The option value is continuous except at expiry, but the A is discontinuous at S = E. It is clear that the optimal exercise boundary is always at S = E\ there is no gain to be made from holding such an option once S has reached the exercise price. Indeed, interest on the payoff is lost if the option is held after S has reached E. Thus, there is no point in hedging for S > E\ looked at another way, A = 0 for S > E. Clearly A > 0 for S < E.

Mathematically, we find that we have two boundary conditions, namely V(0, t) = 0, V(E, t) = B and a payoff condition V(S, T) = 0 for 0 < S < E. There is no point in considering values of S > E, since the option would have been exercised. Unlike the usual American option, where 'spatial' boundary conditions are applied at an unknown value of S, and so an extra condition is needed, both the spatial conditions here are at known values of S. These three conditions therefore give a unique solution of the Black-Scholes equation. This option is one of the few American options to have a useful explicit solution (see Exercise 6 of this chapter).

This option also illustrates very well the idea that the exercise strategy for an American option should maximise its value to the holder. It is particularly clear here that the choice Sj(t) = E gives the largest values of V(S,t) for all S < E, as illustrated in Figure 7.5.

### 7.6 Linear Complementarity Problems

It is clear from the discussion above that the mathematical analysis of American options is more complicated than that of European options. It is almost always impossible to find a useful explicit solution to any given free boundary problem, and so a primary aim is to construct efficient

and robust numerical methods for their computation. This means that we need a theoretical framework within which to analyse free boundary problems in fairly general terms.

Our starting point is the idea that, since it is difficult to deal with free boundaries, it is worth the effort of attempting to reformulate the problem in such a way as to eliminate any explicit dependence on the free boundary. The free boundary does not then interfere with the solution process, and it can be recovered from the solution after the latter has been found. We start by considering a simple example of such a reformulation, called a linear complementarity problem, in the context of the obstacle problem. We then apply the lessons learnt from the obstacle problem to more complicated American options . These problems too have linear complementarity formulations which lectd to efficient and accurate numerical solution schemes with the desirable property of not requiring explicit tracking of the free boundary (the same ideas apply equally well to complicated exotic options). These methods are discussed in detail in Chapter 9. The linear complementarity approach also leads to the idea of a variational inequality, and thence to existence and uniqueness proofs for the solution.

### 7.6.1 The Obstacle Problem

Consider the obstacle problem described in Section 7.2, in which we take the ends of the string to be at x = ±1 and write u(x) for the string displacement and f(x) for the height of the obstacle, both for -1 < x < 1. We assume that /(±1) < 0, and that f(x) > 0 at some points — 1 < x < 1, so that there definitely is a contact region. We also assume, at least initially, that f" < 0, where ' = d/dx, thereby guaranteeing that there is only one contact region. The free boundary is then the set of points, marked as P (x = xp) and Q (x = xq) in Figure 7.2, where the string first meets the obstacle. These are a priori unknown, and have to be determined as part of the solution.

In the contact region, u = f, while where the string is not in contact with the obstacle it is straight, so u" = 0. Normally, one would need just two boundary conditions to determine the straight portions of the string uniquely, and the values of u at the two ends of each straight portion would certainly do; indeed, we do have u(—1) = 0, u(xp) = f(xp) and similar conditions for the other straight portion. However, because P and Q are unknown, we need two more boundary conditions than usual in order to determine these points, and here a physical argument based on a force balance shows that at points such as P and Q, u' must be continuous as well as u. As a free boundary problem we can write the particular example given in Figure 7.2 as the problem of finding u(x) and the points P, Q such that

Given any particular f(x) with the same general shape as in Figure 7.2 it is straightforward in principle to show that u(x), P and Q are uniquely determined by this problem, and to find them. The procedure is tedious, and for all but specially simple /, P and Q must be determined numerically as solutions of an algebraic or transcendental equation. The details are even more complicated when f" is not always less than or equal to zero, because then multiple contact regions can occur, but again, in principle, it can be done.

u(-l) = 0, u" = 0, u(xP) = f(xP), u(x) = f(x), uixo) = f{xq), u" = 0, u'(xQ) = f(xQ),

— 1 < X < Xp, u'{xp) = f'(xp), xp < x < xq, (7.4)

An alternative approach to the problem is to note that the string either lies above the obstacle, u > f, in which case it is straight, u" = 0, or is in contact with the obstacle, u = /, in which case u" — f" < 0. This means that we can write the problem as what is called a linear complementarity problem:4

u" • (u — /) = 0, —u" > 0, (u-f)> 0, (7.5)

subject to the conditions that u(—1) = u(l) = 0, u, u' are continuous. (7.6)

This statement of the problem has a tremendous advantage over the free boundary version (7.4): there is no explicit mention of the free boundary points A and B. They are still present, but only implicitly via the constraint u > /. If we can devise an algorithm to solve the constrained problem, we just have to look at the resulting values of u — /; the free boundaries are where this function switches from being zero to nonzero. One such algorithm is the Projected SOR algorithm described in Chapter 9 in the context of American options; this is an iterative procedure which, starting with an initial guess for u that is certainly above /, produces a sequence of ever more accurate approximations to the true solution. The constraint is simply implemented, for if values of u less than / are generated, they are simply reset to equal /.

It is beyond the scope of this book to prove that the linear complementarity formulation is equivalent to the free boundary formulation (the hard part being to prove that any solution of the former is also a solution of the latter), nor do we show that there is a unique solution to the former. The proofs use techniques of functional analysis, in particular the theory of variational inequalities, but the basic idea is simply minimisation of the appropriate energy functional over the convex space of all suitably smooth functions v(x) that satisfy the constraint v > f.

7.6.2 A Linear Complementarity Problem for the American Put Option

We now extend the analogy between the obstacle problem and the Black-Scholes formulation of the free boundary problem for an American put by

4 In general, a problem of the form

AB = 0, A > 0, B > 0, is called a complementarity problem, and in this example the factors A = u" and

showing that the latter can also be reduced to a linear complementarity problem. In fact, the 'spatial' parts of the two problems are almost identical, and the major difference is that the put gives an evolution problem, in contrast to the static obstacle problem. The analogy also carries over to the numerical methods for the two problems: the 'spatial' part of the put problem is solved by the same Projected SOR algorithm, while the 'temporal', or evolution, part is used to time-step the solution forwards.

We first transform the American put problem from the original (S, t) variables to (x, r) as before. We see in later chapters that this is in some ways better from the numerical point of view, and some of the manipulations are easier. These transformations were given in Chapter 5; the only difference now is that there is an optimal exercise boundary. In original variables this was S = Sf(t), and we write it as x = xf(r); note that because S/(t) < E, x/(t) < 0. Also, the payoff function raax(£ — S, 0) becomes the function g(x,r) = max^-^-e^+^o). (7.7)

u(x,t) = g(x,r) for x<x/(t), with the initial condition u(x, 0) = g(x, 0) = max (e*(*"1)e - e?(k+1)x, o) (7.9)

and the asymptotic behaviour lim u(x, t) = 0. (7.10)

(As x —> —oo, we are in the region where early exercise is optimal, and so u = g.) We also have the crucial constraint u(x, t) > max (e^k~1)x - e*1(fc+1)x, o) (7.11)

and the conditions that u and du/dx are continuous at x = x/(t), all of which follow from the corresponding conditions in the original problem.

In order to avoid technical complications, let us accept that, since we are going to have to restrict any numerical scheme to a finite mesh, we may as well restrict the problem to a finite interval. That is, we consider the problem (7.8)-(7.11) only for x in the interval —x~ < x < x+, where x+ and x are large. This means that we impose the boundary conditions u{x+,t) = 0, u{-x-,T)=g{-x~,T). (7.12)

In financial terms, we assume that we can replace the exact boundary conditions by the approximations that for small values of S, P = E — S, while for large values, P = 0.

The fact that both the obstacle problem and the American put problem satisfy constraints suggests that the latter might also have a linear complementarity formulation, and this is indeed the case. The option problem is very similar to the obstacle problem, but with an obstacle which is time-dependent, that is, the transformed payoff function g(x, r).

We can write (7.8)-(7.11) in the linear complementarity form id a2 \ (7'13)

("M-g(x,T))> 0, with the initial and boundary conditions (7.9) and (7.12), namely u(i,0) = g(x,0), u(-x~,t) = g(-x~,t), u{x+,t) = g(x+,r) = 0, and the conditions that du u(x,t) and — (x,r) are continuous. (7-14)

The two possibilities in this formulation correspond to situations in which it is optimal to exercise the option (u = g) and those in which it is not (u > g).

Once again, the great advantage of this formulation is that the free boundary (or boundaries) need not be tracked explicitly. As with the obstacle problem, a considerable amount of work is necessary to prove that the linear complementarity formulation is equivalent to the original free boundary problem, and that there is a unique solution. Again, the techniques used are those of functional analysis and parabolic variational inequalities, and more details can be found in the books referred to above.

7.7 The American Call with Dividends

(This section may be omitted at a first reading without loss of continuity.)

We now consider some analytical aspects of the model for an American call option on a dividend-paying asset, introduced in Section 6.2. Recall that the value C(S, t) of the call satisfies f + + = 0 (7,5)

so long as exercise is not optimal. The payoff condition is

and because the option can be exercised at any time, we always have

If there is an optimal exercise boundary S = Sf(t) (and we shortly see that there is), then at S = Sf(t),

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