Abstract

We connect the exercisability randomized American option to the penalty method by showing that the randomized American option value is the unique classical solution to the Cauchy problem corresponding to the canonical penalty problem for American options. We also establish a uniform bound for , where is the infinitesimal generator of a geometric Brownian motion.

1. Introduction

The Black-Scholes model is the lingua franca [1], the vehicular language, of option pricing. Yet pricing American options in the Black-Scholes remains the topic of a significant literature because every known method requires significant numerical calculations. For surveys on American option pricing we refer to [24].

Among all methods, Howison et al. [5] argue that, with a quadratic convergence [6], the so-called penalty method for the value of the American option in the Black-Scholes model is “the most efficient numerical approximation methods presently available for American option valuation.” It was used among others in [514]. The penalty method transforms the free boundary problem associated with the price of the American option into a partial differential equation (PDE) of the formwhereis the infinitesimal generator of a risk neutral geometric Brownian motion with volatility and is the risk-free rate and where the penalty term, , is a term which typically is zero when allowing the option to behave like a European option and which drastically pushes the value of the option higher when . Indeed, the “canonical” [5] penalty term isfor some large value of .

In studies [5, 6, 912, 14] where such canonical penalty method was used for approximating the American option value in the Black-Scholes model, the solution to the associated PDE is seen as a viscosity solution or as a weak solution that is the solution to a variational problem. It is well known that, in general, viscosity and weak solutions do not possess the regularity properties of classical solutions which can actually be differentiated in the classical sense to solve the PDE.

In this paper, we connect the randomized American option [15] to the penalty method, showing that not only does its value solve the canonical penalty problem (1), but also it is a classical solution to this Cauchy problem and, for a given maturity, is bounded.

Many of the above cited papers using the canonical penalty method as well as other papers using penalized problems such as [16, 17] were actually concerned not by estimating the value of American options but rather by determining the exact speed of convergence of option values under tree schemes approximations of the Black-Scholes model, a difficult [18] and long lasting problem still unsolved when the maturity is not allowed to float. Indeed, randomized American options can be used as a tool to help determine this exact speed of convergence. It is well known that payoff smoothness drastically affects this rate of convergence. We believe that our result may contribute to solving this problem. Yet the submitted paper answers the very natural question of whether or not the canonical penalty problem has a classical solution.

A randomly exercisable American option is an option which, prior to maturity, can be exercised only at some exercisable times following each other independently after an exponentially distributed waiting time of average . Under the label “option with random intervention time,” randomly exercisable American options were first introduced in Dupuis and Wang [19] for American perpetuities, then in Guo and Liu [20] for American lookback perpetuities, and then in Leduc [15] for American options. Note that the exercisability randomized American option considered in this paper differs from Carr’s maturity randomized option [21] which can be exercised any time up to some random maturity. In contrast, the exercisability randomized option can be exercised only at random times up to a fixed maturity.

We denote by the value of a randomly exercisable American option with maturity and payoff function , when the spot price of the underlying at time is . The value of this randomized American option is given bywhere is the set of exercisable stopping times in and where is the expectation of given that . As shown in [15], is the only solution to the following evolution equation:where, for functions , the expression denotes the discounted expectationIt is also shown in [15] that solveswhereand where is the semigroup associated with ; that is, for functions ,

Recall that a Lipschitz function is absolutely continuous and almost surely differentiable. In a slight abuse of notion, we replace the Lipschitz constant of by so that for every Finally, we denote by the identity function: for every .

Theorem 1. If is a Lipschitz function and , then is the unique classical solution to the Cauchy problem:Furthermore,and for every , there exists a constant depending only on , , , , and such that, for every and ,

The proof of our main result is divided into several steps. In Section 2, we show that is continuous. In Section 3, we prove that is Lipschitz with respect to . In Section 4, we show that is twice continuously differentiable with respect to , and the bounds for and are established. In Section 5, we show that is a classical solution to (11).

2. Continuity

Fix some value and for every function setFurthermore, definewhere . One easily gets from (5) that yieldingwhich is possible only if , showing that is continuous.

3. Lipschitz Property

We show here that is Lipschitz with respect to .

Lemma 2. Under the assumptions of Theorem 1, for every and every ,

Proof. Note first that because is Lipschitz, for every there exists a quantity , with such thatLet and be, respectively, the optimal (exercisable) stopping time for the randomized American option with maturity when and when .
Step 1. Note first that when , the stopping time is suboptimal and therefore for some random variable satisfying . Hence, since is risk neutral, Step 2. By definition, for some random variable satisfying . By suboptimality of when , we conclude that

4. Solution

Defineand for functions and integers and , letwhere is the probability density function of a standard normal random variable. Note that if is bounded thenFor any family of functions , , set

A consequence of Lemma 2 is that since is bounded and Lipschitz,is also a bounded Lipschitz function. It follows from Theorem A.1 in Appendix that is infinitely differentiable and that there exist constants such that Because , it follows from (5) that Using Lemma A.2 in Appendix we getFrom this we obtain

For a fixed , setwhich is Lebesgue almost everywhere since is Lipschitz. Note thatAgain, from Theorem A.1 in Appendix, there exist constants and such thatThe fact that implies thatThis in turn yields As is infinitely differentiable with respect to , function is continuous in and, with (36), dominated convergence gives that is continuous. From Theorem A.1 in Appendix we obtain that, for some constant ,and hence

5. Classical Solution

Recall from (8). Equation (7) can be rewritten asfrom which it follows that, for , ,Letting go to zero, we getwhere is the infinitesimal generator of the GBM. In other words,

6. Conclusion

In this paper, we showed that the exercisability randomized American option is a classical solution (and therefore the only classical solution) to the canonical penalized problem. This relationship can be extended to a broader class of models than the Black-Scholes model. Indeed, the key property is to obtain uniform bounds for in terms of powers of as in Theorem A.1 in Appendix.

Appendix

Theorem A.1. If is a bounded Lipschitz function then for every integer and every there exist real numbers , such thatIf additionally , there exist real numbers , such that

Proof. See [22, Th. 4.1] where the assumption that is polynomially bounded, continuous, and piecewise continuously differentiable with polynomially bounded derivative can be replaced by our assumptions on without any change in the argument, as a Lipschitz function is Lebesgue almost everywhere differentiable and absolutely continuous allowing the required integration by parts.

Lemma A.2. If is a bounded Lipschitz function then for every

Proof. As is Lipschitz it is Lebesgue almost everywhere differentiable and the proof of [22, Lemma ] can be followed without any change.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.