The Stability Analysis of Predictor–Corrector Method in Solving American Option Pricing Model

Abstract

In this paper, a new technique is investigated to speed up the order of accuracy for American put option pricing under the Black–Scholes (BS) model. First, we introduce the mathematical modeling of American put option, which leads to a free boundary problem. Then the free boundary is removed by adding a small and continuous penalty term to the BS model that cause American put option problem to be solvable on a fixed domain. In continuation we construct the method of lines (MOL) in space and reach a non-linear problem and we show that the proposed MOL is more stable than the other kinds. To deal with the non-linear problem, an algorithm is used based on the predictor–corrector method which corresponds to two parameters, $\theta$ and $\phi$. These parameters are chosen optimally using a rational approximation to determine the order of time convergence. Finally in numerical results a second order convergence is shown in both space and time variables.

Appendices

Appendix 1

1.1 Solving an Ordinary Differential Equation by Penalty Method

We consider a simple ordinary differential equation

$$\begin{aligned} \left\{ \begin{array}{ccccccc} &{}u'=-u,\\ &{}u(0)=2, \end{array}\right. \end{aligned}$$

(34)

with the additional constraint that

$$\begin{aligned} u(t)\ge 1. \end{aligned}$$

(35)

The solution to this problem can be computed analytically and is given by

$$\begin{aligned} u(t) =\left\{ \begin{array}{lllllll} 2e^{-t}&{}\quad \text{ for } \;\;t\le ln2,\\ 1&{}\quad \text{ for } \;\;t > ln2. \end{array}\right. \end{aligned}$$

(36)

Suppose, however, that we want to solve the initial-value problem (34)–(35) numerically. Then we would have to check, for each time step, whether the constraint is satisfied or not. Let $u_n$ be a numerical approximation of $u(t_n)$ where $t_n=n\varDelta t$ , and $\varDelta t >0$ is the time step. We compute a numerical solution of the initial-value problem (34)–(35) using an explicit finite-difference scheme:

$$\begin{aligned} u_{n+1}=\max ((1-\varDelta t)u_n,1),\;\;n\ge 0, \end{aligned}$$

(37)

where $u_0=2$. This corresponds to a Brennan-Schwartz type of algorithm for pricing American put options Brennan and Schwartz (1977).

An equation which approximates this property fairly well can be derived by adding an extra term to equation (35). Consider the initial-value problem

$$\begin{aligned} \left\{ \begin{array}{lllllll} &{}v'=-v+\frac{\epsilon }{v+\epsilon -1},\\ &{}v(0)=2, \end{array}\right. \end{aligned}$$

(38)

where $\epsilon >0$ is a small parameter. Note that, initially, $v = 2$, so the penalty term is

$$\begin{aligned} \frac{\epsilon }{v+\epsilon -1}. \end{aligned}$$

For more detail see Nielsen et al. (2002).

Now we use predictor–corrector method developed in Sect. 4 for solving (38). For simplicity, we assume a uniform time step size $\varDelta t$ and $T=1+\varDelta t\theta$, then for the corrector

$$\begin{aligned} Tv_{n+1}=\left( 1-\frac{\varDelta t}{2}\right) v_n+\frac{\varDelta t}{2}\frac{\epsilon }{v_n+\epsilon -1}+\varDelta t(\theta +\phi )\frac{\epsilon }{\widehat{v}_{n+1}+\epsilon -1}, \end{aligned}$$

(39)

the predicted solution $\widehat{v}^{n+1}$ is given by

$$\begin{aligned} T\widehat{v}_{n+1}=(1-\varDelta t(1-\theta ))v_n+\varDelta t\frac{\epsilon }{v_n+\epsilon -1}. \end{aligned}$$

(40)

Table 2, show the maximum absolute error:

$$\begin{aligned} Error=\max _{n}\mid u(t_n) -v_n\mid , \end{aligned}$$

where $u(t_n)$ and $v_n$ are the exact and approximate solution of Eq. (34), respectively.

Table 2 Error of penalty method with $\varDelta t=\frac{\epsilon }{1+\epsilon }$

Appendix 2

1.1 Penalty Method for American Put Option Nielsen et al. (2002)

In this section we derive an implicit and a semi-implicit scheme. For both schemes we assume that

$$\begin{aligned} C > rE. \end{aligned}$$

(41)

Under this mild assumption it turns out that the implicit scheme is stable, whereas the semi-implicit scheme is stable if the additional condition (8) (see Sect. 3) is satisfied.

Using the notation introduced above, we consider forward FD for $\frac{\partial V}{\partial S}$ and obtain the following scheme:

$$\begin{aligned}&\frac{V^{n+1}_j-V^n_j}{\varDelta t} + \frac{1}{2}\sigma ^2 S^2_j\frac{V^n_{j+1}-2V^n_j+V^n_{j-1}}{(\varDelta S)^2}+r S_j \frac{V^n_{j+1}-V^n_j}{\varDelta S}\nonumber \\&\qquad -\,rV^n_j+\frac{\epsilon C}{V^{n+\frac{1}{2}}_j+\epsilon -q_j}=0. \end{aligned}$$

(42)

Here, we put $V^{n+\frac{1}{2} }_j= V^n_j$ in the fully implicit scheme and $V^{n+\frac{1}{2}}_j = V^{n+1}_j$ in the semi-implicit scheme. The scheme (42) can be rearranged as

$$\begin{aligned}&\left( 1+r\varDelta t+\frac{\varDelta t}{(\varDelta S)^2 }S^2_j\sigma ^2+r\frac{\varDelta t}{\varDelta S}S_j\right) V^n_j=V^{n+1}_j\nonumber \\&\quad +\left( \frac{1}{2}\frac{\varDelta t}{(\varDelta S)^2 }S^2_j\sigma ^2+r\frac{\varDelta t}{\varDelta S}S_j\right) V^n_{j+1}+\left( \frac{1}{2}\frac{\varDelta t}{(\varDelta S)^2 }S^2_j\sigma ^2\right) V^n_{j-1}\nonumber \\&\quad +\frac{\epsilon \varDelta tC}{V^{n+\frac{1}{2}}_j+\epsilon -q_j}. \end{aligned}$$

(43)

Our aim is to show that

$$\begin{aligned} V^n_j \ge \max (q_j,0), \forall \; j,n. \end{aligned}$$

(44)

We do this in two steps; first, we show that

$$\begin{aligned} V^n_j \ge q_j \forall j, \end{aligned}$$

(45)

and next that

$$\begin{aligned} V^n_j \ge 0. \end{aligned}$$

(46)

In order to prove (45), we introduce

$$\begin{aligned} u^n_j=V^n_j-q_j. \end{aligned}$$

(47)

By substituting (47) in (43) and that $q_j=E-S_j$, we have

$$\begin{aligned}&\left( 1+r\varDelta t+\frac{\varDelta t}{(\varDelta S)^2 }S^2_j\sigma ^2+r\frac{\varDelta t}{\varDelta S}S_j\right) u^n_j=u^{n+1}_j\nonumber \\&\quad +\left( \frac{1}{2}\frac{\varDelta t}{(\varDelta S)^2 }S^2_j\sigma ^2+r\frac{\varDelta t}{\varDelta S}S_j\right) u^n_{j+1} +\left( \frac{1}{2}\frac{\varDelta t}{(\varDelta S)^2 }S^2_j\sigma ^2\right) u^n_{j-1}\nonumber \\&\quad +\frac{\epsilon \varDelta tC}{u^{n+\frac{1}{2}}_j+\epsilon }-r\varDelta t(E-S_j), \end{aligned}$$

(48)

where $u^{n+\frac{1}{2}}_j = u^n_j$ in the fully implicit case and $u^{n+\frac{1}{2}}_j = u^{n+1}_j$ in the semiimplicit case. Define

$$\begin{aligned} u^n=\min _j u^n_j, \end{aligned}$$

(49)

and let $k$ be an index such that

$$\begin{aligned} u^n_k=u^n. \end{aligned}$$

(50)

For $j = k,$ it follows from (48) that

$$\begin{aligned}&\left( 1+r\varDelta t+\frac{\varDelta t}{(\varDelta S)^2 }S^2_k\sigma ^2+r\frac{\varDelta t}{\varDelta S}S_k\right) u^n \ge u^{n+1}_k\nonumber \\&\quad +\left( \frac{1}{2}\frac{\varDelta t}{(\varDelta S)^2 }S^2_k\sigma ^2+r\frac{\varDelta t}{\varDelta S}S_k\right) u^n +\left( \frac{1}{2}\frac{\varDelta t}{(\varDelta S)^2 }S^2_k\sigma ^2\right) u^n\nonumber \\&\quad +\frac{\epsilon \varDelta tC}{u^{n+\frac{1}{2}}_k+\epsilon }-r\varDelta tE, \end{aligned}$$

(51)

or

$$\begin{aligned} (1+r\varDelta t)u^n \ge u^{n+1}_k+\frac{\epsilon \varDelta t C }{u^{n+\frac{1}{2}}+\epsilon }-r \varDelta tE. \end{aligned}$$

(52)

Let us now consider the fully implicit case. Here (52) takes the form

$$\begin{aligned} (1+r\varDelta t)u^n-\frac{\epsilon \varDelta t C }{u^n+\epsilon }+r \varDelta tE\ge u^{n+1}_k\ge u^{n+1}. \end{aligned}$$

(53)

If we assume that

$$\begin{aligned} u^{n+1} \ge 0, \end{aligned}$$

(54)

then we have

$$\begin{aligned} F(u^n)\ge 0, \end{aligned}$$

(55)

where

$$\begin{aligned} F(x)=(1+r\varDelta t)x-\frac{\epsilon \varDelta t C }{x+\epsilon }+r \varDelta tE. \end{aligned}$$

(56)

Since

$$\begin{aligned} F(0)=\varDelta t(rE-C)\le 0, \end{aligned}$$

(57)

(see (41)), and

$$\begin{aligned} F'(x)=1+r\varDelta t+\frac{\epsilon \varDelta t C }{(x+\epsilon )^2} >0, \end{aligned}$$

(58)

it follows from (55) that

$$\begin{aligned} u^n\ge 0. \end{aligned}$$

(59)

Consequently, by induction on $n$, it follows from (47) that

$$\begin{aligned} V^n_j\ge q_j, \forall j,\quad n=N+1, N,\ldots ,0. \end{aligned}$$

(60)

Next we consider the semi-implicit scheme and we assume that (8) holds. It follows from (52) that

$$\begin{aligned} (1+r \varDelta t)u^n \ge \frac{u^{n+1}_k(u^{n+1}_k+\epsilon )+\epsilon \varDelta tC-r\varDelta tE(u^{n+1}_k+\epsilon )}{u^{n+1}_k+\epsilon }, \end{aligned}$$

(61)

we assume that $u^{n+1}\ge 0$, and thus $u^{n+1}_k \ge 0$.

Let

$$\begin{aligned} G(x)=x(x+\epsilon )+\epsilon \varDelta tC-r\varDelta tE(x+\epsilon ), \end{aligned}$$

(62)

then

$$\begin{aligned} G(0)=\varDelta t\epsilon (C-rE) \ge 0, \end{aligned}$$

(63)

(see (41)), and

$$\begin{aligned} G'(x)=2x+\epsilon -r\varDelta tE, \end{aligned}$$

(64)

so $G'(x) \ge 0$ for $x \ge 0$ provided that (8) holds. Hence, we have

$$\begin{aligned} u^n_j \ge 0, \end{aligned}$$

(65)

and thus by (47)

$$\begin{aligned} V^n_j \ge q_j,\forall j, \quad n = N + 1, N, N - 1,\ldots , 0. \end{aligned}$$

(66)

Next we consider (46), i.e, we want to show that

$$\begin{aligned} V^n_j \ge 0. \end{aligned}$$

(67)

As above, we define

$$\begin{aligned} V^n= \min _jV^n_j, \end{aligned}$$

(68)

and let $k$ be an index such that

$$\begin{aligned} V^n_k= V^n. \end{aligned}$$

(69)

It follows from (43) that

$$\begin{aligned}&\left( 1+r\varDelta t+\frac{\varDelta t}{(\varDelta S)^2 }S^2_k\sigma ^2+r\frac{\varDelta t}{\varDelta S}S_k\right) V^n \ge V^{n+1}_k\nonumber \\&\quad +\left( \frac{1}{2}\frac{\varDelta t}{(\varDelta S)^2 }S^2_k\sigma ^2+r\frac{\varDelta t}{\varDelta S}S_k\right) V^n +\left( \frac{1}{2}\frac{\varDelta t}{(\varDelta S)^2 }S^2_k\sigma ^2\right) V^n\nonumber \\&\quad +\frac{\epsilon \varDelta tC}{V^{n+\frac{1}{2}}_k+\epsilon -q_k}, \end{aligned}$$

(70)

or

$$\begin{aligned} (1+r\varDelta t)V^n\ge V^{n+1}+\frac{\epsilon \varDelta tC }{V^{n+\frac{1}{2}}_k+\epsilon -q_k }. \end{aligned}$$

(71)

Since we have just seen that

$$\begin{aligned} V^{n+\frac{1}{2}}_k\ge q_k, \end{aligned}$$

(72)

both in the fully implicit and in the semi-implicit case, it follows from (71) that

$$\begin{aligned} (1+r\varDelta t)V^n\ge V^{n+1}, \end{aligned}$$

(73)

and then it follows by induction on $n$ that

$$\begin{aligned} V^n_j\ge 0, \quad \forall j, \quad n=N+1, N,\ldots ,0. \end{aligned}$$

(74)

Theorem 2

If (41) holds, the numerical solution computed by the fully implicit scheme (42) satisfies the bound

$$\begin{aligned} V^n_j\ge \max (E-S_j,0), \quad \forall j, \quad n=N+1, N,\ldots ,0. \end{aligned}$$

(75)

Similarly, if (41) and (8) hold, the numerical solution computed by the semi-implicit version of (42) satisfies the bound (75).

We can implement backward and central FDs similar to forward FDs for $\frac{\partial V}{\partial S}$ and obtain similar results.