A Comparison of Iterated Optimal Stopping and Local Policy Iteration for American Options Under Regime Switching

Abstract

A theoretical analysis tool, iterated optimal stopping, has been used as the basis of a numerical algorithm for American options under regime switching (Le and Wang in SIAM J Control Optim 48(8):5193–5213, 2010). Similar methods have also been proposed for American options under jump diffusion (Bayraktar and Xing in Math Methods Oper Res 70:505–525, 2009) and Asian options under jump diffusion (Bayraktar and Xing in Math Fin 21(1):117–143, 2011). An alternative method, local policy iteration, has been suggested in Huang et al. (SIAM J Sci Comput 33(5):2144–2168, 2011), and Salmi and Toivanen (Appl Numer Math 61:821–831, 2011). Worst case upper bounds on the convergence rates of these two methods suggest that local policy iteration should be preferred over iterated optimal stopping (Huang et al. in SIAM J Sci Comput 33(5):2144–2168, 2011). In this article, numerical tests are presented which indicate that the observed performance of these two methods is consistent with the worst case upper bounds. In addition, while these two methods seem quite different, we show that either one can be converted into the other by a simple rearrangement of two loops.

Appendix: Error Bound for Local Policy Iteration with Inexact Inner Solution

In this Appendix, we generalize the result in Theorem 6.1 to include the effect of an approximate solution to Line 5 in Algorithm 6.1. We need only generalize the steps used in [21].

If $V^{n+1}$ is a solution to Eq. (5.2) then

$$\begin{aligned} \displaystyle \max _{Q^{\prime } } \left\{ -\mathcal{A }(Q^{\prime } ) V^{n+1} + \mathcal{B }(Q^{\prime }) V^{n+1} + \mathcal{C }(Q^{\prime }, V^n) \right\} = 0 , \end{aligned}$$

(9.1)

while from Algorithm 6.1, we have

$$\begin{aligned} \displaystyle \max _{Q } \left\{ -\mathcal{A }(Q) (V^{n+1})^{k+1} + \mathcal{B }(Q) (V^{n+1})^k + \mathcal{C }(Q, V^n) \right\} = \mathcal{E }^{k}. \end{aligned}$$

(9.2)

The term $\mathcal{E }^{k}$ in Eq. (9.2) takes into account that that we may not necessarily have the exact solution to Line 5 in Algorithm 6.1, if we use Algorithm 6.2.

Subtracting Eq. (9.1) from Eq. (9.2) we obtain

$$\begin{aligned} \mathcal{E }^{k}&= \displaystyle \max _{Q } \left\{ -\mathcal{A }(Q) (V^{n+1})^{k+1} + \mathcal{B }(Q) (V^{n+1})^k + \mathcal{C }(Q, V^n) \right\} \nonumber \\&\quad - \displaystyle \max _{Q^{\prime } } \left\{ -\mathcal{A }(Q^{\prime } ) V^{n+1} + \mathcal{B }(Q^{\prime }) V^{n+1} + \mathcal{C }(Q^{\prime }, V^n) \right\} \nonumber \\&\le \displaystyle \max _{Q } \left\{ -\mathcal{A }(Q) ((V^{n+1})^{k+1} - V^{n+1}) + \mathcal{B }(Q) ( (V^{n+1})^k - V^{n+1}) \right\} . \end{aligned}$$

(9.3)

If ${\hat{Q}}$ satisfies

$$\begin{aligned} {\hat{Q}} \in \mathop {\text{ arg } \text{ max }}_{Q } \left\{ \mathcal{A }(Q) ( (V^{n+1})^{k+1} - V^{n+1}) + \mathcal{B }(Q) ((V^{n+1})^k - V^{n+1}) \right\} . \end{aligned}$$

(9.4)

then, from Eq. (9.3), we have, $(E^{k+1} = (V^{n+1})^{k+1} - V^{n+1})$

$$\begin{aligned} \mathcal{A }( {\hat{Q}}) E^{k+1} \le \mathcal{B }({\hat{Q}}) E^k - \mathcal{E }^{k}~, \end{aligned}$$

(9.5)

or, since $\mathcal{A }(Q)$ is an $M$ matrix,

$$\begin{aligned} E^{k+1}&\le \mathcal{A }({\hat{Q}})^{-1} \mathcal{B }({\hat{Q}}) E^k - \mathcal{A }({\hat{Q}})^{-1} \mathcal{E }^{k}~ \le C_1 \Vert E^k \Vert _{\infty } \mathbf{{e}} + C_2 \Vert \mathcal{E }^{k} \Vert _{\infty } \mathbf{{e}}\nonumber \\ C_1&= \max _{Q } \Vert \mathcal{A }({Q})^{-1} \mathcal{B }({Q}) \Vert _{\infty }\nonumber \\ C_2&= \max _{Q } \Vert \mathcal{A }({Q})^{-1} \Vert _{\infty } \end{aligned}$$

(9.6)

where $\mathbf{{e}} = [1,1,\ldots ,1]^{\prime }$. Similarly

$$\begin{aligned} \mathcal{E }^{k}&= \displaystyle \max _{Q } \left\{ -\mathcal{A }(Q) (V^{n+1})^{k+1} + \mathcal{B }(Q) (V^{n+1})^k + \mathcal{C }(Q, V^n) \right\} \nonumber \\&- \displaystyle \max _{Q^{\prime } } \left\{ -\mathcal{A }(Q^{\prime } ) V^{n+1} + \mathcal{B }(Q^{\prime }) V^{n+1} + \mathcal{C }(Q^{\prime }, V^n) \right\} \nonumber \\&\ge \displaystyle \min _{Q } \left\{ -\mathcal{A }(Q) ( (V^{n+1})^{k+1} - V^{n+1}) + \mathcal{B }(Q) ((V^{n+1})^k - V^{n+1} ) \right\} . \end{aligned}$$

(9.7)

Hence if

$$\begin{aligned} {\bar{Q}} \in \mathop {\text{ arg } \text{ min }}_{Q \in Z} \left\{ -\mathcal{A }(Q) ( (V^{n+1})^{k+1} - V^{n+1}) + \mathcal{B }(Q) ( (V^{n+1})^k - V^{n+1}) \right\} , \end{aligned}$$

(9.8)

then

$$\begin{aligned} E^{k+1} \ge \mathcal{A }( {\bar{Q}})^{-1} \mathcal{B }({\bar{Q}}) E^k - \mathcal{A }({\hat{Q}})^{-1} \mathcal{E }^{k} \ge - C_1 \Vert E^k \Vert _{\infty } \mathbf{{e}} - C_2 \Vert \mathcal{E }^{k} \Vert _{\infty } \mathbf{{e}}~. \end{aligned}$$

(9.9)

Equations (9.6) and (9.9) then give

$$\begin{aligned} \Vert E^{k+1} \Vert _{\infty } \le C_1 \Vert E^k \Vert _{\infty } + C_2 \Vert \mathcal{E }_{ \max } \Vert _{\infty } \Vert \mathcal{E }_{ \max } \Vert _{\infty } = \max _{k} \Vert \mathcal{E }^{k} \Vert _{\infty }. \end{aligned}$$

(9.10)

From [21], we have that

$$\begin{aligned} C_1 \le \frac{\theta \hat{\lambda } \Delta \tau }{1 + \theta (r + \hat{\lambda })\Delta \tau }; \qquad \hat{\lambda } = \max _j \lambda _j \quad < 1, \end{aligned}$$

(9.11)

and $C_2$ bounded, hence

$$\begin{aligned} \Vert E^{k+1} \Vert _{\infty } \le C_1^{k+1} \Vert E^0 \Vert _{\infty } + C_2 \left( \frac{ 1 - C_1^{k+1} }{ 1 - C_1} \right) \Vert \mathcal{E }_{ \max } \Vert _{\infty }, \end{aligned}$$

(9.12)

and in view of Eq. (9.11), we obtain

$$\begin{aligned} \lim _{k \rightarrow \infty } \Vert E^{k+1} \Vert _{\infty } \le \frac{C_2}{ 1 - C_1} \Vert \mathcal{E }_{ \max } \Vert _{\infty }. \end{aligned}$$

(9.13)

Manipulation of Line 4 in Algorithm 6.2 results in

$$\begin{aligned} \mathcal{A }(Q^m) (U^{m+1} - U^m) = \max _{Q} \biggl \{ - \mathcal{A }(Q) U^m + \mathcal{B }(Q) (V^{n+1})^k + \mathcal{C }(Q, V^n) \biggr \} \end{aligned}$$

(9.14)

Recall that at convergence of Algorithm 6.2, we have

$$\begin{aligned} (V^{n+1})^{k+1} = U^{m+1}. \end{aligned}$$

(9.15)

When Algorithm 6.2 terminates, we have

$$\begin{aligned} \mathcal{E }^k = \max _{ Q } \left\{ - \mathcal{A }(Q) (V^{n+1})^{k+1} + \mathcal{B }(Q) (V^{n+1})^k + \mathcal{C }(Q, V^n) \right\} , \end{aligned}$$

(9.16)

then, from Eqs. (9.14) and (9.16) (with $U^{m+1} = (V^{n+1})^{k+1}$)

$$\begin{aligned} \mathcal{E }^k&= \mathcal{A }(Q^m) (U^{m+1} - U^m) + \max _{ Q } \left\{ - \mathcal{A }(Q) U^{m+1} + \mathcal{B }(Q) (V^{n+1})^k + \mathcal{C }(Q, V^n)\right\} \nonumber \\&- \max _{Q} \left\{ - \mathcal{A }(Q) U^m + \mathcal{B }(Q) (V^{n+1})^k + \mathcal{C }(Q, V^n) \right\} \end{aligned}$$

(9.17)

and then

$$\begin{aligned} | \mathcal{E }^k | \le | \mathcal{A }(Q^m) (U^{m+1} - U^m) | + \max _{ Q } | A(Q) (U^{m+1} - U^m) | ~, \end{aligned}$$

(9.18)

which implies

$$\begin{aligned} \Vert \mathcal{E }^k \Vert _{\infty } \le 2 \cdot \max _{Q} \Vert A(Q) \Vert _{\infty } \Vert U^{m+1} - U^m \Vert _{\infty }. \end{aligned}$$

(9.19)

Since $\mathcal{A }(Q^m)$ is bounded, $\Vert \mathcal{E }^{k} \Vert _{\infty }$ (and hence $\Vert \mathcal{E }_{ \max } \Vert _{\infty }$) can be made arbitrarily small by making $tolerance_{inner}$ small in Algorithm 6.2.