Analysis of Markov chain approximation for Asian options and occupation-time derivatives: Greeks and convergence rates

Abstract

The continuous-time Markov chain (CTMC) approximation method is a powerful tool that has recently been utilized in the valuation of derivative securities, and it has the advantage of yielding closed-form matrix expressions suitable for efficient implementation. For two types of popular path-dependent derivatives, the arithmetic Asian option and the occupation-time derivative, this paper obtains explicit closed-form matrix expressions for the Laplace transforms of their prices and the Greeks of Asian options, through the novel use of pathwise method and Malliavin calculus techniques. We for the first time establish the exact second-order convergence rates of the CTMC methods when applied to the prices and Greeks of Asian options. We propose a new set of error analysis methods for the CTMC methods applied to these path-dependent derivatives, whose payoffs depend on the average of asset prices. A detailed error and convergence analysis of the algorithms and numerical experiments substantiate the theoretical findings.

A Proof of Propositions 2.1 and 2.2 and Lemma 2.1

Proof of Proposition 2.1

The main steps of the proof are similar to that of (Cai et al. 2015, Theorem 1), thus we shall only present the steps where significant changes need to be made. In particular, note that for the case of occupation time functional, we can relax the assumption by not requiring any moment conditions on S. Suppose that f is any bounded solution to the functional equation (6). Define

$$\begin{aligned} M_t:=f(S_t)e^{-\varsigma t-\theta A_t}+\int _0^t e^{-\varsigma u-\theta A_u}du, \quad t\ge 0. \end{aligned}$$

(A.1)

We shall first establish a useful identity:

$$\begin{aligned} \frac{d}{dt}\mathbb {E}^x [f(S_t)e^{-\varsigma t-\theta A_t}]&=-\mathbb {E}^x [e^{-\varsigma t-\theta A_t}]. \end{aligned}$$

(A.2)

The proof of this identity is essentially very similar to that of Cai et al. (2015), and the following steps need adjustment: we have the same decomposition as in (Cai et al. 2015, A3): For any fixed $t\ge 0$,

$$\begin{aligned}&\frac{1}{u}\left( \mathbb {E}^x[f(S_{t+u})e^{-\varsigma (t+u)-\theta A_{t+u}}]-\mathbb {E}^x[f(S_{t})e^{-\varsigma t-\theta A_{t}}]\right) \nonumber \\&\quad =\mathbb {E}^x\left[ \frac{f(S_{t+u}) (e^{-\varsigma (t+u)-\theta A_{t+u}}-e^{-\varsigma t-\theta A_t})}{u}\right] \nonumber \\&\qquad +\mathbb {E}^x\left[ \frac{(f(S_{t+u})-f(S_t))e^{-\varsigma t-\theta A_t}}{u}\right] , \end{aligned}$$

(A.3)

and for the first term of (A.3), we have

$$\begin{aligned}&\left| \frac{f(S_{t+u})(e^{-\varsigma (t+u)-\theta A_{t+u}}-e^{-\varsigma t-\theta A_t})}{u} \right| \nonumber \\&\quad = \left| \frac{f(S_{t+u})\int _0^u (-\varsigma -\theta \mathbb {1}_{\lbrace \mathcal {A}< S_{t+s}<\mathcal {B} \rbrace })e^{ -\varsigma (t+s)-\theta A_{t+s}} ds }{u} \right| \nonumber \\&\quad \le \frac{C}{u} \int _0^u \mid \varsigma +\theta \mathbb {1}_{\lbrace \mathcal {A}< S_{t+s}<\mathcal {B}\rbrace }\mid \cdot \mid e^{-\varsigma (t+s)-\theta A_{t+s}} \mid ds\nonumber \\&\quad \le C(\mid \varsigma \mid +\mid \theta \mid )<\infty , \end{aligned}$$

(A.4)

and we do not need to assume any moment condition and the remaining analysis follows. For the second part of (A.3), just note that $\mid \mathcal {G}f(x)\mid =\mid (\varsigma +\theta )f(x)-1 \mid \le (\mid \varsigma \mid +\mid \theta \mid )C+1$, then we can apply dominated convergence theorem to similarly conclude that the identity (A.2) holds. Then we can establish the remaining results using the lemma similarly as Cai et al. (2015) does. This completes the proof. $\square$

Proof of Proposition 2.2

The proof is similar to that of Cai et al. (2015), and the following key steps need adjustment: For the first corresponding term with the same form as (A.3), when $u\in (0,u_0)$ with any fixed $u_0>0$

$$\begin{aligned}&\left| \frac{f(S_{t+u})(e^{-\varsigma (t+u)-\theta A_{t+u}-\gamma B_{t+u}}-e^{-\varsigma t-\theta A_t-\gamma B_{t}})}{u} \right| \nonumber \\&\quad = \left| \frac{f(S_{t+u})\int _0^u (-\varsigma -\theta \mathbb {1}_{\lbrace \mathcal {A}< S_{t+s}<\mathcal {B} \rbrace }-\gamma S_{t+s}\mathbb {1}_{\lbrace \mathcal {A}< S_{t+s}<\mathcal {B} \rbrace })e^{ -\varsigma (t+s)-\theta A_{t+s}-\gamma B_{t+s}} ds }{u} \right| \nonumber \\&\quad \le \frac{C}{u} \int _0^u \mid \varsigma +\theta \mathbb {1}_{\lbrace \mathcal {A}< S_{t+s}<\mathcal {B} \rbrace }+\gamma S_{t+s} \mathbb {1}_{\lbrace \mathcal {A}< S_{t+s} <\mathcal {B} \rbrace }\mid \cdot \mid e^{-\varsigma (t+s)-\theta A_{t+s}-\gamma B_{t+s}} \mid ds\nonumber \\&\quad \le C(\mid \varsigma \mid +\mid \theta \mid +\mid \gamma S_{t+s}\mid )\nonumber \\&\quad \le C (\mid \varsigma \mid +\mid \theta \mid +\mid \gamma \mid \max \limits _{t\le s \le t+u}S_s)\le C (\mid \varsigma \mid +\mid \theta \mid +\mid \gamma \mid \max \limits _{t\le s \le t+u_0}S_s), \end{aligned}$$

(A.5)

and given that we have assumed the moment condition: $\mathbb {E}^x[S_t^{1+\epsilon }]<\infty$ for some $\epsilon >0$, we have from Doob’s inequality that the right-hand side of (A.5) is bounded. The remaining steps follow similarly from Cai et al. (2015). This completes the proof. $\square$

Proof of Lemma 2.1

Under the condition that $\hbox {Re}(\varsigma ) >\Vert \mathbf{G}-\theta \mathbf{A}-\gamma \mathbf{B}\Vert$, then $\left\| \frac{\mathbf{G}-\theta \mathbf{A}-\gamma \mathbf{B}}{\varsigma }\right\| <1$. From Hom and Johnson (1985, Corollary 5.6.16), we can obtain that

$$\begin{aligned} \left( \mathbf{I}- \frac{\mathbf{G}-\theta \mathbf{A}-\gamma \mathbf{B}}{\varsigma }\right) ^{-1}=\sum ^{\infty }_{k=0}\left( \frac{\mathbf{G}-\theta \mathbf{A}-\gamma \mathbf{B}}{\varsigma }\right) ^k. \end{aligned}$$

(A.6)

Plugging (9) and (A.6) into (7) yields

$$\begin{aligned} \int _0^{\infty } e^{-\varsigma t} \mathbb {E}^{ x}[e^{-\theta A_t-\gamma B_t}]dt \approx \sum ^{\infty }_{k=0}\frac{\mathbf{e}\cdot \left( \mathbf{G}-\theta \mathbf{A}-\gamma \mathbf{B}\right) ^k\cdot \mathbf{1}}{\varsigma ^{k+1}}. \end{aligned}$$

Taking the inverse Laplace transform w.r.t. $\varsigma$ on both hand sides of the above equation, we can obtain

$$\begin{aligned} \mathbb {E}^{x}[e^{-\theta A_t-\gamma B_t}]\approx \sum ^{\infty }_{k=0}\frac{\mathbf{e}\cdot \left( \mathbf{G}-\theta \mathbf{A}-\gamma \mathbf{B}\right) ^k\cdot \mathbf{1}}{k!}t^k=\mathbf{e}\cdot e^{\left( \mathbf{G}-\theta \mathbf{A}-\gamma \mathbf{B}\right) t }\cdot \mathbf{1}. \end{aligned}$$

This completes the proof. $\square$