Asian option as a fixed-point

Abstract

We characterize the price of an Asian option, a financial contract, as a fixed-point of a non-linear operator. In recent years, there has been interest in incorporating changes of regime into the parameters describing the evolution of the underlying asset price, namely the interest rate and the volatility, to model sudden exogenous events in the economy. Asian options are particularly interesting because the payoff depends on the integrated asset price. We study the case of both floating- and fixed-strike Asian call options with arithmetic averaging when the asset follows a regime-switching geometric Brownian motion with coefficients that depend on a Markov chain. The typical approach to finding the value of a financial option is to solve an associated system of coupled partial differential equations. Alternatively, we propose an iterative procedure that converges to the value of this contract with geometric rate using a classical fixed-point theorem.

Appendix A: Proofs

Proof of Lemma 2.1

Define the probability measure $P^*$ equivalent to P via the Radon Nikodym derivative,

$$\begin{aligned} \frac{{\text {d}}P^*}{{\text {d}}P}\mid _{{\mathcal {F}}_T}= {\mathcal {E}}_T \end{aligned}$$

where,

$$\begin{aligned} {\mathcal {E}}_t:=\exp \left( \int _0^t \sigma (Y_u){\text {d}}B_u-\frac{1}{2}\int _0^t \sigma ^2(Y_u){\text {d}}u \right) . \end{aligned}$$

The call option satisfies:

$$\begin{aligned} \begin{aligned} \frac{C(s,x,a,i)}{x}&={\mathbb {E}}_{s,x,a,i}\left[ \frac{e^{-\int _s^T r(Y_u){\text {d}}u} X_T}{x} \left( 1-\frac{1}{T-t_0}\frac{A_T}{X_T}\right) ^+ \right] \\&={\mathbb {E}}_{s,x,a,i}\left[ {\mathcal {E}}_T {\mathcal {E}}^{-1}_s \,e^{-\delta (T-s)} \left( 1-\frac{1}{T-t_0}\frac{A_T}{X_T}\right) ^+ \right] \\&={\mathbb {E}}_{s,x,a,i}^*\left[ e^{-\delta (T-s)}\left( 1-\frac{1}{T-t_0}\frac{A_T}{X_T}\right) ^+ \right] \le 1. \end{aligned} \end{aligned}$$

where the expectation ${\mathbb {E}}^*$ is with respect to $P^*$. The result is now clear. $\square$

Proof of Lemma 2.2

Following up the proof of Lemma 2.1, we have that,

$$\begin{aligned} C^0(s,x,a,i)={\mathbb {E}}_{s,x,a,i}^*\left[ e^{-\delta (T-s)}\left( x-\frac{x}{T-t_0}\frac{A_T}{X_T}\right) ^+\,\mid \, Y_t=i, \forall t\in [s,T] \right] , \end{aligned}$$

and ${\hat{B}}_u=B_u-\int _0^u\sigma (Y_s){\text {d}}s$ is a Brownian motion under $P^*$. Here,

$$\begin{aligned} x\frac{A_T}{X_T} =\frac{x}{X_T}\left( a+\int _s^T X_u\,{\text {d}}u\right) . \end{aligned}$$

The process $(B^*_u)_{s\le u\le T}$, defined by $B^*_u:=B^*_{s}+{\hat{B}}_{T+s-u}-{\hat{B}}_T$ with $B^*_{s}$ a constant, is also a Brownian motion under $P^*$ starting at $B^*_{s}$. Now, conditional on $X_s=x$, $A_s=a$, and $Y_t=i$ for all $t\in [s,T]$,

$$\begin{aligned} \begin{aligned} \frac{x}{X_T}&=\exp \left( \sigma (i)({\hat{B}}_s-{\hat{B}}_T)+\left[ r(i)-\delta +\frac{\sigma ^2(i)}{2}\right] (s-T)\right) \\&{\mathop {=}\limits ^\mathrm{law}}\exp \left( \sigma (i)(B^*_T-B^*_s)+\left[ \delta -r(i)-\frac{\sigma ^2(i)}{2}\right] (T-s)\right) \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} x\int _s^T \frac{X_u}{X_T}&=\int _{s}^T x\exp \left( \sigma (i)({\hat{B}}_u-{\hat{B}}_T)+\left[ r(i)-\delta +\frac{\sigma ^2(i)}{2}\right] (u-T) \right) {\text {d}}u \\&{\mathop {=}\limits ^\mathrm{law}}\int _{s}^T x\exp \left( \sigma (i)(B^*_{T+s-u}-B^*_{s})+\left[ \delta -r(i)-\frac{1}{2}\sigma ^2(i)\right] (T-u) \right) {\text {d}}u \\&=\int _{s}^T x\exp \left( \sigma (i)(B^*_w-B^*_{s})+\left[ \delta -r(i)-\frac{1}{2}\sigma ^2(i)\right] (w-s) \right) {\text {d}}w \end{aligned} \end{aligned}$$

where the third equality is obtained after the change of variable $w=T+s-u$. Therefore, $C^0(s,x,a,i)$ is given by:

$$\begin{aligned} {\mathbb {E}}^*_{s,x,a,i}\left[ e^{-\delta (T-s)} \left( x-\frac{a}{x(T-t_0)}X_T^*-\frac{T-s}{T-t_0}\frac{1}{T-s}\int _s^T X_u^*{\text {d}}u \right) ^+\right] \end{aligned}$$

where the underlying process $X^*$ follows:

$$\begin{aligned} {\text {d}}X^*_t=X^*_t[(\delta -r(i)){\text {d}}t+\sigma (i){\text {d}}B^*_t], \quad X^*_{s}=x, \qquad t\ge s. \end{aligned}$$

Defining the parameters $\lambda =\frac{a}{x(T-t_0)}$ and $\beta =\frac{T-s}{T-t_0}$ the proof is complete.

$\square$

Proof of Lemma 2.3

Part (i). Let $s\le t_0$. Then

$$\begin{aligned} \begin{aligned} C_K(s,x,0,i)&\le {\mathbb {E}}_{s,x,0,i}\left[ e^{-\int _s^T r(Y_u){\text {d}}u}\frac{A_T}{T-t_0}\right] \\&\le \frac{1}{T-t_0}\int _{t_0}^T {\mathbb {E}}_{s,x,0,i}\left[ e^{-\int _s^t r(Y_u){\text {d}}u}X_t\right] {\text {d}}t \\&\le \frac{x}{T-t_0}\int _{t_0}^Te^{-\delta (t-s)}{\text {d}}t \le x. \end{aligned} \end{aligned}$$

Part(ii). Let $s>t_0$. Then

$$\begin{aligned} \begin{aligned} C_K(s,x,a,i)&\le {\mathbb {E}}_{s,x,a,i}\left[ e^{-\int _s^T r(Y_u){\text {d}}u}\left( \frac{a+\int _s^T X_t {\text {d}}t}{T-t_0}\right) \right] \\&\le \frac{a}{T-t_0} +\left( \frac{T-s}{T-t_0}\right) \frac{1}{T-s}\int _{s}^T {\mathbb {E}}_{s,x,a,i}\left[ e^{-\int _s^t r(Y_u){\text {d}}u}X_t\right] {\text {d}}t \\&\le \frac{a}{T-t_0}+\left( \frac{T-s}{T-t_0}\right) x \le \frac{a}{T-t_0}+x. \end{aligned} \end{aligned}$$

$\square$