Option Pricing Under a Stochastic Interest Rate and Volatility Model with Hidden Markovian Regime-Switching

Dong-Mei Zhu; Jiejun Lu; Wai-Ki Ching; Tak-Kuen Siu

doi:10.1007/s10614-017-9754-9

Computational Economics

February 2019, Volume 53, Issue 2, pp 555–586 | Cite as

Option Pricing Under a Stochastic Interest Rate and Volatility Model with Hidden Markovian Regime-Switching

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Dong-Mei Zhu
Jiejun Lu
Wai-Ki Ching
Tak-Kuen Siu

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First Online: 22 September 2017

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Abstract

In this paper we discuss an option pricing problem in a hidden Markovian regime-switching model with a stochastic interest rate and volatility. Regime switches are attributed to structural changes in an hidden economic environment and are described by a continuous-time, finite-state, unobservable Markov chain. The model is then applied to the valuation of a standard European option. By means of the standard separation principle, filtering and option valuation problems are separated. Robust filters for the hidden states of the economy and their robust filtered estimates of unknown parameters from the expectation maximization algorithm are presented based on standard techniques in filtering theory. Then an explicit expression of a conditional characteristic function relevant to option pricing is presented and the valuation of the option is discussed using the inverse Fourier transformation approach. Using the limiting behavior of the conditional characteristic function, an efficient implementation of the transform inversion integral is considered. Numerical experiments are given to illustrate the flexibility of filtering algorithms and the significance of regime-switching in option pricing.

Keywords

Option pricing Hidden Markov model (HMM) Regime-switching Characteristic function Fourier transformation

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Notes

Acknowledgements

A two-page abstract of the paper has been accepted for presentation in 2016 International Congress on Banking, Economics, Finance, and Business, 24–26 June 2016, Sapporo, Japan. This research work was supported by Research Grants Council of Hong Kong under Grant Number 17301214, HKU Strategic Theme on Computation and Information and National Natural Science Foundation of China under Grant Number 71601044.

Appendix

The Proofs of Propositions 6.1–6.4 in “Appendix” follow the methods discussed in the literature (see Elliott et al. 1995). The Proof of Proposition 6.5 follows the basic idea in van Haastrecht et al. (2009).

Proof of Proposition 1

Given that

$$\begin{aligned} H_t=H_0 + \int _0^t \alpha _s ds+\int _0^t \beta '_s dM_s+\int _0^t \delta '_s dW_s \end{aligned}$$

and

$$\begin{aligned} X_t=X_0+\int _0^t A X_sds+M_t, \end{aligned}$$

then

$$\begin{aligned} H_t X_t= & {} H_0 X_0+\int _0^t\alpha _s X_s ds+\int _0^tX_s \beta '_sdM_s\\&+\int _0^t X_s\delta '_sdW_s+\int _0^t H_sAX_sds+\int _0^t H_sd M_s + \sum _{0<s\le t}\beta '_s\Delta X_s\Delta X_s. \end{aligned}$$

Here

$$\begin{aligned} \sum _{0<s\le t}\beta '_s\Delta X_s\Delta X_s= & {} \sum _{i,j=1}^N\int _0^t \left( {\beta '}_s^j-{\beta '}_s^i\right) \langle X_{r-},e_i\rangle \langle e_j, dM_s\rangle ( e_j-e_i)\\&+\,\sum _{i,j=1}^N\int _0^t\langle {\beta '}_s^jX_s-{\beta '}_s^iX_s,e_i\rangle a_{ji}ds ( e_j-e_i). \end{aligned}$$

Set

$$\begin{aligned} \Lambda _t=\exp \left( \int _0^t \langle \gamma ^{-1}\mu _s,\gamma ^{-1}dY_s\rangle -\frac{1}{2}\int _0^t \langle \gamma ^{-1}\mu _s,\gamma ^{-1}\mu _s\rangle ds\right) , \end{aligned}$$

which implies that

$$\begin{aligned} d \Lambda _t=\Lambda _t\langle \gamma ^{-1}\mu _t,\gamma ^{-1}dY_t \rangle . \end{aligned}$$

Then it gives

$$\begin{aligned} \Lambda _tH_tX_t= & {} H_0X_0+\int _0^t \Lambda _s \alpha _s X_s ds+\int _0^t \Lambda _{s-} X_{s-} \beta '_sdM_s\\&+\int _0^t \Lambda _{s-} X_{s-}\delta '_s\gamma ^{-1}dY_s +\int _0^t\Lambda _s H_sAX_sds+\int _0^t\Lambda _{s-} H_d M_s\\&+\sum _{i,j=1}^N\int _0^t \left( {\beta '}_s^j-{\beta '}_s^i\right) \langle \Lambda _{s-}X_{s-},{\varvec{e}}_i\rangle \left\langle {\varvec{e}}_j, dM_s\right\rangle ( {\varvec{e}}_j-{\varvec{e}}_i)\\&+\sum _{i,j=1}^N\int _0^t\langle \Lambda _s{\beta '}_s^jX_s-\Lambda _s{\beta '}_s^iX_s,{\varvec{e}}_i\rangle a_{ji}ds ( {\varvec{e}}_j-{\varvec{e}}_i) \\&+\int _0^t \Lambda _s \langle \gamma ^{-1}\mu _s,\gamma ^{-1}dY_s\rangle H_sX_s. \end{aligned}$$

Denote $\sigma (H_t X_t)=\bar{E}(\Lambda _t H_t X_t|\mathcal {F}_t).$ Under $\bar{P}$, $\gamma ^{-1}Y$ is a n-dimensional standard Brownian motion, thus

$$\begin{aligned} \sigma (H_tX_t)= & {} \sigma (H_0X_0)+\int _0^t \sigma (\alpha _s X_s) ds+\int _0^tA\sigma (H_sX_s)ds\\&+\int _0^t\sum _{i,j=1}^N\langle \sigma ({\beta '}_s^jX_s-{\beta '}_s^iX_s),{\varvec{e}}_i\rangle a_{ji}ds ( {\varvec{e}}_j-{\varvec{e}}_i) +\int _0^t \sigma \left( X_{s}\delta '_s\gamma ^{-1}dY_s\right) \\&+\int _0^t \sigma \left( \langle \gamma ^{-1}\mu _s,\gamma ^{-1}dY_s\rangle H_sX_s\right) . \end{aligned}$$

Since

$$\begin{aligned} \sigma \left( X_{s}\delta '_s\gamma ^{-1}dY_s\right) = \sum _{k=1}^n \left( \gamma ^{-1}dY_s\right) _k\sigma \left( \delta _s^k X_{s}\right) \end{aligned}$$

and

$$\begin{aligned} \sigma \left( \langle \gamma ^{-1}\mu _s,\gamma ^{-1}dY_s\rangle H_sX_s\right) =\text {diag}({\varvec{\mu }}'({\gamma }\gamma ')^{-1}dY_s)\sigma (H_sX_s), \end{aligned}$$

the result follows.

Proof of Lemma 1

Take $H_t={\mathcal J}_t^{ij}$, which implies that

$$\begin{aligned} H_0=0, \alpha _s=\langle X_s,\mathbf{e}_i\rangle a_{ji}, \beta _s=\langle X_s,\mathbf{e}_i\rangle e_{j},\delta _s=0\in \mathcal {R}^n. \end{aligned}$$

Then the Zakai equation for $\sigma \left( {\mathcal J}_t^{ij}X_t\right) $ can be obtained by substituting in Proposition 1.

Take $H_t={\mathcal O}_t^i$, which implies that

$$\begin{aligned} H_0=0, \alpha _s=\langle X_s,\mathbf{e}_i\rangle a_{ji}, \beta _s=0\in \mathcal {R}^N,\delta _s=0\in \mathcal {R}^n. \end{aligned}$$

Then the Zakai equation for $\sigma \left( {\mathcal O}_t^iX_t\right) $ can be obtained by substituting in Proposition 1.

Take $H_t={\mathcal T}_t^{ij}$, which implies that

$$\begin{aligned} H_0=0, \alpha _s=\langle X_s,\mathbf{e}_i\rangle \mathbf {{\varvec{\mu }}}_{ji}, \beta _s=0\in \mathcal {R}^N,\delta _s=\langle X_s,\mathbf{e}_i\rangle (\mathbf {\gamma }_{j-row})'. \end{aligned}$$

Then the Zakai equation for $\sigma \left( {\mathcal T}_t^{ij}X_t\right) $ can be obtained by substituting in Proposition 1.

Proof of Proposition 2

For $i=1,2,\dots ,N$, define

$$\begin{aligned} \phi _i(t):=\exp \left( {{{\varvec{\mu }}}^i}'(\gamma \gamma ')^{-1}Y(t)-\frac{1}{2} {{{\varvec{\mu }}}^i}'(\gamma \gamma ')^{-1}{{\varvec{\mu }}}^it\right) \end{aligned}$$

where ${{\varvec{\mu }}}^i$ represents the i-th column of the matrix ${{\varvec{\mu }}}$. Note that

$$\begin{aligned} d\varPhi _t= & {} \mathrm{diag}(d \phi _1(t), d \phi _2(t),\dots ,d \phi _N(t))\\= & {} \varPhi _t \mathrm{diag}({{\varvec{\mu }}}'(\gamma \gamma ')^{-1}dY(t)) \end{aligned}$$

and

$$\begin{aligned} d\varPhi ^{-1}_t= & {} -\varPhi ^{-2}_td\varPhi _t+ \varPhi ^{-3}_td\varPhi _t^2\\= & {} -\varPhi _t^{-1} \mathrm{diag}({{\varvec{\mu }}}'(\gamma \gamma ')^{-1}dY(t)) + \varPhi _t^{-1}\mathrm{diag}(({\varvec{\mu }}'{{\gamma ^{-1}}}'\mathbf {\gamma }^{-1}{\varvec{\mu }})_{ii})dt. \end{aligned}$$

The transformed quantity is defined by

$$\begin{aligned} \bar{\sigma }(X(t))=\varPhi _t^{-1}\sigma (X(t)) \end{aligned}$$

which satisfies

$$\begin{aligned} d\bar{\sigma }(X(t))= & {} d\varPhi _t^{-1} {\sigma }(X(t))+\varPhi _t^{-1} d{\sigma }(X(t))+d\varPhi _t^{-1} d{\sigma }(X(t)). \end{aligned}$$

The equation for $d\bar{\sigma }(X(t))$ can be obtained by substituting $\varPhi _t^{-1}$, ${\sigma }(X(t))$, $d\varPhi _t^{-1}$ and $d{\sigma }(X(t))$ in to the above function. Equations for $d\bar{\sigma }({\mathcal J}_t^{ij}X(t))$, $d\bar{\sigma }({\mathcal O}_t^i X(t))$ and $d\bar{\sigma }({\mathcal T}_t^{ij} X(t))$ are obtained by applying similar method.

Proof of Proposition 3

For the time interval [0, T], an equidistant time discretization is considered:

$$\begin{aligned} 0=t_0<t_1<\dots<t_k<\dots <t_K=T, \end{aligned}$$

where $\delta =T/K$ and $t_k=k\delta $. Integrating both sides of the equation for $d \bar{\sigma }( X(t))$ in Proposition 2 from $t_k$ to $t_{k+1},$

$$\begin{aligned} \int _{t_k}^{t_{k+1}}d \bar{\sigma }( X(t))=\int _{t_k}^{t_{k+1}}\varPhi _t^{-1}A\varPhi _t\bar{\sigma }( X(t))dt. \end{aligned}$$

The above equation can be transformed to

$$\begin{aligned} \bar{\sigma }( X_{k+1})- \bar{\sigma }( X_{k})=\varPhi _k^{-1}A\varPhi _k \bar{\sigma }( X_k)\delta \end{aligned}$$

where $\varPhi _t^{-1},$$\varPhi _t$ and $ \bar{\sigma }( X(t))$ are assumed to be unchanged during the time interval $\delta $ when $\delta $ is small enough. By substituting $ \bar{\sigma }( X_k)=\varPhi _k^{-1}{\sigma }( X_k), $ then

$$\begin{aligned} \sigma (X_{k+1})=\varPsi _{k,k+1}(I+\delta A)\sigma (X_k). \end{aligned}$$

Here $\varPsi _{k,k+1}=\varPhi _{k+1}\varPhi _k^{-1}.$ The equations for $\sigma \left( {\mathcal J}_{k+1}^{ij}X_{k+1}\right) $ and $\sigma ({\mathcal O}_{k+1}^i X_{k+1})$ are obtained by using the similar method.

To derive the equation for $\sigma ({\mathcal T}^{ij}_t X(t))$, according to Proposition 2,

$$\begin{aligned} \int _{t_k}^{t_{k+1}} d\bar{\sigma }\left( {\mathcal T}^{ij}_t X(t)\right)= & {} \int _{t_k}^{t_{k+1}}\varPhi _t^{-1}A{\sigma }\left( {\mathcal T}_t^{ij}X(t)\right) dt\\&+\int _{t_k}^{t_{k+1}}\varPhi _t^{-1}\langle {\sigma }(X(t)),\mathbf{e}_i\rangle (dY(t))_j\mathbf{e}_i. \end{aligned}$$

Here using the function that

$$\begin{aligned} \int _{t_k}^{t_{k+1}}\varPhi _t^{-1}\langle {\sigma }(X(s)),\mathbf{e}_i\rangle (dY(t))_j\mathbf{e}_i= & {} \langle \varPhi _t^{-1}{\sigma }(X(s)),\mathbf{e}_i\rangle (Y(t))_j\mathbf{e}_i|_{t_k}^{t_{k+1}}\\&- \int _{t_k}^{t_{k+1}}(Y(t))_j\langle \varPhi _t^{-1}A{\sigma }(X(t))dt,\mathbf{e}_i\rangle \mathbf{e}_i\\= & {} \left( \delta \varPhi _t^{-1}\langle A \sigma (X_k),\mathbf{e}_i\rangle +\varPhi _t^{-1}\langle \sigma (X_k),\mathbf{e}_i\rangle \right) (\Delta Y_k)_j\mathbf{e}_i, \end{aligned}$$

then the equation for $\sigma ({\mathcal T}^{ij}_t X(t))$ can be obtained by using similar method in deriving $\sigma (X(t))$.

Proof of Proposition 4

The model we are considering is affine in y(t), $\sigma (t)$ and $\sigma ^2(t)$, that is, in the space as

Open image in new window

We can derive the characteristic function of Eq. (27) by solving the partial differential equation Eq. (28). For convenience, we dropped the explicit (u, T)-dependence for A, B, C, D terms. Then using the assumption, we could write f as follows

$$\begin{aligned} f(t,y,\sigma )=\exp \left[ A(t)+B(t)y(t)+C(t)\sigma (t)+\frac{1}{2}D(t)\sigma ^2(t)\right] , \end{aligned}$$

The partial derivatives for $f=f(t,y,\sigma )$ are given by

Open image in new window

Substituting $f_t,f_{\sigma },f_{\sigma \sigma }$ into f yields

$$\begin{aligned}&(A'(t)+B'(t)y(t)+C'(t)\sigma (t)+\frac{1}{2}D'(t)\sigma ^2(t))+\xi (t)(C(t) +D(t)\sigma (t))\\&+\frac{1}{2}(\sigma ^2(t)+2\rho _{sr}\sigma (t)\gamma _1B_r(t,T) +\gamma _1^2B_r^2(t,T))(B^2(t)-B(t))\\&+(\rho _{s\sigma }\gamma _2\sigma (t)+\rho _{r\sigma }\gamma _1\gamma _2B_ r(t,T))B(t)(C(t)+D(t)\sigma (t))\\&+\frac{1}{2}\gamma _2^2(C^2(t)+D(t)+2C(t)D(t)\sigma (t)+D^2(t)\sigma ^2(t))=0. \end{aligned}$$

Since the equation equals 0, the coefficient of $y(t),\sigma (t),\sigma ^2(t)$ and the constants must be 0. We derive the following four O.D.E for the functions A(t), B(t), C(t) and D(t) as

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(33)

From the first equation in Eq. (33), $B(t)=B$ is a constant. Subject to the boundary condition in Eq. (28), we obtain

$$\begin{aligned} B=iu. \end{aligned}$$

According to the second equation in Eq. (33) and the boundary condition $D(T)=0$, we first derive a Riccati equation with constant coefficients as

$$\begin{aligned} D'(t)&=-(B^2-B)-2\rho _{s\sigma }\gamma _2BD(t)-\gamma _2^2D(t)\\&=:q_0+q_1D(t)+q_2D^2(t). \end{aligned}$$

Substituting $D(t)=\frac{-\sigma '(t)}{q_2\sigma (t)}$ into the Riccati equation then gives the second order linear differential equation with constant coefficients as follows:

$$\begin{aligned} \sigma ''(t)-q_1\sigma '(t)+q_0q_2\sigma (t)=0, \end{aligned}$$

which solution is given by

$$\begin{aligned} \sigma (t)&=k_1\exp [\lambda _+(T-t)]+k_2\exp [\lambda _-(T-t)], \end{aligned}$$

where $\lambda _{\pm }=\frac{-q_1\pm \sqrt{q_1^2-4q_0q_2}}{2}$.

Let $k=\sqrt{q_1^2-4q_0q_2}$, we have

$$\begin{aligned} D(t)=-u(i+u)\frac{1-e^{-2k(T-t)}}{k_1+k_2e^{-2k(T-t)}}, \end{aligned}$$

where

Open image in new window

The third equation in Eq. (33) is actually a first-order linear differential equation of the form $C'(t)+g(t)C(t)+h(t)=0$. Subject to the boundary condition $C(T)=0$, we find a solution for C(t)

$$\begin{aligned} C(t)=\int _t^T h(s)\exp [\int _t^s g(w)dw]ds, \end{aligned}$$

where

$$\begin{aligned} g(w)&=\rho _{s\sigma }\gamma _2B+\gamma _2^2D(w),\\ h(s)&=\rho _{sr}\gamma _1B_r(t,T)(B^2-B)+(\xi (s)+\rho _{r\sigma }\gamma _1 \gamma _2B_r(t,T)B)D(s). \end{aligned}$$

Then we consider the integral over g at the first step. In order to solve the tedious integral over D, we divide the second equation in system (33) by D(t) and rearrange terms to get

$$\begin{aligned} \int g(w)dw&=\int (\rho _{s\sigma }\gamma _2B+\gamma _2^2D(w))dw =\int -\rho _{s\sigma }-\frac{B^2-B}{D(w)}-\frac{D'(w)}{D(w)}dw\\&=\log (k_1e^{k(T-w)}k_2e^{-k(T-w)})+C, \end{aligned}$$

where C is a integration constant. Therefore we have the following formula as:

$$\begin{aligned} \exp \left[ \int _t^s g(w)dw\right] =\frac{k_1e^{k(T-s)}k_2e^{-k(T-s)}}{k_1e^{k(T-t)}k_2e^{-k(T-t)}}, \end{aligned}$$

and hence by a straightforward integration we have

$$\begin{aligned} C(t)&=\frac{1}{k_1e^{k(T-t)}k_2e^{-k(T-t)}}\int _t^T h(s)(k_1e^{k(T-s)}k_2e^{-k(T-s)})ds\\&=\frac{-u(i+u)}{k_1 +k_2 e^{-2k(T-t)}} [k_3-k_4+k_5-k_6+k_7] \end{aligned}$$

where $k,k_1,...,k_7$ are defined in Eq. (29)

Lastly, for the fourth equation of system (33), we derive the solution for A(t) by computing directly

$$\begin{aligned} A(t)= & {} \int _t^T\frac{1}{2}(B^2-B)\gamma _1^2(T-s)^2ds\\&+\int _t^T[(\xi (s)+\rho _{r\sigma }\gamma _1\gamma _2(T-s)B)C(s)+\frac{1}{2}\gamma _2^2(C^2(s)+D(s))]ds\\= & {} -\frac{1}{6}u(i+u)\gamma _1^2(T-t)^3\\&+\int _t^T[(\widehat{\mu }_2(s)+\rho _{r\sigma }(iu-1)\gamma _1\gamma _2(T-s))C(s)+\frac{1}{2} \gamma _2^2(C^2(s)+D(s))]ds. \end{aligned}$$

Here we leave the formula with respect to a differential equation instead of a closed-form expression which would involve hypergeometric function. Practically, it is more efficient to integrate it by numerical integration.

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Dong-Mei Zhu
- 1
Jiejun Lu
- 2
Wai-Ki Ching
- 2
- 3
- 4
Email author
Tak-Kuen Siu
- 5

1.School of Management and EconomicsSoutheast UniversityNanjingChina
2.Advanced Modeling and Applied Computing Laboratory, Department of MathematicsThe University of Hong KongPokfulamHong Kong
3.Hughes HallCambridgeUK
4.School of Economics and ManagementBeijing University of Chemical TechnologyBeijingChina
5.Department of Applied Finance and Actuarial Studies, Faculty of Business and EconomicsMacquarie UniversitySydneyAustralia

Option Pricing Under a Stochastic Interest Rate and Volatility Model with Hidden Markovian Regime-Switching

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