A closed-form pricing formula for variance swaps under a stochastic volatility model with a stochastic mean-reversion level

Abstract

In this paper, we consider the valuation of variance swaps under a modified Heston model with a stochastic mean-reversion level, as a time-dependent mean reversion level for the variance of the Heston volatility model could provide better fit into the term structure of implied volatility and variance swap curve (Byelkina and Levin, in: Sixth World Congress of the Bachelier Finance Society, Toronto 2010; Forde and Jacquier in Appl Math Financ 17:241-259). We present a closed-form pricing formula for discretely sampled variance swaps based on the dimensional reduction technique. The validity of the newly derived formula is demonstrated through the comparison with the Monte Carlo simulation. The influence of introducing the stochastic mean-reversion level on variance swap prices is further investigated with numerical experiments.

Appendices

Appendix A

Here is the proof of Proposition 2.1.

If we make the transformation of $\tau _i=t_i-t$ and $x=\ln (S)$, the PDE system (2.9) can be converted into

$$\begin{aligned} \displaystyle \left\{ \begin{array}{lll} &{}\displaystyle \frac{\partial U_i}{\partial \tau _i}=\frac{1}{2}v\frac{\partial ^{2}U_i}{\partial x^2}+\frac{1}{2}\sigma _{1}^{2}v\frac{\partial ^{2}U_i}{\partial v^2}+\frac{1}{2}\sigma _{2}^{2}\frac{\partial ^{2}U_i}{\partial \theta ^2}+\rho \sigma _1v\frac{\partial ^{2}U_i}{\partial v\partial x}\\ &{}\qquad \qquad \displaystyle +\left( r-\frac{1}{2}v\right) \frac{\partial U_i}{\partial x}+k(\bar{v}+\theta -v)\frac{\partial U_i}{\partial v}\\ &{}\qquad \qquad +\lambda \frac{\partial U_i}{\partial \theta }-rU_i, t_{i-1}\le t\le t_i,\\ &{}\displaystyle U_i(x,v,\theta ,I,0)=\left( \frac{e^{x}}{I}-1\right) ^2\triangleq G(e^x). \end{array} \right. \end{aligned}$$

(A-1)

By applying the generalized Fourier transform, the PDE system (A-1) can be further transformed into

$$\begin{aligned} \displaystyle \left\{ \begin{array}{lll} &{}\displaystyle \frac{\partial \bar{U}_i}{\partial \tau _i}=\frac{1}{2}\sigma _{1}^{2}v\frac{\partial ^{2}\bar{U}_i}{\partial v^2}+\frac{1}{2}\sigma _{2}^{2}\frac{\partial ^{2}\bar{U}_i}{\partial \theta ^2}\\ &{}\qquad \qquad +[k(\bar{v}+\theta -v)+\rho \sigma _1vj\phi ]\frac{\partial \bar{U}_i}{\partial v}+\lambda \frac{\partial \bar{U}_i}{\partial \theta }\\ &{}\qquad \qquad \displaystyle +[(r-\frac{1}{2}v)j\phi -r-\frac{1}{2}v\phi ^2]\bar{U}_i, 0\le \tau _i\le t_i-t_{i-1},\\ &{}\displaystyle \bar{U}_i(\phi ,v,\theta ,I,0)=F[G(e^x)]. \end{array} \right. \end{aligned}$$

(A-2)

Here, $F(\cdot )$ denotes the generalized Fourier transform and $\bar{U}_i=F(U_i)$. Following He and Chen (2021), we assume that the solution to the PDE system (A-2) takes the form of

$$\begin{aligned} \bar{U}_i(\phi ,v,\theta ,I,\tau _i)=e^{C(\phi ;\tau _i)+D(\phi ;\tau _i)v+E(\phi ;\tau _i)\theta }\bar{U}_i(\phi ,v,\theta ,I,0). \end{aligned}$$

(A-3)

Substituting (A-3) into (A-2) yields the following three ordinary differential equations

$$\begin{aligned} \displaystyle \frac{\partial D}{\partial \tau _i}= & {} \frac{1}{2}\sigma _1^2D^2+(\rho \sigma \phi j-k)D-\frac{1}{2}(j\phi +\phi ^2),\\ \frac{\partial E}{\partial \tau _i}= & {} kD,\\ \frac{\partial C}{\partial \tau _i}= & {} \frac{1}{2}\sigma _2^2E^2+\lambda E+k\bar{v}D+r(j\phi -1). \end{aligned}$$

After working out $C(\phi ;\tau _i)$, $D(\phi ;\tau _i)$ and $E(\phi ;\tau _i)$, the solution to $U_i(x,v,\theta ,I,\tau _i)$ can be obtained through the inverse Fourier transform as

$$\begin{aligned} \displaystyle U_i(x,v,\theta ,I,\tau _i)=F^{-1}\{e^{C(\phi ;\tau _i)+D(\phi ;\tau _i)v+E(\phi ;\tau _i)\theta }F[G(e^x)]\}. \end{aligned}$$

It is never an easy task to analytically carry out the inverse Fourier transform. Fortunately, with the following two properties of the generalized Fourier transform

$$\begin{aligned} \displaystyle F(e^{jax})=2\pi \delta _a(\phi ), \end{aligned}$$

and

$$\begin{aligned} \displaystyle \int _{-\infty }^{+\infty }\delta _a(\phi )f(x)dx=f(a), \end{aligned}$$

it is not difficult to find that the pay-off function after the generalized Fourier transform is clearly

$$\begin{aligned} \displaystyle F[G(e^x)]= & {} F\left( \frac{e^{2x}}{I^2}-\frac{2e^x}{I}+1\right) ,\\= & {} \frac{\delta _{-2j}(\phi )}{I^2}-\frac{2\delta _{-j}(\phi )}{I}+\delta _0(\phi ). \end{aligned}$$

Therefore, we can finally arrive at

$$\begin{aligned} U_i(x,v,\theta ,I,\tau _i)&=\int _{-\infty }^{+\infty }e^{jx\phi }e^{C(\phi ;\tau _i)+D(\phi ;\tau _i)v +E(\phi ;\tau _i)\theta }\left[ \frac{\delta _{-2j}(\phi )}{I^2}\right. \\&\quad \left. -\frac{2\delta _{-j}(\phi )}{I}+\delta _0(\phi )\right] d\phi ,\\&=\frac{e^{C(\phi ;\tau _i)+D(\phi ;\tau _i)v+E(\phi ;\tau _i)\theta +jx\phi }}{I^2}|_{\phi =-2j}\\&\quad -\frac{2e^{C(\phi ;\tau _i)+D(\phi ;\tau _i)v+E(\phi ;\tau _i)\theta +jx\phi }}{I}|_{\phi =-j}\\&\quad +e^{C(\phi ;\tau _i)+D(\phi ;\tau _i)v+E(\phi ;\tau _i)\theta +jx\phi }|_{\phi =0}. \end{aligned}$$

This has completed the proof.

Appendix B

Here is the proof of Proposition 2.2.

We assume that the solution to the PDE system (2.13)-(2.14) take the form of

$$\begin{aligned}&U_i(S,v,\theta ,I,\tau _{i-1})\nonumber \\&\quad =e^{{\widetilde{C}}(\tau _{i-1})+{\widetilde{D}}(\tau _{i-1})v+{\widetilde{E}}(\tau _{i-1})\theta }\nonumber \\&\qquad +(e^{-r\Delta t}-2)e^{-r\tau _{i-1}}, \end{aligned}$$

(B-1)

and substitute it into the PDE system. Then it is not difficult to obtain

$$\begin{aligned}&\frac{\partial {\widetilde{C}}}{\partial \tau _{i-1}}+\frac{\partial {\widetilde{D}}}{\partial \tau _{i-1}}v +\frac{\partial {\widetilde{E}}}{\partial \tau _{i-1}}\theta \nonumber \\&\quad =\frac{1}{2}\sigma _1^2v{\widetilde{D}}^2+\frac{1}{2}\sigma _2^2{\widetilde{E}}^2+k(\bar{v}+\theta -v){\widetilde{D}}+\lambda {\widetilde{E}}-r. \end{aligned}$$

(B-2)

By noticing the fact that Eq. (B-2) should hold for arbitrary v and $\theta$, we can obtain the following three ordinary equations

$$\begin{aligned} \frac{\partial {\widetilde{D}}}{\partial \tau _{i-1}}= & {} \frac{1}{2}\sigma _1^2{\widetilde{D}}^2-k{\widetilde{D}},\\ \frac{\partial {\widetilde{E}}}{\partial \tau _{i-1}}= & {} k{\widetilde{D}},\\ \frac{\partial {\widetilde{C}}}{\partial \tau _{i-1}}= & {} \frac{1}{2}\sigma _2^2{\widetilde{E}}^2+k\bar{v}{\widetilde{D}}+\lambda {\widetilde{E}}-r. \end{aligned}$$

It should be mentioned that once we have worked out, D, C and E can be straightforwardly obtained by integrating on both sides of its ODE, respectively, which means that we need to figure out D first. In fact, the ODE governing D can be solved directly with the separation of variables, and thus the result follows. This has completed the proof.