Modeling Tail Dependence Using Stochastic Volatility Model

Abstract

As one can see in many previous well-known papers, an one–factor stochastic volatility model has its limitation to fit the market dynamics. Based on empirical facts that the market volatility can be well explained by the combination of short-term and long-term volatilities, a multi–scale stochastic volatility model that is governed by two factors evolving on different time-scales: a fast mean-reverting factor and a persistent, slow mean-reverting factor is applied to capture the dynamics of two assets in this paper. The validity of the model was tested by calibration against the market return distribution of the S&P 500 and Dow Jones Industrial Average Indices. Based on this multiscale model, an analytically approximate formula, in terms of the Gaussian copula, was obtained for the joint transition density and the parameters of this density were estimated using daily data from the S&P 500 and DAX Indices.

Appendices

Appendix A. Correlation Between $X_t^{(1)}$ and $X_t^{(2)}$ Under the Model (2.2)

From the product rule for Itô Processes and the model dynamics given in (2.1) and (2.2),

$$\begin{aligned} \begin{aligned}&d(X_t^{(1)}X_t^{(2)})=X_t^{(1)}dX_t^{(2)}+X_t^{(2)}dX_t^{(1)}+dX_t^{(1)}dX_t^{(2)}\\&\quad \rightarrow \mathbb {E}[X_t^{(1)}X_t^{(2)}]=\int _0^t \rho \mathbb {E}\bigg [f_1(X_s, Y_s)f_2(X_s, Y_s)\bigg ]ds. \end{aligned} \end{aligned}$$

To derive $\mathbb {E}[f_1(X_s, Y_s)f_2(X_s, Y_s)]$, assume that the OU process as follows:

$$\begin{aligned} dx_t=-\kappa x_t dt+\sqrt{2\kappa }\nu dW_t, \end{aligned}$$

where $\kappa$, $\nu > 0$ and $W_t$ is a standard Brownian motion. Then, it is known that

$$\begin{aligned} x_t=x_0 e^{-\kappa t}+\nu \sqrt{1-e^{-2\kappa t}}Z, \end{aligned}$$

where Z is a standard normal random variable. Therefore, $\alpha Y_t +\beta Z_t$ can be expressed as

$$\begin{aligned} \alpha Y_t + \beta Z_t = \alpha (Y_0 e^{-t/\epsilon }+\nu _1\sqrt{1-e^{2t/\epsilon }}Z_1)+\beta (Z_0 e^{-t \delta }+\nu _2\sqrt{1-e^{2t \delta }}Z_2), \end{aligned}$$

where $Z_1$ and $Z_2$ are standard normal random variables with the correlation $\rho _{YZ}$.

Now, since $\alpha Y_t + \beta Z_t \sim N(\mu ,\sigma ^2)$,

$$\begin{aligned} \mathbb {E}[f_1(X_s, Y_s)f_2(X_s, Y_s)] = \mathbb {E}[e^{(\alpha _1+\alpha _2)X_s+(\beta _1+\beta _2)Y_s}] = e^{\mu (t, \alpha _1, \alpha _2, \beta _1, \beta _2)+\sigma ^2(t, \alpha _1, \alpha _2, \beta _1, \beta _2)/2}, \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} \mu (t, \alpha _1, \alpha _2, \beta _1, \beta _2)&= (\alpha _1+\alpha _2) Y_0 e^{-t/\epsilon } + (\beta _1+\beta _2) Z_0 e^{-t \delta }, \\ \sigma ^2(t, \alpha _1, \alpha _2, \beta _1, \beta _2)&= (\alpha _1+\alpha _2)^2\nu _1(1-e^{-2t/\epsilon })+(\beta _1+\beta _2)^2\nu _2(1-e^{-2t\delta })+(\alpha _1+\alpha _2)(\beta _1+\beta _2)\nu _1\nu _2\sqrt{(1-e^{-2t/\epsilon })(1-e^{-2t\delta })}\rho _{XY}. \end{aligned} \end{aligned}$$

Finally, the correlation Corr between $X_t^{(1)}$ and $X_t^{(2)}$ is derived as

$$\begin{aligned} \begin{aligned} Corr_t&=\frac{\int _0^t\mathbb {E}[f_1(X_s, Y_s)f_2(X_s, Y_s)]ds}{\sqrt{\int _0^t\mathbb {E}[f_1^2(X_s, Y_s)]ds\int _0^t\mathbb {E}[f_2^2(X_s, Y_s)]ds}} \\&=\frac{\int _0^te^{\mu (s, \alpha _1, \alpha _2, \beta _1, \beta _2)+\sigma ^2(s, \alpha _1, \alpha _2, \beta _1, \beta _2)/2}ds}{\sqrt{\int _0^te^{\mu (s, \alpha _1, \alpha _1, \beta _1, \beta _1)+\sigma ^2(s, \alpha _1, \alpha _1, \beta _1, \beta _1)/2}ds\int _0^te^{\mu (s, \alpha _2, \alpha _2, \beta _2, \beta _2)+\sigma ^2(s, \alpha _2, \alpha _2, \beta _2, \beta _2)/2}ds}} \end{aligned} \end{aligned}$$

Appendix B. Proof of Theorem 3.1

Applying the expansion (3.2) with $j=0$ to (3.3) leads to

$$\begin{aligned} \frac{1}{\epsilon }{\mathcal {L}}_0 u_{0,0} + \frac{1}{\sqrt{\epsilon }}\left( {\mathcal {L}}_0 u_{1,0}+{\mathcal {L}}_1 u_{0,0} \right) +\left( {\mathcal {L}}_0 u_{2,0}+{\mathcal {L}}_1 u_{1,0}+{\mathcal {L}}_2 u_{0,0} \right) +\sqrt{\epsilon }\left( {\mathcal {L}}_0 u_{3,0} +{\mathcal {L}}_1 u_{2,0} +{\mathcal {L}}_2 u_{1,0} \right) + \cdots =0. \end{aligned}$$

(5.1)

From the $\mathcal {O} \big ( \frac{1}{\epsilon } \big )$ term in (5.1), we have a differential equation $\mathcal {L}_{0}u_{0,0}=0$ for $u_{0,0}$. Note that the operator $\mathcal {L}_{0}$ is the generator of the OU process $Y_{t}$. If we impose a growth condition on $u_{0,0}$ such that the partial derivative $\frac{\partial u_{0,0}}{\partial y}$ does not grow as much as

$$\begin{aligned} \frac{\partial u_{0,0}}{\partial y} \backsim e^{y^{2}/2}, \text { as } y \rightarrow \infty , \end{aligned}$$

then the solution $u_{0,0}$ must be a constant with respect to the y variable; $u_{0,0}=u_{0,0}(t,x_{1},x_{2},z).$ Next, From the $\mathcal {O} \big ( \frac{1}{\sqrt{\epsilon }} \big )$ terms in (5.1) and the $y-$independence of $u_{0,0}$, we have

$$\begin{aligned} \mathcal {L}_{0}u_{1,0}=0. \end{aligned}$$

(5.2)

Then again if it is assumed that $\frac{\partial u_{1,0}}{\partial y}$ does not grow as much as $e^{y^{2}/2}$ as y goes to infinity, $u_{1,0}$ is also independent with respect to the y variable; $u_{1,0}=u_{1,0}(t,x_{1},x_{2},z).$ Therefore, the first two terms $u_{0,0}$ and $u_{1,0}$ do not depend on the stochastic volatility’s current level y of the fast-scale volatility driving the process $Y_{t}$. Throughout this paper, all the terms of the expansions are assumed to have a reasonable growth condition such that they do not grow as much as $e^{y^{2}/2}$.

Now, the terms of order $1,\sqrt{\epsilon },\epsilon , and \cdots$ can continue to be eliminated. From the $\mathcal {O}(1)$ terms in (5.1), we get $\mathcal {L}_{0}u_{2,0}+\mathcal {L}_{1}u_{1,0}+\mathcal {L}_{2}u_{0,0}=0$. This PDE becomes

$$\begin{aligned} \mathcal {L}_{0}u_{2,0}+\mathcal {L}_{2}u_{0,0}=0 \end{aligned}$$

(5.3)

due to the y-independence of $u_{1,0}$. This is a Poisson equation for $u_{2,0}$ with respect to the operator $\mathcal {L}_{0}$ with the source term $\mathcal {L}_{2}u_{0,0}$. Then, Lemma 1 applied to (5.3) leads to (3.6).

Appendix C. Proof of Theorem 3.2

The $\mathcal {O}(\sqrt{\epsilon })$ terms in (5.1) lead to $\mathcal {L}_{0}u_{3,0}+\mathcal {L}_{1}u_{2,0}+\mathcal {L}_{2}u_{1,0}=0$, which is a Poisson equation for $u_{3,0}$ the centering condition of which is given by

$$\begin{aligned} \langle \mathcal {L}_{1}u_{2,0}+\mathcal {L}_{2}u_{1,0} \rangle =0. \end{aligned}$$

(5.4)

Meanwhile, from PDEs (3.5) and (5.3) we get

$$\begin{aligned} u_{2,0}=-\mathcal {L}_{0}^{-1}(\mathcal {L}_{2}- \langle \mathcal {L}_{2} \rangle )u_{0,0}+c(t,x,z) \end{aligned}$$

(5.5)

for an arbitrary function c(t, x, z) that is independent of the y variable. Plugging (5.5) into (5.4), we derive a PDE for $u_{1,0}$ as follows:

$$\begin{aligned} \langle \mathcal {L}_{2} \rangle u_{1,0} =\mathcal {A} u_{0,0}. \end{aligned}$$

Then, we obtain the result of Theorem 3.2 by direct computation.

Appendix D. Proof of Theorem 3.3

To obtain another first-order correction term, it is necessary to consider another singular perturbation problem (3.4). Expansion (3.2) with $j=0$ and $j=1$ applied to (3.4) leads to

$$\begin{aligned}&\frac{1}{\epsilon }{\mathcal {L}}_0 u_{0,1} + \frac{1}{\sqrt{\epsilon }}\left( {\mathcal {L}}_0 u_{1,1}+{\mathcal {L}}_1 u_{0,1} \right) +\left( {\mathcal {L}}_0 u_{2,1}+{\mathcal {L}}_1 u_{1,1}+{\mathcal {L}}_2 u_{0,1} \right) +\sqrt{\epsilon }\left( {\mathcal {L}}_0 u_{3,1} +{\mathcal {L}}_1 u_{2,1} +{\mathcal {L}}_2 u_{1,1} \right) + \cdots \nonumber \\&\quad =-\frac{1}{\sqrt{\epsilon }}\mathcal {M}_{3} u_{0,0}- (\mathcal {M}_{1} u_{0,0}+\mathcal {M}_{3} u_{1,0} ) - \sqrt{\epsilon } (\mathcal {M}_{1} u_{1,0}+\mathcal {M}_{3} u_{2,0} ) - \cdots . \end{aligned}$$

(5.6)

By multiplying (5.6) by $\epsilon$ and then letting $\epsilon$ go to zero, we find the first two leading-order terms as follows:

$$\begin{aligned} \mathcal {L}_{0}u_{0,1}=0 ~~~~~~\mathrm {and}~~~~~~ \mathcal {L}_{0}u_{1,1}+\mathcal {L}_{1}u_{0,1}=-\mathcal {M}_{3}u_{0,0}. \end{aligned}$$

Because the operator $\mathcal {L}_{0}$ is the generator of the OU process $Y_{t}$, $u_{0,1}$ (the solution of $\mathcal {L}_{0}u_{0,1}=0$) must be a constant with respect to the y variable. Because $\mathcal {M}_{3}$ has a derivative with respect to the y variable and $u_{0,0}$ does not rely on y, we obtain $\mathcal {M}_{3}u_{0,0}=0$. Moreover, because each term of $\mathcal {L}_{1}$ has a derivative with respect to y, $\mathcal {L}_{1}u_{0,1}=0$ holds. Thus, the equation $\mathcal {L}_{0}u_{1,1}+\mathcal {L}_{1}u_{0,1}=-\mathcal {M}_{3}u_{0,0}$ reduces to $\mathcal {L}_{0}u_{1,1}=0$, meaning that $u_{1,1}$ does not depend on the y variable. Hence, the two terms $u_{0,1}$ and $u_{1,1}$ do not depend on the current level y of the fast-scale volatility driving process $Y_{t}$; $u_{0,1}=u_{0,1}(t,x_{1},x_{2},z)$ and $u_{1,1}=u_{1,1}(t,x_{1},x_{2},z)$. In this way, it becomes possible to continue to remove the terms of order 1, $\sqrt{\epsilon }$, $\epsilon$ and others. For the order-1 term, we have $\mathcal {L}_{0}u_{2,1}+\mathcal {L}_{1}u_{1,1} +\mathcal {L}_{2}u_{0,1}=-(\mathcal {M}_{1}u_{0,0}+\mathcal {M}_{3}u_{1,0})$. This PDE becomes $\mathcal {L}_{0}u_{2,1}+\mathcal {L}_{2}u_{0,1}+\mathcal {M}_{1}u_{0,0}=0$ due to the y-independence of $u_{1,0}$ and $u_{1,1}$. This is a Poisson equation for $u_{2,1}$ with respect to the operator $\mathcal {L}_{0}$ in the y variable, which has a solution only if $\mathcal {L}_{2}u_{0,1}+\mathcal {M}_{1}u_{0,0}$ is centered with respect to the invariant distribution of $Y_{t}$. Because $u_{0,0}$ and $u_{0,1}$ do not depend on the variable y, we have

$$\begin{aligned} \langle \mathcal {L}_{2} \rangle u_{0,1}= -\langle \mathcal {M}_{1}\rangle u_{0,0}. \end{aligned}$$

Then, we obtain the result of Theorem 3.3 by direct calculation.