Jump diffusion models, Merton and Bates.

To increase the integral convergence, we add and remove the call price in another model, for instance in the Merton model noted $C_{M}$:

\[ Call(0, S, K, t) = C_{M}(0, S, K, t) + C(t)^{-1} \mathbb{E} [ (S_t - K)_{+} | F_{0}] - C_{M}(0, S, K, t) \]

Simply applying the previously found general formula, we have for $Call(0, S, K, t) $:

$$ \begin{align*} & C_{M}(0, S, K, t) \\ & + S_0 \chi_{Bates}(1) + \left({\frac{K}{C(t)}}\right)^{(-\alpha + 1)} \frac{S_0^{\alpha} }{\pi} \displaystyle \int\limits_{0}^{\infty} Re \left( \frac{e^{-k i u}}{(i u + \alpha) (i u + \alpha -1)} \chi_{Bates} (\alpha + i u) \right) du \\ & - S_0 \chi_{m}(1) - \left({\frac{K}{C(t)}}\right)^{(-\alpha + 1)} \frac{S_0^{\alpha} }{\pi} \displaystyle \int\limits_{0}^{\infty} Re \left( \frac{e^{-k i u}}{(i u + \alpha) (i u + \alpha -1)} \chi_{m} (\alpha + i u) \right) du \\ = & C_{M}(0, S, K, t) \\ & + \left({\frac{K}{C(t)}}\right)^{(-\alpha + 1)} \frac{S_0^{\alpha} }{\pi} \displaystyle \int\limits_{0}^{\infty} Re \left[ \frac{e^{-k i u}}{(i u + \alpha) (i u + \alpha -1)} ( \chi_{Bates} (\alpha + i u) - \chi_{m}(\alpha + i u) ) \right] du \end{align*} $$

The first term is computed using the series expansion found in the section dedicated to the Merton model.

Expression of $\chi_{Bates}(u)$ and $\chi_{m}(u)$ $\forall u \in \mathbb{C}$

Computation of $\chi_{m}(u)$, $\forall u \in \mathbb{C}$

$$ \begin{align*} \forall u \in \mathbb{C}, \chi_m(u) & = \chi_m^d(u) \times \chi_m^c(u) \\ \forall u \in \mathbb{C}, \chi_m^c(u) & = \chi_m^c(u) = \chi_{BS}(u) = e^{ - \frac{\sigma^2 t}{2}(u-u^2)} \\ \forall u \in \mathbb{C}, \chi_{m}^d(u) & = \mathbb{E} [ e^{J_t} ] = e^{ \lambda t ( \chi_a(u) - 1 )} \\ & = \displaystyle e^{ \lambda t \left( e^{u ( \mu_z + \frac{u \sigma_z^2}{2} )} - 1 \right)} \end{align*} $$

With $\chi_a$ the Moment-Generating function of the random variable $A_i$.

We note $\forall u \in \mathbb{C}$, $ \displaystyle \chi_{m}^d(u) = \chi_{Jump}(u) $

Computation of $\chi_{Bates}(u)$, $\forall u \in \mathbb{C}$

$$ \begin{align*} \forall u \in \mathbb{C}, \chi_{Bates}(u) & = \chi_{Bates}^c(u) \times \chi_{Bates}^d(u) \\ \forall u \in \mathbb{C},\chi_{Bates}^d(u) & = \chi_{Jump}(u) = \mathbb{E} [ e^{J_t} ] = exp\left(\lambda T( e^{u \mu_z + u^2 \frac{\sigma_z^2 }{2} } - 1)\right) \\ \forall u \in \mathbb{C}, \chi_{Bates}^c(u) & = \chi_{Heston}(u) \end{align*} $$

First of all, let’s note $ X = log(S) $. Applying Itô formula to $ log(S) $, we have:

$$ dX_t = (r_t + \lambda(1 - mu_a) - \frac{1}{2}V_t )dt + \sqrt{V_t} dW_t^s $$

Let’s note $ \forall t \in \mathbb{R}^{+} $ $ M_t = f(X_t^c, V_t, t ) $ with $ \forall (x,v,t) \in (\mathbb{R}^{+})^3 $ $ f(x, v, t ) = \mathbb{E} [ e^{u X_t^c} |X_t^c = x , V_t = v ] $

For the sake of clarity, let’s introduce $ f = f(X_t^c, V_t, t) $ and let’s apply the Itô formula for continuous processes to $ M_t $.

\[ d M_t = \displaystyle \frac{\partial f }{\partial x } d X_t^c + \frac{\partial f }{\partial v } d V_t + \frac{\partial f }{\partial t } d t + \frac{1}{2} \left[ \frac{\partial^2 f }{\partial x^2 } d[ X^c ]_t + \frac{\partial^2 f }{\partial v^2 } d[ V ]_t + 2 \frac{\partial^2 f }{\partial x \partial v } d[ X^c , V ]_t \right] \]

Let’s compute each terms:

$$ \begin{align*} d X_t^c & = \displaystyle (r_t + \lambda (1 - \mu_A) - \frac{V_t}{2}) dt + \sqrt{V_t} d W_t^s \\ d V_t & = \xi (\eta - V_{t}) dt + \theta \sqrt{V_{t}} d W_t^v \\ d[ X^c ]_t & = V_t dt \\ d[ V ]_t & = \theta^2 V_t dt \\ d[ X^c , V ]_t & = \rho \theta V_t dt \end{align*} $$

Substituting, we have:

$$ \begin{align*} d M_t = & \displaystyle \frac{\partial f }{\partial x } ( (r_t + \lambda (1 - \mu_A) - \frac{V_t}{2}) dt + \sqrt{V_t} d W_t^s ) + \frac{\partial f }{\partial v } ( \xi (\eta - V_{t}) dt + \theta \sqrt{V_{t}} d W_t^v ) + \frac{\partial f }{\partial t } d t \\ & \displaystyle + \frac{1}{2} [ \frac{\partial^2 f }{\partial x^2 } V_t + \frac{\partial^2 f }{\partial v^2 } \theta^2 V_t + 2 \frac{\partial^2 f }{\partial x \partial v } \rho \theta V_t ] dt \\ = & [ \displaystyle \frac{\partial f }{\partial x } (r_t + \lambda (1 - \mu_A) - \frac{V_t}{2}) + \frac{\partial f }{\partial v } \xi (\eta - V_{t}) + \frac{\partial f }{\partial t } \\ & + \frac{1}{2}\frac{\partial^2 f }{\partial x^2 } V_t + \frac{1}{2}\frac{\partial^2 f }{\partial v^2 } \theta^2 V_t + \frac{\partial^2 f }{\partial x \partial v } \rho \theta V_t ] dt + \displaystyle \theta \sqrt{V_{t}} d W_t^v + \sqrt{V_{t}} d W_t^s \end{align*} $$

Moreover, we know that $f(X_{\cdot}^c, V_{\cdot}, \cdot)$ has to be a martingale. The drift must be equals to zero because $f(X_{\cdot}^c, V_{\cdot}, \cdot)$ is continue.

\[ \displaystyle \frac{\partial f }{\partial x } (r_t + \lambda (1 - \mu_A) - \frac{V_t}{2}) + \frac{\partial f }{\partial v } \xi (\eta - V_{t}) + \frac{\partial f }{\partial t } + \frac{1}{2}\frac{\partial^2 f }{\partial x^2 } V_t + \frac{1}{2}\frac{\partial^2 f }{\partial v^2 } \theta^2 V_t + \frac{\partial^2 f }{\partial x \partial v} \rho \theta V_t = 0 \]

with the terminal condition $ f(x,v,t) = e^{ux} $

We suppose the solution can be written as follow: $ f(x, v, t) = e^{ A(T-t) + v B(T-t) + u x } $ with initial condition $ A(0) = 0 $ and $ B(0) = 0 $. We have:

$$ \begin{align*} \displaystyle \frac{\partial f }{\partial x } & = u f \\ \frac{\partial f }{\partial v } & = B(T-t) f \\ \frac{\partial f }{\partial t } & = - \frac{\partial A }{\partial t }(T-t) + -V_t \frac{\partial B }{\partial t }(T-t) \\ \frac{\partial^2 f }{\partial x^2 } & = u^2 f \\ \displaystyle \frac{\partial^2 f }{\partial v^2 } & = B(T-t)^2 f \\ \frac{\partial^2 f }{\partial x \partial v } & = - u B(T-t) f \end{align*} $$

The equation can be written by substituting:

$$ \begin{align*} & \displaystyle u f (r_t + \lambda (1 - \mu_A) - \frac{V_t}{2}) + B(t) f \xi (\eta - V_{t}) - (\frac{\partial A }{\partial t }(t) + V_t \frac{\partial B }{\partial t }(t) ) \\ & + \frac{1}{2}u^2 f V_t + \frac{1}{2}B(t)^2 f \theta^2 V_t - u B(t) f \rho \theta V_t = 0 \end{align*} $$ $$ \begin{align*} & \displaystyle u(r_t + \lambda (1 - \mu_A)) + B(t)\xi \eta - \frac{\partial A }{\partial t }(t) \\ & + \left[ - \frac{\partial B }{\partial t }(t) - B(t)\xi + \frac{1}{2}u^2 + \frac{1}{2}B(t)^2 \theta^2 - u B(t) \rho \theta - \frac{u}{2} \right] V_t = 0 \end{align*} $$

The two solutions of our problem are:

$$ \begin{align*} \displaystyle \frac{\partial A }{\partial t }(t) & = B(t) \xi \eta + u (r_t + \lambda (1 - \mu_A)) \\ \displaystyle \frac{\partial B }{\partial t }(t) & = \frac{1}{2}\theta^2 B(t)^2 - \xi B(t) - \rho \theta u B(t) + \frac{1}{2}u^2 - \frac{1}{2}u = \frac{1}{2}\theta^2 B(t)^2 - ( \rho \theta u + \xi ) B(t) - \frac{u}{2}( 1 - u ) \end{align*} $$

The first one is an integral computation. The second one is a Riccati equation.

The solutions of our two equations can be written as follow:

$$ \begin{align*} \displaystyle A(t) & = u \left( \frac{1}{log ( C(t))} + t \lambda ( 1 - \mu_A) \right) + \frac{\xi \eta t (\xi + \rho \theta u)}{\theta^2} - \frac{2 \xi \eta}{\theta^2} log \left( cosh ( \frac{\gamma t}{2} ) - \frac{\xi + \rho \theta u}{\gamma} sin ( \frac{\gamma t}{2} ) \right) \\ \displaystyle B(t) & = - \frac{u - u^2}{\gamma coth ( \frac{\gamma t}{2} ) - \xi - \rho \theta u } \end{align*} $$

with $ \gamma = \displaystyle \sqrt{\theta^2 (u - u^2) + (\xi + \rho \theta u)^2} $

We have the log characteristic function:

$$ \begin{align*} log(\chi_{Bates}^c(u)) &= u \left( \frac{1}{log ( C(t) )} + t \lambda ( 1 - \mu_A)\right) + \frac{\xi \eta t (\xi + \rho \theta u)}{\theta^2}\\ &- \frac{2 \xi \eta}{\theta^2} log \left(cosh ( \frac{\gamma t}{2}) - \frac{\xi + \rho \theta u}{\gamma} sin ( \frac{\gamma t}{2} ) \right) - V_0 \frac{u - u^2}{\gamma coth ( \frac{\gamma t}{2} ) - \xi - \rho \theta u } + u X_0 \end{align*} $$

and the call price in the Bates model is given by:

$$ \begin{align*} Call(0, S, K, t) &= C_M(0, S, K, t)\\ & - \frac{K}{\pi C(t)} \displaystyle \int\limits_{0}^{\infty} Re \left[ \frac{e^{-k( \alpha + i u )} \chi_J (\alpha + i u) }{(i u + \alpha) (i u + \alpha -1)} \left( \chi_{Heston}(\alpha + i u) - \chi_{BS}(\alpha + i u) \right) \right] du \end{align*} $$

Control variate choice

To compute the Bates’ call price, we can choose the Merton, Black-Scholes, or no control variate to increase the accuracy.

Remark: Using a Black-Scholes or Merton call price as control variate, we have to choose the volatility $ \sigma $ parameter.

• Let's identify $\sigma$ by choosing $\theta = 0$. We have $ \forall t \in \mathbb{R}^{+*} $: \begin{align*} \displaystyle d( e^{\xi t} V_t ) & = V_t \xi e^{\xi t} dt + e^{\xi t} dV_t = V_t \xi e^{\xi t} dt + e^{\xi t} \xi (\eta - V_{t}) dt =e^{\xi t} \xi \eta dt \\ \displaystyle e^{\xi t} V_t - V_0 & = \int\limits_0^t e^{\xi t} \xi \eta dt \iff V_t = e^{-\xi t} (V_0 - \eta) + \eta \end{align*}
• $ \sigma $ can be chosen as the squared root of the terminal expected variance: \[ \sigma = \displaystyle \sqrt{e^{-\xi t} (V_0 - \eta) + \eta} \]
• $\sigma$ can be chosen as the squared root of the integrated variance without noise: \[ \sigma^2 = \displaystyle \int\limits_0^t V_s ds = \frac{1}{t} [ \frac{(e^{-\xi t} - 1)(\eta - V_0)}{\xi} +\eta t ] \iff \sigma = \sqrt{\frac{(e^{-\xi t} - 1)(\eta - V_0)}{\xi t} +\eta} \]

Calibration methods

We develop two calibration methods for the Bates model:

• In one step and using weight functions
• In two steps: Firstly a set of parameters is found using Merton and Heston calibration and secondly a local optimisation is performed

One step method

We minimise the imply/model volatilities in $L^2$ norm, as explain in the Merton calibration section. Let’s define a weight function applied respectively to strikes and to maturities.

• Weight function for strikes:

  $ \forall x \in \mathbb{R}^{+*}, $ $ f(x) = \displaystyle \frac{ e^{ - \left( \frac{(x - \mu)}{\sigma} \right)^2 }}{\sigma \sqrt{2 \pi}} $ with $\mu = S_0 C(t) $, $ \sigma = scale \times \sqrt{ \frac{1}{2}(k_{min}^2 + k_{max}^2) - \frac{1}{4} (k_{min} + k_{max} )^2 } $

  $ k_{min} $ minimum strike and $ k_{max} $ maximum strike for a fixed $ T \in \mathbb{R}^{+*} $ and $ scale \in \mathbb{R}^{+*} $ a scale factor

• Weight function for maturities:

  $ \forall x \in \mathbb{R}^{+*}, $ $ f(x) = \displaystyle (\frac{x}{x_0})^{p x_0} e^{ p(x-x_0) } $ with $ x_0 $ the centered maturity and $p$ the bandwidth

Two steps method

We inspect a first solution in the set of Bates parameters.

For the height parameters, we have:

• For $\lambda$, $\mu_z$, $\sigma_z$ and $\sigma$:

  We calibrate those parameters in the Merton model for maturities lower than one year. Remark that $\sigma$ is the volatility in the Merton model and not a Bates parameter

• For $V_0$, $\xi$, $\nu$, $\theta$ and $\rho$:

  We calibrate those parameters in the Heston model for maturities greater than one year. Remark that $v_0$ can be taken equals to $ \sigma^2 $

The second step use the previously found parameters as initial solution for a local minimisation.

Parameters impact

To expose the parameters influence, we expose the implied volatility surface in function of the model parameters. Don’t hesitate to expand this section

Conclusion

We presented in this post some theoretical aspects on discontinuous processes. We also exposed a generic formulation for jump diffusion model using Lévy process. We investigated two specific models: Merton and Bates models. The dynamic, call pricing, calibration and influence of parameters are exposed for each models. Numerical experiments are provided.

This methodology has been chosen to be implemented in LexiFi Apropos.