3
$\begingroup$

I am investigating time changed Levy models. As far as I have seen, these models are usually directly described under the risk neutral measure $\mathbb{Q}$. However, I'm interested in first modelling the dynamics under the physical measure $\mathbb{P}$ and then using some change of measure to find an EMM. I work in the following framework: First I model the stock price $S(t)$ under the physical measure $\mathbb{P}$ as

$$ S(t):=S(0)\exp(X(t)) $$ with an arbitrary Levy process $X(t)$. The time change I use is given by an integrated CIR process, i.e. $$ T_t = \int_0^t\lambda(s)\mathrm{d}s $$ with $$ \mathrm{d}\lambda(t)=\kappa(\theta-\lambda(t))\mathrm{d}t+\eta\sqrt{\lambda(t)}\mathrm{d}W^1(t),\quad \lambda(0)=\lambda_0. $$ In order to capture the leverage effect, we can allow the Brownian motion $W^1(t)$ driving the intensity $\lambda(t)$ to be correlated with the Brownian motion part of $X(t)$. The stock price process under the time change is then given by $$ \tilde S(t):=S(T_t). $$

To start, I consider the following simple case: $$ X(t)=\mu t+\sigma \rho W^1(t) + \sigma\sqrt{1-\rho^2}W^2(t) $$ where $W^1(t)$ and $W^2(t)$ are two independent Brownian motions. In this case $\rho$ controls the leverage. The time changed price process is then given by $$ \tilde S(t)=S(0)\exp\left(\mu\int_0^t\lambda(s)\mathrm{d}s+\sigma\rho W^1\left(\int_0^t\lambda(s)\mathrm{d}s\right)+\sigma\sqrt{1-\rho^2} W^2\left(\int_0^t\lambda(s)\mathrm{d}s\right)\right). $$ In order to exclude arbitrage the discounted stock price process $$ \exp(-rt)\tilde S(t) = S(0)\exp\left(-rt+\mu\int_0^t\lambda(s)\mathrm{d}s+\sigma\rho W^1\left(\int_0^t\lambda(s)\mathrm{d}s\right)+\sigma\sqrt{1-\rho^2} W^2\left(\int_0^t\lambda(s)\mathrm{d}s\right)\right). $$ must be a martingale. By the Dubins-Schwarz theorem it holds $$ W^i\left(\int_0^t \lambda(s)\mathrm{d}s\right)=\int_0^t\sqrt{\lambda(s)}\mathrm{d}W^i(s), \quad i=1,2. $$ Hence, I can write $$ \exp(-rt)\tilde S(t)=S(0)\exp\left(-rt+\mu\int_0^t\lambda(s)\mathrm{d}s+\sigma\rho\int_0^t\sqrt{\lambda(s)}\mathrm{d}W^1(s)+\sigma\sqrt{1-\rho^2}\int_0^t\sqrt{\lambda(s)}\mathrm{d}W^2(s)\right) $$ In order to find a change of measure, I write $$ \exp(-rt)\tilde S(t)=S(0)\exp\left(-rt+\mu\int_0^t\lambda(s)\mathrm{d}s-\sigma\rho\int_0^t\gamma(s)\sqrt{\lambda(s)}\mathrm{d}s-\sigma\sqrt{1-\rho^2}\int_0^t\delta(s)\sqrt{\lambda(s)}\mathrm{d}s\\ +\sigma\rho\int_0^t\sqrt{\lambda(s)}\mathrm{d}\left(W^1(s)+\int_0^s\gamma(r)\mathrm{d}r\right)\\ +\sigma\sqrt{1-\rho^2}\int_0^t\sqrt{\lambda(s)}\mathrm{d}\left(W^2(s)+\int_0^s\delta(r)\mathrm{d}r\right)\right) $$ It must now hold that $$ -rt+\mu\int_0^t\lambda(s)\mathrm{d}s-\sigma\rho\int_0^t\gamma(s)\sqrt{\lambda(s)}\mathrm{d}s-\sigma\sqrt{1-\rho^2}\int_0^t\delta(s)\sqrt{\lambda(s)}\mathrm{d}s \\ = -\frac{\sigma^2}{2}\int_0^t\lambda(s)\mathrm{d}s $$ Differentiating both sides gives $$ -r+\mu\lambda(t)+\sigma\rho\gamma(t)\sqrt{\lambda(t)}-\sigma\sqrt{1-\rho^2}\delta(t)\sqrt{\lambda(t)}=-\frac{\sigma^2}{2}\lambda(t). $$ We set $\gamma(t):=\mu\sqrt{\lambda(t)}$ with arbitrary $\mu\in\mathbb{R}$. Hence, it must hold $$ \delta(t) = -\frac{r}{\sigma\sqrt{1-\rho^2}}+\frac{(\mu-\sigma\rho+\sigma^2/2)\sqrt{\lambda(t)}}{\sigma\sqrt{1-\rho^2}} $$ The risk neutral measure can then be defined via $$ \frac{\mathrm{d}\mathbb{Q}}{\mathrm{d}\mathbb{P}}:=\mathcal{E}(-\gamma(t)\bullet W^1(t)-\delta(t) \bullet W^2(t)). $$ The dynamics of the intensity under $\mathbb{Q}$ are then given by $$ \mathrm{d}\lambda(t)=\kappa^*(\theta^*-\lambda(t))\mathrm{d}t+\sigma\sqrt{\lambda(t)}\mathrm{d}W^1(t) $$ with $\kappa^*=\kappa+\eta\mu$ and $\theta^*=\frac{\kappa\theta}{\kappa+\eta\mu}$. Now my questions:

  • Is the whole argument sound?
  • In the part where I used the Dubins-Schwarz theorem, can I really use the same Brownian on both sides of the equation? In the theorem it says that there is some Brownian motion such that it holds. But if it isn't the same brownian motion I have no clue on how to proceed.
  • With respect to which filtration is the discounted stock price a martingale under $\mathbb{Q}$? I guess it must be $\mathcal{G}_t:=\sigma\left(W^i(T_s), T_s; 0\leq s\leq t\right)$ . How can I make this rigorous?
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.