# Bond PDE under an Affine Jump Diffusion model

Under the Jump extended Vasicek model, the dynamics of the short rate are as follow : $$dr_t=\kappa(\theta-r_t)dt+\sigma\sqrt{r_t}\,dW_t+d\left(\sum\limits_{i=1}^{N_t}\,J_i\right)$$ where $$N_t$$ represents a Poisson process with constant intensity rate $$\lambda>0$$ and $$\{J_i\}_{i=1}^{\infty}$$ denotes the magnitudes of jump, which are assumed to be i.i.d. random variables with distribution $$f_J$$ independent of $$W_t$$ and $$N_t$$. Moreover,$$W_t$$ is assumed to be independent of $$N_t$$. In addition the jump sizes $$\,J_i$$ has an exponential distribution with density: $${{f}_{J}}(\chi )=\left\{ \begin{matrix} \eta {{e}^{-\eta\,\chi}}\,,\,\,\chi >0\, \\ 0\,\,\,\,\,\,\,\,,\,\,\,\,o.w. \\ \end{matrix} \right.$$ where $$\eta > 0$$ is an constant. Can some one explain how to find the following parabolic partial integro differential equation for an arbitrage-free price at time $$t$$ of of a ZC bond of maturity $$T$$ ? : $$\frac{\partial P}{\partial t}+\frac{1}{2}{{\sigma }^{2}}r\frac{{{\partial }^{2}}P}{\partial {{r}^{2}}}+\kappa (\theta -r)\frac{\partial P}{\partial r}-rP+\lambda \int_{-\infty }^{\infty }{(P(t,r+\chi ,T)-P(t,r,T)d\chi =0}$$ with boundary condition $$P(T,r,T)=1$$.

Thank you

• Your equation has an error. See the derivation below. Oct 26, 2018 at 20:25

Let $$P(t, r_t, T)$$ be the bond price at time $$t$$, where $$0 \leq t \leq T$$. Then, by Ito's formula, \begin{align*} &\ P(t, r_t, T) \\ =& P(0, r_0, T) + \int_0^t\partial_s P(s, r_s, T) ds + \int_0^t\partial_r P(s, r_{s-}, T) dr_s + \frac{1}{2}\sigma^2 \int_0^t r_s\partial_{rr} P(s, r_s, T)ds\\ & \quad +\sum_{s \leq t}\big[P(s, r_s, T) - P(s, r_{s-}, T) - \partial_r P(s, r_{s-}, T)\Delta r_s\big] \quad (\mbox{where } \Delta r_s=r_s - r_{s-})\\ =& P(0, r_0, T) + \int_0^t\partial_s P(s, r_s, T) ds + \int_0^t\partial_r P(s, r_s, T) dr_s^c + \frac{1}{2}\sigma^2 \int_0^t r_s\partial_{rr} P(s, r_s, T)ds\\ & \quad +\sum_{s \leq t}\big[P(s, r_s, T) - P(s, r_{s-}, T) \big] \quad (\mbox{where } dr_t^c = \kappa(\theta - r_t)dt + \sigma \sqrt{r_t} d W_t)\\ =& P(0, r_0, T) + \int_0^t\partial_s P(s, r_s, T) ds + \int_0^t\partial_r P(s, r_s, T) dr_s^c + \frac{1}{2}\sigma^2 \int_0^t r_s\partial_{rr} P(s, r_s, T)ds\\ & \quad +\int_0^t \int_{\mathbb{R}}\big[ P(s, r_{s-}+y, T) - P(s, r_{s-}, T)\big]\mu(ds, dy) \quad (\mbox{where } \mu = \sum_{i=1}^{\infty} \delta_{\tau_i, J_i})\\ =& P(0, r_0, T) + \int_0^t\partial_s P(s, r_s, T) ds + \int_0^t\partial_r P(s, r_s, T) dr_s^c + \frac{1}{2}\sigma^2 \int_0^t r_s\partial_{rr} P(s, r_s, T)ds\\ &\quad +\int_0^t \int_{\mathbb{R}}\big[P(s, r_{s-}+y, T) - P(s, r_{s-}, T)\big](\mu(ds, dy) - ds v(dy)) \\ &\quad +\int_0^t ds\int_{\mathbb{R}}\big[ P(s, r_s+y, T) - P(s, r_s, T)\big]\lambda f_J(y)dy, \end{align*} where $$v(dy) = \lambda f_J(y)dy$$. Here \begin{align*} M_t = \int_0^t \int_{\mathbb{R}}\big[ u(X_{s-} + y, s) - u(X_{s-}, s))\big](\mu(ds, dy) - ds v(dy)) \end{align*} is a martingale. Since $$P(t, r_t, T) e^{-\int_0^t r_s ds}$$ is a martingale, and \begin{align*} d\Big(P(t, r_t, T) e^{-\int_0^t r_s ds}\Big) &= e^{-\int_0^t r_s ds}\big[-r_t P(t, r_t, T) dt + dP(t, r_t, T)\big], \end{align*} we obtain that \begin{align*} &-r_t P(t, r_t, T) + \partial_t P(t, r_t, T) + \kappa(\theta-r_t)\partial_r P(t, r_t, T) + \frac{1}{2}\sigma^2 r_t\partial_{rr} P(t, r_t, T) \\ & \qquad\qquad + \int_{\mathbb{R}}\big[ P(t, r_t+y, T) - P(t, r_t, T\big]\lambda f_J(y)dy = 0. \end{align*} That is, \begin{align*} & \partial_t P(t, r_t, T) + \kappa(\theta-r_t)\partial_r P(t, r_t, T) + \frac{1}{2}\sigma^2 r_t\partial_{rr} P(t, r_t, T) -(r_t+\lambda)P(t, r_t, T)\\ & \qquad\qquad + \lambda \int_{\mathbb{R}} P(t, r_t+y, T) f_J(y)dy = 0. \end{align*}
• Thank you for your answer. You claim that $P(t, r_t, T) e^{-\int_0^t r_s ds}$ is a martingale, but under which measure ? Nov 5, 2018 at 16:41