3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Your equation has an error. See the derivation below. $\endgroup$ – Gordon Oct 26 '18 at 20:25
2
$\begingroup$

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*}

$\endgroup$
  • $\begingroup$ 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 ? $\endgroup$ – Younes S Nov 5 '18 at 16:41
  • $\begingroup$ It is under the risk-neutral measure. $\endgroup$ – Gordon Nov 5 '18 at 17:27
  • $\begingroup$ So, shouldn't be a market price of risk parameter in your equation ? $\endgroup$ – Younes S Nov 6 '18 at 9:17
  • $\begingroup$ We assume that all models are under the risk-neutral measure. Note that, the short rate is not a trading asset. Then, as long as the bond price is specified properly, the deflated bond price is a martingale, under the risk-neutral measure. See also discussions in this question. $\endgroup$ – Gordon Nov 6 '18 at 15:03

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.