8
$\begingroup$

I am struggling in this question:

Let $P(t,T)$ denote the price of a zero-coupon bond (with marturity at time $T$) at time $t \in [0,T]$.

As usual, at time $t$ for maturity $T$, the forward rate is defined by

$$f(t,T)= - \frac{\partial}{\partial T} \log P(t,T)$$.

Consider a short interest rate process $(r_t)$ satisfying the following dynamics: \begin{equation} dr_t = a(r_t) \,dt + b(r_t) \, dW_t \end{equation} for two smooth functions $a$ and $b$.

Let the function $G: [0,T] \times \mathbb{R} \rightarrow \mathbb{R}$ satisfy the following integral-differential equation: \begin{equation} \frac{\partial G}{\partial t} (t,r) = a(r) \frac{\partial G}{\partial r} (t,r) + \frac{ b(r)^2}{2} \frac{{\partial}^2 G}{\partial r^2} (t,r) - b(r)^2 \frac{\partial G}{\partial r} (t,r) \int_0^t \frac{\partial G}{\partial r} (s,r) \,ds, \end{equation} with initial condition $G(0,r)=r$.

We want to show that there is no arbitrage if the forward rate function is defined by $f(t,T) = G(T-t, r_t)$.

The main problem I encounter is the fact that $ \frac{1}{G(t,r)} b(r)^2 \frac{\partial G}{\partial r} (t,r) \int_0^t \frac{\partial G}{\partial r} (s,r)\,ds $ isn't a function of $r$ only. Therefore, I don't know how to apply Feynman-Kac in this situation.

Any suggestions on how to transform this to Feynman-Kac?

$\endgroup$
2
$\begingroup$

Integrate the integral-differential equation from 0 to $T-t$, we obtain that \begin{align*} G(T-t, r) - r &= a(r) \int_0^{T-t}\frac{\partial G}{\partial r} (s,r)ds + \frac{ b(r)^2}{2}\int_0^{T-t} \frac{{\partial}^2 G}{\partial r^2} (s,r)ds\\ &\qquad - b(r)^2 \int_0^{T-t}\frac{\partial G}{\partial r} (u,r) \int_0^u\frac{\partial G}{\partial r} (s,r) \,ds du\\ &=a(r) \int_0^{T-t}\frac{\partial G}{\partial r} (s,r)ds + \frac{ b(r)^2}{2}\int_0^{T-t} \frac{{\partial}^2 G}{\partial r^2} (s,r)ds\\ &\qquad - \frac{ b(r)^2}{2}\left(\int_0^{T-t}\frac{\partial G}{\partial r} (s,r) \,ds\right)^2.\tag{1} \end{align*}

Let $$\widehat{P}(t,T) = e^{-\int_0^{T-t} G(s, r_t)ds}.$$ Then \begin{align*} \frac{\partial \widehat{P}}{\partial t} &= \widehat{P}\, G(T-t, r_t),\\ \frac{\partial \widehat{P}}{\partial r} &=-\widehat{P}\int_0^{T-t}\frac{\partial G(s, r)}{\partial r}ds,\\ \frac{\partial^2 \widehat{P}}{\partial r^2}&=\widehat{P}\left(\int_0^{T-t}\frac{\partial G(s, r)}{\partial r}ds \right)^2 - \widehat{P}\int_0^{T-t}\frac{\partial^2 G(s, r)}{\partial r^2}ds. \end{align*} Moreover, from $(1)$, \begin{align*} &\ \frac{b(r)^2}{2}\frac{\partial^2\widehat{P}}{\partial r^2}+a(r)\frac{\partial \widehat{P}}{\partial r}+\frac{\partial \widehat{P}}{\partial t}-r\widehat{P}\\ =&\ \widehat{P}\Bigg[\frac{b(r)^2}{2}\left(\int_0^{T-t}\frac{\partial G(s, r)}{\partial r}ds \right)^2 - \frac{b(r)^2}{2}\int_0^{T-t}\frac{\partial^2 G(s, r)}{\partial r^2}ds \\ &\qquad - a(r)\int_0^{T-t}\frac{\partial G(s, r)}{\partial r}ds + G(T-t, r_t) -r\Bigg]. \end{align*} That is, $$\frac{b(r)^2}{2}\frac{\partial^2\widehat{P}}{\partial r^2}+a(r)\frac{\partial \widehat{P}}{\partial r}+\frac{\partial \widehat{P}}{\partial t}-r\widehat{P}=0. $$ Now, you can apply the Feynman-Kac formula to obtain that \begin{align*} \widehat{P}(t, T) = E\left(e^{-\int_t^T r_s ds} \mid \mathcal{F}_t \right). \end{align*} Therefore, $\widehat{P}(t, T) = P(t, T)$ is the zero-coupon bond price. Consequently, \begin{align*} f(t, T) &= - \frac{\partial}{\partial T} \ln \widehat{P}(t,T)\\ &=G(T-t, r_t). \end{align*}

$\endgroup$

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.