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Suppose that interest rate $r(t)$ follows some short-rate models, say Vasicek, so that$dr = a(b-r) dt + \sigma dZ$, with constants $a,b,\sigma$.

It is well known that the price of zero-coupon bond $P(r,t)$ at current time $t$ maturing at $T$ with face value 1 follows (for example, see McDonald's Derivatives Markets, 3rd ed, p.758): $$\frac{\sigma^2}{2} \frac{\partial^2 P}{\partial r^2} + a(b-r) \frac{\partial P}{\partial r} + \frac{\partial P}{\partial t} - r P=0$$ with boundary condition $P(r,T) = 1$. Note that we could write $P(r,t) = \mathbb{E}^\mathbb{Q} \Big[ e^{- \int_t^T r(u) du} \big| F_t \Big]$ for all $t \leq T$.

Trying to generalize, for some smooth condition of $h(r,T)$ depending only on $r$ at $T$: If we define $Q(r,t) = \mathbb{E}^\mathbb{Q} \Big[ e^{- \int_t^T r(u) du} h(r,T) \big| F_t \Big]$ for all $t \leq T$, does the following PDE $$\frac{\sigma^2}{2} \frac{\partial^2 Q}{\partial r^2} + a(b-r) \frac{\partial Q}{\partial r} + \frac{\partial Q}{\partial t} - r Q=0$$ with boundary condition $Q(r,T) = h(r,T)$ hold?

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  • $\begingroup$ you mean $Q(t,T) = \mathbb{E}^{\mathbb{Q}}\left[e^{-\int_t^T r(u) du} h(r_T)\right]$ ? $\endgroup$ Sep 8, 2016 at 12:03
  • $\begingroup$ Hi, MJ73550, yes I mean $h(r, T)$, to be terminal condition for $Q$. Thanks. $\endgroup$
    – quant123
    Sep 8, 2016 at 13:15
  • $\begingroup$ My question is whether $h $ depends only on the final value of $r$ or the whole path. $\endgroup$ Sep 9, 2016 at 5:32
  • $\begingroup$ $h$ depends only on final value $r$ at $T$. $\endgroup$
    – quant123
    Sep 9, 2016 at 7:09

1 Answer 1

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Note that \begin{align*} M(r_t, t) &\equiv Q(r_t, t) e^{-\int_0^t r_u du} \\ &=E\left(e^{-\int_0^T r_u du} h(r_T, T) \mid \mathscr{F}_t \right) \end{align*} is a martingale. Moreover, \begin{align*} dM &= \frac{\partial M}{\partial t}dt + \frac{\partial M}{\partial r} dr_t + \frac{1}{2}\frac{\partial^2 M}{\partial r^2}d\langle r, r\rangle_t\\ &=\left[\frac{\partial Q}{\partial t} e^{-\int_0^t r_u du} - r_t Q(r_t, t) e^{-\int_0^t r_u du}\right]dt+e^{-\int_0^t r_u du}\frac{\partial Q}{\partial r}dr+ \frac{\sigma^2}{2}\frac{\partial^2 Q}{\partial r^2}e^{-\int_0^t r_u du}dt\\ &=e^{-\int_0^t r_u du}\left[\left(\frac{\sigma^2}{2}\frac{\partial^2 Q}{\partial r^2} +a(b-r_t)\frac{\partial Q}{\partial r} -r_t Q(r_t, t) + \frac{\partial Q}{\partial t} \right)dt + \sigma \frac{\partial Q}{\partial r} dZ \right]. \end{align*} Therefore, \begin{align*} \frac{\sigma^2}{2}\frac{\partial^2 Q}{\partial r^2} +a(b-r_t)\frac{\partial Q}{\partial r} -r_t Q(r_t, t) + \frac{\partial Q}{\partial t}=0. \end{align*}

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