# Extended Hull White Interest Rate Model for Zero Coupon Bond

Let's take the following three SDEs:

$$dr=u(r,t)dt + w(r,t)dX$$ $$u(r,t)=a(t)-br$$ $$w(r,t)=c$$

where $b$ and $c$ are constants and $a(t)$ an arbitrary function of time $t$.

If Zero Coupon Bond $Z(r,T,T)=1$ for this model has the form

$$Z(r,t,T)=e^{(A(t,T)-B(t,T)r)}$$

How do you find $A$ and $B$?

I have derived the PDE for this model using no arbitrage condition. Substituting this in the PDE is not giving the right answers.

• This is pretty standard, and you can find it in any book that talks about interest rate models.In addition, please use latex to make it more readable. Oct 30, 2015 at 17:28
• Ok. Can you please give me some online link ? I have found for other IR models but not this one. Oct 30, 2015 at 18:15
• Those books are not openly available. If possible, check the book by Brigo. On this site, there may already have some discussions. Oct 30, 2015 at 19:07
• This linke math.nyu.edu/~benartzi/Slides10.3.pdf may be helpful. Oct 30, 2015 at 19:19

Here is a solution without using the PDE technique, which is preferred as we do not need to assume the affine form of a zero-coupon price from the start.

we assume that, under the risk-neutral measure, \begin{align*} dr_t = (\theta(t)-a r_t) dt + \sigma dW_t, \end{align*} where $a$ and $\sigma$ are constants, $a(t)$ is a deterministic function, and $W_t$ is a standard Brownian motion. We seek to compute the zero-coupon bond price defined by \begin{align*} P(t, T) &= E\left(e^{-\int_t^T r_s ds} \mid \mathcal{F}_t \right), \end{align*} where $\mathcal{F}_t$ is the information set up to time $t$. Note that \begin{align*} d\left(e^{at} r_t\right) &= be^{at}r_t dt + e^{at} dr_t\\ &=\theta(t)e^{at} dt + \sigma e^{at} dW_t. \end{align*} Then, for $s \geq t \geq 0$, \begin{align*} e^{as} r_s = e^{at} r_t + \int_t^s \theta(u)e^{au} du + \int_t^s \sigma e^{au} dW_u. \end{align*} That is, \begin{align*} r_s = e^{-a(s-t)} r_t + \int_t^s \theta(u)e^{-a(s-u)} du + \int_t^s \sigma e^{-a(s-u)} dW_u. \end{align*} We then have the integral \begin{align*} &\ \int_t^T r_s ds \\ =&\ r_t \int_t^T e^{-a(s-t)} ds + \int_t^T\!\!\!\!\int_t^s \theta(u)e^{-a(s-u)} du ds + \int_t^T\!\!\!\!\int_t^s\sigma e^{-a(s-u)} dW_u ds\\ =&\ \frac{1}{a}\Big(1-e^{-a(T-t)} \Big) r_t + \int_t^T\!\!\!\!\int_u^T \theta(u)e^{-a(s-u)} ds du + \int_t^T\!\!\!\!\int_u^T \sigma e^{-a(s-u)} ds dW_u\\ =&\ \frac{1}{a}\Big(1-e^{-a(T-t)} \Big) r_t + \int_t^T\!\! \frac{\theta(u)}{a}\Big(1-e^{-a(T-u)} \Big)du + \int_t^T \!\!\frac{\sigma}{a}\Big(1-e^{-a(T-u)} \Big)dW_u. \end{align*} Let \begin{align*} B(t, T) = \frac{1}{a}\Big(1-e^{-a(T-t)} \Big). \end{align*} Then, \begin{align*} \int_t^T r_s ds &= B(t, T) r_t + \int_t^T \theta(u) B(u, T) du + \int_t^T \sigma B(u, T) dW_u. \end{align*} Moreover, the zero-coupon bond price is then given by \begin{align*} P(t, T) &= E\left(e^{-\int_t^T r_s ds} \mid \mathcal{F}_t \right)\\ &=\exp\left(-B(t, T) r_t - \int_t^T \theta(u) B(u, T) du + \frac{1}{2}\int_t^T \sigma^2 B(u, T)^2 du\right). \end{align*} Note that \begin{align*} \int_t^T \sigma^2 B(u, T)^2 du &= \frac{\sigma^2}{a^2}\int_t^T \left(1 - 2e^{-a(T-u)} + e^{-2a(T-u)}\right) du\\ &=\frac{\sigma^2}{a^2}\left(T-t-\frac{2}{a}\Big(1-e^{-a(T-t)}\Big) +\frac{1}{2a} \Big(1-e^{-2a(T-t)}\Big) \right)\\ &= \frac{\sigma^2}{a^2}\left(T-t -\frac{1}{2a}\Big(1-e^{-a(T-t)}\Big)^2-\frac{1}{a}\Big(1-e^{-a(T-t)}\Big)\right)\\ &= -\frac{\sigma^2}{a^2}\big(B(t, T) -T+t\big)-\frac{\sigma^2}{2a}B(t, T)^2. \end{align*} Then \begin{align*} P(t, T) &= A(t, T) e^{-B(t, T) r_t}, \end{align*} where \begin{align*} A(t, T) &= \exp\left(- \int_t^T \theta(u) B(u, T) du -\frac{\sigma^2}{2a^2}\big(B(t, T) -T+t\big)-\frac{\sigma^2}{4a}B(t, T)^2\right). \end{align*}

See http://www.math.nyu.edu/~benartzi/Slides10.3.pdf for another derivation using the PDE approach.

• Thanks for this.The another is martingale approach. Both the methods including PDE are clear now. Nov 2, 2015 at 18:13