I am having some trouble to understand the derivation of the parameters of zero coupon pricing formula using Hull White. Specifically I am trying to understand how to get
--[1]
where is the continuously compounded zero spot rate:
and is the initial instantaneous forward rate curve
I understand the process to get below:
--[2]
and
---[3]
Can someone please help me with deriving [1] from [2] or [3]?
Thanks very much in advance.
Under the Hull-White interest rate model, the short rate $r_t$ satisfies a risk-neutral SDE of the form \begin{align*} dr_t = (\theta(t)-a r_t)dt+ \sigma dW_t. \end{align*} The price at time $t$ of a zero-coupon bond with maturity $T$ and unit face value is then given by \begin{align*} P(t, T) &= A(t, T) e^{-B(t, T) r_t}, \end{align*} where \begin{align*} B(t, T) &= \frac{1}{a}\Big(1-e^{-a(T-t)} \Big),\\ 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), \tag{1}\\ \theta(t) &= af^M(0, t) +\frac{\partial f^M(0, t)}{\partial t}+\frac{\sigma^2}{2a}\left(1-e^{-2at}\right), \\ \end{align*} and \begin{align*} f^M(0, t) &= -\frac{\partial \ln P(0, t)}{\partial t}. \end{align*} Note that, \begin{align*} \ln P(0, T) = -\int_0^T f^M(0, u) du. \end{align*} Moreover, we define the yield-to-maturity $R(t, T)$ by \begin{align*} R(t, T) &=\frac{-\ln P(t, T)}{T-t}. \end{align*} We show that \begin{align*} \ln A(t, T) &= \left[tR(0, t)-TR(0, T) \right] + B(t, T)f^M(0, t) -\frac{\sigma^2}{4a}\left(1-e^{-2at} \right)B(t, T)^2.\tag{2} \end{align*}
Note that, \begin{align*} & \ \int_t^T \frac{\partial f^M(0, u)}{\partial u}B(u, T) \\ =&\ f^M(0, u)B(u, T)\, \big|_t^T - \int_t^T f^M(0, u)\frac{\partial B(u, T)}{\partial u}du\\ =&\ -f^M(0, t)B(t, T) + \int_t^T f^M(0, u) e^{-a(T-u)} du\\ =&\ -f^M(0, t)B(t, T) - a\int_t^T f^M(0, u) B(u, T) du + \int_t^T f^M(0, u)du. \end{align*} That is, \begin{align*} &\ a\int_t^T f^M(0, u) B(u, T) du + \int_t^T \frac{\partial f^M(0, u)}{\partial T}B(u, T) \\ =&\ -f^M(0, t)B(t, T)+\int_t^T f^M(0, u)du\\ =&\ -f^M(0, t)B(t, T)- \ln P(0, T)+\ln(0, t)\\ =&\ -f^M(0, t)B(t, T)+ TR(0, T)-tR(0, t). \end{align*} Then \begin{align*} &\ \int_t^T \frac{\sigma^2}{2a}\left(1-e^{-2au}\right) B(u, T) du \\ =&\ \frac{\sigma^2}{2a^2}\int_t^T \left(1-e^{-2au}\right)\left(1-e^{-a(T-u)} \right)du\\ =&\ \frac{\sigma^2}{2a^2}\int_t^T \left(1-e^{-2au} - e^{-a(T-u)} + e^{-a(T+u)}\right)du\\ =&\ \frac{\sigma^2}{2a^2}\left[T-t+\frac{1}{2a}\left(e^{-2aT}-e^{-2at} \right) + \frac{1}{a}\left(1-e^{-a(T-t)} \right) -\frac{1}{a}\left(e^{-2aT}-e^{-a(T+t)}\right) \right] \\ =&\ \frac{\sigma^2}{2a^2}\left[\big(T-t- B(t, T)\big) + \frac{1}{2a}e^{-2at}\left(-e^{-2a(T-t)}-1 +2e^{-a(T-t)} \right)\right]\\ =&\ \frac{\sigma^2}{2a^2}\big(T-t- B(t, T)\big) - \frac{\sigma^2}{4a}e^{-2at}B(t, T)^2. \end{align*} Therefore, \begin{align*} &\ \ln A(t, T) \\ =&\ - \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\\ =&\ f^M(0, t)B(t, T)+ tR(0, t)- TR(0, T)- \frac{\sigma^2}{2a^2}\big(T-t- B(t, T)\big) + \frac{\sigma^2}{4a}e^{-2at}B(t, T)^2\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad -\frac{\sigma^2}{2a^2}\big(B(t, T) -T+t\big)-\frac{\sigma^2}{4a}B(t, T)^2\\ =&\ \big(tR(0, t)- TR(0, T)\big)+f^M(0, t)B(t, T) - \frac{\sigma^2}{4a}\left(1-e^{-2at} \right)B(t, T)^2, \end{align*} which is the required Claim $(2)$ above.
