3
$\begingroup$

I'm studying the following paper on Hull-White model calibration: Hull-White paper

In this paper they study the general form of the HW model with time-dependent mean reversion and volatility: $$dr(t) = (\theta(t) - a(t)r(t))dt + \sigma(t)dW(t)$$ I am trying to prove their formula for the solution $\theta(t)$ which matches a prescribed discount rate at each maturity (on p. 36 of the paper). I have not seen this function derived under such generality.

The solution in the paper is as follows: $$\theta(t) = \frac{\partial}{\partial t} f(0,t) + a(t) f(0,t) + \frac{1}{2} \left(\frac{\partial^2}{\partial t^2} V(0,t) + a(t) \frac{\partial}{\partial t} V(0,t) \right)$$ Where: $$f(0,t) = \text{ instantaneous forward rate at time } t$$ $$V(t,T) = \int_t^T \sigma(u,T)^2 du$$ $$\sigma(u,T) = \sigma(u)B(u,T)$$ $$B(t,T) = E(t) \int_t^T \frac{du}{E(u)}$$ $$E(t) = e^{\int_0^t a(u) du}$$ I imagine one can follow a similar method done in simpler versions of the model and:

  1. Compute $r(t)$ via integration

  2. Compute the bond prices

  3. Take derivative of log of the bond price to compute the forward price

  4. Somehow isolate $\theta(t)$ after possibly taking another derivative of the previous equation?


Update: Here is what I got so far. Apply Itô's Lemma to the function: $$F(r(t), t) = E(t) r(t)$$ We get: $$F_t = a(t) E(t) r(t) \hspace{10pt} F_r = E(t) \hspace{10pt} F_{rr} = 0$$ Resulting in: $$\begin{align*} dF(t) &= F_t dt + F_r dr \\ &= E(t)\theta(t)dt + E(t)\sigma(t)dW(t) \end{align*}$$ Integrating from $t$ to $s$ one gets: $$\begin{align*} F(s) - F(t) &= \int_t^s E(u)\theta(u)du + \int_t^s E(u)\sigma(u)dW(u) \\ E(s)r(s) &= E(t)r(t) + \int_t^s E(u)\theta(u)du + \int_t^s E(u)\sigma(u)dW(u) \\ r(s) &= \frac{E(t)}{E(s)}r(t) + \frac{1}{E(s)}\int_t^s E(u)\theta(u)du + \frac{1}{E(s)}\int_t^s E(u)\sigma(u)dW(u) \end{align*}$$ We now compute the bond prices by integrating this short rate from $t$ to $T$ and applying the definition of the bond price. First we integrate and make use of Fubini's theorem: $$\begin{align*} \int_t^T r(s) ds &= \int_t^T \frac{E(t)}{E(s)}r(t) ds + \int_t^T \int_t^s \frac{E(u)}{E(s)}\theta(u) du ds + \int_t^T \int_t^s \frac{E(u)}{E(s)}\sigma(u) dW(u) ds \\ &= \int_t^T \frac{E(t)}{E(s)}r(t) ds + \int_t^T \int_u^T \frac{E(u)}{E(s)}\theta(u) ds du + \int_t^T \int_u^T \frac{E(u)}{E(s)}\sigma(u) ds dW(u) \\ &= r(t)\underbrace{E(t) \int_t^T \frac{1}{E(s)} ds}_{B(t,T)} + \int_t^T \theta(u) \underbrace{E(u) \int_u^T \frac{1}{E(s)} ds}_{B(u,T)} du + \int_t^T \sigma(u) \underbrace{E(u) \int_u^T \frac{1}{E(s)} ds}_{B(u,T)} dW(u) \\ &= r(t) B(t,T) + \int_t^T \theta(u) B(u,T) du + \int_t^T \sigma(u) B(u,T) dW(u) \end{align*}$$ Now we can compute the bond prices via: $$\begin{align*} P(t,T) &= \mathbb{E}_t \left[e^{-\int_t^T r(s) ds} \right] \\ &= e^{-r(t)B(t,T) - \int_t^T\theta(u)B(u,T)du} \mathbb{E}_t \left[ e^{-\int_t^T \sigma(u) B(u,T) dW(u)} \right] \\ &= e^{-r(t)B(t,T) - \int_t^T\theta(u)B(u,T)du} e^{\frac{1}{2} \int_t^T \sigma(u)^2 B(u,T)^2 du} \end{align*}$$ Here we used that $r(t)$ is known at $t$ and the formula for the expectation for a lognormal variable. We can now take a log and derivative to produce a formula for the instantaneous forward rate: $$\begin{align*} f(0,T) &= - \frac{\partial}{\partial T} \ln P(0,T) \\ &= \frac{\partial}{\partial T} \left[ r(0) B(0,T) + \int_0^T \theta(u) B(u,T) du - \frac{1}{2} \int_0^T \sigma(u)^2 B(u,T)^2 du \right] \end{align*}$$ Here we will use a couple properties of $B(t,T)$ which follow directly from its definition: $$\frac{\partial}{\partial T} B(t,T) = \frac{E(t)}{E(T)} \hspace{5pt} \text{ and } \hspace{5pt} B(T,T) = 0$$ $$\begin{align*} f(0,T) &= r(0) \frac{E(0)}{E(T)} + \int_0^T \theta(u) \frac{E(u)}{E(T)} du - \frac{1}{2} \int_0^T \sigma(u)^2 2B(u,T)\frac{E(u)}{E(T)} du \end{align*}$$ Taking one more derivative with respect to $T$ will result in two instances of $\theta(t)$ in the formula. I don't see a way to isolate this function by itself to recover the formula cited in the paper above.

$\endgroup$
  • 1
    $\begingroup$ Yes, as you said, it is done in the steps you outlined. What is your question then? $\endgroup$ – Gordon May 1 '18 at 19:49
  • $\begingroup$ Thanks Gordon. Even knowing that I'm trying the right method is helpful. I'm just getting stuck somewhere along the way. I'll keep trying and add some of my work to my question-post so that people can suggest where I went wrong and others can use it as a reference. $\endgroup$ – RyanM May 1 '18 at 19:53
6
$\begingroup$

We assume that the process $\{r(t), \, t \ge 0\}$ satisfies an SDE of the form \begin{align*} dr(t) = \big( \theta(t) - a(t) r(t) \big)dt + \sigma(t) dW_t, \quad t > 0, \end{align*} where $\{W_t, \, t \ge 0 \}$ is a standard Brwonian motion. Note that \begin{align*} d\left(e^{\int_0^t a(u) du}r(t) \right) &=a(t) e^{\int_0^t a(u) du}r(t) dt + e^{\int_0^t a(u) du} dr(t)\\ &= \theta(t) e^{\int_0^t a(u) du} dt + \sigma(t)e^{\int_0^t a(u) du}dW_t. \end{align*} Then, for $v \ge t$, \begin{align*} r(v) = r(t) e^{-\int_t^v a(u) du} + \int_t^v \theta(s)e^{-\int_s^v a(u)du} ds+\int_t^v \sigma(s)e^{-\int_s^v a(u)du} dW_s. \end{align*} For $0 \le t \le T$, let \begin{align*} P(t, T) = E\left(e^{-\int_t^T r_s ds}\,|\, \mathcal{F}_t \right), \end{align*} where $E$ is the expectation operator under the risk-neutral measure and $\mathcal{F}_t$ is the information set up to time $t$, be the value of a zero-coupon bond with maturity $T$ and unit notional amount. Moreover, let \begin{align*} B(t, T) &= \int_t^T e^{-\int_t^v a(u) du} dv,\\ \sigma(t, T) &= \sigma(t) B(t, T),\\ V(t, T) &= \int_t^T \sigma^2(u, T) du. \end{align*} From the above \begin{align*} P(t, T) &= E\left(e^{-\int_t^T r_v dv}\,|\, \mathcal{F}_t \right) \\ &=E\left(e^{-\int_t^T \left(r(t) e^{-\int_t^v a(u) du} + \int_t^v \theta(s)e^{-\int_s^v a(u)du} ds\right)dv - \int_t^T \left(\int_t^v \sigma(s)e^{-\int_s^v a(u)du} dW_s\right) dv}\,|\, \mathcal{F}_t \right) \\ &=e^{-r(t) B(t, T)}E\left(e^{-\int_t^T \left(\int_t^v \theta(s)e^{-\int_s^v a(u)du} ds\right)dv - \int_t^T \left(\int_t^v \sigma(s)e^{-\int_s^v a(u)du} dW_s\right) dv}\,|\, \mathcal{F}_t \right) \\ &=e^{-r(t) B(t, T)}E\left(e^{-\int_t^T \left(\int_s^T \theta(s)e^{-\int_s^v a(u)du} dv\right)ds - \int_t^T \left(\int_s^T \sigma(s)e^{-\int_s^v a(u)du} dv\right) dW_s}\,|\, \mathcal{F}_t \right) \\ &=e^{-r(t) B(t, T)}E\left(e^{ -\int_t^T \theta(s)B(s, T)ds - \int_t^T \sigma(s)B(s, T) dW_s}\,|\, \mathcal{F}_t \right) \\ &=e^{-r(t) B(t, T)-\int_t^T \theta(s)B(s, T)ds + \frac{1}{2}V(t, T)}. \end{align*} Then, \begin{align*} f(0, t)&=-\frac{\partial}{\partial t}\ln P(0, t)\\ &=r(0)\frac{\partial}{\partial t}B(0, t) + \theta(t)B(t, t) + \int_0^t\theta(s)\frac{\partial}{\partial t}B(s, t)ds - \frac{1}{2}\frac{\partial}{\partial t}V(0, t)\\ &=r(0)e^{-\int_0^t a(u) du} + \int_0^t\theta(s)e^{-\int_s^t a(u) du}ds- \frac{1}{2}\frac{\partial}{\partial t}V(0, t). \end{align*} and \begin{align*} \frac{\partial}{\partial t} f(0, t) &= -a(t)r(0)e^{-\int_0^t a(u) du} + \theta(t) - a(t)\int_0^t \theta(s)e^{-\int_s^t a(u) du}ds-\frac{1}{2}\frac{\partial^2}{\partial t^2}V(0, t)\\ &=-a(t)\left(f(0, t) + \frac{1}{2}\frac{\partial}{\partial t}V(0, t) \right) + \theta(t) -\frac{1}{2}\frac{\partial^2}{\partial t^2}V(0, t). \end{align*} That is, \begin{align*} \theta(t) = \frac{\partial}{\partial t} f(0,t) + a(t) f(0,t) + \frac{1}{2} \left(\frac{\partial^2}{\partial t^2} V(0,t) + a(t) \frac{\partial}{\partial t} V(0,t) \right). \end{align*}

$\endgroup$
  • 2
    $\begingroup$ Thank you for the detailed proof, Gordon! I see what I was missing now. I want to up-vote, but I suppose I have to wait until I'm not a newbie (no reputation points yet). Cheers! $\endgroup$ – RyanM May 2 '18 at 12:45

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.