Duality of callable bond price

Question

I am trying to understand the relationship between two methods of pricing callable bonds in the risk-neutral pricing framework.

Problem statement

Let's consider zero-coupon bond with face value 1, expiring in 1 year and callable at 6 months with strike $K =0.4$. I am going to work in the Ho-Lee model in which the spot rate follows the generalized Brownian motion $$ dr(t) = \theta(t) dt + \sigma d\hat{w}(t). $$ For the purposes of obtaining a numerical result, I will assume $\sigma = 0.16$ and $r(0) = 0.3$ and choose the fitting function to be $\theta(t) = 0$ (unrealistic values in practice).

I can show numerically that the price obtained from the finite-difference method (Method 2) converges to the analytical price (Method 1). I would like to obtain the analytical result directly from the PDE in Method 2.

Method 1

The zero-coupon discount curve is given by $ P(t,T) = e^{-r(t) (T-t) + \frac{\sigma^2}{6}(T-t)^3} $. Applying Ito's lemma we obtain $$ dP(t,T) = P(t,T) r(t) dt - P(t,T)(T-t)\sigma d\hat{w}(t). $$ It follows that a European call option with expiration $T_i$ on the ZCB with maturity $T_j>T_i$ can be priced using the Black pricing formula using a time-depedent volatility, \begin{align} \mathbf{CZCB}(t,T_i,T_j,K) & = P(t,T_i)\mathbb{E}^{Q_i}_t\left[\max(G(T_i)-K,0)\right] \\ & = P(t,T_i)\left[G(t)N(y+\bar{\sigma}\sqrt{T_i-t})-KN(y)\right] \end{align} where \begin{equation} G(t) = \frac{P(t,T_j)}{P(t,T_i)}, \quad \quad y = \frac{\log\left(\frac{G(0)}{K}\right)-\frac{1}{2}\bar{\sigma}^2(T_i-t)}{\bar{\sigma}\sqrt{T_i-t}}, \quad \quad \bar{\sigma} =\sigma(T_j-T_i). \end{equation} Thinking of the callable bond as a straight bond minus the call option we obtain \begin{equation} P(0,T_e)-\mathbf{CZCB}(0,T_c,T_e,K) = \$ \, 0.344467, \end{equation} where I have set $T_e = 1$, $T_c = 0.5$, and $K = 0.4$

Method 2

Thinking of the callable bond as a spot-rate contingent claim we see that the bond price satisfies the PDE, \begin{equation} \frac{\partial g}{\partial t} + \frac{1}{2}\sigma^2 \frac{\partial^2 g}{\partial r^2} - rg(r,t) = 0. \end{equation} In a fully explicitly discretization with $t_n = n \Delta t$ ($0 \leq n \leq N$) we have \begin{equation} g_n^j = h_n^j , \quad \quad h_n^j = \frac{1}{1+r_j\Delta t} \bigl[p g_{n+1}^{j+1}+(1-2p) g_{n+1}^{j} + p g_{n+1}^{j-1}\bigr], \quad \quad p = \frac{\sigma^2\Delta t}{2\Delta r^2}. \end{equation} with the following boundary conditions in the $t$-direction (assuming $N$ is even) \begin{align} g^j_N & = 1 \\ g^j_{N/2+1} & = \max(K,h^j_{N/2+1}) \end{align} where the second condition accounts for the possibility of exercise at the 6 month mark.

Answer 1

Transformation to Heat PDE

First define $T_1$ to be the call date and $T_2$ to be the maturity date. We start by making a change of variables. Let

\begin{equation} \tau = T_1 - t, \quad x = r - \frac{1}{2} \sigma^2 \tau^2, \quad g(t, r) = \exp \left\{ -r \tau + \frac{1}{6} \sigma^2 \tau^3 \right\} h(\tau, x). \end{equation}

Then

\begin{eqnarray} \frac{\partial g}{\partial t} & = & \exp \left\{ -r \tau + \frac{1}{6} \sigma^2 \tau^3 \right\} \left( \left( r - \frac{1}{2} \sigma^2 \tau^2 \right) h(\tau, r) - \frac{\partial h}{\partial \tau} + \sigma^2 \tau \frac{\partial h}{\partial x} \right),\\ \frac{\partial g}{\partial r} & = & \exp \left\{ -r \tau + \frac{1}{6} \sigma^2 \tau^3 \right\} \left( -\tau h(\tau, r) + \frac{\partial h}{\partial x} \right),\\ \frac{\partial^2 g}{\partial r^2} & = & \exp \left\{ -r \tau + \frac{1}{6} \sigma^2 \tau^3 \right\} \left( \tau^2 h(\tau, x) - 2 \tau \frac{\partial h}{\partial x} + \frac{\partial^2 h}{\partial x^2} \right) \end{eqnarray}

and we get

\begin{equation} \frac{\partial h}{\partial \tau} = \frac{1}{2} \sigma^2 \frac{\partial^2 h}{\partial x^2} \end{equation}

subject to the initial condition

\begin{eqnarray} h(0, x) & = & \min \left\{ K, P \left( T_1, T_2; x \right) \right\}\\ & = & P \left( T_1, T_2; x \right) - \max \left\{ P \left( T_1, T_2; x \right) - K, 0 \right\} \end{eqnarray}

where $P(t, T; r)$ is defined as in the question.

Greens Function Solution

The fundamental solution is the heat kernel given by

\begin{equation} \phi(\tau, x) = \frac{1}{\sqrt{2 \pi \sigma^2 \tau}} \exp \left\{ -\frac{x^2}{2 \sigma^2 \tau} \right\}. \end{equation}

We obtain $h(\tau, x)$ through the convolution

\begin{eqnarray} h(\tau, x) & = & \int_{-\infty}^\infty h(0, y) \phi(\tau, x - y) \mathrm{d}y\\ & = & \underbrace{\int_{-\infty}^\infty \exp \left\{ -y \left( T_2 - T_1 \right) + \frac{1}{6} \sigma^2 \left( T_2 - T_1 \right)^3 \right\} \phi(\tau, x - y) \mathrm{d}y}_{h_1(\tau, x)}\\ & & - \underbrace{\int_{-\infty}^\infty \left( \exp \left\{ -y \left( T_2 - T_1 \right) + \frac{1}{6} \sigma^2 \left( T_2 - T_1 \right)^3 \right\} - K \right)^+ \phi(\tau, x - y) \mathrm{d}y}_{h_2(\tau, x)}. \end{eqnarray}

Some tedious computations show that for some $\alpha, \beta \in \mathbb{R}$,

\begin{equation} \int_{-\infty}^\alpha e^{\beta y} \phi(\tau, x - y) \mathrm{d}y = \exp \left\{\beta x + \frac{1}{2} \beta^2 \sigma^2 \tau \right\} \mathcal{N} \left( \frac{\alpha - x - \beta \sigma^2 \tau}{\sigma \sqrt{\tau}} \right). \end{equation}

Solution of $h_1(\tau, x)$

Thus, with $\alpha = \infty$ and $\beta = -\left( T_2 - T_1 \right)$, the first integral evaluates to

\begin{eqnarray} h_1(\tau, x) & = & \exp \left\{ -x \left( T_2 - T_1 \right) + \frac{1}{2} \sigma^2 \left( T_2 - T_1 \right)^2 \tau + \frac{1}{6} \sigma^2 \left( T_2 - T_1 \right)^3 \right\} \end{eqnarray}

Reversing the change of variables yields

\begin{eqnarray} g_1(t, r) & = & \exp \left\{ -r \left( T_2 - t \right) + \frac{1}{6} \sigma^2 \left( T_2 - t \right)^3 \right\}\\ & = & P \left( t, T_2; r \right) \end{eqnarray}

This is, as expected, just the time $t$ price of a zero coupon bond with maturity in $T_2$. However, explicitly computing it servers as a good sanity check if something went wrong on the way.

Solution of $h_2(\tau, x)$

The integrand in $h_2(\tau, x)$ is positive when

\begin{equation} y < \frac{-\ln (K) + \frac{1}{6} \sigma^2 \left( T_2 - T_1 \right)^3}{T_2 - T_1} =: \alpha \end{equation}

Thus, the integral evaluates to

\begin{eqnarray} h_2(\tau, x) & = & h_1(\tau, x) \mathcal{N} \left( \frac{\alpha - x + \sigma^2 \left( T_2 - T_1 \right) \tau}{\sigma \sqrt{\tau}} \right) - K \mathcal{N} \left( \frac{\alpha - x}{\sigma \sqrt{\tau}} \right). \end{eqnarray}

Reversing the change of variables and after a lot of tedious algebra, you find

\begin{eqnarray} g_2(t, r) & = & P \left( t, T_2; r \right) \mathcal{N} \left( d_+ \right) - P \left( t, T_1; r \right) K \mathcal{N} \left( d_- \right), \end{eqnarray}

where

\begin{equation} d_\pm = \frac{1}{\sigma \left( T_2 - T_1 \right) \sqrt{\tau}} \left( \ln \left( \frac{P \left( t, T_2; r \right)}{P \left( t, T_1; r \right) K} \right) \pm \frac{1}{2} \sigma^2 \left( T_2 - T_1 \right)^2 \tau \right). \end{equation}

Putting everything together, you obtain the same expression as the one you provided in the question.

Answer 2

Can you try to plug your analytical solution obtained Method 1 into the PDE and show that it satisfies the PDE with boundary conditions ?