3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

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.