A Steady State Solution to a Mortgage Pricing Problem

TL;DR

This paper derives an analytical steady-state solution for a mortgage pricing model with prepayment options, assuming CIR interest rate dynamics, using variational methods to solve a free boundary problem.

Contribution

It provides the first analytical solution to a mortgage pricing problem with prepayment under CIR interest rate dynamics using variational methods.

Findings

Analytical solution to the free boundary problem is obtained.

The free boundary is characterized by a transcendental algebraic equation.

The model captures the impact of CIR interest rates on mortgage prepayment pricing.

Abstract

This paper considers a mortgage contract where the borrower pays a fixed mortgage rate and has the choice of making prepayment. Assume the market interest follows the CIR model, a free boundary problem is formulated. Here we focus on the infinite horizon problem. Using variational method, we obtain an analytical solution to the problem, where the free boundary is implicitly given by a transcendental algebraic equation.

Equations

$$\frac{dM(t)}{dt}=-m+cM(t).$$

$$M(t)=\frac{m}{c}(1-e^{(-c(T-t))}).$$

$$\displaystyle\left\{\begin{array}[]{ll}\displaystyle\textbf{L}(V)=m,&$for$\quad x>h(t),t>0\vskip 6.0pt plus 2.0pt minus 2.0pt\\ \displaystyle V=\frac{m}{c}[1-e^{-ct}],&$for$\quad x\leq h(t),t>0\vskip 6.0pt plus 2.0pt minus 2.0pt\\ \displaystyle\frac{\partial V}{\partial x}(h(t),t)\equiv 0\vskip 6.0pt plus 2.0pt minus 2.0pt\\ V(x,0)=0,&$for all$\quad x\geq 0\vskip 6.0pt plus 2.0pt minus 2.0pt\\ h(0)=c\end{array}\right.$$

$$\text{L}(V)=\frac{\partial V}{\partial t}-\frac{{\sigma }^2}{2}x\frac{{\partial }^{2}V}{\partial x^2}-k(\theta -x)\frac{\partial V}{\partial x}+xV$$

$$\displaystyle\left\{\begin{array}[]{ll}\displaystyle xV^{\prime\prime}+(\frac{2k\theta}{\sigma^{2}}-\frac{2k}{\sigma^{2}}x)V^{\prime}-\frac{2}{\sigma^{2}}xV=-\frac{2c}{\sigma^{2}},\qquad x\in(R^{\ast},~{}\ \infty)\\ V(R^{\ast})=1\\ V_{x}(R^{\ast})=0\\ V(x=\infty)=0\end{array}\right.$$

$$V(x)$$

$$z$$

$$\lambda$$

$$\displaystyle\left\{\begin{array}[]{ll}\displaystyle zu^{\prime\prime}+(\frac{2k\theta}{\sigma^{2}}-z)u^{\prime}-\frac{k\theta}{\sigma^{2}}(1-\frac{k}{\sqrt{k^{2}+2\sigma^{2}}})u=-\frac{c}{\sqrt{k^{2}+2\sigma^{2}}}e^{(\frac{1}{2}-\frac{k}{2\sqrt{k^{2}+2\sigma^{2}}})z},z\in(z^{\ast},\infty)\\ u(z^{\ast})=e^{(\frac{1}{2}-\frac{k}{2\sqrt{k^{2}+2\sigma^{2}}})z^{\ast}}\\ u^{\prime}(z^{\ast})=(\frac{1}{2}-\frac{k}{2\sqrt{k^{2}+2\sigma^{2}}})e^{(\frac{1}{2}-\frac{k}{2\sqrt{k^{2}+2\sigma^{2}}})z^{\ast}}\\ e^{\frac{\lambda}{p}z}u(z)=0,\qquad z\rightarrow\infty\end{array}\right.$$

$$u_1=M(\frac{k\theta }{{\sigma }^2}(1-\frac{k}{\sqrt{k^2+2{\sigma }^2}}),\frac{2k\theta }{{\sigma }^2},z)$$

$$u_2=U(\frac{k\theta }{{\sigma }^2}(1-\frac{k}{\sqrt{k^2+2{\sigma }^2}}),\frac{2k\theta }{{\sigma }^2},z)$$

$$W(M,U)(z)=-\frac{\Gamma (\frac{2k\theta }{{\sigma }^2})}{\Gamma (\frac{k\theta }{{\sigma }^2}(1-\frac{k}{\sqrt{k^2+2{\sigma }^2}}))}{z}^{-\frac{2k\theta }{{\sigma }^2}}{e}^z.$$

$$\begin{array}{rl}& u(z)=c_{1}M(\alpha ,\gamma ,z)+c_{2}U(\alpha ,\gamma ,z)\\ & +{\int }_z^{\infty }\frac{M(\alpha ,\gamma ,\xi )U(\alpha ,\gamma ,z)-U(\alpha ,\gamma ,\xi )M(\alpha ,\gamma ,z)}{-\frac{\Gamma (\gamma )}{\Gamma (\alpha )}{\xi }^{-\gamma }}(-\frac{c}{\sqrt{k^2+2{\sigma }^2}})e^{-(\frac{1}{2}+\frac{k}{2\sqrt{k^2+2{\sigma }^2}})\xi }d\xi \end{array},$$

$$\begin{array}{rl}& V(x)=e^{\lambda x}\{c_{1}M(\alpha ,\gamma ,px)+c_{2}U(\alpha ,\gamma ,px)\\ & -{\int }_z^{\infty }\frac{M(\alpha ,\gamma ,\xi )U(\alpha ,\gamma ,px)-U(\alpha ,\gamma ,\xi )M(\alpha ,\gamma ,px)}{-\frac{\Gamma (\gamma )}{\Gamma (\alpha )}{\xi }^{-\gamma }}[\frac{c{e}^{-(\frac{1}{2}+\frac{k}{2\sqrt{k^2+2{\sigma }^2}})\xi }}{\sqrt{k^2+2{\sigma }^2}}]d\xi \}\end{array}$$

$$\alpha$$

$$\gamma$$

$$p$$

$$M(\alpha ,\gamma ,z)$$

$$U(\alpha ,\gamma ,z)$$

$$u(z^{\ast })$$

$$u^{\prime }(z^{\ast })$$

$$\begin{array}{rl}& \frac{U(\alpha ,\gamma ,z^{\ast })}{U(\alpha +,\gamma +1,z^{\ast })}=\\ & \frac{e^{(\frac{1}{2}-\frac{k}{2\sqrt{k^2+2{\sigma }^2}})z^{\ast }}+\frac{c}{\sqrt{k^2+2{\sigma }^2}}\frac{{\Gamma }^2(\alpha )}{{\Gamma }^2(\gamma )}{\int }_{z^{\ast }}^{\infty }\frac{M(\alpha ,\gamma ,\xi )U(\alpha ,\gamma ,z^{\ast })-U(\alpha ,\gamma ,\xi )M(\alpha ,\gamma ,z^{\ast })}{{\xi }^{-\gamma }{e}^{(\frac{1}{2}+\frac{k}{2\sqrt{k^2+2{\sigma }^2}})\xi }}d\xi }{e^{(\frac{1}{2}-\frac{k}{2\sqrt{k^2+2{\sigma }^2}})z^{\ast }}+\frac{c}{\sqrt{k^2+2{\sigma }^2}}\frac{{\Gamma }^2(\alpha )}{{\Gamma }^2(\gamma )}{\int }_{z^{\ast }}^{\infty }\frac{M(\alpha ,\gamma ,\xi )U(\alpha +1,\gamma +1,z^{\ast })-U(\alpha ,\gamma ,\xi )M(\alpha +1,\gamma +1,z^{\ast })}{{\xi }^{-\gamma }{e}^{(\frac{1}{2}+\frac{k}{2\sqrt{k^2+2{\sigma }^2}})\xi }}d\xi }\end{array}$$

Full text

1. Formulation of the problem

We consider a mortgage contract where the borrower pays a fixed rate of $c$ $(year^{-1})$ to the lender. In reality this mortgage rate is implicitly represented by a continuous payment of $m$ \/year$. At each time$t$when the contract is effect, the borrower has two choices: to continue the mortgage by paying$mdt$for the next$dt$period or to close the mortgage by paying off all the loan balance$M(t)$, where the loan balance$M(t)$ is determined by

[TABLE]

When the contract duration $T$ is given and $M(T)=0$ specified, the above ODE has a unique soultion

[TABLE]

Here we assume the borrower always has sufficient amount of capital. The borrower chooses not to pay $M(t)$ even though he is financially capable to do so if the expected future market return from an equal amount of investment is higher enough. On the other hand, if the expected future market return from an equal amount of investment is lower enough, he should choose to settle $M(t)$. From the lender’s point of view, the value of the contract $V$, as a function of time $t$ and market interest rate $x$, is determined by the market interest rate $x$. The higher the market interest rate $x$ is, the lower the contract value $V$, but $V$ shall never be lower than 0. The lower the market interest rate $x$ is, the higher the contract value $V$, but $V$ shall never be exceed $M(t)$ since the borrower has the choice to settle the loan once $V$ reaches $M(t)$. From standard mathematical finance theory, one can find the value of the contract $V(x,t)$ and the optimal level of market interest $x=h(t)$ at which the borrower should make prepayment of $M(t)$ by solving the following free boundary problem:

[TABLE]

where $t$, for mathematical convenience, is defined to be the time to expiry of the contract, $h(t)$ is the unknown free boudary to be determined together with $V$, and the differential operator L is defined as

[TABLE]

The differential operator in the the system, referred as Kolmogorov equation, which can also be derived from Feymann-Kac Theorem [13, 3, 23]. Because of the important role played by the mortgage securities in real economy, there exists a considerable literature (see [6, 7, 19, 8, 11], for instance) dedicated to the topic, mostly of which have studied the problem from option-theoretical point with relatively less rigorous mathematical proofs. In recent development, a free boundary approach was introduced in [5, 10] to a similar problem. In particular D. Xie et al. (2007) have formulated the integral representations of the problems and proposed a Newton’s iteration scheme under the assumption that the underlying interest rate follows the Vasicek model [5]. In this paper, we shall study the same type of mortgage contract with market interest rate following CIR model instead of Vasicek model. We use CIR model because it is observed that Vasicek model allows negative interest rate, which contradicts the empirical statistics from market [1]. In this paper, we focus on the infinite horizon (steady state) of the problem.

2. The Infinite Horizon Problem

Without loss of generality, we assume $m=c$, thus we derive the following infinite horizon problem of the original system.

[TABLE]

where $R^{\ast\text{ }}$ is to be determined together with $V.$ To solve this free boundary problem, we let

[TABLE]

The above infinite horizon problem is transformed into the following one.

[TABLE]

We first solve the ODE in $z$ , disregarding any boundary conditions.  The homogeneous equation, as a standard confluent hypergeometric equation, has two linearly independent solutions

[TABLE]

one condition that both $\frac{k\theta}{\sigma^{2}}(1-\frac{k}{\sqrt{k^{2}+2\sigma^{2}}})$ and $\frac{2k\theta}{\sigma^{2}}$ are strictly positive, which is certainly satisfied for this problem. And the Wronskian of these two linearly independent solutions is calculated to be

[TABLE]

By the standard variational method, we can find one particular solution to the inhomogeneous equation, and together with above two linearly independent solutions, we obtain the general solution of the ODE for $u$

[TABLE]

and correspondingly, the general solution to the ODE in (2.5) is given by

[TABLE]

where

[TABLE]

Next we need to decide, by the conditions at free boundary and infinity, the values of two constants in the general solution as well as the unknown infinite horizon z${}^{\ast}.$ We firstly convince ourselves that $c_{1}=0$ by investigation of the asymptotic behavior of the $\ M(\alpha,\gamma,z)$ and $U(\alpha,\gamma,z)$ for large z. Recall that

[TABLE]

If we let $z\rightarrow\infty$ in the expression of $u(z)$, the definite integral term vanishes, i.e., the $U$ part contribution goes zero, but the $M$ part go to $e^{z}.$ If we multiply them with $e^{\frac{\lambda}{p}z},$ the magnitude contributed by the definite integral or the $U$ part still go zero, which is the desired, because $\lambda<0.$  But now the contribution of $M$ will be $e^{\frac{\lambda}{p}z}e^{z}$ multiplying a nontrivial polynomial in $z$, which clearly goes to infinity. This implies that $M(\alpha,\gamma,z)$ should not be included in the general solution. Now we only have two unknowns $c_{1\text{ }}$and $z^{\ast}.$ It is easy to use the fact that

[TABLE]

to get the implicit solution of $z^{\ast}\,:$

[TABLE]

And the $x^{\ast}$ in original system (1.8) can be simply recovered by $x^{\ast}=pz^{\ast}.$