Given that $dr=(\eta-\gamma r)dt+\sqrt{\alpha r+\beta}dW$

Let $Z(r,t)=e^{A(t;T)-rB(t;T)}$,

\begin{matrix} \frac{dA}{dt}=\eta B-\frac{1}{2}\beta {{B}^{2}} \\ \frac{dB}{dt}=\frac{1}{2}\alpha {{B}^{2}}+\gamma B-1 \\ \end{matrix}

How can we solve for $B$ in this case?

$B$ is a Riccati Equation. The answer for B is given below:


where $\Psi_1=\sqrt{\gamma^2+2\alpha}$


As you noted, this is a Riccati type ODE and it can thus be simplified using the standard transformations for this class - see e.g. Wikipedia. We start by defining

\begin{equation} C(t, T) = \frac{1}{2} \alpha B(t, T) \qquad \Rightarrow \qquad C_t(t, T) = \frac{1}{2} \alpha B_t(t, t) \end{equation}

and get

\begin{eqnarray} C_t(t, T) & = & C^2(t, T) + \gamma C(t, T) - \frac{1}{2} \alpha. \end{eqnarray}

We now set

\begin{eqnarray} C(t, T) = -\frac{D_t(t, T)}{D(t, T)} \qquad \Rightarrow \qquad C_t(t, T) & = & -\frac{D_{tt}(t, T)}{D(t, T)} + \frac{D_t^2(t, T)}{D^2(t, T)}\\ & = & \frac{-D_{tt}(t, T)}{D(t, T)} + C^2(t, T) \end{eqnarray}

and get

\begin{eqnarray} \frac{D_{tt}(t, T)}{D(t, T)} & = & C^2(t, T) - C_t(t, T)\\ & = & -\gamma C(t, T) + \frac{1}{2} \alpha\\ & = & \gamma \frac{D_t(t, T)}{D(t, T)} + \frac{1}{2} \alpha \end{eqnarray}


\begin{equation} D_{tt}(t, T) = \gamma D_t(t, T) + \frac{1}{2} \alpha D(t, T). \end{equation}

This is a homogeneous second order linear ODE with constant coefficients and can be solved using standard methods. We note that $T$ has been fixed and make another substitution by defining $\tau = T - t$ such that $E(\tau) = D(t, T)$. We get

\begin{equation} E_{\tau \tau}(\tau) + \gamma E_\tau(\tau) - \frac{1}{2} \alpha E(\tau) = 0. \end{equation}

The characteristic equation is

\begin{equation} r^2 + \gamma r - \frac{1}{2} \alpha = 0 \end{equation}

with solutions

\begin{equation} r_{1, 2} = -\frac{1}{2} \gamma \pm\frac{1}{2} \sqrt{\gamma^2 + 2 \alpha} := \beta \pm \lambda. \end{equation}

Note that $\lambda = \frac{1}{2} \psi_1$ in your notation. We thus have the general solution

\begin{equation} E(\tau) = c_1 e^{(\beta + \lambda) \tau} + c_2 e^{(\beta - \lambda) \tau} \end{equation}


\begin{equation} E_\tau(\tau) = (\beta + \lambda) e^{(\beta + \lambda) \tau} + (\beta - \lambda) c_2 e^{(\beta - \lambda) \tau} \end{equation}

and for some constants $c_1$ and $c_2$ that have to be determined. We obtain the solution to the Riccati ODE by substituting back

\begin{equation} B(t, T) = \frac{2 C(t, T)}{\alpha} = -\frac{2 D_t(t, T)}{\alpha D(t, T)} = \frac{2 E_\tau(\tau)}{\alpha E(\tau)}. \end{equation}

Applying the terminal condition yields

\begin{eqnarray} B(T, T) = 0 \qquad & \Leftrightarrow & \qquad E_\tau(0) = 0\\ & \Leftrightarrow & \qquad c_1 = -c_2 \frac{\beta - \lambda}{\beta + \lambda}. \end{eqnarray}


\begin{equation} E(\tau) = \frac{c_2}{\beta + \lambda} e^{\beta \tau} \left( (\beta + \lambda) e^{-\lambda \tau} - (\beta - \lambda) e^{\lambda \tau} \right) \end{equation}


\begin{equation} E_\tau(\tau) = (\beta - \lambda) c_2 e^{\beta \tau} \left( e^{-\lambda \tau} - e^{\lambda \tau} \right). \end{equation}

We finally get

\begin{eqnarray} B(t, T) & = & \frac{2 (\beta^2 - \lambda^2) \left( e^{-\lambda \tau} - e^{\lambda \tau} \right)}{\alpha \left( (\beta + \lambda) e^{-\lambda \tau} - (\beta - \lambda) e^{\lambda \tau} \right)}\\ & = & \frac{\left( e^{2 \lambda \tau} - 1 \right)}{(\beta + \lambda) - (\beta - \lambda) e^{2 \lambda \tau}}\\ & = & \frac{2 \left( e^{\psi_1 (T - t)} - 1 \right)}{\left( \gamma + \psi_1 \right) \left( e^{\psi_1 (T - t)} - 1 \right) + 2 \psi_1}. \end{eqnarray}


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.