Cox-Ingersoll-Ross Stochastic Clocks

In this post, we consider stochastic clocks whose instantaneous rate of activity follows a square root process. We present all intermediate steps in the derivation of the characteristic function of the corresponding time change. The corresponding steps closely resemble those in the derivation of the zero-coupon bond price in the Cox et al. (1985) model for the short-term interest rate.

In a previous post we introduced the general concept of stochastic clocks. Here, we consider the case where the instantaneous rate of activity process has the dynamics

    \[ \mathrm{d}y_t = \kappa \left( \eta - y_t \right) \mathrm{d} t + \lambda \sqrt{y_t} \mathrm{d}W_t, \]

where W = \left\{ W_t : t \in \mathbb{R}_{+, 0} \right\} is a standard one-dimensional Brownian motion and \kappa, \eta, \lambda \in \mathbb{R}_+ are constants. The corresponding time-change is given by

    \[ Y(t, T) = \int_t^T y_u \mathrm{d}u. \nonumber \]

Laplace Transform of the Time-Change

We fix a T \in \mathbb{R}_+ and start by computing the Laplace transform

    \[ \mathcal{L} \left\{ Y(t, T) \right\} (\alpha) = \mathbb{E} \left[ \left. \exp \left\{ -\alpha \int_t^T y_u \mathrm{d}u \right\} \right| \mathfrak{F}_t \right]. \]

When \alpha = 1, we recognize this as the time t price of a zero-coupon bond with maturity in T in the Cox et al. (1985) model; see e.g. Musiela and Rutkowski (2005). Since solutions to stochastic differential equations have the Markov property, it follows that

    \[ \mathcal{L} \left\{ Y(t, T) \right\} (\alpha) = \mathbb{E} \left[ \left. \exp \left\{ -\alpha \int_t^T y_u \mathrm{d}u \right\} \right| \sigma \left( y_t \right) \right], \]

where \sigma \left( y_t \right) is the \sigma-algebra generated by y_t. Now, since \mathcal{L} \left\{ Y(t, T) \right\} (\alpha) is adapted to the filtration \sigma \left( y_t \right), it follows that there exists a function f: [0, T] \times \mathbb{R}_+ such that \mathcal{L} \left\{ Y(t, T) \right\} (\alpha) = f \left(  t, y_t; \alpha \right).

Note that for s \in [0, t], we have

    \begin{eqnarray*} & & \mathbb{E} \left[ \left. \exp \left\{ -\alpha \int_0^t y_u \mathrm{d}u \right\} f \left( t, y_t; \alpha \right) \right| \mathfrak{F}_s \right]\\ & = & \mathbb{E} \left[ \left. \exp \left\{ -\alpha \int_0^t y_u \mathrm{d}u \right\} \mathbb{E} \left[ \left. \exp \left\{ -\alpha \int_t^T y_u \mathrm{d}u \right\} \right| \mathfrak{F}_t \right] \right| \mathfrak{F}_s \right]\\ & = & \mathbb{E} \left[ \left. \exp \left\{ -\alpha \int_0^T y_u \mathrm{d}u \right\} \right| \mathfrac{F}_s \right]\\ & = & \exp \left\{ -\alpha \int_0^s y_u \mathrm{d}u \right\} \mathbb{E} \left[ \left. \exp \left\{ -\alpha \int_s^T y_u \mathrm{d}u \right\} \right| \mathfrac{F}_s \right]\\ & = & \exp \left\{ -\alpha \int_0^s y_u \mathrm{d}u \right\} f \left( s, y_s; \alpha \right) \end{eqnarray*}

It follows that the process X = \left\{ X_t : t \in \mathbb{R}_{0, +} \right\} defined as

    \[ X_t = \exp \left\{ -\alpha \int_0^t y_u \mathrm{d}u \right\} f \left( t, y_t; \alpha \right). \]

is a \left( \mathfrak{F}_t \right)_{t \in \mathbb{R}_{0, +}}-martingale. Using the Itō product rule, we find that its differential is given by

    \begin{eqnarray*} \mathrm{d}X_t & = & \exp \left\{ -\alpha \int_0^t y_u \mathrm{d}u \right\} \left( \frac{\partial f}{\partial t} \mathrm{d}t + \frac{\partial f}{\partial y} \mathrm{d}y_t + \frac{1}{2} \frac{\partial^2 f}{\partial y^2} \mathrm{d} \langle y \rangle_t - \alpha y_t f \left( t, y_t; \alpha \right) \mathrm{d}t \right)\\ & = & \exp \left\{ -\alpha \int_0^t y_u \mathrm{d}u \right\} \left( \left( \frac{\partial f}{\partial t} + \kapa \left( \eta - y_t \right) \frac{\partial f}{\partial y} + \frac{1}{2} \lambda^2 y_t \frac{\partial^2 f}{\partial y^2} - \alpha y_t f \left( t, y_t; \alpha \right) \right) \mathrm{d}t \right.\\ & & \left. + \lambda \sqrt{y_t} \frac{\partial f}{\partial y} \mathrm{d}W_t \right). \end{eqnarray*}

Here, we used that the first factor of X is a process of bounded variation. Furthermore, we assume that f \in \mathcal{C}^{1, 2} so that the Itō formula can be applied. It follows from the martingale property that the drift term in the above differential has to vanish. We obtain the PDE

    \[ \frac{\partial f}{\partial t} + \kappa (\eta - y) \frac{\partial f}{\partial y} + \frac{1}{2} \lambda^2 y \frac{\partial^2 f}{\partial y^2} - \alpha y f(t, y; \alpha) = 0 \]

with terminal condition f(T, y; \alpha) = 1 for all y. We postulate a solution of the form

    \[ f(t, y; \alpha) = \exp \left\{ m(t, T) - \alpha y n(t, T) \right\} \]


    \begin{align*} \frac{\partial f}{\partial t} & = \left( m_t(t, T) - \alpha y n_t(t, T) \right) f(t, y; \alpha),\\ \frac{\partial f}{\partial y} & = -\alpha n(t, T) f(t, y; \alpha)\\ \frac{\partial^2 f}{\partial y^2} & = \alpha^2 n^2(t, T) f(t, y; \alpha). \end{align*}

We also require that m(T, T) = n(T, T) = 0 such that f(t, y; \alpha) = 1 for all y and the terminal condition is satisfied. Substituting into the PDE yields

    \begin{align*} 0 & = \left( \left( m_t(t, T) - \alpha y n_t(t, T) \right) - \alpha \kappa (\eta - y) n(t, T) + \frac{1}{2} \lambda^2 \alpha^2 y n^2(t, T) - \alpha y \right) f(t, y; \alpha)\\ & = m_t(t, T) - \alpha \kappa \eta n(t, T) - \alpha y \left( n_t(t, T) - \kappa n(t, T) - \frac{1}{2} \lambda^2 \alpha n^2(t, T) + 1 \right). \end{align*}

Here, the second step follows from f(t, y; \alpha) being strictly positive. Since this PDE has to hold for all values of y, the term that multiplies it has to be equal to zero and we obtain the first ODE

    \[ n_t(t, T) = \kappa n(t, T) + \frac{1}{2} \lambda^2 \alpha n^2(t, T) - 1 \]

with terminal condition n(T, T) = 0. Setting this term equal to zero in the original PDE yields the second ODE

    \[ m_t(t, T) = \alpha \kappa \eta n(t, T) \]

with terminal condition m(T, T) = 0.

Solving the ODEs

The first ODE is of the Riccati type and can thus be simplified using the standard transformations for this class. We start by defining

    \[ a(t, T) = \frac{1}{2} \lambda^2 \alpha n(t, T) \]

such that

    \[ a_t(t, T) = \frac{1}{2} \lambda^2 \alpha n_t(t, T) \]

and get

    \[ a_t(t, T) = a^2(t, T) + \kappa a(t, T) - \frac{1}{2} \lambda^2 \alpha. \]

We now set

    \[ a(t, T) = -\frac{b_t(t, T)}{b(t, T)} \]

such that

    \begin{eqnarray*} a_t(t, T) & = & -\frac{b_{tt}(t, T)}{b(t, T)} + \frac{b_t^2(t, T)}{b^2(t, T)}\\ & = & -\frac{b_{tt}(t, T)}{b(t, T)} + a^2(t, T) \end{eqnarray*}

and get

    \begin{eqnarray*} \frac{b_{tt}(t, T)}{b(t, T)} & = & a^2(t, T) - a_t(t, T)\\ & = & -\kappa a(t, T) + \frac{1}{2} \lambda^2 \alpha\\ & = & \kappa \frac{b_t(t, T)}{b(t, T)} + \frac{1}{2} \lambda^2 \alpha \end{eqnarray*}


    \[ b_{tt}(t, T) = \kappa b_t(t, T) + \frac{1}{2} \lambda^2 \alpha b(t, T). \]

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 c(\tau) = b(t, T) with c_\tau = -b_t(t, T) and c_{\tau \tau}(\tau) = b_{tt}(t, T). We get

    \[ c_{\tau \tau}(\tau) + \kappa c_\tau(\tau) - \frac{1}{2} \lambda^2 \alpha c(\tau) = 0. \]

The characteristic equation is

    \[ r^2 + \kappa r - \frac{1}{2} \lambda^2 \alpha = 0 \]

with roots

    \[ r_{1, 2} = -\frac{1}{2} \kappa \pm \frac{1}{2} \sqrt{\kappa^2 + 2 \lambda^2 \alpha} := \beta \pm \gamma. \]

We thus have the general solution

    \[ c(\tau) = c_1 e^{(\beta + \gamma) \tau} + c_2 e^{(\beta - \gamma) \tau} \]


    \[ c_\tau(\tau) = (\beta + \gamma) c_1 e^{(\beta + \gamma) \tau} + (\beta - \gamma) c_2 e^{(\beta - \gamma) \tau} \]

and for some constants c_1 and c_2 to be determined. We obtian the solution to the Riccati ODE by back substituting

    \[ n(t, T) = \frac{2 a(t, T)}{\lambda^2 \alpha} = -\frac{2 b_t(t, T)}{\lambda^2 \alpha b(t, T)} = \frac{2 c_\tau(\tau)}{\lambda^2 \alpha c(\tau)}. \]

By the terminal condition n(T, T) = 0, it follows that

    \[ n(T, T) = 0 \qquad \Leftrightarrow \qquad c_\tau(0) = 0 \qquad \Leftrightarrow \qquad c_1 = -c_2 \frac{\beta - \gamma}{\beta + \gamma}. \]


    \begin{align*} c(\tau) & = \frac{c_2}{\beta + \gamma} e^{\beta \tau} \left( (\beta + \gamma) e^{-\gamma \tau} - (\beta - \gamma) e^{\gamma \tau} \right)\\ & = \frac{c_2}{\beta + \gamma} e^{\beta \tau} \left( \beta \left( e^{-\gamma \tau} - e^{\gamma \tau} \right) + \gamma \left( e^{-\gamma \tau} + e^{\gamma \tau} \right) \right)\\ & = \frac{2 c_2}{\beta + \gamma} e^{\beta \tau} \left( \beta \sinh(-\gamma \tau) + \gamma \cosh(-\gamma \tau) \right) \end{align*}


    \begin{align*} c_\tau(\tau) & = (\beta - \gamma) c_2 e^{\beta \tau} \left( e^{-\gamma \tau} - e^{\gamma \tau} \right)\\ & = 2 (\beta - \gamma) c_2 e^{\beta \tau} \sinh(-\gamma \tau). \end{align*}

We get

    \begin{align*} n(t, T) & = \frac{2 \left( \beta^2 - \gamma^2 \right) \sinh(-\gamma \tau)}{\lambda^2 \alpha \left( \beta \sinh(-\gamma \tau) + \gamma \cosh(-\gamma \tau) \right)}\\ & = -\frac{\sinh(-\gamma \tau)}{\beta \sinh(-\gamma \tau) + \gamma \cosh(-\gamma \tau)}\\ & = \frac{\sinh(\gamma \tau)}{\gamma \cosh(\gamma \tau) - \beta \sinh(\gamma \tau)}\\ & = \frac{1}{\gamma \coth(\gamma \tau) - \beta}. \end{align*}

To solve for m(t, T), we first note that

    \[ m_\tau(t, T) = -\alpha \kappa \eta n(t, T) \]

which can be solved by integration

    \begin{align*} m(t, T) & = c_3 - \alpha \kappa \eta \int_0^{T - t} n(T - s, T) \mathrm{d}s\\ & = c_2 - \frac{2 \kappa \eta}{\lambda^2} \int_0^\tau \frac{c_\tau(s)}{c(s)} \mathrm{d}s\\ & = c_3 - \left. \frac{2 \kappa \eta}{\lambda^2} \ln \left( c(s) \right) \right|_{s = 0}^{s = \tau}\\ & = c_3 - \frac{2 \kappa \eta}{\lambda^2} \ln \left( \frac{c(\tau)}{c(0)} \right)\\ & = c_3 - \frac{2 \kappa \eta}{\lambda^2} \ln \left( \frac{e^{\beta \tau}}{\gamma} \left( \beta \sinh(-\gamma \tau) + \gamma \cosh(-\gamma \tau) \right) \right). \end{align*}

By the terminal condition m(T, T) = 0, it follows that c_3 = 0 and we get

    \begin{align*} m(t, T) & = \frac{2 \kappa \eta}{\lambda^2} \ln \left( \frac{\gamma e^{-\beta \tau}}{\gamma \cosh(\gamma \tau) - \beta \sinh(\gamma \tau)} \right)\\ & = -\frac{2 \kappa \eta}{\lambda^2} \left( \ln \left( \cosh(\gamma \tau) - \frac{\beta}{\gamma} \sinh(\gamma \tau) \right) + \beta \tau \right). \end{align*}

Putting Everything Together

Putting everything together, we find that the Laplace transform of the integrated time-change is given by

    \[ \mathcal{L} \left\{ Y(t, T) \right\} (\alpha) = \frac{\exp \left\{ \kappa^2 \eta \tau / \lambda^2 - 2 \alpha y_t / (2 \gamma \coth(\gamma \tau) + \kappa) \right\}}{\left( \cosh(\gamma \tau) + \kappa \sinh(\gamma \tau) / (2 \gamma) \right)^{2 \kappa \eta / \lambda^2}}. \]

The characteristic function of Y_t = Y(0, t) is

    \[ \phi_{Y_t}(\omega) = \frac{\exp \left\{ \kappa^2 \eta t / \lambda^2 - 2 \mathrm{i} \omega y_0 / (2 \gamma \coth(\gamma \tau) + \kappa) \right\}}{\left( \cosh(\gamma \tau) + \kappa \sinh(\gamma \tau) / (2 \gamma) \right)^{2 \kappa \eta / \lambda^2}}, \]


    \[ \gamma = \frac{1}{2} \sqrt{\kappa^2 - 2 \mathrm{i} \omega \lambda^2}. \]

This expression is equivalent to the ones given in Carr et al. (2003) and Schoutens (2003), who both define \gamma slightly differently.


Carr, Peter, Hélyette Geman, Dilip B. Madan, and Marc Yor (2003) “Stochastic Volatility for Lévy Processes”, Mathematical Finance, Vol. 13, No. 3, pp. 345-382

Cox, John C., Jonathan Ingersoll Jr., and Stephen A. Ross (1985) “A Theory of the Term Structure of Interest Rates”, Econometrica, VOl. 53, No. 2, pp. 385-407

Musiela, Marek and Marek Rutkowski (2005) Martingale Methods in Financial Modelling: Springer

Schoutens, Wim (2003) Lévy Processes in Finance: John Wiley & Sons

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