Deriving a Two-Factor Bond Pricing PDE from a Discounted Martingale
Summary
The answer derives the pricing equation for a zero-coupon bond in a model with two state variables: the short rate and a second factor. It represents the bond price as a conditional expectation of discounted face value, then discounts the price back to the present and identifies the resulting process as a martingale. Applying Itô’s lemma and setting the drift to zero yields a two-factor partial differential equation. Replacing calendar time with time to maturity changes the sign of the time derivative.
The derivation includes mean-reverting drift terms, state-dependent diffusion for the short rate, and square-root diffusion for the second factor. It states that the remaining step to a Riccati equation is straightforward, but does not show that step or solve the equation. The result depends on the assumed risk-neutral dynamics and bond pricing representation; the excerpt does not discuss parameter restrictions, boundary conditions, or model calibration.
Key ideas
- A discounted conditional expectation gives the zero-coupon bond price in the stated model.
- The discounted bond price is treated as a martingale under the pricing measure.
- Applying Itô’s lemma and eliminating the drift produces the pricing PDE.
- Writing the equation in time to maturity reverses the sign of its time derivative.
- The excerpt does not derive or solve the resulting Riccati equation.
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# Problem of stochastic differential equation (SDE)
# Problem of stochastic differential equation (SDE)
Please help to answer this stochastic differential equation (SDE). Thank you very much.
## Answer by Gordon (score 4)
https://quant.stackexchange.com/a/49264
We assume that the price at time $t$ of a zero-coupon bond, with maturity $u$ and unit face value, is of the form \begin{align*} f(u-t, r_t, x_t) = E\left(e^{-\int_t^u r_s ds}\mid \mathcal{F}_t\right). \end{align*} Note that \begin{align*} M(t, r_t, x_t) &\equiv f(u-t, r_t, x_t) e^{-\int_0^t r_s ds} \\ &=E\left(e^{-\int_0^u r_s ds} \mid \mathcal{F}_t \right) \end{align*} is a martingale. Moreover, \begin{align*} dM &= - r f e^{-\int_0^t r_s ds}dt + e^{-\int_0^t r_s ds}df\\ &= e^{-\int_0^t r_s ds}\bigg[- r f dt + \frac{\partial f}{\partial t}dt + \frac{\partial f}{\partial r} dr_t + \frac{\partial f}{\partial x} dx_t\\ &\qquad\qquad\qquad + \frac{1}{2}\frac{\partial^2 f}{\partial r^2}d\langle r, r\rangle_t + \frac{1}{2}\frac{\partial^2 f}{\partial x^2}d\langle x, x\rangle_t + \frac{\partial^2 f}{\partial r\partial x}d\langle r, x\rangle_t \bigg]\\ &=e^{-\int_0^t r_s ds}\bigg[\frac{\partial f}{\partial t} - r f + \kappa_r(x-r)\frac{\partial f}{\partial r} + \kappa_x(\theta - x) \frac{\partial f}{\partial x} \\ &\qquad\qquad\qquad\qquad\qquad\qquad + \frac{1}{2}(\alpha + \beta r)\frac{\partial^2 f}{\partial r^2} + \frac{1}{2}\sigma^2 x \frac{\partial^2 f}{\partial x^2} \bigg]dt\\ & \quad +e^{-\int_0^t r_s ds}\left[ \sqrt{\alpha + \beta r_t}\frac{\partial f}{\partial r}dB_r(t) + \sigma \sqrt{x} \frac{\partial f}{\partial x}dB_x(t)\right]. \end{align*} Therefore, \begin{align*} \frac{\partial f}{\partial t} - r f + \kappa_r(x-r)\frac{\partial f}{\partial r} + \kappa_x(\theta - x) \frac{\partial f}{\partial x} + \frac{1}{2}(\alpha + \beta r)\frac{\partial^2 f}{\partial r^2} + \frac{1}{2}\sigma^2 x \frac{\partial^2 f}{\partial x^2}=0. \end{align*} In term of $\tau = u-t$, \begin{align*} -\frac{\partial f}{\partial \tau} - r f + \kappa_r(x-r)\frac{\partial f}{\partial r} + \kappa_x(\theta - x) \frac{\partial f}{\partial x} + \frac{1}{2}(\alpha + \beta r)\frac{\partial^2 f}{\partial r^2} + \frac{1}{2}\sigma^2 x \frac{\partial^2 f}{\partial x^2}=0. \end{align*} The remaining derivation of the Ricaati equation is then straightforward.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.