Exact Gaussian Transitions for a Separable HJM Short-Rate Model
Summary
This note describes a finite-dimensional representation of an HJM interest-rate model when its deterministic volatility is separable across time and maturity. The short rate is written using a vector state process whose drift includes a deterministic forcing term and a mean-reversion matrix. The forcing term is itself a deterministic matrix that can be characterized by an explicit integral or an ordinary differential equation.
For simulation, the state at the next time point is conditionally Gaussian given its current value. The note gives the conditional mean and proposes drawing the next state using that mean, a square root of its conditional covariance, and a standard Gaussian vector. Its question concerns numerical evaluation of the mean integral when model functions are piecewise constant on the time grid. It considers using the explicit expression or evolving the auxiliary matrix with a discretized ODE, alongside a right-endpoint Riemann approximation. No numerical comparison or preferred scheme is supplied, so discretization accuracy remains open.
Key ideas
- A separable deterministic volatility structure permits a finite-dimensional state representation of the HJM short rate.
- The state vector has a conditionally Gaussian transition over each simulation interval.
- A deterministic matrix in the drift can be computed from an integral or its associated ODE.
- Piecewise-constant inputs simplify the scheme, but the note does not establish which approximation is most accurate.
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# How to deal with the deterministic $y$ in the d-dimensional gaussian model
# How to deal with the deterministic $y$ in the d-dimensional gaussian model
Suppose that under the risk-neutral measure $\mathbf{Q}$ we have an HJM framework dynamics for the instantaneous forward rate $$df_{t,T} = \left(\ldots\right) dt + {}^t \sigma_f (t,T) d W^{Q}_t$$ where $\sigma_f \in\mathbf{R}^d$ is deterministic and where $W^{\mathbf{Q}} \in\mathbf{R}^d$ is a Brownian Motion.
If we suppose that $\sigma_f$ is separable, that is, that you can write $$\sigma_f (t,T) = g(t) h(T)$$ for $g$ a $d\times d$ deterministic matrix function and $h$ a $d$-dimensional deterministic vector function, you can develop a whole theory. Namely, if you set $$H(t) = diag (h(t))$$ and $$\chi (t) = - H'(t) H(t)^{-1}$$ (assuming differentiability and invertibility) you can show that $r_t = f_{0,t} + x_{1,t} + \ldots x_{d,t}$ where the $x_{i,t}$ are the coefficient of $d$-dimensional random vector $x_t$ satisfying the following SDE : $$ dx_t = (y(t)\mathbf{1} - \chi(t) x_t) dt + {}^t \sigma_x (t) dW^{\mathbf{Q}}_t$$ with $x_0 = 0$ where $\mathbf{1}$ is the $d$-dimensional vector with all coefficients equal to $1$, $\sigma_x (t) \equiv g(t) h(t)$ and where $y(t)$ is a $d\times d$ deterministic matrix equal to $$H(t) \left( \int_0^t {}^t g(s) g(s) \right) H(t).$$
One can also show that $y(t)$ satisfies the following ODE : $$y'(t) = H(t) {}^t g(t) g(t) H(t) - \chi (t) y(t) - y(t) \chi(t)$$ with $y(0) = 0$.
Now if we want to simulate $x$ at a time discretization $0 = t_0 < t_1 < \ldots < t_d$ one can tactically take advantage of the fact that $x_{t_{i+1}}$ is, conditionally to $x_{t_i}$, a gaussian vector with computable mean and variance, namely : $$(E)\;\;\;\;\;\;\;\;\mathbf{E}^{\mathbf{Q}}\left[ x_{t_{i+1}} \left| x_{t_i} \right.\right] = e^{-\int_{t_i}^{t_{i+1}} \chi (u)du} x_{t_i} + \int_{t_i}^{t_{i+1}} e^{-\int_s^{t_{i+1}} \chi (u)du} y(s) \mathbf{1} ds.$$
and write that $$x_{t_{i+1}} = e^{-\int_{t_i}^{t_{i+1}} \chi (u)du} x_{t_i} + \int_{t_i}^{t_{i+1}} e^{-\int_s^{t_{i+1}} \chi (u)du} y(s) \mathbf{1} ds + \sqrt{\mathbf{Var}^{\mathbf{Q}}\left[ x_{t_{i+1}} \left| x_{t_i} \right.\right]} Z_i$$ where $\sqrt{\mathbf{Var}^{\mathbf{Q}}\left[ x_{t_{i+1}} \left| x_{t_i} \right.\right]}$ is a square-root of the variance matrix (Cholesky for instance) and $Z_1,\ldots,Z_d$ a sequence of independent and identically distributed standard $d$-dimensional gaussian vectors.
Fine. So we need to treat numerically the integral $$\int_{t_i}^{t_{i+1}} e^{-\int_s^{t_{i+1}} \chi (u)du} y(s) \mathbf{1} ds$$ from equation (E).
How do we do that under the hypothesis that all deterministic functions are piecewise constant on the given discretization ? Do we explicitely compute $y(s)$ on $[t_i, t_{i+1}]$ from $y$'s explicit formula or do we use the ODE satisfied by $y$ and if so, how ? Or do we simply say that $$\int_{t_i}^{t_{i+1}} e^{-\int_s^{t_{i+1}} \chi (u)du} y(s) \mathbf{1} ds \simeq (t_{i+1} - t_i) y(t_{i+1}) \mathbf{1}$$ (Riemann right sum) and calculate recursively the $y(t_{i+1})$'s from a discretized version $$y(t_{i+1}) = y(t_i) + (t_{i+1} - t_i) \left(H(t_i) {}^t g(t_i) g(t_i) H(t_i) - \chi(t_i) y(t_i) - y(t_i) \chi(t_i)\right)$$ of the ODE satisfied by $y$ ?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.