1
$\begingroup$

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 $y$ ?

$\endgroup$
1
  • $\begingroup$ This is in @piterbarg and Andersen's Interest Rates Modeling book, volume 2. $\endgroup$
    – 11house
    Commented May 21 at 10:37

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.