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Quadratic Variation in a One-Factor Cheyette Interest Rate Model

Article Quant Q&A · Author: Jason

Summary

The document asks how two representations of a one-factor Cheyette interest-rate model relate. One uses state variables for the short-rate factor and an auxiliary variance-like process, while the other expresses discounted zero-coupon bond prices using the factor and its quadratic variation. The central question is how the second bond-price formula can recover the first model’s auxiliary-state dynamics, including its mean-reversion term.

The post supplies the factor dynamics, bond-price expressions, and a relationship between their time-dependent loading functions, but provides no answer or derivation. It therefore serves as a focused problem statement rather than a worked explanation. Resolving it would require checking the change of variables and applying Itô calculus to the transformed factor and quadratic variation; the document itself offers no evidence establishing that the formulas coincide or identifying conditions under which they do.

Key ideas

  • The post compares two formulations of a one-factor Cheyette interest-rate model.
  • One formulation tracks an auxiliary state with drift tied to mean reversion and factor volatility.
  • The alternate bond-price expression uses the factor and its quadratic variation.
  • A relationship between the model loading functions is stated, but the requested derivation is absent.

Tags

Full text
# Where is the Quadratic Variation Coming from in this One-Factor Cheyette Model?


# Where is the Quadratic Variation Coming from in this One-Factor Cheyette Model?












I am having difficulty switching from a general interest rate model (the quasi-gaussian or cheyette model) and a specific version of this model. In particular, I assume the following instantaneous forward vol separation:

$$\sigma_f(t, T, \omega) = g(t,\omega)h(T)$$

and define $$h(t) = \exp\left(\int_0^t \kappa(u)\,du\right) \hspace{1cm} G(t,T) = \int_t^Te^{\int_t^s \kappa(u)\,du} \,ds \hspace{1cm} \sigma_r(t,\omega) = g(t,\omega) h(t)$$

The model is driven by stochastic processes $x$ and $y$ with dynamics given by

$$\,dx_t = (y_t - \kappa(t)x_t) \,dt + \sigma_r(t, \omega) \,dW_t$$ $$\,dy_t = (\sigma_r(t,\omega)^2 - 2\kappa(t)y_t)\,dt$$ and $x(0) = y(0) = 0$. From here I know that we can recover discounted ZCB prices as $$P_1(t,T, x_t, y_t) = \frac{P(0,T)}{P(0,t)} \exp \left(-G(t,T)x_t - \frac{1}{2}G(t,T)^2y_t\right)$$

This is all well and good, my problem is that I have a specification of the model with $\,dx(t) = \sigma_r(t,\omega) \,dW_t$ (no problem) but with discounted bond prices as $$P_2(t,T, x_t, \langle x \rangle_t) = \frac{P(0,T)}{P(0,t)} \exp \left( -(H_T - H_t) x_t -\frac{1}{2} (H_T^2 - H_t^2)\langle x \rangle_t\right)$$ where $H_t = \int_0^t e^{\int_0^s \kappa(u)\,du} \,ds$. My question is, how are we recovering the dynamics of $y_t$ when $\,d\langle x \rangle_t = \sigma_r^2\,dt$ and the factor of $2\kappa(t)y(t)$ is removed? ie. how do we go from $P_2$ to $P_1$?

We have that $G(t,T) = (H_T - H_t)e^{\int_0^t \kappa(s)\,ds}$. At this point I have spent enough time on this problem that I feel silly and hope I am just missing something obvious. Help is very much appreciated.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.