Decomposing a Mean-Reverting Stochastic Interest Rate

Summary

The document shows how to express a stochastic interest rate with deterministic time-varying drift as the sum of a random component and a deterministic component. The rate follows a mean-reverting diffusion with constant mean-reversion and volatility parameters, while its drift level is a nonstochastic function of time. Applying an integrating factor gives an explicit solution: the random part is an exponentially weighted Brownian integral, and the deterministic part combines the decayed initial rate with an integral of the drift function.

The answer verifies that the random component satisfies its own mean-reverting stochastic differential equation, driven by the same Brownian motion, while the remaining component is deterministic. This decomposition can help separate stochastic fluctuations from the time-dependent level in interest-rate modeling. The result assumes deterministic drift and constant coefficients as stated; the thread does not discuss calibration, alternative rate models, or empirical performance.

Key ideas

• An integrating factor solves the linear stochastic differential equation.
• The rate can be decomposed into a stochastic fluctuation and a deterministic level component.
• The stochastic component is an exponentially weighted integral of Brownian increments.
• The deterministic component includes the decayed initial value and accumulated time-varying drift.
• The decomposition relies on the specified deterministic drift and constant model parameters.

Tags

# stochastic interest rate $r_t=x_t+y_t$


# stochastic interest rate $r_t=x_t+y_t$












Let $$dr_t=(\alpha(t)-\beta r_t)dt+\sigma dW_t$$ where $\alpha$ is non stochastic process and $\beta$ and $\sigma$ are constant. Can we write process $r_t$ in the form $$r_t=x_t+y_t$$ where the process $x_t$ satisfies $$dx_t=-\beta x_t dt+\sigma dW_t$$ and $y_t$ be a deterministic function. I used Ito's lemma but was not useful.

Thanks in advanced.

## Answer by Gordon (score 2, accepted)

https://quant.stackexchange.com/a/27983

By the usual integrating factor method, \begin{align*} r_t = r_0e^{-\beta t} + \int_0^t \alpha(s) e^{-\beta(t-s)}ds +\sigma \int_0^t e^{-\beta(t-s)}dW_s. \end{align*} Let \begin{align*} x_t &=\sigma \int_0^t e^{-\beta(t-s)}dW_s, \textrm { and}\\ y_t &=r_0e^{-\beta t} + \int_0^t \alpha(s) e^{-\beta(t-s)}ds. \end{align*} Then $r_t = x_t + y_t$, moreover, \begin{align*} dx_t &= d\left(\sigma e^{-\beta t} \int_0^t e^{\beta s}dW_s \right)\\ &=-\beta \left(\sigma e^{-\beta t} \int_0^t e^{\beta s}dW_s\right)dt + \sigma dW_t\\ &=-\beta x_t dt + \sigma dW_t, \end{align*} and $y_t$ is a deterministic function.

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