Verifying the Exponential Solution to a Mean-Reverting Rate ODE

Summary

The document considers the deterministic rate equation in which the rate moves toward a long-run level at a speed set by a positive adjustment parameter. It asks whether the proposed exponential expression satisfies the differential equation and its initial condition. The response verifies this directly: differentiate the proposed function, then substitute the function into the equation’s right-hand side; the expressions agree.

This substitution check is a straightforward alternative to deriving the solution by separation of variables. It confirms the stated formula and its behavior at the initial time, but the exchange does not discuss stochastic rate models, parameter restrictions, or applications to pricing. The example is useful as a compact method for checking a candidate solution to a deterministic mean-reversion equation.

Key ideas

• The proposed rate path approaches a constant long-run level exponentially.
• Differentiate the candidate function to check whether it satisfies the rate equation.
• Substitute the candidate into the equation’s right-hand side to verify the derivative.
• The verification concerns a deterministic ordinary differential equation, not a stochastic rate model.

Tags

# Proof verification : risk free rate


# Proof verification : risk free rate












I want to prove that $$r_t = \theta + (r_0 -\theta)e^{-kt}$$ satisfies $$dr_t = k(\theta-r_t)dt, \ r(0) = r_0$$ I have \begin{split}\frac{1}{\theta - r_t} dr_t = kdt \Rightarrow & \int_0^t \frac{1}{\theta - r_s} dr_s = \int_0^t kds\\ \Rightarrow &- \ln|\theta - r_s|\big\lvert_0^t = ks\big\lvert_0^t\\ \Rightarrow &-\ln|\theta - r_t| + \ln|\theta - r_0| = kt\\ \Rightarrow &\ln|\theta - r_t| - \ln|\theta - r_0| = -kt\\ \Rightarrow &\ln\left|\frac{\theta - r_t}{\theta - r_0}\right|= -kt\\ \Rightarrow &\frac{r_t - \theta}{r_0-\theta} = e^{-kt}\\ \Rightarrow & r_t = \theta + (r_0 -\theta)e^{-kt} \end{split} Is this the way to proceed? Thanks!

## Answer by Daneel Olivaw (score 3, accepted)

https://quant.stackexchange.com/a/68910

You have: $$r(t):=\theta+(r_0-\theta)e^{-kt}\tag{1}$$ Then: $$r^\prime(t)=-k(r_0-\theta)e^{-kt}\tag{2}$$ which is clearly equal to: $$k(\theta-r(t))\tag{3}$$ based on $(1)$.

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