The Vasicek Model As A Mean-Reverting Linear SDE

From a Linear Drift to the Vasicek Model

The Vasicek model describes a short rate whose drift pulls it toward a fixed level while Brownian noise continually perturbs it. Its defining feature is not simply that the drift is linear: the coefficient of the rate must have the sign that makes the drift mean-reverting. Once the equation is written in that form, an integrating factor gives the full rate process as a deterministic relaxation of its initial value plus a stochastic convolution.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ relative to the filtration. The equation below is an Itô SDE under $\mathbb{P}$. A solution is adapted to this filtration, and each stochastic integral $\int_0^t H_s\,dB_s$ used here is an Itô integral with progressively measurable integrand satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$ for every finite $T$.

Definition: The Vasicek short-rate model is the SDE $$ dr_t=\kappa(\theta-r_t)\,dt+\sigma\,dB_t,\qquad r_0\text{ given}, $$ where $\kappa>0$ is the mean-reversion speed, $\theta\in\mathbb{R}$ is the long-run level, and $\sigma\ge0$ is the volatility. The drift coefficient is $\kappa(\theta-r)$; the diffusion coefficient is the constant $\sigma$.

The sign of the drift explains the mean-reversion terminology. If $r_t>\theta$, then $\kappa(\theta-r_t)<0$; if $r_t<\theta$, then $\kappa(\theta-r_t)>0$. Thus the drift points toward $\theta$. The parameter $\kappa$ controls the rate of this pull, while $\sigma$ controls the size of the Brownian perturbation. The model permits real-valued rates: it does not impose a positivity constraint.

Identifying the Parameters

A common starting point is a linear drift written with an intercept and a rate coefficient. Its relation to the Vasicek parameters follows from matching coefficients, not from assigning the intercept itself as the long-run level.

Proposition: For $\beta>0$, the equation $$ dr_t=(\alpha-\beta r_t)\,dt+\sigma\,dB_t $$ is the Vasicek model with $\kappa=\beta$ and $\theta=\alpha/\beta$. Conversely, the Vasicek drift has intercept $\alpha=\kappa\theta$ and rate coefficient $\beta=\kappa$.

Proof. Since $\beta>0$, division by $\beta$ is defined. Factoring the drift gives $\alpha-\beta r=\beta(\alpha/\beta-r)$, which is $\kappa(\theta-r)$ when $\kappa=\beta$ and $\theta=\alpha/\beta$. Conversely, expanding $\kappa(\theta-r)$ gives $\kappa\theta-\kappa r$, so $\alpha=\kappa\theta$ and $\beta=\kappa$. The strict condition $\beta>0$ is essential for mean reversion in this parameterization; $\beta=0$ gives constant drift, and $\beta<0$ gives a drift that pushes away from its equilibrium level. $\square$

Worked Example: Reading the Mean-Reversion Parameters

Write $$ dr_t=(0.18-0.6r_t)\,dt+0.12\,dB_t $$ in Vasicek form and interpret the parameters.

Matching the coefficient of $r_t$ gives $\kappa=0.6$. The intercept is $\alpha=0.18$, so the long-run level is $\theta=\alpha/\beta=0.18/0.6=0.3$. The diffusion coefficient is $\sigma=0.12$. Therefore the equation is $$ dr_t=0.6(0.3-r_t)\,dt+0.12\,dB_t. $$

Check: expanding the drift gives $0.6(0.3-r_t)=0.18-0.6r_t$, the original drift. At $r_t=0.4$, it equals $0.6(0.3-0.4)=-0.06$, pointing downward toward $0.3$; at $r_t=0.2$, it equals $0.6(0.3-0.2)=0.06$, pointing upward.

Explicit Solution by an Integrating Factor

The linear equation is solved by multiplying the rate by $e^{\kappa t}$. This calculation uses Itô’s formula. For a twice differentiable function $f(t,x)$ and an Itô process $X_t$, the second-order term is $\tfrac12 f_{xx}$ times the squared diffusion coefficient and $dt$; in differential notation $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$.

Theorem: Suppose $\kappa>0$, $\theta\in\mathbb{R}$, $\sigma\ge0$, and $r_0$ is $\mathcal{F}_0$-measurable with $\mathbb{E}[r_0^2]<\infty$. The Vasicek SDE has the strong solution $$ r_t=\theta+(r_0-\theta)e^{-\kappa t} +\sigma e^{-\kappa t}\int_0^t e^{\kappa s}\,dB_s,\qquad t\ge0. $$ On every finite time interval, this solution is pathwise unique.

Proof. Apply Itô’s formula to $f(t,x)=e^{\kappa t}x$ and $x=r_t$. Its derivatives are $f_t=\kappa e^{\kappa t}x$, $f_x=e^{\kappa t}$, and $f_{xx}=0$. The correction term is explicitly $\tfrac12 f_{xx}\sigma^2dt=\tfrac12\cdot0\cdot\sigma^2dt=0$. Hence $$ d(e^{\kappa t}r_t) =\kappa e^{\kappa t}r_t\,dt +e^{\kappa t}\big(\kappa(\theta-r_t)\,dt+\sigma\,dB_t\big) =\kappa\theta e^{\kappa t}\,dt+\sigma e^{\kappa t}\,dB_t. $$ For fixed finite $T$, the integrand $s\mapsto e^{\kappa s}$ is deterministic and progressively measurable, and $\mathbb{E}\int_0^T e^{2\kappa s}\,ds=(e^{2\kappa T}-1)/(2\kappa)<\infty$. Thus its Itô integral is defined. Integrating from $0$ to $t$, and using $\int_0^t\kappa\theta e^{\kappa s}\,ds=\theta(e^{\kappa t}-1)$, gives $$ e^{\kappa t}r_t=r_0+\theta(e^{\kappa t}-1)+\sigma\int_0^t e^{\kappa s}\,dB_s. $$ Multiplication by $e^{-\kappa t}$ yields the displayed formula. Its initial value is $r_0$; it is adapted and continuous, and its stochastic integral is well-defined on each finite interval. This constructs a strong solution, meaning a solution built on the given space and driven by its given Brownian motion. For uniqueness, let $r$ and $\widetilde r$ be two solutions with the same initial value and Brownian motion. Their difference $D_t=r_t-\widetilde r_t$ satisfies $dD_t=-\kappa D_t\,dt$, $D_0=0$. The ordinary product rule gives $d(e^{\kappa t}D_t)=0$, so $e^{\kappa t}D_t=0$ and $D_t=0$ for all $t$. This is pathwise uniqueness. It is a statement about solutions driven by the same Brownian motion; it is distinct from uniqueness in law, which concerns equality of distributions. $\square$

The coefficient functions $\mu(t,x)=\kappa(\theta-x)$ and $\sigma(t,x)=\sigma$ also satisfy the usual global Lipschitz and linear growth conditions. In fact, $|\mu(t,x)-\mu(t,y)|=\kappa|x-y|$, and $|\sigma-\sigma|=0$. Also, $|\mu(t,x)|\le\kappa|\theta|+\kappa|x|$, so the drift grows at most linearly, while the diffusion is bounded. Here the explicit construction and uniqueness argument establish the solution directly. A strong solution under $\mathbb{P}$ should not be silently transferred to a different measure: if a model is instead specified under a risk-neutral measure $\mathbb{Q}$, its Brownian motion and parameters must be interpreted under that measure.

Interpreting the Solution

The formula separates the three contributions: $\theta$, the initial displacement $(r_0-\theta)e^{-\kappa t}$, and the stochastic convolution. Since $\kappa>0$, the initial displacement is multiplied by a factor that decays as time increases. The stochastic convolution records the accumulated Brownian shocks, with earlier shocks discounted by the same exponential relaxation. The stochastic integral here is an Itô integral; it is not a Stratonovich integral, and no ordinary chain rule is being used to derive the solution.

Worked Example: Writing the Rate at a Fixed Time

Let $r_0=0.04$, $\kappa=0.5$, $\theta=0.06$, and $\sigma=0.02$. Express $r_2$ using the explicit solution.

First, $e^{-\kappa t}=e^{-0.5\cdot2}=e^{-1}$. The initial displacement is $r_0-\theta=0.04-0.06=-0.02$. Substitution gives $$ r_2=0.06-0.02e^{-1} +0.02e^{-1}\int_0^2 e^{0.5s}\,dB_s. $$

The integral is admissible because its deterministic integrand is progressively measurable and $$ \mathbb{E}\int_0^2 (e^{0.5s})^2\,ds =\int_0^2 e^s\,ds=e^2-1<\infty. $$ Check: setting $t=0$ in the general solution makes the exponential $1$, the integral has zero length, and the two deterministic terms give $\theta+(r_0-\theta)=r_0=0.04$. Thus the initial condition is preserved.

Worked Example: The Zero-Volatility Limit

Take $r_0=0.07$, $\kappa=2$, $\theta=0.03$, and $\sigma=0$. Find $r_{0.5}$ and check the initial direction of motion.

With $\sigma=0$, the stochastic term vanishes. At $t=0.5$, $\kappa t=2\cdot0.5=1$, so $$ r_{0.5}=0.03+(0.07-0.03)e^{-1}=0.03+0.04e^{-1}. $$ The initial derivative of this deterministic solution is $-\kappa(r_0-\theta)=-2(0.07-0.03)=-0.08$. The SDE drift at the initial rate gives $2(0.03-0.07)=-0.08$ as well.

Check: differentiating $0.03+0.04e^{-2t}$ gives $-0.08e^{-2t}$, while the drift evaluated at this rate is $2(0.03-(0.03+0.04e^{-2t}))=-0.08e^{-2t}$. The deterministic formula therefore satisfies the drift equation, including its initial value.

Verification and a Common Parameter Error

The explicit formula can also be checked directly in the SDE. Put $J_t=\int_0^t e^{\kappa s}\,dB_s$, which is well-defined by the finite-interval integrability calculation above, and write $r_t=\theta+(r_0-\theta)e^{-\kappa t}+\sigma e^{-\kappa t}J_t$. The deterministic factor $e^{-\kappa t}$ has finite variation, so its quadratic covariation with $J_t$ is zero. Since $dJ_t=e^{\kappa t}dB_t$, the product rule gives $$ d(e^{-\kappa t}J_t)=-\kappa e^{-\kappa t}J_t\,dt+dB_t. $$ Consequently, $$ dr_t=-\kappa(r_t-\theta)\,dt+\sigma\,dB_t =\kappa(\theta-r_t)\,dt+\sigma\,dB_t, $$ as required. This substitution check confirms both the sign of the drift and the coefficient of the noise.

Common mistake: in $dr_t=(0.12-0.4r_t)dt+0.03\,dB_t$, taking the long-run level to be $0.12$ confuses the intercept with the equilibrium. The correct values are $\kappa=0.4$ and $\theta=0.12/0.4=0.3$. The incorrect drift $0.4(0.12-r_t)$ expands to $0.048-0.4r_t$, not $0.12-0.4r_t$. The corrected drift $0.4(0.3-r_t)$ expands to $0.12-0.4r_t$, matching the given equation.

The Vasicek form is useful because it makes the direction and speed of mean reversion explicit while retaining an exact solution. Its real-valued state space is also a modeling limitation: unlike a model designed to preserve nonnegative rates, the Vasicek equation can take negative values. The equation is a specification of rate dynamics under the stated probability measure, not a guarantee that every rate path is positive.

Check Your Understanding

Use the parameter matching rule and the explicit solution to answer the following.

Rewrite $dr_t=(0.15-0.5r_t)dt+0.02\,dB_t$ in Vasicek form and identify each parameter.
For $\kappa=1$, $\theta=0.04$, and $r_0=0.05$, write the solution at time $t$ as a deterministic term plus an Itô integral.
Explain why $\beta>0$ is needed for the equation $dr_t=(\alpha-\beta r_t)dt+\sigma dB_t$ to have a mean-reverting drift.
For the explicit solution, verify the initial value by setting $t=0$.
State the difference between pathwise uniqueness and uniqueness in law for this SDE.