The Vasicek Short-Rate Model
The Vasicek model describes a short rate that is pulled toward a long-run level while also receiving random shocks. Its linear drift makes it possible to solve the stochastic differential equation exactly. The solution reveals not only the expected future rate but its full conditional distribution, and it also makes the expected accumulated rate over a time interval explicit.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\). For the calculations, let \(\mathbb{Q}\) be an equivalent risk-neutral measure and let \(B_t\) be Brownian motion under \(\mathbb{Q}\), with the filtration satisfying the usual conditions. The model is specified under \(\mathbb{Q}\): for constants \(\kappa>0\), \(\theta\in\mathbb{R}\), and \(\sigma\ge0\),
Here \(\kappa\) is the speed of mean reversion, \(\theta\) is the long-run level, and \(\sigma\) is the diffusion coefficient. The drift is \(\kappa(\theta-r_t)\). All conditional expectations and laws below are under \(\mathbb{Q}\). If the same SDE is instead specified under \(\mathbb{P}\) with a \(\mathbb{P}\)-Brownian motion, the calculations give its \(\mathbb{P}\)-conditional law; changing measures does not by itself preserve the drift specification.
Every stochastic integral below is an Ito integral. Its integrand is adapted and progressively measurable, and belongs to \(\mathcal{L}^2\) under \(\mathbb{Q}\): for each finite horizon \(T\), \(\mathbb{E}^{\mathbb{Q}}\int_0^T H_u^2\,du<\infty\). The deterministic exponential integrands used here are bounded on finite intervals, so they satisfy this condition.
Explicit Solution and Conditional Law
Proof. Multiply the SDE by the deterministic integrating factor \(e^{\kappa t}\). The Ito product rule has no quadratic-covariation term with this finite-variation factor, and gives \(d(e^{\kappa t}r_t)=\kappa\theta e^{\kappa t}\,dt+\sigma e^{\kappa t}\,dB_t\). Integrating from \(s\) to \(t\), then multiplying by \(e^{-\kappa t}\), yields $$ r_t=e^{-\kappa(t-s)}r_s+\theta(1-e^{-\kappa(t-s)}) +\sigma\int_s^t e^{-\kappa(t-u)}\,dB_u, $$ which is the stated representation.
For a deterministic square-integrable function \(g\), the Ito integral \(\int_s^t g(u)\,dB_u\) is normal with mean zero and variance \(\int_s^t g(u)^2\,du\). To justify this fact, first take \(g\) to be a step function: the integral is a sum of independent centered normal Brownian increments, hence is normal with variance equal to the integral of \(g^2\). For general deterministic \(g\in L^2([s,t])\), approximate it in \(L^2\) by step functions. The Ito isometry gives convergence of the integrals in \(L^2(\mathbb{Q})\); their characteristic functions therefore converge to that of the centered normal with variance \(\int_s^t g(u)^2\,du\). The same argument applies conditionally on \(\mathcal{F}_s\), because Brownian increments after \(s\) are independent of \(\mathcal{F}_s\). Thus the stochastic integral in the solution is conditionally centered Gaussian, with variance $$ \sigma^2\int_s^t e^{-2\kappa(t-u)}\,du =\frac{\sigma^2}{2\kappa}\left(1-e^{-2\kappa(t-s)}\right). $$ The remaining terms are \(\mathcal{F}_s\)-measurable. This proves the conditional mean, variance, and Gaussian law, including the stated zero-variance cases. \(\square\)
Worked Example: One-Year Conditional Rate Distribution
Suppose \(\kappa=2\), \(\theta=0.06\), \(\sigma=0.03\), and \(r_s=0.04\). Find the conditional law one year later, meaning \(t-s=1\).
The conditional mean is $$ m_{s,t}=0.06+(0.04-0.06)e^{-2} =0.06-0.02e^{-2}. $$ The conditional variance is $$ v_{s,t}=\frac{(0.03)^2}{2(2)}(1-e^{-2(2)(1)}) =0.000225(1-e^{-4}). $$ Consequently \(r_t\mid\mathcal{F}_s\) is normal with this mean and variance. Both quantities are conditional on the observed current rate \(r_s=0.04\); they are not unconditional constants if \(r_s\) is itself random.
Check: if the horizon is zero, the mean formula becomes \(0.06+(0.04-0.06)=0.04\), and the variance becomes \(0.000225(1-1)=0\), as required for \(r_s\) given \(\mathcal{F}_s\).
Accumulated Short Rate
For \(h\ge0\), define the integrated short rate over \([s,s+h]\) by the ordinary time integral \(I_{s,h}=\int_s^{s+h}r_u\,du\). It is the accumulated rate over the interval; it is not itself a discount factor. The explicit rate solution also gives its conditional mean and variance.
Proof. Let \(N_u=\int_s^u e^{-\kappa(u-v)}\,dB_v\) for \(s\le u\le s+h\). Its integrand is deterministic, progressively measurable, and has finite squared integral on this finite interval, so this Ito integral is well-defined in \(\mathcal{L}^2\). The rate representation gives \(r_u=\theta+(r_s-\theta)e^{-\kappa(u-s)}+\sigma N_u\). Integrating the first two terms in time gives \(\theta h+(r_s-\theta)(1-e^{-\kappa h})/\kappa\).
For the noise term, the defining integral for \(N_u\) solves \(dN_u=-\kappa N_u\,du+dB_u\), with \(N_s=0\). Integrating this identity from \(s\) to \(s+h\) yields $$ N_{s+h}=-\kappa\int_s^{s+h}N_u\,du+(B_{s+h}-B_s), $$ and therefore $$ \int_s^{s+h}N_u\,du =\frac{1}{\kappa}\left((B_{s+h}-B_s)-N_{s+h}\right) =\frac{1}{\kappa}\int_s^{s+h}\left(1-e^{-\kappa(s+h-v)}\right)dB_v. $$ The last identity follows by substituting the defining expression for \(N_{s+h}\). The final integrand is deterministic, progressively measurable, and square-integrable on the finite interval. Thus \(I_{s,h}\) is a conditionally Gaussian variable, with the stated conditional mean and with variance $$ \frac{\sigma^2}{\kappa^2}\int_0^h(1-e^{-\kappa u})^2\,du =\frac{\sigma^2}{\kappa^2}\left[ h-\frac{2(1-e^{-\kappa h})}{\kappa} +\frac{1-e^{-2\kappa h}}{2\kappa} \right]. $$ The variance is zero if \(h=0\) or \(\sigma=0\); otherwise the conditional Gaussian law follows from the deterministic-integrand result proved above. \(\square\)
Worked Example: Expected Integrated Rate Over Two Years
Take \(\kappa=1\), \(\theta=0.05\), \(r_s=0.03\), \(\sigma=0.02\), and \(h=2\). Compute the conditional mean and variance of the two-year integrated rate.
Substituting into the mean formula gives $$ \mathbb{E}^{\mathbb{Q}}[I_{s,2}\mid\mathcal{F}_s] =0.05(2)+(0.03-0.05)(1-e^{-2}) =0.1-0.02(1-e^{-2}). $$ For the variance, \(\sigma^2=0.0004\) and \(\kappa=1\), so $$ \operatorname{Var}^{\mathbb{Q}}(I_{s,2}\mid\mathcal{F}_s) =0.0004\left[2-2(1-e^{-2})+\frac{1-e^{-4}}{2}\right] =0.0004\left[\frac12+2e^{-2}-\frac12e^{-4}\right]. $$ Thus the integrated rate has a conditional Gaussian distribution with the displayed mean and variance.
Check: letting \(\sigma=0\) in the variance calculation gives zero. In that case the rate follows its deterministic mean-reverting path and the integrated rate equals \(0.1-0.02(1-e^{-2})\), agreeing with the conditional mean.
Worked Example: A One-Half-Life Horizon
Let \(\kappa=1\), \(\sigma=2\), and \(h=\ln 2\), so \(e^{-\kappa h}=1/2\). The current rate \(r_s\) and long-run level \(\theta\) may be any fixed values. Compute the rate variance, integrated-rate variance, and their conditional covariance.
The rate variance is $$ \operatorname{Var}^{\mathbb{Q}}(r_{s+h}\mid\mathcal{F}_s) =\frac{2^2}{2(1)}(1-e^{-2\ln2}) =2\left(1-\frac14\right)=\frac32. $$ The integrated-rate variance is $$ \operatorname{Var}^{\mathbb{Q}}(I_{s,h}\mid\mathcal{F}_s) =4\left[\ln2-2\left(1-\frac12\right) +\frac{1-\frac14}{2}\right] =4\ln2-\frac52. $$ To obtain the covariance, use the stochastic parts of the two representations. Their deterministic integrands are \(2e^{-(s+h-v)}\) for the terminal rate and \(2(1-e^{-(s+h-v)})\) for the integrated rate. The covariance of deterministic Brownian integrals is the integral of the product of their integrands; hence $$ \operatorname{Cov}^{\mathbb{Q}}(r_{s+h},I_{s,h}\mid\mathcal{F}_s) =4\int_0^{\ln2}e^{-u}(1-e^{-u})\,du =4\left[\frac12-\frac38\right]=\frac12. $$ The terminal rate and accumulated rate are therefore jointly Gaussian conditionally on \(\mathcal{F}_s\), since each is an affine function of deterministic Brownian integrals over the same future interval.
Check: the covariance is positive because both stochastic integrands are nonnegative on this interval. The integrated-rate variance is also positive: its bracket is \(\ln2-\tfrac58>0\), as follows directly from its equivalent integral form \(\int_0^{\ln2}(1-e^{-u})^2du\), whose integrand is positive for \(u>0\).
Interpreting the Formulas and Avoiding a Common Error
The conditional mean moves from \(r_s\) toward \(\theta\) at the exponential rate \(e^{-\kappa h}\). The conditional variance depends on the diffusion coefficient and horizon, not on the current rate or long-run level. As \(h\) increases, the terminal-rate variance approaches \(\sigma^2/(2\kappa)\); the integrated-rate variance instead records shocks accumulated throughout the interval.
Common mistake: using Brownian variance without the mean-reverting kernel. A wrong calculation treats the rate noise as \(\sigma(B_t-B_s)\), and consequently assigns terminal variance \(\sigma^2(t-s)\). For the one-year parameters in the first example, that gives \((0.03)^2(1)=0.0009\). The actual Vasicek variance is \((0.03)^2(1-e^{-4})/4=0.000225(1-e^{-4})\), because shocks received earlier are damped by mean reversion. The factor \(e^{-\kappa(t-u)}\) in the solution must be retained when computing variance.
Finally, the conditional expected accumulated rate is not the conditional expected discount factor. In general, the expectation of an exponential is not the exponential of an expectation. The Gaussian law derived here supplies the moments needed to distinguish those quantities, while pricing a short-rate bond requires the separate expectation of the discount factor under \(\mathbb{Q}\).
Check Your Understanding
Use the conditional formulas and keep track of which quantities are conditioned on \(\mathcal{F}_s\).
- For \(\kappa=3\), \(\theta=0.04\), and \(r_s=0.07\), write the conditional mean of \(r_{s+h}\) for an arbitrary \(h\ge0\).
- What is the conditional variance of \(r_{s+h}\) when \(\sigma=0\), and what does this imply about its conditional law?
- For general parameters, derive the conditional mean of \(I_{s,h}\) by integrating the conditional mean of \(r_{s+u}\) over \(0\le u\le h\).
- Explain why the conditional variance of the terminal rate does not depend on \(r_s\).
- For \(h>0\) and \(\sigma>0\), explain why the integrated-rate variance is strictly positive using its integral representation.