Solving the Vasicek Short-Rate Model for Its Mean and Variance
Summary
The document derives the conditional distribution moments for a Vasicek short-rate process with mean-reversion rate, long-run average rate, and constant diffusion coefficient. Rewriting the drift in terms of the long-run level makes the process a mean-reverting equation. Multiplication by an exponential integrating factor yields an explicit solution: a deterministic component that approaches the long-run rate, plus a stochastic integral of the driving process.
For Brownian noise, the conditional mean is the deterministic component, and Ito isometry gives the variance as the integral of the squared exponential kernel. The answer also cautions that an extended Vasicek model can use a time-varying mean-reversion level or drift, so the constant-level derivation does not by itself solve that broader specification. The response supplies the standard calculation but leaves the relevant extension to a separate derivation.
Key ideas
- The Vasicek drift can be expressed as mean reversion toward a long-run short-rate level.
- An exponential integrating factor gives an explicit solution consisting of deterministic and stochastic terms.
- The conditional mean follows from the deterministic part when the stochastic integral has zero mean.
- Ito isometry yields the conditional variance of the stochastic integral.
- A time-varying extended Vasicek specification requires additional drift treatment beyond this constant-level result.
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# Vasicek model and spot interest rate parametrised by reversion rate
# Vasicek model and spot interest rate parametrised by reversion rate
By solving an SDE I want to derive the analytical results for mean and variance of the process of extended Vasicek model.
$$ dr(t) = \left(\eta - \gamma r(t) \right)dt + c dX(t) $$
where $\gamma$ is the reversion rate and $s = \frac{\eta}{\gamma}$ is the average short rate.
How can I set $X(t) = r(t) - s$ and solve by integration over both sides of the SDE with the help of the integrating factor $e^{yt}$ and in a second step derive the mean and variance?
## Answer by Magic is in the chain (score 0, accepted)
https://quant.stackexchange.com/a/46251
Take your equation,
$ dr(t) = \left(\eta - \gamma r(t) \right)dt + c \, dX(t)$
and rearrange it as you suggested:
$dr(t) = \gamma \left(\frac{\eta}{\gamma} - r(t) \right)dt + c \, dX(t)$
$dr(t) = \gamma \left(s - r(t) \right)dt + c \, dX(t)$
Now if you multiply through by the integrating factor $e^{\gamma t}$ as you mentioned, you should get this expression after a little bit of manipulation:
$d \left( e^{\gamma t} r_{t} \right) = e^{\gamma t}\gamma \, s \,dt + e^{\gamma t} c \, d X(t)$
And then integrate from 0 to t to get:
$r_{t} = r_{0}e^{-\gamma t}+ s\left( 1-e^{-\gamma t }\right) +c\int_{0}^{t} {e^{-\gamma \left( t- u \right) } d X(u)}$
So the mean is just the deterministic term, and you can determine the variance via Ito isometry.
$V \left[ r_{t} \mid r_0 \right]={c}^{2} \int_{0}^{t} {e^{-2 \gamma \left(t-u \right)} du}=\frac{{c}^2}{2{\gamma}} {\left( 1- e^{-2 \gamma t} \right)}$
With the above steps in sight, could you clarify which particular part you are after?
You also use the 'Extended Vasicek', which is different in the sense that the mean is a function of time (in its simplest form). If that's what you are after then please google the derivation of the drift of the Extended Vasicek. The info here will also be useful. How to get set the theta function in the Hull-White model to replicate the current yield curveShown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.