Vasicek Short-Rate Mean-Reversion Half-Life
Summary
The document asks how to measure the time for a Vasicek short rate to move halfway toward its long-run mean. It separates the deterministic mean-reverting part of the stochastic differential equation from its random component and relates the decay of a deviation to an exponential function. For a deviation that decays at speed k, solving for when it is half its initial size gives a half-life of ln(2)/k.
The answer instead identifies 1/k as the exponential time constant and calls it the half-time. These are related but not equal: after 1/k, about 1/e of the initial deviation remains, rather than one half. The discussion also connects the process to the Ornstein–Uhlenbeck model and radioactive decay. It gives no empirical rate data; the result is mathematical and depends on the constant-speed mean-reversion assumption.
Key ideas
- The deterministic Vasicek dynamics make deviations from the long-run mean decay exponentially.
- The exponential decay rate is the mean-reversion parameter k.
- The half-life is ln(2)/k, while 1/k is the exponential time constant.
- The stochastic term adds uncertainty around the mean-reverting path.
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Full text
# Half-life of short rate
# Half-life of short rate
The SDE for the short rate r(t) in the Vasicek model is given by:
$$ d(r) = k(r^* - r)dt + \sigma dW $$
The deterministic part of the above SDE is the following ODE
$$ d(r) = k(r^* - r)dt, $$
where $k$ is the mean reversion speed and $r^*$ is a constant representing the long-term average mean.
We have to use this to find the half-life of the above short rate. Half-life is the time it takes for the interest rate to move half the distance towards its long-term average. Any ideas?
## Answer by SimoPape (score 1, accepted)
https://quant.stackexchange.com/a/77025
Okay, I try a non-traditional approach. You know that the state variable in the Vasicek (or Ornstein Uhlenbeck) process $$ dx_{t} = \alpha(\gamma - x_{t})dt + \sigma dW_{t} $$ is a normally distributed random variable $$ x_{t} \sim \mathcal{N}\left(\gamma(x_{0}-\gamma)e^{-\alpha t},\frac{\sigma^2}{2\alpha}[1 - e^{-2\alpha t}]\right) $$ Now, you know that (at whatever time point) $x(s)$ you are at if you wait for a time $t \to \infty$ the random variable $x(t)$ will be a normal centered on the long-run average parameter.
Now, your question is centered on how long the time interval has to be to get to a stationary situation $$ x_{\infty} \sim \mathcal{N} \left(\gamma, \frac{\sigma^2}{2\alpha}\right) $$
You read this value from the term $e^{-\alpha t}$ and the average time with which the exponential is half-timed is $$ \frac{1}{\alpha}$$
This more generally applies to all transient processes (the deterministic equation for vasicek corresponds to the model for radioactive decay). In all these cases the differential model is linear of the type $$ dx = -\alpha x dt $$ and thus leads to a negative exponential solution.
I hope I have correctly interpreted your question and that this can help you :)Shown 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.