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?

  • 1
    $\begingroup$ What approaches have you already tried? $\endgroup$ Commented Oct 16, 2023 at 13:02

1 Answer 1


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 :)


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