Difference between Ornstein-Uhlenbeck, Vasicek and Geometric Mean Reversion

Question

I am currently reading the following 3 paper:

• Paper 1: Decision Support for IT Investment Projects
• Paper 2: A Stochastic Processes Toolkit for Risk Management
• Paper 3: Calibration of the exponential Ornstein–Uhlenbeck process when spot prices are visible through the maximum log-likelihood method. Example with gold prices

Each of them is terming what to me seems the same stochastic process differently. Paper 1 p. 387 Equation 4 is terming the process Geometric Mean Reversion (GMR) and Equation 5 the exponential version Exponential Mean Reversion (EMR), Paper 2 p. 27 Equation 47 is terming it Vasicek Model and on p. 31/32 Equation 62 Exponential Vasicek Model and Paper 3 p. 4 Equation 3 is terming it Exponential Ornstein-Uhlenbeck process.

Am I correct, that all of these 3 essentially are the same stoachstic process?

Also, for the exponential version two paper seem to offer a closed form solution, where I spotted a small difference. Paper 1 p. 387 Equation 6 has the following solution:

$$S_{T,i}=\exp\left(\ln(S_0)e^{-\alpha T}+\left(\theta-\frac{\sigma^2}{4\alpha}\right)(1-e^{-\alpha T})+\sqrt{(1-e^{-2\alpha T})\frac{\sigma^2}{2\alpha}}\epsilon_i\right)$$

Paper 3 p. 5 Equation 9 on the other hand has the following solution:

$$\ln(X_{t})=\ln(X_{t-1})e^{-\theta \Delta t}+\left(\mu-\frac{\sigma^2}{2\alpha}\right)(1-e^{-\theta \Delta t})+\sigma\sqrt{\frac{1}{2\alpha}(1-e^{-2\theta \Delta t})}\epsilon_i$$

Now different variables aside, the first equation has $\frac{\sigma^2}{4\alpha}$ and the second equation has $\frac{\sigma^2}{2\alpha}$. Can anyone tell me which one is correct?

Answer 1

Short Answer:

• Ornstein-Uhlebeck = Vasicek (only difference being that Vasicek can sometimes have (deterministic) non-constant reversion speed parameter, whilst Ornstein-Uhlenbeck has all parameters constant by definition).

• Exponetial Vasicek = EMR.

• The term $4 \alpha$ in the denominator looks like a typo, should be $2 \alpha$

Long Answer:

Ornstein-Uhlenbeck process is defined as (I use $\theta$ instead of your $\alpha$, otherwise the notation is identical):

$$\tag{1} X_t:=X_0+\int_{h=0}^{h=t}\theta(\mu- X_h)dh + \int_{h=0}^{h=t}\sigma dW_h$$

Or in "short-hand" notation we can write:

$$dX_t=\theta (\mu-X_t)dt+\sigma dW_t$$

In an answer to another question on Math SE here, I show how the Ornstein-Uhlenbeck can be solved step-by-step (the bottom section of my answer titled "Why not to use Short-hand notation")

The final step I don't show is to complete the Riemann integral which solves as follows:

$$\int_{h=0}^{h=t}\left(e^{\theta(h-t)}\theta\mu\right)dh=\mu(1-e^{\theta t})$$

So the full solution is:

$$X_t=X_0e^{-\theta t}+\mu(1-e^{\theta t})+\sigma\int_{h=0}^{h=t} e^{\theta(h-t)} dW_h$$

Ornstein-Uhlenbeck and Vasicek (and GMR) processes are the same thing (except that Vasicek process can have the parameter $\theta$ non-constant, it can be a deterministic function of $t$).

In the first reference you provide, the author introduces the EMR process (or Exponential Vasicek) for $S_t$ (i.e. what I call $X_t$) as:

$$dS_t=\alpha(\theta - ln(S_t))S_tdt + \sigma_t S_t dW_t$$

Or in long-hand notation:

$$S_t=S_0+\int_{h=0}^{h=t}\alpha(\theta - ln(S_h))S_hdh+\int_{h=0}^{h=t}\sigma_hS_s dW_h$$

We can see right away that this is not the same process as the Ornstein-Uhlenbeck defined in (1).

As for your last question, the $4\alpha$ in the denominator looks like a typo, should be $2\alpha$ I believe (indeed the author who has $4 \alpha$ doesn't derive the solution, he just quotes Brigo. Also see the Wikipedia link at the bottom for why it should be $2 \alpha$).

Last but not least, note that Ito integral of any deterministic function is a Normally distributed random variable. Variance of the term $\sigma\int_{h=0}^{h=t} e^{\theta(h-t)} dW_h$ is derived here on Wikipedia, and this variance and the Normal distribution property of the Ito Integral are used to rewrite the solution to the EMR process in the form you show in the two equations you quote.