# FxPaul

Math in finance or vice versa

## Process definition

In this article we deduce the closed-from solution of the modified version of Ornstein-Uhlenbeck process:
$dS = \theta(\mu - S) dt + \sigma S dW_t$
where $\theta$ – mean reversion parameter, $\mu$ – mean and $\sigma$ – volatility.

## Integrating factor approach

There exists a general approach to non-linear stochastic differential equations of the form:
$dX_t = f(t,X_t) dt + c(t) X_t dW_t \, , \, X_0 = x$
where $f$ and $g$ are given continuous and deterministic functions.

The method consists of:

1. Define the integrating factor:
$F_t = F_t(\omega) = \exp\left( - \int_0^t {c(s) dW_s} + \frac{1}{2}\int_0^t {c^2(s) dW_s} \right)$
2. So the original equation could be written as $d\left(F_t X_t\right) = F_t f(t,X_t) dt$
3. Now define
$Y_t(\omega) = F_t(\omega) X_t(\omega)$
so that $X_t = F^{-1}_t Y_t$
4. And it yields the deterministic differential equation for each $\omega\in\Omega$
$\frac{dY_t(\omega)}{dt} = F_t(\omega) f\left(t, F^{-1}_t(\omega) Y_t(\omega)\right); \, Y_0 = x$

We can therefore solve it with $\omega$ as a parameter to find $Y_t(\omega)$ and then obtain $X_t(\omega)$

## Modified Ornstein-Uhlenbeck process solution

Let’s apply the described approach to the process. Thus we’ve got in notation of the method:
$f(S, t) = \theta(\mu - S)$
$c = \sigma$

Integrating factor transforms to:
$F_t = \exp{\left( -\sigma W_t+\frac{1}{2}\sigma^2t\right)}$
$Y_t = F_t S_t$
and ODE for it is:
$\frac{dY(t)}{dt} = \theta\exp{\left( -\sigma w+\frac{1}{2}\sigma^2t\right)} \left(\mu - Y(t) \exp{\left( \sigma w- \frac{1}{2}\sigma^2t\right)} \right)$
and initial conditions are:
$Y(0) = S_0$

Thus the solution $Y(t)$ is:
$Y(t) ={\frac { 2\theta \mu {\exp\left(-\sigma W_t+\frac{1}{2}\sigma^2 t \right)}}{{\sigma}^{2}+2\,\theta}}+\exp\left(-\theta\,t\right) \left( S_0 - 2 \frac {\theta\mu \exp(-\sigma W_t)}{\sigma^2+2\theta} \right)$

And recovering $S(t)$ solution:
$S(t) = F_t^{-1} Y_t = \exp{\left(\sigma W_t - \frac{1}{2}\sigma^2t \right)} Y(t)$
$S(t) = S_0\exp{\left( - \alpha t + \sigma W_t\right)} + \frac{\theta\mu}{\alpha}\left( 1 - \exp\left( - \alpha t\right) \right)$
where $\alpha = \theta + \frac{1}{2}\sigma^2$

UPD: Fixed signs in 2 last equations.

Written by fxpaul

May 27, 2011 at 08:39

### 5 Responses

1. Thanks for sharing the solution. I have been looking everywhere for this :) I have a few questions though.
In the expression for S(t), shouldn’t the second term be (1-exp(-alpha t)) instead of (1+exp(-alpha t)). Just feel weird thats its so different from the non-stochastic version.

It seems that the variance either converge to infinity or zero. I wonder if there is there any lognormal process that has bounded variance? (like the OU process)

Thanks!

Boon Teik Ooi

July 14, 2013 at 22:02

• You are right! Thanks for the proof reading of the solution. I’ve just made an update.

I think that the variance converge to something like OU process with $\frac{\theta\mu}{\alpha}$ as mean and $\alpha$ as mean-reversion parameter.

fxpaul

July 16, 2013 at 08:43

2. Excellent stuff! I was looking for this problem.

quant3

December 24, 2013 at 16:14

3. Hi,

I was also looking for the solution of this problem, but i am afraid what you derive for S(t) is not correct. Nevertheless the integrating factor technique is interesting, but i am not sure it works this way in this case.

As far as i know in the literature it is known as Generalized OU process, or sometimes Nelson’s diffusion process, or even continuous GARCH(1,1) process.

Anyhow, i think the correct solution is more complicated, it should depend on the full path of the Brownian motion up to time t, not just on the value W(t). I think Eq. (6) in this paper (https://www.tu-braunschweig.de/Medien-DB/stochastik/lindner5.pdf) is the correct solution. Also, see Eq. (9) and the text below that.

Cheers,
andras

Andras

December 26, 2013 at 21:33

• Hi Andras,

Sorry for very long reply. I’ve got too much work over this Christmas.

There is an error in the derivation of $F_t$.
It should be $F_t = \exp{\left( -\sigma W_t+\frac{1}{2}\sigma^2 W_t\right)}$.

I will fix it over this weekend.

Thanks a lot!

fxpaul

February 8, 2014 at 22:02