# What is the mean and the standard deviation for Geometric Ornstein-Uhlenbeck Process?

I am uncertain as to how to calculate the mean and variance of the following Geometric Ornstein-Uhlenbeck process.

$$d X(t) = a ( L - X_t ) dt + V X_t dW_t$$

Is anyone able to calculate the mean and variance of this process as well as include the calculations for the solution?

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Wait guys, he's not talking about the normal OU process, he's looking at the Geometric version. I couldn't find easily the answer on google for that. –  SRKX Jan 7 '13 at 7:53

An OU process

$dx_t = \theta(\mu - x_t)dt + \sigma dW_t$

where W follows standard Wiener process then mean is $\mu$ and variance is $\sigma^2/2\theta$ . You can substitute your factors.

Using mathematica

mean = Mean[OrnsteinUhlenbeckProcess[$\mu,\sigma,\theta$]]

variance = Variance[OrnsteinUhlenbeckProcess[$\mu,\sigma,\theta$]]

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He's apparently not looking for the classic Ornstein-Uhlenbeck process, but the Geometric version. –  SRKX Jan 7 '13 at 8:00
Any comments for downvote ?! –  ash Jan 7 '13 at 10:42
Your answer is wrong; I mean it's not the right process. –  SRKX Jan 7 '13 at 10:54

Okay so I'll take Jase answer and format it properly so that it answers your question and it will be useful for users in the future.

For clarity, let me restate the dynamics of the Modified Ornstein-Uhlenbeck model using the more common notation:

$$dS_t = \theta (\mu-S_t)dt + \sigma S_t dW_t$$

This blog post provides a closed form solution:

$$S_t = S_0 \exp(- \alpha t + \sigma W_t) + \frac{\theta \mu}{\alpha} (1+ \exp(-\alpha t))$$

where $\alpha=\theta+\frac{1}{2} \sigma^2$.

So first, the expectation:

$$\mathbb{E}[S_t] = \mathbb{E}[ S_0 \exp(- \alpha t + \sigma W_t) + \frac{\theta \mu}{\alpha} (1+ \exp(-\alpha t))]$$

$$\mathbb{E}[S_t] = \mathbb{E}[ S_0 \exp(- \alpha t + \sigma W_t)] + \mathbb{E}[\frac{\theta \mu}{\alpha} (1+ \exp(-\alpha t))]$$

$$\mathbb{E}[S_t] = S_0 \mathbb{E}[\exp(- \alpha t + \sigma W_t)] + \frac{\theta \mu}{\alpha} (1+ \exp(-\alpha t))$$

Now, note that $\exp(- \alpha t + \sigma W_t)$ can be expressed as $\exp(- \alpha t + \sigma \sqrt{t} Z)$ (with $Z \sim \mathcal{N}(0,1)$) hand is log-normally distributed: $\sim \ln \mathcal{N} (-\alpha t, \sigma^2 t)$, so you can apply the formulas of the log-normal distribution for mean and variance.

$$\mathbb{E}[\exp(- \alpha t + \sigma \frac{1}{\sqrt{t}} Z)]= \exp(- \alpha t + \frac{1}{2} \frac{\sigma^2}{t})$$

So, we get

$$\mathbb{E}[S_t] = S_0 \exp(- \alpha t + \frac{1}{2} \sigma^2 t) + \frac{\theta \mu}{\alpha} (1+ \exp(-\alpha t))$$

Now for the variance:

$$Var[S_t]= Var[S_0 \exp(- \alpha t + \sigma W_t) + \frac{\theta \mu}{\alpha} (1+ \exp(-\alpha t))]$$

As the second term is constant, we get:

$$Var[S_t]= Var[S_0 \exp(-\alpha t + \sigma W_t)]$$ $$Var[S_t]= S_0^2 Var[\exp(- \alpha t + \sigma W_t)]$$

Using again the log-normal formula $$Var[S_t]= S_0^2 (\exp(\sigma^2 t)-1) \exp(-2 \alpha t+\sigma^2t)$$

To sum up, and substituting $\alpha$ with $\theta+\frac{1}{2} \sigma^2$ we get:

$$\mathbb{E}[S_t] = S_0 \exp(- \theta t) + \frac{\theta \mu}{\theta+\frac{1}{2} \sigma^2} (1+ \exp(- (\theta+\frac{1}{2} \sigma^2) t))$$

$$Var[S_t]= S_0^2 (\exp(\sigma^2 t)-1) \exp(-2 \theta t)$$

Shouldn't it be $dS_t$ instead of $dX_t$ in the first formula? –  vonjd Jan 18 '13 at 12:17
In the expression for $S_t$, shouldn’t the second term be $\frac{\theta\mu}{\alpha}(1-exp(- \alpha t))$ instead of $\frac{\theta\mu}{\alpha}(1+exp(-\alpha t))$? –  Boon Teik Ooi Jul 14 '13 at 21:27