I'm new to the forum so I hope this first question goes well.

Let the Ornstein-Uhlenbeck process be defined as: $$ dV_t = - \beta V_t dt + \sigma dW_t $$

with $V_0 = v$, where $W_t$ is a Wiener Process (or Brownian Motion) started at $0$.

I have spent quite a while trying to prove if this process has independent increments, i.e., $Cov(V_t - V_s, V_s - V_0) =0$. The result I'm obtaining is different than zero, which would imply that the increments are not independent, but I am unsure about the accuracy of my calculations.

Intuitively, I would say I've gone wrong, because if the process is driven by a Brownian Motion, and the latter has independent increments, one could perhaps argue that then the O-U process must have independent increments too.

This leads me to ask two questions:

  1. Is there an intuitive reasoning to argue if $V_t$ has (or has not) independent increments?
  2. Should we take the stationary distribution of $V_t$ (letting $t \rightarrow \infty$) such that $V_t \sim N\left(0, \frac{\sigma^2}{2\beta}\right)$, what would happen to the increments?.

P.S: I can guess that if we consider the non-stationary distribution of $V_t$, the increments are non-stationary, given that $V_t \sim N\left(e^{-\beta t}v, \frac{\sigma^2}{2\beta}(1-e^{-2\beta t})\right)$ and thus they would always depend on $t$. Perhaps you can prove me wrong here too.

  • $\begingroup$ If it had independent increment and were stationary, it would be a Brownian motion + drift (as it would be a continuous Lévy process). The whole point of havind a PDE is that - in this case - there is some "tension" driving the process toward zero. $\endgroup$
    – D. Thomine
    Commented May 15, 2014 at 11:53

2 Answers 2

  1. Intuitively, @D.Thomine already explained why the increments are not independent nor stationary. Another way to see it is by noting that the Ornstein-Uhlenbeck is characterized by its "return to mean" property. What I mean by that is that when $V_t$ is far from zero, the term $-\beta V_t\,\mathrm dt$ has a tendency to drag back the process to $0$ since (again heuristically) $\mathbb E[\Delta V_t]=-\beta V_t\Delta t$. It thus seems clear, at least intuitively, that the increments are not independent.

  2. We can do the calculations. I am not $100\%$ sure of these, so be sure to check if something seems out of the ordinary.

The unique solution of the Ornstein-Uhlenbeck stochastic differential equation is $$ V_t=\nu e^{-\beta t}+\sigma\int_0^te^{-\beta(t-u)}\,\mathrm dW_u, $$ since an application of Itô's formula yields $$ \mathrm dV_t=-\beta\nu e^{-\beta t}\,\mathrm dt-\sigma\beta e^{-\beta t}\left(\int_0^te^{\beta u}\,\mathrm dW_u\right)\,\mathrm dt+\sigma\,\mathrm dW_t =-\beta V_t\,\mathrm dt+\sigma\,\mathrm dW_t. $$ As you hint in OP, $(V_t)$ is a Gaussian process, thus it has independent increments if and only if $\mathrm{Cov}\,(V_{t+s}-V_s,V_s)=0$ for all $t,s\ge 0$. This happens if and only if $$ \mathbb E\left[\left(\int_0^{t+s}e^{-\beta(t+s-u)}\,\mathrm dW_u-\int_0^se^{-\beta(s-u)}\,\mathrm dW_u\right)\left(\int_0^se^{-\beta(s-u)}\,\mathrm dW_u\right)\right]=0. $$ Each of the terms can be calculated according to the Itô isometry formula: \begin{align*} &\mathbb E\left[\left(\int_0^{t+s}e^{-\beta(t+s-u)}\,\mathrm dW_u\right)\left(\int_0^se^{-\beta(s-u)}\,\mathrm dW_u\right)\right]\\ &=e^{-\beta(t+2s)}\mathbb E\left[\left(\int_0^se^{\beta u}\,\mathrm dW_u\right)^2\right]+\mathbb E\left[\left(\int_s^{t+s}e^{-\beta(t-u)}\,\mathrm dW_u\right)\left(\int_0^se^{-\beta(s-u)}\,\mathrm dW_u\right)\right]\\ &=e^{-\beta(t+2s)}\frac1{2\beta}\left(e^{2\beta s}-1\right), \end{align*} and similarly, $$ \mathbb E\left[\left(\int_0^se^{-\beta(s-u)}\,\mathrm dW_u\right)\left(\int_0^se^{-\beta(s-u)}\,\mathrm dW_u\right)\right]=e^{-2\beta s}\frac1{2\beta}\left(e^{2\beta s}-1\right). $$ Thus, $$ \mathrm{Cov}\,(V_{t+s}-V_s,V_s)=-\frac{\sigma^2}{2\beta}\left(1-e^{-2\beta s}\right)\left(1-e^{-\beta t}\right). $$ At this point, we note that the covariance is negative, as expected. Additionally, if $t$ is fixed, and $s$ goes to infinity, then heuristically, the covariance of the increments tend to a covariance which depends on $t$.

Thus, the increments are negatively correlated, even as $s$ goes to infinity.

Lastly, note that if we start from a distribution $\nu\sim N(0,\frac{\sigma^2}{2\beta})$ which is independent from the Wiener process $(W_t)$, then $(V_t)$ has a constant distribution, $\forall t\ge 0, V_t\sim N(0,\frac{\sigma^2}{2\beta})$. In this case, \begin{align*} \mathrm{Cov}\,(V_{t+s}-V_s,V_s)&=\mathbb E\left[\left(\nu e^{-\beta(t+s)}-\nu e^{-\beta s}\right)\nu e^{-\beta s}\right]-\frac{\sigma^2}{2\beta}\left(1-e^{-2\beta s}\right)\left(1-e^{-\beta t}\right)\\ &=\frac{\sigma^2}{2\beta}\left(e^{-\beta(t+2s)}-e^{-2\beta s}\right)-\frac{\sigma^2}{2\beta}\left(1-e^{-2\beta s}\right)\left(1-e^{-\beta t}\right)\\ &=-\frac{\sigma^2}{2\beta}\left(1-e^{-\beta t}\right), \end{align*} so here also the increments remain of negative covariance, and incidentally, this is the limiting covariance of the initially considered process. Even if the distribution of the process is stationary, the increments remain negatively correlated.

We can also look at the distribution of $V_{t+s}-V_s$. It is clearly Gaussian, of mean $0$, and the variance is given by \begin{align*} &\mathrm{Var}\,(V_{t+s}-V_s)\\ &=\frac{\sigma^2}{2\beta}\left(\left(e^{-\beta(t+s)}-e^{-\beta s}\right)^2+\left(\left(1-e^{-2\beta(t+s)}\right)+\left(1-e^{-2\beta s}\right)-2e^{-\beta(t+2s)}\left(e^{2\beta s}-1\right)\right)\right)\\ &=\frac{\sigma^2}{\beta}\left(1-e^{-\beta t}\right). \end{align*} Hence the increments are stationary when the distribution of the process is stationary, and $$ V_{t+s}-V_s\sim N\left(0,\frac{\sigma^2}{\beta}\left(1-e^{-\beta t}\right)\right). $$


No formulas required. Since the mean of the increment depends on $V_t$, and since the previous increment modifies $V_t$, the increments cannot be independent.

As for sampling from the unconditional distribution, what we can say is that this is the same as sampling from an infinitely long sample. Since the sample never gets anywhere, the average change from a randomly sampled point must also be zero.


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