I am trying to prove the continuity of Ornstein-Uhlenbeck process which is a stationary Gaussian process with covariance kernel $k(x,y) = \exp(-|x-y])$.

Let $(X_t)_{t\ge 0}$ be an Ornstein-Uhlenbeck process with $0$ mean. The increment $(X_{t+h} - X_t)$ follows a Normal distribution with mean $0$ and variance $2(1-\exp(-h))$. In order to prove that the process is continuous, I need to show

$$ \Bbb{P}(\sup_{h<\delta} |X_{t+h} - X_t|> \epsilon)\xrightarrow[\delta \to 0]{} 0$$

Edit 1

Alternatively, one can show $$P(\lim_{h \to 0} X_{t+h} = X_t) = 1$$

It is enough to show that \begin{equation} P(\lim_{n \to \infty} X_{t+1/n} = X_t) = 1 \end{equation}

Now, $\sum_{n=1}^\infty P(|X_{t+1/n}-X_t|> \epsilon) < \sum_{n=1}^\infty \frac{var(X_{t+1/n}-X_t)}{\epsilon^2} = \sum_{n=1}^\infty \frac{(1-e^{-1/n})}{\epsilon^2}$

Hence, if we can bound the last term, then by Borel-Cantelli's lemma, the desired result will hold.

Are the details of the proof correct?

  • $\begingroup$ Please take a look at this question to understand what it means for a stochastic process to be continuous. math.stackexchange.com/questions/1355012/… $\endgroup$
    – Calculon
    Jul 10, 2015 at 8:47
  • $\begingroup$ I cannot see any straightforward way of using that definition here. $\endgroup$ Jul 10, 2015 at 9:16
  • 1
    $\begingroup$ I posted it so you would see where your reasoning goes wrong. When looking at path properties, you fix $\omega$. So a statement like "$X_{t+h} - X_t$ is unbounded because it is normal" doesn't fit here. $\endgroup$
    – Calculon
    Jul 10, 2015 at 9:20
  • $\begingroup$ Okay, I will edit the question. However, do you have any idea how I can continue that proof. $\endgroup$ Jul 10, 2015 at 9:22

2 Answers 2


If $X_t$ is given as $X_t = e^{-\lambda t} \left(X_0 + \int _0^t e^{\lambda u} dW_u\right)$ for some $\mathcal{F}_0$ measurable r.v. $X$ and a standard Brownian motion (BM) $W$ defined on the same space and with respect to the same filtration as $X_0$, then you call $X$ the Ornstein-Uhlenbeck process.

You can view the $X_0$ part as a constant stochastic process so that part is continuous. The Ito integral $\int _0^t e^{\lambda u} dW_u$ is continuous as well since $W$ is continuous. I must say though that the proof of the latter statement is rather deep.

  • $\begingroup$ I know this proof, but this isn't the proof that I want. $\endgroup$ Jul 10, 2015 at 9:17

It is not enought to check continuity for each $t$. That is, the proof in the Edit 1 part is incorrect for two reasons,

1) It does not provide the continuity result, as you can see by a counter example. The poisson point process satisfies the condition $$ P(\lim_{h \to 0} X_{t+h} = X_t) = 1$$ and it is discontinuous almost surely (the reason this fails to prove continuity is that you have to consider non denumerable sets, one for each $t$)

2) The term

$$\sum_{n=1}^\infty P(|X_{t+1/n}-X_t|> \epsilon) < \sum_{n=1}^\infty \frac{var(X_{t+1/n}-X_t)}{\epsilon^2} = \sum_{n=1}^\infty \frac{(1-e^{-1/n})}{\epsilon^2} = \infty$$

So you can't apply Borel Cantelli.

Continuity for the Orstein-Uhlembeck(O-U) must be proved by other means. For instance, you could use Kolmogorov lemma to check if you can find a modification that has the same finite dimensionals and is continuous almost surely. Or you could find a sequence of random variables that converge to the O-U and satisfy the continuity condition.

$$ sup_n P(\sup_{h<\delta} |X^n_{t+h} - X^n_t|> \epsilon)\xrightarrow[\delta \to 0]{} 0$$

Calculon answer is also very good. Once you construct the O-U via Stochastic integration you immediately obtain continuous processes.

Ps: take a look at https://en.wikipedia.org/wiki/Kolmogorov_continuity_theorem


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