# Ornstein-Uhlenbeck stochastic process and gaussian process

The solution of a Ornstein-Uhlenbeck stochastic process is the following: $$X_t= m +(x_0-m)e^{-\lambda t }+ \sigma \int_0^te^{-\lambda(t-s)} \, dBs$$

we know that $$\sigma \int_0^te^{-\lambda(t-s)}\, dBs$$ is a gaussian process. I also know that $$X_t$$ is a gaussian process and i thought that this is an immediate consequence of the fact that $$\sigma \int_0^te^{-\lambda(t-s)}\, dBs$$ is a gaussian process and that adding to this term $$m +(x_0-m)e^{-\lambda t }$$ does not cause any problem. But actually it does, but I'm not able to understand why. So, why is not immediately visible that $$X_t$$ is a gaussian process applying the definition ( vec$$(X_{t_1},.. X_{t_n})$$ is multivariate normal for every $$t_n$$)?

• You seem to be asking a variant of "why isn't a true statement true?" Can you be more clear/specific about your question, and also give some context for why you're asking it? (My guess: this was homework and you don't know why your instructor counted your answer wrong) Jan 24 '20 at 18:40
• @BrianMoehring I edited Jan 25 '20 at 9:30
• Okay... let me rephrase the second part of my comment: You literally state "But actually it does". Why do you think it causes a problem? Note I'm not asking why it causes a problem (which you state you don't know) but rather what is your reason for claiming it causes a problem. Absent that, I don't know how to determine whether any given answer is appropriate. For instance, one could simply note that a non-random function of $t$ added to a Gaussian process always makes a Gaussian process since it only changes the mean of your multivariate normal. This seems "immediately visible" to me. Jan 25 '20 at 9:44
• I'm saying that it causes a problem because my professor said so during the lecture. So it is instead directly visible that it is a gaussian process? Jan 25 '20 at 11:56

Depending on the precise context, we might be able to use the following theorem:

If $$f : [0,T] \to \mathbb{R}$$ is nonrandom and $$Y_t$$ is a Gaussian process for $$0 \leq t \leq T$$, then $$f(t)+Y_t$$ is a Gaussian process for $$0 \leq t \leq T$$.

For the proof, let $$0 \leq t_1 < t_2 < \cdots < t_n \leq T$$ and $$a_1, a_2, \ldots, a_n \in \mathbb{R}$$. Then $$\sum_{k=1}^n a_k\left(f(t_k)+Y_{t_k}\right) = \sum_{k=1}^na_kf(t_k) + \sum_{k=1}^na_kY_{t_k} = \text{constant } + \text{ normal random variable }$$ which is normal.

In your case, it seems that $$f(t) = m + (x_0 - m)e^{\lambda t}$$ is nonrandom and $$Y_t = \sigma \int_0^te^{-\lambda(t-s)}\, dB_s$$ was already proven to be a Gaussian process, so the theorem is immediately applicable.

Whether this result is "immediately visible" could be debated, I suppose, but since it's just a "throw-the-definition-at-it" proof, I don't know what problem your professor might be referencing.

I want to make one final comment: I haven't studied SDEs in a Bayesian context. If your class is Bayesian, then it would have been prudent for you to say so, and $$f : [0,T] \to \mathbb{R}$$ would likely need to be replaced by the random $$f : [0,T] \times \Omega \to \mathbb{R}$$. In this broader context, I have no clue if your $$X_t$$ even is a Gaussian process (my gut would say it isn't), much less how to prove it.

• Okey, if you wish you could give a look at this my question about the proof of $Y_t = \int_0^tf_sdB_s$ to be gaussian Jan 25 '20 at 21:21