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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 $t_n$)?

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    $\begingroup$ 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) $\endgroup$ Jan 24 '20 at 18:40
  • $\begingroup$ @BrianMoehring I edited $\endgroup$
    – Buddy_
    Jan 25 '20 at 9:30
  • $\begingroup$ 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. $\endgroup$ Jan 25 '20 at 9:44
  • $\begingroup$ 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? $\endgroup$
    – Buddy_
    Jan 25 '20 at 11:56
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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.

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  • $\begingroup$ 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 $\endgroup$
    – Buddy_
    Jan 25 '20 at 21:21

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