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 $t_n$)?

  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.