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A stochastic process $X(t)$ by definition is gaussian iff all its finite-dimensional joint probability density functions are multivariate gaussian. Namely iff given the times $(t_1,t_2,...,t_n)$ , the random variables $(X(t_1),X(t_2),...,X(t_n))$ are jointly gaussian (n is arbitrary).

My intent is understanding why the solution of the following simple stochastic differential equation is a gaussian process:

$$dX(t)=a(t)dt+b(t)dW(t)$$

where W(t) is a Wiener process with diffusion coefficient D and the initial condition is $X(t_0)=X_0$, where $X_0$ is a random variable that is independent of W(t). a(t) and b(t) are deterministic "regular" functions.

The explicit solution is:

$$X(t)=X_0+ \int_{t_0}^{t}a(s)ds+\int_{t_0}^{t}b(s)dW(s)$$

If the integral $\int_{t_0}^{t}b(t)dW(t)$ were a gaussian process (I'm not sure about this) and the initial condition were a gaussian random variable, than at fixed t, the random variable $X(t)$ would be the sum of 3 independent gaussian random variables ( $\int_{t_0}^{t}a(t)dt$ is in fact degenerate) and would be therefore gaussian. But what about the multiple-times joint densities of $(X(t_1),X(t_2),...,X(t_n))$?

Antonio19932806
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    You were correct to point out that my answer didn't solve your question, I deleted my answer and will wait for someone else to answer. – Jan Stuller Feb 25 '21 at 17:30

1 Answers1

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You can get rid of the drift applying the Girsanov's theorem (see for instance Oksendal's introductory book on SDEs, or "Introduction to stochastic integration" by H.H. Kuo), so you end up with $X(t)$ being a Wiener integral (plus the initial condition, but since it's independent and Gaussian I will ignore it hereafter ).

As you may know the Wiener integrals are gaussian random variables, and by the linearity of the integral any linear combination of Wiener integrals will be in turn a Wiener integral and hence gaussian. This allows to conclude that variables $X(t_1),...X(t_n)$ are jointly gaussian.

Chaos
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  • Could you please link a source where the Grisanov's theorem you are referring to is stated and possibly demonstrated? If you do, I can accept the answer. – Antonio19932806 Feb 28 '21 at 23:01
  • @Antonio19932806 Girsanov's theorem for "change of measure" is a quite standard result in stochastic calculus, I provided two good references where you can find a proof. – Chaos Mar 01 '21 at 07:34
  • Unfortunately my level of understanding (i'm a physics student) is not sufficient to understand neither statement of the theorem (I woulnd't say it's a standard one), nor the way you suggested to use it for showing that X(t) is a Wiener integral. I think this is the result, when in university advanced arguments (such as EDS) are treated, without having the minimal basis for that...I have faith that the answer is correct and undestrendable for a better prepared person. – Antonio19932806 Mar 10 '21 at 08:38
  • @Antonio19932806 I apologize if my statement was kinda vague in this sense.All the discussion in the body of your question wasn't trivial and I assumed you knew this stuff. I will try yo explain it in brief detail here in the comments, latter I can put together a better answer – Chaos Mar 10 '21 at 09:08
  • @Antonio19932806 basically the Girsanov's theorem states that a brownian motion (under measure P) with an absolutely continuous drift can be seen as a brownian motion (without the drift) under some probability $Q$ which is absolutely continuous wrt $P$. The Radon Nikodym derivative will be the so called Doolean Dade exponential – Chaos Mar 10 '21 at 09:10
  • @Antonio19932806 Then, the first deterministic integral can be seen as this drift, and "absorbed" by this change of measure (now you work under $Q$). Then at this point you only have a Wiener integral, since the drift term disappeared. Then the conclusion follows – Chaos Mar 10 '21 at 09:12