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I know that the above statement is true, but I can't demonstrate it. It's a pretty powerful theorem, here is its mathematical formulation:

Theorem: The only continuous martingales with stationary increments are Brownian motions

(1) being for all u, e, X(u+e) - X(u) = X(e) (equal in distribution)

I would greatly appreciate if you could point me in the right direction or link me to some material :)


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This is quite standard and any decent textbook on continuous time stochastic processes should have this, so I don't really think this site is really suitable for this kind of request. However, I do have a proof of this on my blog, if that helps.… – George Lowther Feb 23 '11 at 12:35
Actually, I thought you were assuming independence but, retracing your question, I see this is not the case. Then, your statement is false. Consider the integral of a stationary process with respect to an indepenent Brownian motion. – George Lowther Feb 23 '11 at 12:39
@George Lowther: I think that your comments are really worth to be posted as an answer. – Andrey Rekalo Feb 23 '11 at 14:13
Apparently, the OP is referring to the following paper (access to the content is restricted to subscribers): – Shai Covo Feb 23 '11 at 15:59
@Albert: The theorem you link to is not clear. There seems to be information missing in from the bit you posted, as it does not metion stationary increments, and refers to "(1)" which is not in the snapped image. – George Lowther Feb 24 '11 at 3:29

This question keeps getting bumped up, so (at long last) I'll convert my comments above to an answer.

It is true that every continuous martingale $X$ with stationary independent increments is a Brownian motion or, to be precise, $X=X_0+\sigma B_t$ for a standard Brownian motion $B$ and constant $\sigma$. This is because any such process is a Lévy process, and Brownian motions (possibly with drift) are the only continuous Lévy processes. This is standard, and most decent book on continuous-time stochastic processes should show this. I also have a proof of this on my blog here.

However, you do not mention the necessary condition that the increments of $X$ are independent. So, the statement in the question is false. There do in fact exist continuous martingales with stationary increments which are not Brownian motions. You can take $$ X_t=\int_0^t Y\\,dB $$ for a standard Brownian motion $B$ and independent stationary process $Y$. Then $X$ has stationary increments, but is not a Brownian motion unless $Y$ is constant. For example, $Y$ could be an Ornstein-Uhlenbeck process started in its stationary distribution.

[Finally, the link to the theorem in the question is not very useful, as it is not complete and references some undefined "(1)" that the process is supposed to follow.]

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@Shai yes I am referring to that article, but the proofs in there are not clear to me. Any other material that covers the same topic?

@George I read your blog, but isn't your result much more general than what we're trying to demonstrate here?

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@Albert: I don't have access to that article. – Shai Covo Feb 23 '11 at 17:52
@Shai the document is on Scribd: – Albert Feb 23 '11 at 18:04
@Albert: Well, on my blog page I was looking at general d-dimensional processes with independent increments. You are only looking at 1-dimensional martingales with stationary increments (also independent, otherwise the result doesn't hold, but you don't say that in the question). But the idea is the same. Actually the theorem you link to seems even more specialized, as it only refers to integrals w.r.t. Brownian motion but, again, you don't say that in the question. – George Lowther Feb 24 '11 at 3:28

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