Login

+1 vote

In the notes they mention that Lévy gives an alternative way of characterising a brownian motion. That is, if \(\{W\}_{t\geq0}\) is some process and \(Cov(W_t,W_s)=min(s,t)\) then \(W_t\) is a Brownian motion.

Now I know this is wrong because you can take a counterexample of a poisson process. For a poisson process, \(N_t\) with rate parameter \(\lambda, Cov(N_t,N_s)=min(\lambda s,\lambda t) \) so in the case that \(\lambda=1\) this matches what is given in the notes and is obviously wrong.

What I think the notes are missing is that if you have a process with continuous sample paths that is also a

martingale and \(Cov(W_t,W_s)=min(s,t)\) then the process is a brownian motion. I am not sure though.

I have seen alternative characterisations being that if you have a martingale process \(W_t\) with quadratic variation \(t\) then the process is a Brownian motion. Or if \(W_t\) is a martingale such that \(W_t^2-t\) is also a martingale then \(W_t\) is a brownian motion.

Would these definitions be equivalent to saying that if you have a process with continuous sample paths that is also a martingale and \(Cov(W_t,W_s)=min(s,t)\) then the process is a brownian motion? I seem to have a bit of a proof but I am not sure if its rigorous.

+2 votes

Best answer

The requirement is that the process is also Gaussian. i.e. - normal as per the unique characteristic function. I think that is assumed under their statement - maybe check again.

As to your other assertions, I am not sure that they will always be the case although both seem likely to be true. I would be reluctant to say this for certain, for the purposes of your course your understanding should be fine.

As you can see the more general form of Levy's theorem revolves around the uniqueness of the characteristic function. I would focus on this distinction.

If you think about it, saying \(W_t^2 - t\) is martingale is equivalent to saying the quadratic variation is t, you can take expectations and use the linearity of expectations to show this. So both of those statements are equivalent.

I guess you could have a discrete time process that also has these properties, but not continuous sample paths, so continuous sample paths and initialisation at zero are pre-conditions. Given that both of these hold then your alternative definitions are correct.

Thanks, that does make sense. Saying that \(t\) is such that \(W^2_t-t\) is martingale \(\Rightarrow [W]_t=t \) only holds if \(W_t\) is a continuous martingale itself though in anyway. That definition is for the predictable quadratic variation which is the same as the usual quadratic variation if \(W_t\) is also a continuous martingale. So I suppose it would have to be shown that \(W_t\) is a martingale itself also.

Thanks for the help!

- All categories
- BUS 1003H - Introduction to Financial Risk (52)
- BUS 2016H - Financial Mathematics (55)
- BUS 3018F - Models (74)
- BUS 3024S - Contingencies (61)
- BUS 4028F - Financial Economics (39)
- BUS 4027W - Actuarial Risk Management (54)
- BUS 4029H - Research Project (5)
- Mphil (1)
- Calculus and Pure Mathematics (4)
- Statistics (16)

...

Thanks. The notes unfortunately don't say anything more than the covariance requirement. I will take the covariance requirement along with the Gaussian requirement to be the right way. I just wanted to know because these alternative definitions make proofs a bit easier. The definition with it being a martingale with quadratic variation t is used in tutorial 5 q2 to prove Girsanov's theorem.