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in BUS 4028F - Financial Economics by (820 points)
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. 

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+2 votes
by (820 points)
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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.

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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.

by (820 points)

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. 

by (820 points)

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.

by (820 points)

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!

by (740 points)

I think the notes should just have added the bit about continuity as well. From the covariance being equal to min(t,s) it is easy to prove independent increments. Knowing that Var(X_s)=Covar(X_s,X_s)=s and Var(X_t)=Covar(X_t,X_t)=t (with t>s) plus the independent increments gives Var(X_t-X_s)=t-s. So, we have independent increments and the variance of an increment is the time length of the increment. Defining a series of partitions that gets very dense should prove the normality. So, I reckon they should just have added the bit about continuity.

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