Standard Brownian motion, , is defined to be a realvalued process satisfying the following properties.
 .
 is normally distributed with mean 0 and variance ts independently of , for any .
 B has continuous sample paths.
As always, it only really matters is that these properties hold almost surely. Now, to apply the techniques of stochastic calculus, it is assumed that there is an underlying filtered probability space , which necessitates a further definition; a process B is a Brownian motion on a filtered probability space if in addition to the above properties it is also adapted, so that is measurable, and is independent of for each . Note that the above condition that is independent of is not explicitly required, as it also follows from the independence from . According to these definitions, a process is a Brownian motion if and only if it is a Brownian motion with respect to its natural filtration.
The property that has zero mean independently of means that Brownian motion is a martingale. Furthermore, we previously calculated its quadratic variation as . An incredibly useful result is that the converse statement holds. That is, Brownian motion is the only local martingale with this quadratic variation. This is known as Lévy’s characterization, and shows that Brownian motion is a particularly general stochastic process, justifying its ubiquitous influence on the study of continuoustime stochastic processes.
Theorem 1 (Lévy’s Characterization of Brownian Motion) Let X be a local martingale with . Then, the following are equivalent.
 X is standard Brownian motion on the underlying filtered probability space.
 X is continuous and is a local martingale.
 X has quadratic variation .
This result carries directly through to the ddimensional situation. A ddimensional Brownian motion on a filtered probability space is a continuous adapted process with such that, for any , is independent of and multivariate normal with zero mean and covariance matrix .
Theorem 2 Let be a ddimensional local martingale with . Then, the following are equivalent.
 X is a Brownian motion on the underlying filtered probability space.
 X is continuous and is a local martingale for .
 X has quadratic covariations for .
Here, is the Kronecker delta. Theorems 1 and 2 are both special cases of the following characterization of multidimensional centered Gaussian processes with independent increments. Note that, as processes are continuous whenever their quadratic variation is continuous, statement 3 of Theorems 1 and 2 automatically imply continuity of X.
Theorem 3 Let be a ddimensional continuous local martingale with . Also, let be symmetric real matrices such that and is continuous and increasing for all . Then, the following are equivalent.
 is independent of and normally distributed with mean zero and covariance matrix , for all .
 is a local martingale for .
 X has quadratic covariations .
Proof: Suppose that 1 holds and set , for some . Then, , are joint normal with zero mean and covariance , independently of . So
Setting gives , so M is a martingale and 2 holds.
Now suppose that 2 holds, choose any , and set . Then, is a local martingale, as is . It follows that is also a local martingale and, being the difference of continuous increasing processes, V is also an FV process. So, V is constant, giving . By the polarization identity,
giving property 3.
Finally, suppose that 3 holds, choose and set , so that . The process
is bounded by . Applying Ito’s lemma for continuous semimartingales to f gives
By preservation of the local martingale property, M is a local martingale. As it is also bounded over finite intervals, M is a proper martingale. So,
This is the characteristic function of the multivariate normal, independently of , with mean zero and covariance matrix , as required.
Theorem 1 can also be used to characterize those processes which can be expressed as a stochastic integral with respect to a Brownian motion. Other than the local martingale property, it is necessary for the quadratic variation to have absolutely continuous sample paths. A function is said to be absolutely continuous if it is continuous with finite variation and for all bounded measurable functions with . Equivalently, the signed measure is absolutely continuous with respect to the Lebesgue measure which, by the RadonNikodym theorem, is equivalent to f being expressible as an integral for some Lebesgueintegrable function . In the stochastic setting, FV processes with absolutely continuous sample paths can be expressed as an integral of a predictable process.
Lemma 4 Let V be an FV process such that (almost surely) for each and bounded predictable process satisfying (almost surely).
Then, for some predictable process satisfying (almost surely) for each .
Proof: Let us first suppose that V has integrable variation and let denote the predictable sigmaalgebra. Define the following measures on
for all nonnegative bounded and predictable . Here is a (nonnegative) sigmafinite measure and is a finite signed measure.
By the condition of the lemma, whenever so, by the RadonNikodym theorem, there exists a predictable process so that for all bounded predictable . The process is continuous, has integrable variation over all bounded time intervals and satisfies
for bounded predictable . So, W is a martingale as well as a continuous FV process and must be constant. This gives as required.
Now, for general V satisfying the conditions of the lemma, define stopping times
Then the stopped processes have variation bounded by n and, by the above argument, there are predictable processes such that . The result now follows by taking .
Any process of the form for a Brownian motion B has quadratic variation , which is absolutely continuous. We prove that the converse statement also holds. That is, if X is a local martingale with absolutely continuous quadratic variation, then it can be expressed as a stochastic integral with respect to a Brownian motion. Before doing this, there is one technical issue to consider. On any probability space it is always possible to define the process , which is identically zero everywhere. This trivially has zero quadratic variation and we would like to write where . However, this is only possible if there is at least one Brownian motion B defined on the underlying filtered probability space. This is not a major problem though, as it is always possible to enlarge the space to add a Brownian motion. For example, if is the original filtered probability space and is any other space with a Brownian motion defined on it, then
defines a new, enlarged, filtered probability space which contains a Brownian motion and to which all processes and random variables on our original space can be lifted. The statement of the representability of processes as stochastic integrals with respect to Brownian motion is as follows.
Theorem 5 Let X be a local martingale such that [X] has absolutely continuous sample paths. Also, suppose that there exists at least one Brownian motion defined on the filtered probability space.
Then, for a Brownian motion B and Bintegrable process .
Proof: Applying Lemma 4, we can write . Then, as quadratic variations are increasing, will be almosteverywhere nonnegative. Set , so .
The integral
shows that is Xintegrable. So, picking any Brownian motion W defined on the filtered probability space, we can define
By preservation of the local martingale property, B is a local martingale. Its quadratic variation satisfies
so, by Lévy’s characterization, B is a Brownian motion. As , is Bintegrable and,

(1) 
However, is a local martingale with quadratic variation . So, and (1) gives the result.
Hi, I have a little question about the proof of theorem 3.
You claim there : “Then,, are joint normal with zero mean and covariance , independently of .”.
Well it is not obvious to me why they should be normal independently of ?
I’m not sure about that but maybe this should be added to the proof or added as an hypothesis in 1 (in this last case this additional property should be proven in the last implication of 3 implies 1).
Best regards
Comment by TheBridge — 30 October 11 @ 5:14 PM 
Yes, I think you’re right. This should have been included among the hypotheses of statement (1). I updated it, thanks.
Also, the proof of (3) ⇒ (1) is ok as it is. As the characteristic function of X_{t}  X_{s} conditioned on is not random, it is automatically independent of .
Comment by George Lowther — 30 October 11 @ 10:18 PM 
wonder if you can make PDF versions of these. Would be easier to print and read!
Comment by Yufan — 20 January 12 @ 11:18 AM 
Still in the proof of theorem 3, didn’t you mean Y=a’X and Z=b’X (I write b’ for the transposition of b)?
Because you are mentioning the covariance of Y and Z with Z defined as b’Y but then have not the same dimension (Y is in R and Z has 1 line and d columns).
Let me know what you think!
Thanks
Comment by Pito — 5 December 13 @ 10:02 AM 