Hi,I couldn't find answer to this easy question. If W(t) is a standard Brownian motion, then apparently W(t) is a Martingale. Is W(t)^3 also a Martingale?Thanks.

Since X^3 is a measurable function, W^3 is measurable w.r.t. the filtration.Therefore, you should only check the expectation property, i.e. E(W^3 (t+s) |F_t) ) = W^3(t).This is not the case because the third moment of a normal depends on its variance, i.e. LHS here depends on s.Therefore, it's not a martingale and you can't make it a martingale by "centering" it.

just computed(W_t^3)using ito's lemma. if the drift is non-zero, it cant be a martingale

Indeed. One thing I like about is that it's an counterexample to the following: every martingale M_t satisfies E(M_t) = 0. Is the converse true? No E()=0.

...every martingale M_t satisfies E(M_t) = 0....Shisah, you may want to refesh your definition of a martingale....

One thing I like about is that it's an counterexample to the following: every martingale M_t satisfies E(M_t) = 0. Is the converse true? No E()=0. ++++++++++++++++++++You don't have to search for a counterexample at all.If what you said were true it would mean another definition of a martingale which is of coursenot equivalent to the original one.

Clopinette, please read what I said again. If still don't see why what you said is crap, then brush up on your basic reasoning skills. Hint: If I say that every human has two eyes then that's not the same as saying that anything with two eyes is a human.Nikkei, for me it's obvious that one can't define a martingales by saying that E(M_t) = 0 because of counterexamples like the one I have pointed out. But maybe you can explain to me why the definitions are not equivalent without the use of counterexamples.

Because the original definition uses such words as "measurability w.r.t. filtration", "conditional expectation", etc.and the "alternative" definition just says that any process with zero mean is a martingale.If it could be that simple, the "alternative" definition would be what our professors give us in the first place - common sense.

Nikkei, sorry but in mathematics things are not the way they are just because "professors give". Things are the way they are because it's the way it makes _sense_ to define things. Mathematics isn't a religion: "the professor giveth and the professor taketh away". Take the example of conditional expectation. There are (at least) two perfectly sensible ways to define it for general random variables. One is using the Radon-Nikodym theorem. The other one is using orthogonal projections in the L^2 spaces of Random variables. For random variables with E(X^2) < infinity the two definitions are equivalent. But at FIRST SIGHT they look completely different. Moreover you professor is likely to give you only one of them (at least initially). So by your logic you would conclude that other one is wrong. By my logic I would try to find a counterexample, I would fail, then I could proceed to convince myself that the two definitions are in fact equivalent.

Shisha,What Clopinette wants to say is that a martingale can have a mean different from 0. If you need an example, take M_t=a+W_t, where W_t is a standard brownian motion. Then M_t is a martingale: indeed E(M_t|F_s)=a+E(W_t|F_s)=a+W_s=M_s, with s<t, but E(M_t)=a, which can take any value you want.Regards,

Time to eat humble pie for me! Sorry folks. What I had in mind was E(M_t - M_s | F_s) = 0, but that's not what I said. And then when Clopinette kindly pointed out my error I still didn't spot it. Oh dear. I knew it was better to choose a nickname for the forum that has nothing to do with a real name.