## B_t and martingale

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### B_t and martingale

Dr. J and Fellow users,

I have a question on the "bounded" condition on a martingale.As we all know, for a stochastic process to be martingale, it needs to satisfy the "bounded" criterion i.e E(|X_t|)<inf for all t

Now, B_t where B is a BM is a martingale and B_t^3 is not.

Question is this: By direct integration, one can obtain

E(|B_t|)=(constant terms) * sqrt(t)
E(|B_t^3|)=(constant terms) * sqrt(t^3)

From the above results, it's not clear why E(|B_t|) is bounded as it should tend to inf (although at a slower rate) as t -->inf

What am I missing?

Regards,
SR
sraks

Posts: 30
Joined: Tue Feb 16, 2010 3:31 am

### Re: B_t and martingale

the reason B_{t}^{3} is not a martingale is not lack of boundedness.

It is that the conditional expectations are wrong.

The easiest way to see this is to compute d(B_{t}^{3}) and observe that the drift is not zero.
mj

Posts: 1380
Joined: Fri Jul 27, 2007 7:21 am

### Re: B_t and martingale

Dr. J,

d(B_t^3)=(...) dB+ 3*B_t*dt

so,
E[d(B_t^3)/B_t]=(...)*0+ non-zero

The dt term is non-zero because B_t is known (informally ) wrt the filtration. Is that the correct argument?

Regards,
SR
sraks

Posts: 30
Joined: Tue Feb 16, 2010 3:31 am

### Re: B_t and martingale

essentially yes
mj

Posts: 1380
Joined: Fri Jul 27, 2007 7:21 am

### Re: B_t and martingale

Thanks Dr. J.
sraks

Posts: 30
Joined: Tue Feb 16, 2010 3:31 am