In one of the lecture notes that I am reading up, I read that the integral vanishes because of 'martingle property'. Can some one help me understand this?Thanks

It might help to compare this martingale to some other ones you might be more familiar with, like the normal 4*dB and lognormal 4*B*dB . How will the process behave differently for B between 0 and 1; how about above 1?

Hi !In the same spirit, what is

A stochastic integral is a martingale, so that its expected value is 0.If the text is calculating an expectation, the expected value of the stoch int vanishes.Is it the case?

QuoteOriginally posted by: lolilloA stochastic integral is a martingale, so that its expected value is 0.If the text is calculating an expectation, the expected value of the stoch int vanishes.Is it the case?Yes. Actually the exercise is to find It assumesThen it finds the total variation of Zthen it integrates the both sides and claims the first one vanishes.Any help (in more detail?)Thanks

The expected value of an integral is the integral of the expected value.Since the term B(t) must be valued before dBt is observed, then E[B(t)^3*dBt]=B(t)^3*E[dBt]=0Not a formal proof at all, but hope it helps.

SPAAGG: The integral you gave is technically not a stochastic integral (in the Ito sense). The variable you're integrating against has bounded variation on any compact subset of $\R$, and hence such an integral can be given a pathwise definition. Lucetios: Stochastic integrals are only martingales if they are defined in the Ito sense (but don't worry, that's your case). Other stochastic integrals such as Stratonavich do not possess the martingale property. When you're integrating against a Brownian motion, the integral certainly don't vanish! However, it is only when the expectation are taken that makes it disappear. The reason is that Ito integrals are martingales and hence their expectation must be constant. But \int_0^t . dW_s = 0 a.s. when t = 0, which forces the expectation of all Ito integrals to have zero expectation.

You are right SU2, thanks.Often in stoch calculus we use Ito-isometry and Watanabe lemma. Check this out

Thanks evrey one.

A remind to SPAAG:the question which is interesting for the integral:is not the expected value (thats relatively easy).But this integral is a random variable with a normal distribution (little bit harder to proof). But what is the variance?I think t/3. Is it true?

QuoteOriginally posted by: ddrdoubleA remind to SPAAG:the question which is interesting for the integral:is not the expected value (thats relatively easy).But this integral is a random variable with a normal distribution (little bit harder to proof). But what is the variance?I think t/3. Is it true?Shouldn't be zero? I see the ds^2 coming out of the variance and it should be the same as zero . Am I wrong?

It isn't that hard, is it? Here it goes:Now write:which is a sum of independent increments. Calculate the inner expectation and the two other integrals and I think the result is (if I made no mistakes): .Can anyone confirm this?Correction: one can directly write that ,so that we get the integral over a square, [0,t]^2, of the minimum of s and u. This can be evaluated quickly by observing that it equals twice the integral over the lower triangle under the line s=u. This should give , which is different from what I got in my first attempt Cheers, Micha

another ay to compute this variance:

QuoteOriginally posted by: ddrdoubleBut this integral is a random variable with a normal distribution (little bit harder to proof).This follows from partial integration. One can write the integral as a sum of a stochastic integral w.r.t. a BM and a BM itself. Both are normally distributed and the sum of normally distributed things is normally distributed.

I am sorry for sure it is , now comes my way to derive it:Write the integral just as Riemann Integral :But this sum, can be written as: , but this is an stochastic integral.To compute the value of: just use Ito-isometry, this gives:the normal distribution follows from the convolution of a normal distributed random variable.