I have to find a $F_t$ such that $B_T^3=E[B_T^3]+\int_0^T F_t dB_t$. I have shown by ito formula that $B_T^3=\int_0^T 3 B_s^2 dB_s+\int_0^T 3 B_s ds$. Could you please help me?
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1$\begingroup$ Please do not delete your questions unless there is a really good reason for it. Its kind of frustrating if you write an answer and the question is deleted some seconds before you finished writing it. $\endgroup$– sazCommented Apr 19, 2016 at 17:46
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First of all, note that $\mathbb{E}(B_T^3)=0$, i.e. we have to find $F_t$ such that
$$B_T^3 = \int_0^T F_s \, dB_s. \tag{1}$$
It follows from Itô's formula that
$$B_t^3 = 3 \int_0^t B_s^2 \, dB_s + 3 \int_0^t B_s \, ds \tag{2}$$
(please compare this with what you wrote in your question; there are so many typos in there that I'm really not sure whether you know what you are doing). The first term on the right-hand side looks already pretty good, but we have to rewrite the second one. By the integration by parts formula we have
$$t B_t = \int_0^t B_s \, ds + \int_0^t s \, dB_s,$$
i.e.
$$\int_0^t B_s \, ds = t B_t - \int_0^t s \, dB_s = t \int_0^t \, dB_s - \int_0^t s \, dB_s. \tag{3}$$
Plugging this into $(2)$ and evaluating at $t=T$, we get
$$B_T^3 = 3 \int_0^T B_s^2 \, dB_s + 3 \left( T \int_0^T \, dB_s - \int_0^T s \, dB_s \right) = 3\int_0^T (B_s^2+T-s) \, dB_s.$$
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$\begingroup$ Could you please explain why E[B_T^3]=0 $\endgroup$ Commented Apr 18, 2016 at 18:22
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1$\begingroup$ @Mathfreak $(B_t)_{t \geq 0}$ is a Brownian motion, and this means in particular that $B_T$ is Gaussian with mean $0$ and variance $T$... and it is well-known that the odd moments of a centered Gaussian random variable are $0$ (this is a direct consequence of the fact that the density of a centered Gaussian random variable is symmetric). $\endgroup$– sazCommented Apr 18, 2016 at 18:28
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$\begingroup$ Do you also know how it works if I have $cos(B_T)$ instead of $B_T^3$. I tried the same argument as you showed me. $\endgroup$ Commented Apr 18, 2016 at 20:28
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$\begingroup$ @Mathfreak There is a similar question about $\sin(B_T)$: math.stackexchange.com/q/1275966/36150 The approach I suggested there should also work for $\cos$. $\endgroup$– sazCommented Apr 19, 2016 at 5:35