Find the Itô representation of the following random variable

Question

Let $(B_t)_{t \geq 0}$ be a Brownian motion and $T \geq 0$. Find a constant $z_T \in \mathbb{R}$ and $(s,\omega) \in \mathcal{V}([0,T])$ such that $F_T(\omega) = z_T + \int_{0}^{T} \phi_T(s,\omega)dB_s(\omega)$ when $F_T(\omega) = \int_0^T B^3_s(\omega) ds$.

The idea is to rewrite $\int_0^T B^3_s ds$ using Itô's lemma, but it seems like I'm stuck in a loop. We have

\begin{equation*} \begin{split} d(tB_t^3) &= B_t^3dt + 3tB_t^2dB_t + 3t B_t d \langle B \rangle_t \iff \\ \int_0^T B_t^3 dt&= TB_T^3 - \int_0^T 3tB_t^2 d B_t - \int_0^T 3t B_t dt \end{split} \end{equation*}

Now two new terms appeared, $TB_T^3$ and $\int_0^T 3t B_t dt$. If we plug in

$$ TB_T^3 = \int_0^T B_t^3 dt + \int_0^T 3tB_t^2 d B_t + \int_0^T 3t B_t dt $$

everything cancels, yielding $0=0$. Otherwise we can proceed by using Itô on $\int_0^T 3t B_t dt $ to see what we get, but I'm not quite sure if that's gonna give anything. Any hints?

Edit, solution:

We have that

\begin{equation*} \begin{split} &\int_0^T 3t B_t dt = \frac{3}{2}T^2B_T - \int_0^T \frac{3}{2}t^2 dB_t \\ &TB_T^3 = \int_0^T 3T(B_t^2 +T -t) dB_t \\ &\frac{3}{2}T^2B_T = \frac{3}{2}T^2 \int_0^T dB_t \end{split} \end{equation*}

by some Itô engineering, yielding

$$ \int_0^T B^3_t dt = \int_0^T (T-t)(3T + 3B_t^2 + \frac{3}{2}(T+t))dB_t. $$

Answer 1

Seems I obtain something different which cannot be since the representation should be unique.

• By Ito, $$\tag{1} B_t^3=\textstyle 3\int_0^tB_s^2\,dB_s+3\int_0^tB_s\,ds\,. $$
• By integration by parts, $$ \textstyle\int_0^sB_u\,du=\textstyle\int_0^s(s-u)\,dB_u\,.\tag{2} $$
• By stochastic Fubini, \begin{align} \textstyle\int_0^t\int_0^sB_u\,du\,ds&\stackrel{(2)}=\textstyle\int_0^t\int_0^s(s-u)\,dB_u\,ds\stackrel{stoch. Fub.}=\int_0^t\int_u^t(s-u)\,ds\,dB_u\,.\tag{3}\\ \end{align}
• Then, using (1), \begin{align} F_t&=\textstyle\int_0^tB_s^3\,ds\stackrel{(1)}=3\int_0^t\int_0^sB_u^2\,dB_u\,ds+3\int_0^t\int_0^sB_u\,ds\,du\\[2mm] &\stackrel{stoch. Fub.}=\textstyle3\int_0^t\int_u^tB_u^2\,ds\,dB_u+3\int_0^t\int_0^sB_u\,ds\,du\\[2mm] &\stackrel{(3)}=\textstyle3\int_0^tB_u^2(t-u)\,dB_u+3\int_0^t\int_u^t(s-u)\,ds\,dB_u\\[2mm] &=\textstyle3\int_0^tB_u^2(t-u)+\frac{t^2-u^2}{2}-u(t-u)\,dB_u\,.\tag{4} \end{align}