# Problem with a Brownian motion, Ito's formula and an indicator function

so I have done the first part of this question (it is at the bottom), but I have no clue how to do the second part. I think I understand the theory, but I do not know how to apply it. Any help would be really appreciated, thanks a lot! :)

Let $W(t)$, $t\ge 0$ be standard Brownian motion.

Part B Consider the class of general integrands: $$H:=\{{}(h_t)_{t≥0}:h_t\text{ is adapted, }E\int_{0}^{\infty}h^2_t \text{d}t<\infty\},$$ the indicator functions $$1_{[0,T]}(t):=1\text{ if }t\in[0,T]$$ $$1_{[0,T]}(t):=0\text{ if }t\notin[0,T]$$

Question: Show that $B(t)1_{[0,T]}(t)\in H$ and that $$E[sin^2(W(t))]=\int_{0}^{t}E[A(s)]\text{d}s$$

Background Question which might be useful

Part A: Using Ito’s formula, write the stochastic differential of $\sin^2(W(t))$, i.e. find $A(t)$, $B(t)$ such that $\text{d}\sin^2(W(t))=A(t)\text{d}t+B(t)\text{d}W(t)$.

Ito's lemma for Brownian motion is:

$$\text{d}f(W_t,t)=\partial_w f(W_t,t)+\frac{1}{2}\partial^2_w f(W_t,t) \text{d}t+\partial_t f(W_t,t)\text{d}t$$

Let $f(W_t,t)=f(w,t)$. Then $f(w,t)=\sin^2(w)$ $$\text{d}_w f(w,t)=2\sin{w}\cos{w}=\sin(2w)$$ Now, $$\frac{1}{2}\text{d}^2_w f(w,t)=\frac{1}{2}2\cos(2w)=\cos(2w)$$ and $$\text{d}_t f(w,t)=0$$ Hence, we have $$\text{d}f(W_t,t)=\text{d}\sin^2(W(t))=\sin(2W(t))\text{d}W_t+\cos(2W(t))\text{d}t$$

Therefore, $A(t)=\cos(2W(t))$ and $B(t)=sin(2(W(t))$.

Information provided:

Fubini’s theorem: under suitable conditions, for a stochastic process $h_t$ we have

$$\int_{0}^{T}E[h_t]\text{dt}=E\left[\int_{0}^{T}h_t\text{dt}\right]$$

The theorem holds in these two cases:

• When $h_t\ge 0$.
• When $h_t$ is bounded, i.e. $|h_t(\omega)|\le C$ for some $C>0$, for all $t$, $\omega$. In the sequel, $W_t$ is a standard Brownian motion.

Hints:

By definition, to show that $B(s){\bf 1}_{[0,T]}\in H$, you want to prove that $B(t){\bf 1}_{[0,T]}$ is $\mathcal{F}_t$-measurable and $\mathbb E[\int_{0}^{t}B^2(s){\bf 1}_{[0,T]}\mathrm ds] < \infty$.

1. Take measurability, for instance. For $0\leqslant t\leqslant T$, $$~B(t){\bf 1}_{[0,T]} = B(t) = \sin(2W_t).$$ Well, what does this imply for $B(t){\bf 1}_{[0,T]}$ in terms of $\mathcal{F}_t$-measurability? Why? Also, for $T<t$, $$~B(t){\bf 1}_{[0,T]} = 0.$$ What does this imply for $B(t){\bf 1}_{[0,T]}$ in terms of $\mathcal{F}_t$-measurability? Why?

2. From the information provided to you, since $B^2(t){\bf 1}_{[0,T]}\geqslant 0$, then $$\mathbb E[\int_{0}^{t}B^2(s){\bf 1}_{[0,T]}\mathrm ds] = \int_{0}^{t}\mathbb E[B^2(s){\bf 1}_{[0,T]}]\mathrm ds = \ldots <\infty$$ Conclude, therefore, that $B(s){\bf 1}_{[0,T]}\in H.$

3. Finally, why is $$\mathbb E[\int_{0}^{t} B(s) \mathrm d W_s] = 0?$$ If you know why, you are done because, by part(a) of the question and the information provided, $$\mathbb E[\sin^2(W_t)] = \mathbb E[\int_{0}^{t} B(s) \mathrm d W_s + \int_{0}^{t} A(s) \mathrm ds] = \mathbb E[\int_{0}^{t} B(s) \mathrm d W_s] + \mathbb E[\int_{0}^{t} A(s) \mathrm ds] = \mathbb E[\int_{0}^{t} A(s) \mathrm ds] = \int_{0}^{t} \mathbb E[A(s)] \mathrm ds\,.$$

The first part of this is fairly straight forward. Since $B(t)=sin(2W(t))$ is bounded, and in fact $|B(t)|\le1$, then $(B(t)1_{[0,t]}(t))^2\le1$.

Obviously, $E[(B(t)1_{[0,t]}(t))^2]\le{T}$. Thus, $B(t)1_{[0,t]}$ is in $H$.

For the second part, integrate the Stochastic Differential Equation to obtain $$\int_0^tdf(u)du=\int_0^tA(u)du+\int_0^tB(u)dW_u$$ which yields $$f(t)-f(u)=\int_0^tA(u)du+\int_0^tB(u)dW_u$$Now, take expectations on both sides to find $$E[f(t)]=E[\int_0^tA(u)du]+E[\int_0^tB(u)dW_u]$$$$=\int_0^tE[A(u)]du$$since the expected value of both $f(0)$ and $dW_t$ are zero.