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


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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.