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Let $B=(B_t)_{t\geq0}$ be a standard Brownian motion started at $0$. Then how do we work out $$E[(B_t-1)^2\int^{t}_0B_s ds]$$ am I right in saying this is the same as $$E[(B_s-1)^2\int^{s}_0B_s ds]$$ by using the following fact $E[E[X|A]]=E[X]$, where A is some set contained in $\Omega$?

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    The reason why you think "this is the same" is mysterious. And what is $s$ in the second formula? And how can $s$ be at the same time a bound of the integral and the variable of integration? // Hint: Can you compute, for every $0\leqslant s\leqslant t$, $$E((B_t-1)^2B_s)\ ?$$ – 2017-01-20
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    I was thinking using $E[E[(B_t-1)^2\int^t_0B_s ds|\mathcal{F}_s]]$, if this doesn't work how should i approach this? – 2017-01-20
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    Again, what is $s$ in $\mathcal F_s$? For an approach that works, please see my first comment. – 2017-01-20
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    s is $0\leq s\leq t$, oh yes I see my error. So how should I go about it, is the hint you gave equal to 0? – 2017-01-20

1 Answers 1

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Let $t \in \mathbb{R_{+}^{*}}$,

$$E[(B_t-1)^2\int^{t}_0B_s ds]=E[\int_{0}^{t}{B_s(B_t-1)^2}ds]$$

Let define the process $X_s$

$$X_s=B_s(B_t-1)^2$$

where $s \leq t$

In order to apply Fubini's theorem , we must verify that

$$E[\int_{0}^{t}{|X_s|}ds]< +\infty$$

Given that $|X_s| \geq 0$, Tonelli's theorem gives $$E[\int_{0}^{t}{|X_s|}ds]=\int_{0}^{t}E({|X_s|})ds$$

Furthermore, $$E({|X_s|})= |E({|X_s|})|=\left|E({|B_s|(B_t-1)^2})\right|$$

By applying Cauchy-Schwarz inequality, we have

$$\left|E({|B_s|(B_t-1)^2})\right|\leq \sqrt{E(B_s^2)E((B_t-1)^4))}$$

We have that $$E(B_s^2)=var(B_s)=s$$

For the second term , we introduce the process $Y_t=(B_t-1)^4=f(B_t)$ , and $f(x)=(x-1)^4$.

By differentiating $f$ twice, we have $f''(x)=12(x-1)^2 \geq 0$

Using Ito's lemma on $Y_t$, and Tonelli's theorem, we have that $$E(Y_t)=Y_0+E\left(\int_{0}^{t}{\frac{1}{2}f''(B_u)du}\right)$$ $$=1+\int_{0}^{t}{\frac{1}{2}E(f''(B_u))du}=1+6\int_{0}^{t}{(u+1)}du=3t^2+6t+1$$

Finally, $$E[\int_{0}^{t}{|X_s|}ds] \leq \int_{0}^{t}{\sqrt{E(B_s^2)E((B_t-1))^4)}}=\int_{0}^{t}{\sqrt{s(3t^2+6t+1)}}<\infty$$

We can apply Fubini's theorem and we have $$E(\int_0^{t}{X_s}ds)=\int_0^{t}{E(X_s)ds}$$

$$E(X_s)=E(B_s(B_t-1)^2)=E(B_s(B_t-B_s+B_s-1)^2)$$ $$=E(B_s((B_t-B_s)^2+2(B_s-1)(B_t-B_s)+(B_s-1)^2))$$

$$=E(B_s(B_t-B_s)^2)+2E(B_s(B_s-1)(B_t-B_s))+E(B_s(B_s-1)^2)$$

$B_t-B_s$ is independent of $\mathcal{F_s}$, where $\mathcal{F}$ is the natural filtration of $B$, and $E(B_s)=E(B_t-B_s)=0$, therefore $$E(X_s)=E(B_s(B_s-1)^2)$$

Finally, $$E(X_s)=E(B_s(B_s-1)^2)=E(B_s^3)-2E(B_s^2)+E(B_s)=-2s$$

and $$E[(B_t-1)^2\int^{t}_0B_s ds]=\int_{0}^{t}{-2s}=-t^2$$