Quadratic variation of a product

Question

I have been struggling to figure out the quadratic variation of the following product: $$e^{3B(t)}\int_0^tB(s)ds$$ I know that the integral of the Brownian motion w.r.t time is finite variation so its quadratic variation is zero. But the $[e^{3B(t)}]=\int_0^t9e^{6B(s)}ds$. Using the Ito formulas, I converted it to $$\left(1+\int_0^t3e^{B(s)}dB(s)+\int_0^t 9/2 e^{B(s)}ds\right)\cdot \int_0^tB(s)ds$$ where among lots of members (almost all of them cancel out) there will be $$\left[\int_0^te^{3B(s)}dB(s)\cdot\int_0^tB(s)ds\right]$$ and I was given the suggestion that it equals to $$\int_0^te^{3B(s)}B(s)d[B(s),s]$$ which is zero — but I think this is a misuse of the differenial forms. I would appreciate any help or advice!

Answer 1

Welcome on Math.stackexchange! Here is my take on it:

Using Itô's lemma, one finds

$$e^{3B}=e^{3B}\cdot3B+\frac{1}{2}e^{3B}\cdot[3B]$$

Its quadratic variation is

$$[e^{3B}]=9e^{6B}\cdot id$$

Using stochastic integration by parts, one obtains $$e^{3B}(B\cdot id)=e^{3B}\cdot (B\cdot id)+(B\cdot id)\cdot e^{3B}+[e^{3B},B\cdot id]$$

Now, one can calculate the quadratic variation as follows:

$$[e^{3B}(B\cdot id)]=[(B\cdot id)\cdot e^{3B}]=(B\cdot id)^2\cdot[e^{3B}]=9e^{6B}(B\cdot id)^2\cdot id$$

I've used the notation "$\cdot$" for the stochastic integral, therefore the quadratic variation would be

$$9\int_0^te^{6B(s)}\Big(\int_0^sB(r)dr\Big)^2\ ds$$

Answer 2

To keep track of things more easly, write $X_t=e^{3Bt}$ and $Y_t=\int_0^t B_s ds$. Then, as you note, $$ dX_t = 3X_t dB_t. $$ Therefore $$ d(X_tY_t) = 3X_tY_t dB_t + V_t, $$ where $V$ is an unimportant (for present purposes) process of finite variation. From this it follows that the quadratic variation of $XY$ is $$ 9\int_0^t X_s^2Y_s^2 ds. $$