The quadratic variation of the following process...

Question

Let $B$ denotes a Brownian motion, and a stochastic process $X$ is definied as follows: $$X_{t}=e^{3B_{t}}+\int_{0}^{t}B_{s}ds.$$

What is the quadratic variation of $X^2$?

I got the following result: $$\int_{0}^{t}36e^{12B_{s}}ds+2\int_{0}^{t}B_{s}ds\cdot\int_{0}^{t}36e^{9B_{s}}ds.$$

Is it a good result? I calculated $X_{t}^{2}$, then I tried to find the quadratic variation.

Answer 1

I'm using "$\cdot$" as notation for the stochastic integral.

$$X^2=\exp(6B)+2\exp(3B)(B\cdot id)+(B\cdot id)^2$$

Using ito's lemma:

$$[X]=[\exp(3B)]=9\exp(6B)\cdot id$$

Using integration by parts

$$X^2=X_0^2+2X\cdot X+[X]$$

$$[X^2]=4X^2\cdot[X]$$

Calculation:

$$ \begin{align} [X^2]&=4(\exp(6B)+2\exp(3B)(B\cdot id)+(B\cdot id)^2)\cdot (9\exp(6B)\cdot id)\\ &=36(\exp(12B)\cdot id+2\exp(9B)(B\cdot id)\cdot id + \exp(6B)(B\cdot id)^2\cdot id )\\ \end{align} $$

Therefore, I find

$$[X^2]_t=36\ \bigg(\int_0^t\exp(12B_s)ds\ +2\int_0^t\exp(9B_s)\Big(\int_0^sB_r\ dr\Big)\ ds\ + \int_0^t\exp(6B_s)\Big(\int_0^sB_r\ dr\Big)^2\ ds\ \bigg)$$