How to solve for the expectation of the Ito Integral: $\int_0^4 B_t^2 dB_t$?

Question

I would like to find the expectation of the Ito Integral: $\int_0^4 B_t^2 dB_t$. My strategy is to use Ito's general formula with:

$$ f(t, B_t) = f(0,0) + \int_0^t \frac{df}{dx}(s, B_s) dB_s + \int_0^t \frac{df}{dt}(s, B_s) ds + \frac{1}{2}\int_0^t \frac{d^2f}{dx^2}(s, B_s) ds $$

Then, my candidate function is $f(t, B_t) = \frac{B_t^3}{3}$. However, the problem I am having is in the derivative $\frac{df}{dt}(s, B_s)$. Since I want to have my integration go up to $4$, then I necessarily have:

$$ \int_0^4 \frac{df}{d4}(s, B_s) ds = \int_0^4 \frac{\frac{B_t^3}{3}}{d4} ds $$

I know that there is probably a bad/stupid mistake here, and that I should just take the derivative with respect to $s$, but I cannot figure out why exactly we are having the $dt$ MATCH the upper limit of the integral, $t$. Can anyone shed some light here? Thanks!

Answer 1

Choose $$f(t, B_t) = \frac{B_t^3}{3}$$

so that we have

$$ \frac{B_t^3}{3} = \int_0^t B_s^2 dB_s + \int_0^t 0 ds + \frac{1}{2}\int_0^t 2B_s ds $$

$$ \to E[\frac{B_t^3}{3}] = E[\int_0^t B_s^2 dB_s + \int_0^t 0 ds + \frac{1}{2}\int_0^t 2B_s ds] $$

$$ \to E[\frac{B_t^3}{3}] = E[\int_0^t B_s^2 dB_s] + \frac{1}{2}\int_0^t 2E[B_s] ds $$

$$ \to E[\frac{B_t^3}{3}] - \frac{1}{2}\int_0^t 2E[B_s] ds = E[\int_0^t B_s^2 dB_s] $$

$$ \to E[\frac{B_t^3}{3}] = E[\int_0^t B_s^2 dB_s] $$

$$ \to \frac{E[B_t^3]}{3} \stackrel{?}{=} \frac{0}{3} = E[\int_0^t B_s^2 dB_s] $$

Answer 2

Let $f(x) = \frac{x^3}{3}$ and for reference, note that $f'(x) = x^2$ and $f''(x) = 2x$. It\^o's formula gives us that \begin{eqnarray*} f(B_t) - f(0) &=& \int_0^t f'(B_s) dB_s + \frac{1}{2} \int_0^t f''(B_s) ds. \end{eqnarray*} Therefore, we see that \begin{eqnarray*} \frac{1}{3} B_t^3 &=& \int_0^t B_s^2 dB_s + \int_0^t B_s ds \\ \therefore \ \mathbb{E} \left( \frac{1}{3} B_t^3 \right) &=& \mathbb{E} \left( \int_0^t B_s^2 dB_s + \int_0^t B_s ds \right) \\ \therefore \ \frac{1}{3} \mathbb{E} (B_t^3) &=& \mathbb{E} \left( \int_0^t B_s^2 dB_s \right) + \mathbb{E} \left( \int_0^t B_s ds \right) \ \ \ \text{(linearity of expectation)}, \\ 0 &=& \mathbb{E} \left( \int_0^t B_s^2 dB_s \right) + \mathbb{E} \left( \int_0^t B_s ds \right) \\ \therefore \ \mathbb{E} \left( \int_0^t B_s^2 dB_s \right) &=& - \mathbb{E} \left( \int_0^t B_s ds \right) = - \int_0^t \mathbb{E}(B_s) ds \ \ \ \text{(Fubini)} \\ &=& 0, \end{eqnarray*} where the last line follows from the fact that the Brownian motion is normally distributed with mean zero.