How to find the variance $Var[\int_0^t(aB_s+bs)^2dB_s],$ where $B_s$ are standard Brownian motion.
My thoughts:
$Var[\int_0^t(aB_s+bs)^2dB_s] = E[(\int_0^t(aB_s+bs)^2dB_s)^2] - E[\int_0^t(aB_s+bs)^2dB_s]^2.$
For the first term, it seems that we need to use Ito's isometry. Then $E[(\int_0^t(aB_s+bs)^2dB_s)^2] = E[\int_0^t (aB_s+bs)^4ds].$ I don't know how to do the next step. Do I need to expand the integrand $(aB_s+bs)^4$ and calculate the integral term by term? If so, How to calculate $\int_0^t s^i B_s^j dB_s$ ? (Let $f(t,x)=t^ix^j$ and then use ito's formula?)
For the second term, do we just expand the integrand to calculate the integral and then calculate the expectation?
Am I on the right track to solve this problem? Are there an easier way to calculate the variance?
Notice that for each $s \geq 0$, $a B_s + bs \sim N(bs, a^2s)$. Thus, its fourth moment is given by $$\mathbb{E}[(aB_s + bs)^4] = (bs)^4 + 6(bs)^2(a^2s) + 3(a^2s)^2 = b^4 s^4 + 6a^2b^2s^3 + 3a^4s^2$$
Since an Itô integral has mean zero, the variance of the integral is simply its second moment. By using Itô's isometry and Tonelli's theorem, we get:
$$\begin{align*} Var \left( \int_0^t (aB_s + bs)^2 dB_s \right) &= \mathbb{E} \left( \int_0^t (aB_s + bs)^2 dB_s \right)^2 \\ &= \mathbb{E} \left( \int_0^t (aB_s + bs)^4 ds \right)\\ &= \int_0^t\mathbb{E} \left( (aB_s + bs)^4 \right)ds \\ &= \int_0^t b^4 s^4 + 6a^2b^2s^3 + 3a^4s^2 ds \\ &= \frac{b^4t^5}{5} + \frac{6a^2 b^2 t^4}{4} + a^4t^2 \end{align*}$$