expectations of Brownian motions

Question

Let $B_t$ be a standard Brownian motion started at zero, and let $M_t$ be a stochastic process defined by $M_t=3\int_0^{t^{1/9}} s^4dB_s$

Compute $E\left[1+\int_0^t(1+M_s)^4 dM_s\right]$.

Compute $E\left[{M_t}^2\int_0^t M_s dM_s\right]$.

I have already shown that $M_t$ is a standard Brownian motion by showing $\langle M,M\rangle =t$.

Answer 1

1. Since $(M_t)_{t \geq 0}$ is a Brownian motion, we know that $(s,\omega) \mapsto (1+M_s(\omega))^4$ is integrable with respect to $L^2(\lambda_T \otimes \mathbb{P})$ for any $T>0$. Consequently, the stochastic integral $$X_t := \int_0^t (1+M_s)^4 \, dM_s$$ is a martingale; in particular it has constant expectation.
2. Using Itô's formula, it is not difficult to show that $$\int_0^t M_s \, dM_s = \frac{M_t^2-t}{2}.$$ Therefore, $$\mathbb{E} \left(M_t^2 \cdot \int_0^t M_s \, dM_s \right) = \frac{1}{2} \left( \mathbb{E}(M_t^4) - t \cdot \mathbb{E}(M_t^2) \right).$$ As $M_t \sim N(0,t)$ for each $t \geq 0$, the expectations on the right hand side are known.