Obviously, we can write
$$M_T = \mathbb{E} \left[ \left( \int_0^T 1 \, dW_t \right) \cdot \left( \int_0^T t \, dW_t \right) \right],$$
i.e. we are interested in the expectation of a product of stochastic integrals. Using the identity
$$a \cdot b = \frac{1}{4} ((a+b)^2-(a-b)^2) \tag{1}$$
for
$$a := \int_0^T 1 \, dW_t \qquad \quad b := \int_0^T t \, dW_t$$
we get
$$M_T = \frac{1}{4} \mathbb{E} \left[ \left(\int_0^T (1+t) \, dW_t \right)^2 \right] - \frac{1}{4} \mathbb{E} \left[ \left(\int_0^T (1-t) \, dW_t \right)^2 \right].$$
Applying Itô's isometry yields
$$\begin{align*} M_T &= \frac{1}{4} \mathbb{E} \left( \int_0^T (1+t)^2 \, dt \right) - \frac{1}{4} \mathbb{E} \left( \int_0^T (1-t)^2 \, dt \right) \\ &= \int_0^T \frac{1}{4} ((1+t)^2-(1-t)^2) \, dt \\ &\stackrel{(1)}{=} \int_0^T t \, dt. \end{align*}$$
In fact, we have shown the following (more general) statement:
Let $f,g \in L^2([0,T] \otimes \mathbb{P})$ be progressively measurable. Then $$\mathbb{E} \left[ \left( \int_0^T f(t) \, dW_t \right) \cdot \left( \int_0^T g(t) \, dW_t \right) \right] = \mathbb{E} \int_0^T g(t) \cdot f(t) \, dt.$$
Note that for $f=g$ this is the (standard version of) Itô's isometry.
Yes, you can prove it using the multidimensional Levy theorem. This is, checking that the process $((X_t^1, X_t^2))_{t \in (0,\infty)}$ is a 2-dimensional local martingale and that
- $[X^{1}]_t=[X^{2}]_t = t;$
- $[X^{1},X^{2}]_t = 0.$
Best Answer
For part 2:
Is a square-integrable continuous local martingale a true martingale?
I think 'lemma 3' in the first answer tells you how to solve question 2. It shows the stochastic integral inside your expectation is a true martingale, which means the expectation is 0.
For part 3:
The integral is $(M_t^2-t)/2$, where $M$ is a Brownian motion. Then you just cube this and use the fact $E[M_t^n] = t^{n/2}(n-1)!$ for $n$ is even, $E[M_t^n]=0$ for $n$ odd. In fact we could have used it for part 2 but this would be too much hassle...