3rd moment of Stochastic Integral of Simple Process Consider a filtered probabilty space $(\Omega, \mathscr{F},\{\mathscr{F}_t\}_{t \geq 0}, \mathbb{P})$. Let $A_t$ be a simple process defined on $[0,T]$, i.e.:
$$
A_t = \sum_{k=1}^n Y_{t_{k-1}}\chi_{[t_{k-1},t_k)} 
$$
where here, each $Y_{t_{k-1}}$ is a random variable that is $\mathscr{F}_{t_{k-1}}$ measurable. Assume that almost surely, for every $t$, we have that $0 < |A_t| \leq C < \infty$. Iis it true that
$$
\mathbb{E}\left[\left(\int_0^TA_t \mathrm{d}W_t\right)^3\right] = \mathbb{E} \left[\left(\sum_{k =1}^n Y_{t_{k-1}}(W_{t_k} - W_{t_{k-1}}) \right)^3\right] = 0
$$
for any simple process satisfying the above? I was able to prove the claim for processes where each $Y_{t_k}$ is independent of the Brownian motion, but am not sure if it is true otherwise. Does anyone have any ideas?
 A: Here is a proposed solution of mine, i.e., it is the calculations for what J.G. proposed.
Consider the process with coefficients given by $Y_{t_k} = W_{t_k} \wedge C$ and consider a step function that is a single period of time, so the integral has the form:
    $$
W_{t_1}(W_{t_2} - W_{t_1})
$$
    Note that we need only consider this case, as $W_{t_1} < C$ with nonzero probability, hence, if we are able to show that the expectation is nonzero in this case, it will follow in the general case, as $\mathbb{E}[ (C(W_{t_2} - W_{t_1}))^3] = 0$ in general as it is a rescaled odd moment of a symmetrically distributed random variable.
    \newline 
    \newline 
    Let us compute:
    $$
\mathbb{E} [(W_{t_1}W_{t_2} - W_{t_1}^2)^3]
$$
    The cube expands to:
    $$
-W_{t_1}^6 + W_{t_1}^3W_{t_2}^3 - 3W_{t_1}^5W_{t_2} + 3W_{t_1}^4W_{t_2}^2
$$
    Let us compute the expectation of the cross terms. We have:
    $$
\mathbb{E}[W_{t_1}^5W_{t_2}] = \mathbb{E}[\mathbb{E}[W_{t_1}^5W_{t_2}|\mathscr{F}_{t_1}]] = \mathbb{E}[W_{t_1}^5\mathbb{E}[W_{t_2}|\mathscr{F}_{t_1}] = \mathbb{E}[W_{t_1}^6]
$$
    Thanks to the martingale property.For the term with the fourth power, we calculate:
    $$
    \mathbb{E}[W_{t_1}^4W_{t_2}^2] = \mathbb{E}[\mathbb{E}[W_{t_1}^4W_{t_2}^2|\mathscr{F}_{t_1}]] = \mathbb{E}[W_{t_1}^4\mathbb{E}[W_{t_2}^2|\mathscr{F}_{t_1}]]
$$
    The conditional variance inside is computed as:
    $$
\mathbb{E}[W_{t_2}^2 | \mathscr{F}_{t_1}] = \mathbb{E}[W_{t_2}^2 - 2W_{t_2}W_{t_1} + W_{t_1}^2 + 2W_{t_2}W_{t_1} - W_{t_1}^2| \mathscr{F}_{t_1}]
$$
    Completing the square, indepence of increments, and martingality gives:
    $$
(t_2 - t_1) + 2W_{t_1}^2 - W_{t_1}^2 = (t_2 - t_1) + W_{t_1}^2
$$
    Finally, taking the unconditional expectation simplifies this:
    $$
(t_2 - t_1) + t_1^2
$$
    Finally, for the term with the third power, bound it below using Jensen's inequality:
    $$
\mathbb{E}[W_{t_1}^3W_{t_2}^3] \geq \mathbb{E}[W_{t_1}W_{t_2}]^3 = t_1^3
$$
    Our final expression thus reads:
    $$
-\mathbb{E}[W_{t_1}^6] + \mathbb{E}[W_{t_1}^3W_{t_2}^3] - 3(E[W_{t_1}^6]) + 3(t_2 - t_1 + t_1^2) 
$$
    And Jensen's inequality from before bounds this below by:
    $$
    -\mathbb{E}[W_{t_1}^6] + t_1^3 - 3(\mathbb{E}[W_{t_1}^6]) + 3(t_2 - t_1 + t_1^2) 
$$
    Take $t_2$ sufficiently large, such that the expression above is greater than or equal to $0$. It follows that our expectation is positive, as we desire.
