Can the Martingale Representation theorem be used to assume that the integral with respect to Brownian motion, $B(t,\omega)$,
$$X=\int^{T}_{0}B^{4}(t,\omega)dB(t,\omega)$$
is a square integrable random variable with $\mathbb{E}[X]=0$, since the process $B^{4}(t,\omega)$ is an adapted and predictable process with respect to the natural filtration?
In other words, does the implication of the Martingale Representation theorem hold both ways?
Best Answer
You need to be a little careful that the stochastic integral of the process is a true martingale instead of just a local martingale. In general, if $H(s,\omega)$ is a predictable process with respect to your filtration you only have that the process $X(t,\omega) = \int_0^t H(s,\omega)dB(s,\omega)$ is a local martingale. If you have an extra condition on $H$ such as $\mathbb{E}[\int_0^T (H(s,\omega))^2ds] < \infty$ then the process $X(t,\omega)$ defined above is a true martingale so you would indeed have $\mathbb{E}[X(T,\omega)]=0$ and $\mathbb{E}[(X(T,\omega))^2] < \infty$.
In the example you posted, we can check
$$ \mathbb{E}\left[\int_0^T (B^4(t,\omega))^2 dt\right] = \int_0^T \mathbb{E}[B^8(t,\omega)]dt = \int_0^T 105 t^4 dt < \infty$$
so $X$ is a square integrable random variable with $\mathbb{E}[X] = 0$.