How to get the mean, skewness of an Itō integral?

Question

If $B_t$ denotes a standard Brownian motion, and let $X_t = \int f(s)dB_s$, $f(s)$ is a deterministic integrand. I know $B_t$ is a martingale. Is $X_t$ also a martingale? And how can I get the formula of moment statistics of $X$, say, $E(X_t^2)$, $E(X_t^3)$?

Answer 1

For the martingale part, this follows by the Ito construction increments $B_{t_{i}}-B_{t_{i-1}}$. That implies that the sum of them is also a martingale $I(f^{n}) = \int_{0}^{t} f^{n}_u dB_u$ where $f^{n}$ are simple processes.

Then finally we use that the $L_1$-limit of martingales is still martingale eg. see here When is the limit of Martingales a Martingale?

To get the formulas for the moments we use the Gaussianity. Since $f$ is deterministic, we get that $X_{t}$ is Gaussian with variance $\mathbb{E}\int_0^t f(s)^2 ds$. see here

The integral of a progressively measurable process $f$ is a limit of the integrals $I(f^{n}) = \int_{0}^{t} f^{n}_u dB_u$ where $f^{n}$ are simple processes and these integrals are Gaussian by definition.

Then as mentioned there we just use

Let $X_n$ be a sequence of normally distributed random variables with mean zero and variances $\sigma_n^2 \in [0,\infty)$. Suppose $X_n \to X$ in distribution. Then $\sigma_n^2$ converges to some $\sigma^2 \in [0,\infty)$, and $X$ is normally distributed with mean zero and variance $\sigma^2$.

In fact, this process is just a time changed Brownian motion see here