# Ito Moments

Use Ito’s formula to show that if $$\sigma(t,\omega)$$ is a bounded nonanticipating functional, $$|\sigma|\leq M$$, then for the stochastic integral $$I(t,\omega)=\int_0^t \sigma(s,\omega) dB(s,\omega)$$ we have the moment estimates
$\mathbf E\{ |I(t)|^{2p}\}\leq 1\cdot 3\cdot 5 \cdots (2p-1) (M^2 t)^p$
for $$p=1,2,3,…$$. Follow the steps below to achieve this result.

1. First assume that
$\mathbf E \int_0^t |I_s|^{k} \sigma_s dB_s =0$
for all positive , integer $$k$$. Under this assumption, prove the result. Why don’t we know a priori that this expectation is zero ?
2. Now define  for positive $$L$$, define $$\chi^{p}_L(x)$$ as $$x^p$$ for $$|x| < L$$ and 0 if $$|x| > L+1$$ and connected monotonically in between such that the whole function is  smooth. Now define
$\psi^{(p)}_L(x) = p\int_0^x \chi^{(p-1)}_L(y) dy\qquad\text{and}\qquad\phi^{(p)}_L(x)= p\int_0^x \psi^{(p-1)}_L(y) dy$
Observe that  all three of the functions are globally bounded for a given $$L$$. Apply Ito’s formula to $$\phi^{(p)}_L(I_t)$$ and the fact that Fatou’s lemma implies that
$\mathbf E |I_t|^{2p} \leq \lim_{L \rightarrow \infty} \mathbf E \phi^{(2p)}_L(I_t)$
to prove the estimate started at the start by following a similar induction step as used above.