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.

- 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 ? - 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.