Assume that $H\_t$ is a progressively measurable process such that with probability one $|H\_t| \le k$ for all $t$. Let$$Z\_t = \int\_0^t H\_s\,dB\_s.$$How do I see that for all $s < t$, $\lambda \in \mathbb{R}$, that$$\textbf{E}[\text{exp}\{\lambda(Z\_t - Z\_s)\}] \le \text{exp}\left\{{{k^2\lambda^2}\over2}(t - s)\right\}?$$I know how to show this using Itō's lemma, but I was wondering if anyone could supply a proof or a reference to one that does not use it. Many thanks in advance.



Bound on expectation, not a really simple process, circumvent using Itō's lemma?
You can prove it going back to the definition of the stochastic integral. If $H$ is constant on the interval $[s,t)$ then conditionally to $\mathcal F\_s$ (the underlying filtration at time $s$), $Z\_t-Z\_s$ is a centerer gaussian with variance $H\_s^2 (t-s)$, and hence we have $$\textbf{E}[\text{exp}\{\lambda(Z\_t - Z\_s)\} | \mathcal F\_s] =\text{exp}\left\{{{H\_s^2 \lambda^2}\over 2}(t - s)\right\} \le \text{exp}\left\{{{k^2\lambda^2}\over2}(t - s)\right\}.$$
If $H\_s$ is a simple process (ie piecewise constant), then the same inequality $\textbf{E}[\text{exp}\{\lambda(Z\_t - Z\_s)\} | \mathcal F\_s] \le \text{exp}\left\{{{k^2\lambda^2}\over2}(t - s)\right\}$ holds by applying the preceding on each interval on which $H$ is constant. It therefore holds for every progressively measurable process by density of the simple processes.


