Mean of log-normal variable when exponent is replaced by runnung maximum of Ito-integral

Question

Let $W=\{W_t\}_{t\in[0;1]}$ be a real-valued Brownian motion, $\{F_t\}_{t\in [0;1]}$ the filtration generated by $W$, augmented with the nullsets. Let $\{\sigma_t\}_{t\in[0;1]}$ be a continuous and bounded Ito process (EDIT: w.r.t. $W$) with bounded drift and volatility coefficients (I could also live with more restrictions). I want to prove that \begin{align*} A := \mathbb E\bigg( \exp\bigg( \sup_{t\in[0;1]}\int^t_0 \sigma_s \mathrm dW_s\bigg) \bigg) < \infty. \end{align*}

If $\sigma$ was constant, this would be obvious since the distribution of the running maximum of the Brownian motion is known and the above expectation can be computed in closed integral form.

Question: Any ideas how to bound $A$ in a more general case than "$\sigma$ constant"? Or can such a result be found in the literature? Thank you!

This question has been posted on math.stackexchange a few days ago, but got no answers so far. (The "partially solved" in the title refers to the fact that meanwhile I managed to bound $B$ (see below), which was not the case when I asked the question on math.stackexchange originally. Thus, I have edited the question there. The main problem to bound $A$ is still unsolved.)

What I have tried: As an easier exercise, first I tried to prove the following: \begin{align*} B := \mathbb E\bigg( \exp\bigg( \int^1_0 \sigma_s \mathrm dW_s\bigg) \bigg) < \infty. \end{align*} This is also easy since the stochastic exponential is a martingale and then \begin{align*} 1 =& \mathbb E\bigg( \exp\bigg( \int^1_0 \sigma_s \mathrm dW_s - \frac 1 2 \int^1_0 (\sigma_s)^2 \mathrm ds \bigg) \bigg) \\\ge& \mathbb E\bigg( \exp\bigg( \int^1_0 \sigma_s \mathrm dW_s \bigg) \bigg) e^{- \frac 1 2 \Vert \sigma\Vert^2} \\=& B e^{- \frac 1 2 \Vert \sigma\Vert^2}. \end{align*}

Moreover, I thought we could $L^2$-approximate the Ito integral by \begin{align*} \sum_{k=1}^n \sigma_{(k-1)/n} ( W_{k/n} - W_{(k-1)/n} ), \end{align*} but I don't know how to approximate the supremum and anyway we cannot pull $L^2$-convergence into the $\exp$ function because $\exp$ increases faster than $\operatorname{id}^2$.

Answer 1

This edit reflects the actual question asked, and corrects an earlier answer.

You can rewrite the process $\int_0^t \sigma_s dW_s$ as a time change of Brownian motion, where the time change is given by $\tau(t)=\int_0^{t} \sigma^2(s) ds$. If $\sup_{t\in [0,1]} |\sigma_t|<R$ a.s. for some deterministic $R$, then $\tau(1)\leq R^2$. Then $$ E(e^{\sup_{t\in [0,1]}\int_0^t W_s ds}) =E(e^{\sup_{t\in [0,1]} B_{\tau(t)} }) \leq E(e^{\sup_{u\in [0,R^2]} B_{u}})<\infty$$ by the reflection principle.

In the more general case, where $\tau(1)$ is not uniformly bounded, define $A_t= {\int_0^t \sigma_s dW_s}$ and $B_t={\frac12 \int_0^t \sigma_s^2 ds}$. Fix $\alpha\in (0,1)$. If the condition $E(e^{2\tau(1)/\alpha^2})=E(e^{4 B_1/\alpha^2})<\infty$ holds then $$M_t= e^{2A_t/\alpha -(4/\alpha^2)B_t}$$ is a (positive) martingale, by Novikov's criterion, up to $t=1$, and in particular $E(\sup_{t\in [0,1]} M_t^{\alpha})<\infty$ by Doob's inequality. Now, \begin{eqnarray} E(\sup_{t\in[0,1]} e^{A_t})&\leq & E(\sup_{t\in [0,1]} e^{A_t-2B_t/\alpha} e^{2B_1/\alpha})\\ &\leq& (E(\sup_{t\in [0,1]} M_t^{\alpha}))^{1/2} (E(e^{4B_1/\alpha}))^{1/2}<\infty, \end{eqnarray} where the second inequality used Cauchy-Schwarz and the definition of $M_t$.