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](https://math.stackexchange.com/questions/4477916) 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 $\operatorname{id}^2$.



Mean of log-normal variable when exponent is replaced by runnung maximum of Ito-integral
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 $M\_t$.


