2
$\begingroup$

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 $\operatorname{id}^2$.

$\endgroup$
2
  • $\begingroup$ Your computation is suspicious since $\|\sigma\|_2^2$ is a random variable.... $\endgroup$ Commented Jun 24, 2022 at 8:48
  • $\begingroup$ By $\Vert \sigma\Vert \in \mathbb R$, I mean a uniform (in $(\omega,s)$) bound on the value of $\sigma_s(\omega)$. $\endgroup$
    – Kolodez
    Commented Jun 24, 2022 at 9:08

1 Answer 1

2
$\begingroup$

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 $M_t$.

$\endgroup$
3
  • $\begingroup$ What is the "tail" of a stochastic process? And by "endpoint" you mean the random variable corresponding to the largest possible time index? $\endgroup$
    – Kolodez
    Commented Jun 24, 2022 at 9:12
  • 1
    $\begingroup$ Tail is $P(X>x)$ for $x$ large, which is all you care about when computing $E(e^X)$. Recall that the tail of the distribution of the maximum of BM on an interval $[0,T}$ is just twice that of the distribution of the BM at time $T$ (as I wrote, this follows from the reflection principle). $\endgroup$ Commented Jun 24, 2022 at 12:26
  • 1
    $\begingroup$ If (as you asked) $\tau(1)$ is bounded, say by a deterministc $A$, then $\sup_{t\in[0,1]} B_{\tau(t)}\leq \sup_{s\in [0,A]} B_s$. $\endgroup$ Commented Jun 24, 2022 at 19:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.