0
$\begingroup$

Let $W$ be a standard one-dimensional Brownian motion and $0<T<+\infty$. Then $$\lim_{\beta\to+\infty}\sup_{0\le t\le T}|e^{-\beta t}\int_0^te^{\beta s}\mathrm{d}W_s|=0\quad \text{a.s.}$$

Could anyone give some hints or proofs?

$\endgroup$
  • $\begingroup$ If I have done it correctly, stochastic integration by parts gives $$e^{-\beta t} \int_0^t e^{\beta t}\,dW_s = W_t - \int_0^t W_s \beta e^{-\beta(t-s)}\,ds$$ which seems to show this is not true, as the second term on the right side goes to 0 as $\beta \to \infty$. $\endgroup$ – Nate Eldredge Dec 28 '14 at 19:37
  • $\begingroup$ @Nate Eldredge - that was my initial reaction too. However the integrand on the RHS is not bounded by an integrable function as $\beta \to \infty$ ($\beta e^{-\beta t} \ge \beta(1 - \beta t) \ge \beta/2$ for $t \le 1/2\beta$, in particular any (uniform in $\beta$) upper bound $g$ satisfies $g(x) \ge 1/4x$). On the other hand Ito isometry yields $$ E\left[ \left( e^{-\beta t} \int_0^t e^{\beta s} dW_s \right)^2 \right] = \frac{1}{2\beta}(1 - e^{-2\beta t} ) \to 0 $$ as $\beta \to \infty$. Also I did a quick simulation in R. I think the OP's statement is correct. $\endgroup$ – P.Windridge Dec 28 '14 at 21:16
  • 4
    $\begingroup$ Is this a homework problem? $\endgroup$ – Matthias Ludewig Dec 28 '14 at 22:18
  • $\begingroup$ This is an Orenstein-Uhlenbeck process with high damping (solution to $dX_t=-\beta X_t dt+dW_t$, starting at 0). Of course it goes to $0$ as $\beta\to\infty$.... $\endgroup$ – ofer zeitouni Dec 29 '14 at 7:50
1
$\begingroup$

As Nate Eldredge says, integration by parts implies $$ e^{ - \beta t } \int_0^t e^{ \beta s } dW_s = W_t - \beta \int_0^t e^{ - \beta ( t - s ) } W_s ds \\ = \beta \int_0^t e^{ - \beta ( t - s ) } ( W_t - W_s ) ds + e^{ -\beta t }W_t. $$ The supremum of the two terms above tend to zero. The second one, for example, follows because the maximum of the function $ t \mapsto e^{-\beta t } t$ is attained at $\beta^{-1}$, and $\sup_{t \in (0,T )} t^{-1}|W_t|$ is in $L^1$. The first term converges uniformly to zero from the fact that brownian motion is uniformly Holder of order $\alpha\in(0,1/2)$ and because $$ \lim_{ \beta \to \infty }\sup_{ t \in (0,T) } \beta \int_0^te^ss^\alpha ds = 0. $$ The last fact can be proved from the standard integration by parts theorem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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