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?

  • $\begingroup$ 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 $\beta \to \infty$. $\endgroup$ Commented Dec 28, 2014 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. $\endgroup$ Commented Dec 28, 2014 at 21:16
  • 4
    $\begingroup$ Is this a homework problem? $\endgroup$ Commented Dec 28, 2014 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 $\beta\to\infty$.... $\endgroup$ Commented Dec 29, 2014 at 7:50

1 Answer 1


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.


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.