# A question on Ito integral

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?

• 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 $\beta \to \infty$. – Nate Eldredge Dec 28 '14 at 19:37
• @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. – P.Windridge Dec 28 '14 at 21:16
• Is this a homework problem? – Matthias Ludewig Dec 28 '14 at 22:18
• 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 $\beta\to\infty$.... – ofer zeitouni Dec 29 '14 at 7:50

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.