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?
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$\begingroup$ 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 EldredgeCommented Dec 28, 2014 at 19:37
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$\begingroup$ @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.WindridgeCommented Dec 28, 2014 at 21:16
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4$\begingroup$ Is this a homework problem? $\endgroup$– Matthias LudewigCommented Dec 28, 2014 at 22:18
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$\begingroup$ 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 zeitouniCommented Dec 29, 2014 at 7:50
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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.
