Let $W_t$ be a one-dimensional Brownian motion and I would like to prove

$$\lim_{\beta\rightarrow+\infty}\sup_{0\leq t\leq T}\left|e^{-\beta t} \int_0^te^{\beta s}\mathrm dW_s\right|=0$$

This is an exercise in Chapter 3 of Karatzas&Shreve, but I don't think this proposition is true since

$$\sup_{0\leq t\leq T}\left|e^{-\beta t} \int_0^te^{\beta s}\mathrm dW_s\right|\geq \left|e^{-\beta T} \int_0^Te^{\beta s}\mathrm dW_s\right|:=|G|$$

where $G$ is a centered Gaussian random variable of variance $(1-e^{-2\beta T})/2\beta$, which implies the precedent proposition can not be true. Could someone tell me where I am wrong or this proposition is not true? Thanks a lot

  • 1
    $\begingroup$ Why do you think that what you've stated implies the result does not hold? (The variance of your $G$ goes to zero as $\beta$ increases.) $\endgroup$ – cardinal Jul 16 '13 at 21:59
  • 1
    $\begingroup$ This question has been answered on MO: mathoverflow.net/questions/120438/limit-of-a-wiener-integral $\endgroup$ – Ben Derrett Jul 17 '13 at 8:19
  • $\begingroup$ Very nice solution, thanks so much $\endgroup$ – Higgs88 Jul 17 '13 at 18:39

It has been answered on Math Overflow. The key was an integration by parts argument.


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.