6
$\begingroup$

It seems strange to me that all we can find about Schilder's theorem in the literature is on a finite interval of Brownian path.

If we equip the space of continuous function starting from $0$, defined on $\mathbb{R}_+$ with the topology of uniform convergence on compacts. Can we have the similar large deviation principle? With of course the rate function

$$I(f) = \frac{1}{2}\int_0^{\infty} \dot{f}(t)^2 dt.$$

I don't see any objection so far but I don't have any reference to confirm my guess. If it is wrong can you tell me why?

The same question for the Cameron-Martin formula. If we write $\mu$ the Wiener measure on $C_0([0,\infty))$, and $\mu^h$ the measure of $B+h$ where $B \sim \mu$. Can we have, for $h$ of finite functional value $I(h) < \infty$, that $\mu^h$ is absolutely continuous with respect to $\mu$, of density

$$\frac{d\mu^h}{d\mu} = \exp(\int_0^{\infty} \dot{h}_t dx_t - \frac{1}{2}\int_0^{\infty} \dot{h}_t^2 dt),$$

where $\int_0^{\infty} \dot{h}_t dx_t$ is defined in $L^2(\mu)$ by Wiener Integral.

Does that make any sense?

Thank you in advance.

$\endgroup$

1 Answer 1

4
$\begingroup$

It is not in the uniform topology but with a topology tapered off at infinity it is correct. It is done that way in the book of Deuschel and Stroock on large deviations.

$\endgroup$
1
  • $\begingroup$ Thank you very much for that reference, it is really helpful. Sorry for my ignorance about their book. $\endgroup$
    – yilin wang
    Commented Oct 3, 2015 at 9:33

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.