Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $ 0 < r < 1$, fix $x > 1$ and consider the integral

$$ I_{r}(x) = \int_{1}^{\infty} \exp\left( - \frac{x^2}{2y^{2r}} - \frac{y^2}{2}\right) \frac{dy}{y^r}.$$

In the investigation of iterating two Brownian motions the integral above arises in some form or another in the paper titled "Iterated Random Walk" L. Turban 2004 Europhys. Lett. 65 627. or a more recent paper "Fractional diffusion equations and processes with randomly varying time" Enzo Orsingher, Luisa Beghin

The result we have been trying to prove below can be thought of as an analysis of localizing around the maximum of the function $f(y) = \frac{x^2}{2y^{2r}} + \frac{y^2}{2}$ in the exponential of the integrand and then applying appropriate estimates along with integration by parts. I have always thought of this as a localization similar to what is used in Laplace's method but my advisor has explained to me that from a probabilistic perspective we are just applying a standard idea from large deviations and with that in mind much of our analysis has assumed that we are taking $x \gg 1$ much larger than one.

Question 1: Suppose that $p(x),q(x)>0$ are polynomials, $k(r)$ a constant depending on $r$ and that $c(r) = \frac{(r^{-\frac{r}{r+1}} +r^\frac{1}{r+1}) }{2}$ then is it true that $I_r (x) \leq k(r) \frac{p(x)}{q(x)} \exp(-c(r) x^{\frac{2}{1+r}}) $

We have a proof of this result under more strict hypothesis that is rather boring (in the sense it only uses calculus) and involves essentially integration by parts applied to the integral rewritten as $I_r(x) = \int_{1}^{\infty} \frac{y^{-r}}{f'(y)} f'(y) \exp(-f(y)) dy$. The draw back of our proof is it is not uniform in the sense that it only holds for all $x > x_0$ for some $x_0(r) \gg 1$ some constant much greater than $1$. We would like to relax this assumption and try to find a proof that hold for any $x>1$.

share|cite|improve this question
Do you have some specific $k(r)$, or are you asking if one exists such that the statement in question 1 is true? Also, thanks for providing the motivation. – Alex Becker Jan 4 '12 at 6:33
@AlexBecker We do not have a specific $k(r)$ we are asking if one exits such that the statement in question 1 is true but the important thing for us is the identification of the constant $c(r)$ in exponential which gives us our large deviation statement. At the end of the day we will be ignoring everything of polynomial growth or lower so $k(r)$ will not be necessary to identify precisely, only to show its existence. – user7980 Jan 4 '12 at 6:40
What is $H$? Did I miss it somewhere in the exposition? – cardinal Jan 14 '12 at 18:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.