Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

My question is: How do I find sharp upper bounds on $P(|q|\leq \epsilon)$ uniformly over a set of gaussian polynomes $q$ of degree two.

Notations and definitions (to make the question rigorous)

  • Let me define $\mathcal{X}_{2}^*$ as the set of real random variables $q$ that can be written $$q=c+\sum_{i\geq 1}\beta_i (\xi_i^2-1)+\alpha_i\xi_i$$ with $c\in \mathbb{R}$, $\beta=(\beta_i)_i\in l_2(\mathbb{N})$, $\alpha=(\alpha_i)_i\in l^2(\mathbb{N})$ and $(\xi_i)_{i\in \mathbb{N}}$ a sequence of iid gaussian random variables with mean $0$ and variance $1$. This set is also known as gaussian polynomial of degree two (see for example the book of Bogachev 1998).

  • Let $q\in \mathcal{X}_2^*$ given by $q=c+\sum_{i\geq 0}\alpha_i\xi_i+\sum_{i}\beta_i(\xi_i^2-1)$, I use the notation $$n_2(q)=\max_i |\beta_i|, \;\; \sigma(q)=\left (\sum_{i\geq 0}2\beta_i^2+\alpha_i^2\right )^{1/2}$$.

  • I propose to use the following sets of polynomes: $$ \Gamma_{2}(c)= ( q\in \mathcal{X}_{2}^*\;:\sigma(q)\geq c),\; \Gamma_{\infty}(c)=(q\in \mathcal{X}_{2}^*\;:n_2(q)\geq c) $$ and $$ \Gamma_{1}(c)=(q\in \mathcal{X}_{2}^*\;:|\mathbb{E}(q)|\geq c). $$


Motivation for this problem

It is worth noticing that this problem appears when one wants to check the so called Noise condition in an infinite dimensional gaussian classification problem.... Anyway, I called it small crown, even if it is not always a crown ... should be easy for expert of "small ball probabilities" ?


What I have so far

  1. There exists $C(c_0)>0$ such that $\forall \epsilon>0\; \sup_{q\in \Gamma_1(c_0)} P(|q|\leq \epsilon)\leq C(c_0)\epsilon^{2/7}$
  2. There exists $C'(c_0)>0$ such that $\forall \epsilon>0$ $\sup_{q\in \Gamma_2(c_0)} P(|q|\leq \epsilon)\leq C'(c_0)\epsilon^{1/3}$.
  3. Let $q\in \mathcal{X}_{2}^*$, for all $\epsilon> 0$, $P(|q|\leq \epsilon) \leq \sqrt{\frac{1}{\pi}\frac{\epsilon}{n_2(q)}}$.

Comments Point 3 is easy but point 1 and 2 are less easy. I can provide a link to the proof if desired. If $n_2(q)=\max_{i}|\beta_i|>c_0$, bound of point $3$ is optimal in the sense that if $\beta=(1,0,\dots)$, $c=1$ and $\alpha=0$ we get $P(|q|\leq \epsilon)=P(|\xi^2|\leq \epsilon)\sim C\epsilon^{1/2}$ (for a constant $C$ that can be calculated explicitly). I have problem for case 1 and 2 ....

My analysis and conjecture: When $\|\beta\|2 \rightarrow 0$ ($l^2$ norm) the behaviour of $P(|q|\leq \epsilon)$ tends to be the same behaviour that $P(|\|\alpha\|_{l^2} \mathcal{N}(0,1)-c|\leq \epsilon)\sim C'(c_0)\epsilon$. Also, it is possible to conjecture that point $1$ and $2$ of the Theorem can be improved (in order to obtain exponent $1/2$ instead of $2/7$ and $1/3$). The difficult cases to study are those with $\|\beta\|_{\infty}\rightarrow 0$ but $\|\beta\|2$ doesn't tends to zero (there, in the proof of points 1 and 2 I use a gaussian approximation of q). Notice that when you restrict yourself to $\beta=0$ the answer to point 2 is that the best exponent is 1! hence a gap between linear and quadratic form...

Some of the ideas I have tryed (without success :( )

  1. Using an explicit formulae of the density (if the density of $q$ is uniformly in $L^p$ for a good $p$ then we are done.. ) using characteristic funtions.

  2. Using optimal Young inequality (we have an infinite number of convolutions to build q)

share|improve this question

2 Answers 2

Here is a solution for problem 2, with power $1/2$, using your idea 1. First some computations. Let A, B real numbers, let z ~ N(0,1), and X=$B(z^2-1) + Az$. The Fourier transform of the distribution of X (i.e.: the characteristic function of X with some $\pi$) is

$\; \; \; \;\; \; \; \;E(exp(-2\pi i\xi X) = \frac{e^{2\pi i \xi B - \frac{2\pi^2 A^2 \xi^2}{1+4\pi i B \xi}}}{\sqrt{1+4\pi i\xi B}}$

where the square root is the one with positive real part. Then

$\; \; \; \;\; \; \; |E(exp(-2\pi i\xi X)|^2 = \frac{e^{- \frac{4\pi^2 A^2 \xi^2}{1+16\pi^2 B^2 \xi^2} }} {1+16\pi^2 \xi^2 B^2}$

We also have for any pair of real numbers $a, b\ge 0$

$\; \; \; \;\; \; \; \ln(\frac{1+8b + 4a}{1+16b}) < \ln(1+\frac{4a}{1+16b})\le \frac{4a}{1+16b}$

In particular,

$\; \; \; \;\; \; \; -\frac{4a}{1+16b} - \ln(1+16b)<-\ln(1+4(2b + a))$

thus, with $a=\pi^2\xi^2 A^2$ and $b=\pi^2\xi^2 B^2$

$\; \; \; \;\; \; \; |E(exp(-2\pi i\xi X)|^2 \le \frac{1} {1+4\pi^2 \xi^2 (2B^2+A^2)}.$

It follows that the Fourier transform of a Gaussian polynomial of degree two $q$ satisfies, with notation as in your definition:

$\; \; \; \;\; \; \; |E(exp(-2\pi i\xi q)|^2 \le \prod_{j}{\frac{1} {1+4\pi^2 \xi^2 (2\beta_j^2+\alpha_j^2)}}\le\frac{1}{1+4\pi^2\xi^2\sigma(q)^2}.$

Then, the density of $q$ is bounded in $L^2$ , uniformly in $\Gamma_2(c_0)$, and so there exists a constant $C'$ such that for all $\epsilon>0$

$\; \; \; \;\; \; \;\sup_{q\in \Gamma_2(c_0)} P(|q|\leq \epsilon)\leq C'\epsilon^{1/2}.$

share|improve this answer
    
Thanks Victor, don't I much time to redo the calculation myself but it seems you have found the good way to handle the majoration of the fourrier transform ! cool :) –  robin girard Jul 4 '11 at 21:09

Here is a contribution to problem 1, with power arbitrary close to $1/2$. First, we may as well assume $c_0 = 1$, and $0<\epsilon<.5$, since the general case reduces to this case. Then, for a quadratic gaussian polynomial $q = 1 + q_0$ with $E(q_0)=0$,

$\;\;\;\;\;\;P(|q| \le\epsilon) \le P( |q_0|\ge(1-\epsilon))\le \frac{E(|q_0|^p)}{(1-\epsilon)^p}\le$ $ C_1(p)\sigma(q_0)^{p}\le C_2\epsilon^\lambda$

provided $\;\;\\sigma(q_0)^{p}\le\epsilon^\lambda$. But, from the solution to 2,

$\;\;\;\;\;\;P(|q| \le\epsilon) \le C_3\(\frac{\epsilon}{\sigma(q_0)})^{1/2}\le C_3 \epsilon^{\frac{1}{2}-\frac{\lambda}{p}}$

provided $\;\;\sigma(q_0)^{p}>\epsilon^\lambda$. Taking $\lambda=\frac{1}{2+\frac{2}{p}}$ both cases are bound by the same power of $\epsilon$, and we get

$\;\;\;\;\;\;P(|q| \le\epsilon) \le C_4 \epsilon^{\frac{1}{2+(2/p)}}$

whether $\sigma(q_0)$ is large or small.

share|improve this answer
    
Using Chevyshev's inequality with power p is too blunt. Using large deviation ideas one can get much better bounds: as above, with $c_0=1$ and $0<\epsilon<.5$ one can show $P(|q|\le \epsilon)\le e^{-\frac{1}{16\sigma(q)^2}}$ Thus, from problem 2: $P(|q|\le \epsilon)\le $$C_5 \min ( \frac{\epsilon}{2\sigma(q)})^{1/2}, e^{-\frac{1}{16\sigma(q)^2}} ) \le C_6 \frac{\epsilon^{1/2}}{G^{-1}(\epsilon^2)}$ where $G(t)=t^4e^{-\frac{1}{t^4}}$. $\;G^{-1}$ approaches 0 very slowly, so with a little algebra one can get: $P(|q|\le \epsilon)\le C_7 (\epsilon |ln(\epsilon)|)^{1/2} .$ –  Victor Jul 1 '11 at 20:18

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.