# Small crown probabilities (and infinite dimensional margin assumption)

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)

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 $\epsilon>0$

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

• 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 :) 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.

• 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: $P(|q|\le \epsilon)\le C_7 (\epsilon |ln(\epsilon)|)^{1/2} .$ Jul 1 '11 at 20:18