When the a contrario approach becomes generative

Abstract

The a contrario approach is a statistical, hypothesis testing based approach to detect geometric meaningful events in images. The general methodology consists in computing the probability of an observed geometric event under a noise model (null hypothesis) $H_0$ and then declare the event meaningful when this probability is small enough. Generally, the noise model is taken to be the independent uniform distribution on the considered elements. Our aim in this paper will be to question the choice of the noise model: What happens if we “enrich” the noise model? How to characterize the noise models such that there are no meaningful events against them? Among them, what is the one that has maximum entropy? What does a sample of it look like? How is this noise model related to probability distributions on the elements that would produce, with high probability, the same detections? All these questions will be formalized and answered in two different frameworks: the detection of clusters in a set of points and the detection of line segments in an image. The general idea is to capture the perceptual information contained in an image, and then generate new images having the same visual content. We believe that such a generative approach can have applications for instance in image compression or for clutter removal.

Appendix 1

Proof of Lemma 3

Proof

It is based on the fact that we consider here a symmetric function $H_p$ (i.e. symmetric in the sense that it is invariant under any permutation of $q_1,\ldots ,q_n$) under a constraint that is also symmetric. We first notice that a point of local maximum of $H_p(q_1,\ldots ,q_n)=-\sum _{i=1}^n q_i \log \frac{q_i}{p} + (1-q_i) \log \frac{1-q_i}{1-p}$ under the constraint $C(q_1,\ldots ,q_n)\ge \eta$ (where $\eta \in (0,1/2]$ and $C(q_1,\ldots ,q_n)=\mathbb {P}[K\ge k_0]$, with $K$ following a Poisson binomial distribution of parameters $q_1,\ldots ,q_n$) is necessarily achieved when $C(q_1,\ldots ,q_n)=\eta$. Indeed if it is not the case then we will have a point $(q_1,\ldots ,q_n)$ of local maximum of $H_p$ with $C(q_1,\ldots ,q_n)>\eta$. Now, at least one the $q_i$ is not equal to $p$ (because $B(n,k_0,p)<\eta$ by hypothesis) and therefore we can slightly modify it, still satisfying the constraint $C(q_1,\ldots ,q_n)\ge \eta$ and increasing $H_p$.

Let us now prove that $(\overline{q},\ldots ,\overline{q})$ with $\overline{q}:=B_{n,k_0}^{-1}(\eta )$ is a point of local maximum of $H_p$ under the constraint $C(q_1,\ldots ,q_n)=\eta$. We consider smooth (at least $C^2$) curves $t\mapsto q_i(t)$ defined for $t$ real in a neighbourhood $I$ of $0$, and such that $\forall t\in I$, $C(q_1(t),\ldots ,q_n(t))=\eta$ and $q_1(0)=\ldots = q_n(0)=\overline{q}$. Then we define for all $t\in I$, $h(t) = H_p(q_1(t),\ldots ,q_n(t))$ and we will compute $h'(0)$ and $h''(0)$. A simple computation leads to

$$\begin{aligned} h'(0) = - \left( \log \frac{\overline{q}}{1-\overline{q}} - \log \frac{p}{1-p}\right) \sum _{i=1}^n q'_i(0) \end{aligned}$$

(13)

and

$$\begin{aligned} h''(0)= & {} - \left( \log \frac{\overline{q}}{1-\overline{q}} - \log \frac{p}{1-p}\right) \sum _{i=1}^n q''_i(0) \nonumber \\&- \frac{1}{\overline{q}(1-\overline{q})} \sum _{i=1}^n q'_i(0)^2 . \end{aligned}$$

(14)

Now, since $C(q_1(t),\ldots ,q_n(t))=\eta$ for all $t$, and since $C$ is symmetric, we get

$$\begin{aligned} \frac{\partial C}{\partial q_1}(\overline{q},\ldots ,\overline{q}) \sum _{i=1}^n q'_i(0) = 0 \end{aligned}$$

(15)

and

$$\begin{aligned}&\frac{\partial C}{\partial q_1}(\overline{q},\ldots ,\overline{q}) \sum _{i=1}^n q''_i(0)\nonumber \\&\quad + \frac{\partial ^2 C}{\partial q_1 \partial q_2}(\overline{q},\ldots ,\overline{q}) \sum _{i,j=1, i\ne j}^n q'_i(0) q'_j(0) \nonumber \\&\quad + \frac{\partial ^2 C}{\partial q_1^2}(\overline{q},\ldots ,\overline{q}) \sum _{i=1}^n q'_i(0)^2 = 0 . \end{aligned}$$

(16)

To compute the partial derivatives of $C$, we go back to its definition. Indeed $C$ is defined by: $C(q_1,\ldots ,q_n)=\mathbb {P}[Y_1+\ldots +Y_n\ge k_0]$ where the $Y_i$ are independent Bernoulli random variables with respective parameter $q_i$. We can then develop the probability term and rewrite $C$ as

$$\begin{aligned} C(q_1,\ldots ,q_2)= & {} q_1 ( \mathbb {P}[Y_2+\ldots +Y_n\ge k_0-1] \\&- \mathbb {P}[Y_2+\ldots +Y_n\ge k_0])\\&+ \mathbb {P}[Y_2+\ldots +Y_n\ge k_0] \end{aligned}$$

and also

$$\begin{aligned} C(q_1,\ldots ,q_2)= & {} q_1 q_2 ( P_2 - 2 P_1 + P_0 ) \\&+ (q_1+q_2) (P_1 - P_0) + P_0 , \end{aligned}$$

where $P_2:=\mathbb {P}[Y_3+\ldots +Y_n\ge k_0-2]$ ; $P_1:=\mathbb {P}[Y_3+\ldots +Y_n\ge k_0-1]$ and $P_0:=\mathbb {P}[Y_3+\ldots +Y_n\ge k_0]$. These functions depend only on $q_3, \ldots , q_n$. This allows us to easily compute the partial derivatives of $C$ at the point $(\overline{q},\ldots ,\overline{q})$ and get

$$\begin{aligned}&\frac{\partial C}{\partial q_1}(\overline{q},\ldots ,\overline{q}) = \frac{(n-1)!}{(k_0-1)! (n-k_0)!} \overline{q}^{k_0-1} (1-\overline{q})^{n-k_0} \, \, ,\\&\frac{\partial ^2 C}{\partial q_1^2} = 0 \end{aligned}$$

and

$$\begin{aligned} \frac{\partial ^2 C}{\partial q_1 \partial q_2}= & {} \frac{(n-2)!}{(k_0-1)! (n-k_0)!} \overline{q}^{k_0-2} (1-\overline{q})^{n-k_0-1} (k_0\\&-1 - (n-1)\overline{q} ) . \end{aligned}$$

Then from (15), we deduce that

$$\begin{aligned} \sum _{i=1}^n q'_i(0) = 0 , \end{aligned}$$

and as a first consequence we thus have, from (13), that $h'(0)=0$. Then using (16), the values of the partial derivatives of $C$ and the fact that $\sum _{j\ne i} q'_j(0) = -q'_i(0)$, we get that (14) gives:

$$\begin{aligned} h''(0)= & {} - \frac{\overline{q}}{1-\overline{q}} \left( 1 + (\frac{k_0-1}{n-1} -\overline{q}) (\log \frac{\overline{q}}{1-\overline{q}}\right. \\&\left. - \log \frac{p}{1-p})\right) \sum _{i=1}^n q'_i(0)^2 . \end{aligned}$$

By definition of $\overline{q}$, we have $B(n,k_0,\overline{q})=\eta \le \frac{1}{2}$ (by hypothesis). Then by Lemma 1, and more precisely by Equation (7) in its proof, we get

$$\begin{aligned}&1 + \left( \frac{k_0-1}{n-1} -\overline{q}\right) \left( \log \frac{\overline{q}}{1-\overline{q}} - \log \frac{p}{1-p}\right) \\&\quad \ge 1 + \frac{1}{n+1} \log \frac{p}{1-p} > 0 \end{aligned}$$

by hypothesis. Finally, we have obtained that

$$\begin{aligned} h'(0)=0 \text { and } h''(0) < 0 , \end{aligned}$$

showing that the point $(\overline{q},\ldots ,\overline{q})$ is a local maximum of $H_p$ under the constraint $C\ge \eta$. $\square$