Inconsistency of Template Estimation with the Fréchet Mean in Quotient Space

We tackle the problem of template estimation when data have been randomly transformed under an isometric group action in the presence of noise. In order to estimate the template, one often minimizes the variance when the influence of the transformations have been removed (computation of the Fréchet mean in quotient space). The consistency bias is defined as the distance (possibly zero) between the orbit of the template and the orbit of one element which minimizes the variance. In this article we establish an asymptotic behavior of the consistency bias with respect to the noise level. This behavior is linear with respect to the noise level. As a result the inconsistency is unavoidable as soon as the noise is large enough. In practice, the template estimation with a finite sample is often done with an algorithm called max-max. We show the convergence of this algorithm to an empirical Karcher mean. Finally, our numerical experiments show that the bias observed in practice cannot be attributed to the small sample size or to a convergence problem but is indeed due to the previously studied inconsistency.

1. 1.

   Note that in this article, \(g\cdot x\) is the result of the action of g on x, and \(\cdot \) should not to be confused with the multiplication of real numbers noted \(\times \).

2. 2.

   \(d_Q\) is called a pseudometric because \(d_Q([x],[y])\) can be equal to zero even if \([x]\ne [y]\). If the orbits are closed sets then \(d_Q\) is a distance.

3. 3.

   The code used in this Section is available at http://loic.devilliers.free.fr/ipmi.html.

4. 4.

   Indeed we know that \(x\in \mathbb {R}^+\mapsto x^2-2bx+c\) reaches its minimum at the point \(x=b^+\) and \(f(b^+)=c-(b^+)^2\).

5. 5.

   Note that we remove the positive part and the square because \(\text {argmax}\, h=\text {argmax}\, (h^+)^2\) since h takes a non negative value (indeed \(h(v)\ge \mathbb {E}(\left\langle v,\phi \cdot t_0+\epsilon \right\rangle )=\left\langle v,\mathbb {E}(\phi \cdot t_0)\right\rangle \) and this last quantity is non negative for at least one \(v\in S\)).

Authors and Affiliations

1. Université Côte d’Azur, Inria, France

   Loïc Devilliers & Xavier Pennec

2. Université Paris Descartes, INSERM UMRS 1138, Centre de Recherche des Cordeliers, Paris, France

   Stéphanie Allassonnière

Corresponding author

Correspondence to Loïc Devilliers .

Editors and Affiliations

1. University of North Carolina, Chapel Hill, North Carolina, USA

   Marc Niethammer

2. University of North Carolina, Chapel Hill, North Carolina, USA

   Martin Styner

3. Kitware Inc., Carrboro, North Carolina, USA

   Stephen Aylward

4. University of North Carolina, Chapel Hill, North Carolina, USA

   Hongtu Zhu

5. University of Pennsylvania, Philadelphia, Pennsylvania, USA

   Ipek Oguz

6. University of North Carolina, Chapel Hill, North Carolina, USA

   Pew-Thian Yap

7. University of North Carolina, Chapel Hill, North Carolina, USA

   Dinggang Shen

A Proof of Theorem 1

Proof

In the proof, we note by S the unit sphere in H. In order to prove that \(K>0\), we take x in the support of \(\epsilon \) such that x is not a fixed point under the action of G. It exists \(g_0\in G\) such that \(g_0\cdot x\ne x\). We note \(v_0=\frac{g_0\cdot x}{\Vert x\Vert }\in S\), we have \(\left\langle v_0,g_0\cdot x\right\rangle =\Vert x\Vert >\left\langle v_0,x\right\rangle \) and by continuity of the dot product it exists \(r>0\) such that: \( \forall y\in B(x,r)\quad \left\langle v_0,g_0\cdot y\right\rangle >\left\langle v_0,y\right\rangle \) as x is in the support of \(\epsilon \) we have \(\mathbb {P}(\epsilon \in B(x,r))>0\), it follows:

Thanks to Inequality (7) and the fact that \(\sup _{g\in G} \left\langle v_0,g\cdot \epsilon \right\rangle \ge \left\langle v_0,\epsilon \right\rangle \) we have:

Using the Cauchy-Schwarz inequality: \(K\le \sup _{v\in S} \mathbb {E}(\Vert v\Vert \times \Vert \epsilon \Vert )\le \mathbb {E}(\Vert \epsilon \Vert ^2)^{\frac{1}{2}}=1\). We now prove Inequalities (3). The variance at \(\lambda v\) for \(v\in S\) and \(\lambda \ge 0\) is:

Indeed \(\Vert g\cdot Y\Vert =\Vert Y\Vert \) thanks to the isometric action. We note \(x^+=\max (x,0)\) the positive part of x and \(h(v)=\mathbb {E}(\sup _{g\in G}\left\langle v,g \cdot Y\right\rangle )\). The \(\lambda \ge 0\) which^{Footnote 4} minimizes (8) is \(h(v)^+ \) and the minimum value of the variance restricted to the half line \(\mathbb {R}^+v\) is \(F(h(v)^+ v)=\mathbb {E}(\Vert Y\Vert ^2)- (h(v)^+)^2\). To find \([m_\star ]\) the Fréchet mean of [Y], we need to maximize \((h(v)^+)^2\) with respect to \(v\in S\): \(m_\star =h(v_\star )v_\star \) with^{Footnote 5} \(v_\star \in \text {argmax}_{v\in S} \,h(v)\). As we said in the sketch of the proof we are interested in getting a piece of information about the norm of \(\Vert m_\star \Vert \) we have: \( \Vert m_\star \Vert =h(v_\star )=\sup _{v\in S} h. \) Let \(v\in S\), we have: \(-\Vert t_0\Vert \le \left\langle v,g\phi \cdot t_0\right\rangle \le \Vert t_0\Vert \) because the action is isometric. Now we decompose \(Y=\phi \cdot t_0+\sigma \epsilon \) and we get:

By taking the biggest value in these inequalities with respect to \(v\in S\), by definition of K we get:

Thanks to (9) and to (5), Inequalities (3) are proved.   \(\square \)

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