Inconsistency of Template Estimation with the Fréchet Mean in Quotient Space

Abstract

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.

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:

$$\begin{aligned} \mathbb {P}\left( \underset{g\in G}{\sup } \left\langle v_0,g\cdot \epsilon \right\rangle>\left\langle v_0,\epsilon \right\rangle \right) >0. \end{aligned}$$

(7)

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:

$$\begin{aligned} K=\underset{v\in S}{\sup }\, \mathbb {E}\left( \underset{g\in G}{\sup } \left\langle v,g \cdot \epsilon \right\rangle \right) \ge \mathbb {E}\left( \underset{g\in G}{\sup } \left\langle v_0,g \cdot \epsilon \right\rangle \right) >\mathbb {E}(\left\langle v_0,\epsilon \right\rangle )=\left\langle v_0,\mathbb {E}(\epsilon )\right\rangle =0. \end{aligned}$$

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:

$$\begin{aligned} F(\lambda v)=\mathbb {E}\left( \underset{g\in G}{\inf } \Vert \lambda v-g \cdot Y\Vert ^2\right) =\lambda ^2-2\lambda \mathbb {E}\left( \underset{g\in G}{\sup } \left\langle v,g \cdot Y\right\rangle \right) +\mathbb {E}(\Vert Y\Vert ^2). \end{aligned}$$

(8)

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 $\lambda \ge 0$ which⁴ 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 $v\in S$: $m_\star =h(v_\star )v_\star$ with⁵ $v_\star \in \text {argmax}_{v\in S} \,h(v)$. 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:

$$\begin{aligned} h(v)= & {} \mathbb {E}\left( \underset{g\in G}{\sup } \left\langle v,g\cdot Y\right\rangle \right) = \mathbb {E}\left( \underset{g\in G}{\sup } \left( \left\langle v,g\cdot \sigma \epsilon \right\rangle +\left\langle v,g\phi \cdot t_0 \right\rangle \right) \right) \\ h(v)\le & {} \mathbb {E}\left( \underset{g\in G}{\sup } \left( \left\langle v,g \cdot \sigma \epsilon \right\rangle +\Vert t_0\Vert \right) \right) = \sigma \mathbb {E}\left( \underset{g\in G}{\sup } \left\langle v,g \cdot \epsilon \right\rangle \right) +\Vert t_0\Vert \\ h(v)\ge & {} \mathbb {E}\left( \underset{g\in G}{\sup } \left( \left\langle v,g\cdot \sigma \epsilon \right\rangle \right) -\Vert t_0\Vert \right) = \sigma \mathbb {E}\left( \underset{g\in G}{\sup } \left\langle v,g \cdot \epsilon \right\rangle \right) -\Vert t_0\Vert . \end{aligned}$$

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

$$\begin{aligned} -\Vert t_0\Vert +\sigma K \le \Vert m_\star \Vert \le \Vert t_0\Vert +\sigma K. \end{aligned}$$

(9)

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