Sampling Minimal Subsets with Large Spans for Robust Estimation

Abstract

When sampling minimal subsets for robust parameter estimation, it is commonly known that obtaining an all-inlier minimal subset is not sufficient; the points therein should also have a large spatial extent. This paper investigates a theoretical basis behind this principle, based on a little known result which expresses the least squares regression as a weighted linear combination of all possible minimal subset estimates. It turns out that the weight of a minimal subset estimate is directly related to the span of the associated points. We then derive an analogous result for total least squares which, unlike ordinary least squares, corrects for errors in both dependent and independent variables. We establish the relevance of our result to computer vision by relating total least squares to geometric estimation techniques. As practical contributions, we elaborate why naive distance-based sampling fails as a strategy to maximise the span of all-inlier minimal subsets produced. In addition we propose a novel method which, unlike previous methods, can consciously target all-inlier minimal subsets with large spans.

Appendices

Appendix 1: Proof of Proposition 2 for TLS with non-minimal subsets

From Sect. 2.1, the weight of a non-minimal subset $\nu$ is proportional to $|\mathbf{X }(\nu )^T \mathbf{X }(\nu )|$ which, from (36), is equal to

$$\begin{aligned} |\mathbf{X }(\nu )^T \mathbf{X }(\nu )|&= |\mathbf{V }\mathbf{S }^T_m\mathbf{U }_m(\nu )^T\mathbf{U }_m(\nu ) \mathbf{S }_m \mathbf{V }^T |\end{aligned}$$

(65)

$$\begin{aligned}&= |\mathbf{U }_m(\nu )^T \mathbf{U }_m(\nu )| |\mathbf{S }_m \mathbf{V }^T|^2. \end{aligned}$$

(66)

Similarly,

$$\begin{aligned} |\mathbf{Z }(\nu )^T \mathbf{Z }(\nu )| = |\mathbf{U }_m(\nu )^T \mathbf{U }_m(\nu )| |\tilde{\mathbf{S }}_m \mathbf{V }^T|^2. \end{aligned}$$

(67)

Therefore, $|\mathbf{X }(\nu )^T \mathbf{X }(\nu )| = \alpha |\mathbf{Z }(\nu )^T \mathbf{Z }(\nu )|$ where $\alpha$ is a constant which does not depend on $\nu$. This proves that

$$\begin{aligned}&|\mathbf{X }(\nu _1)^T \mathbf{X }(\nu _1)| > |\mathbf{X }(\nu _2)^T \mathbf{X }(\nu _2)|\end{aligned}$$

(68)

$$\begin{aligned}&\implies |\mathbf{Z }(\nu _1)^T \mathbf{Z }(\nu _1)| > |\mathbf{Z }(\nu _2)^T \mathbf{Z }(\nu _2)|. \end{aligned}$$

(69)

Appendix 2: Proof of Propositions 1 and 2 for mixed OLS-TLS

We aim to prove that Proposition 1 holds for the mixed OLS-TLS problem (Sect.3.3), i.e., the solution to the OLS problem

$$\begin{aligned} \underset{{\varvec{\beta }}}{\mathrm{arg~min }}~\Vert \mathbf{y }- \hat{\mathbf{y }} \Vert ^2~~~\mathrm{s.t.}~~~\mathbf{Z }{\varvec{\beta }}= \hat{\mathbf{y }} \end{aligned}$$

(70)

coincides with the mixed OLS-TLS estimate

$$\begin{aligned} \breve{{\varvec{\beta }}} = \left( \mathbf{X }^T\mathbf{X }- \sigma _{m_2 + 1}^2 \mathbf L \right) ^{-1} \mathbf{X }^T \mathbf{y }= (\mathbf{X }^T \mathbf{Z })^{-1} \mathbf{X }^T \mathbf{y }\end{aligned}$$

(71)

where

$$\begin{aligned} \mathbf{Z }:= \mathbf{X }- \sigma _{m_2 + 1}^2(\mathbf{X }^T)^\dagger \mathbf L . \end{aligned}$$

(72)

See Sect. 3.3 for the definition of the other symbols involved.

Let $\tilde{{\varvec{\beta }}}$ be the solution to (70). Then

$$\begin{aligned} \tilde{{\varvec{\beta }}} = (\mathbf{Z }^T\mathbf{Z })^{-1} \mathbf{Z }^T \mathbf{y }\end{aligned}$$

(73)

which, following the proof of Proposition 1, can be rearranged to become

$$\begin{aligned} \mathbf{X }^T\mathbf{Z }\tilde{{\varvec{\beta }}}&= \mathbf{X }^T\mathbf{y }+ \sigma _{m+1}^2 \mathbf L ^T (\mathbf{X }^T)^{\dagger T}(\mathbf{Z }\tilde{{\varvec{\beta }}} - \mathbf{y }). \end{aligned}$$

(74)

As shown in the proof of Proposition 1, the column spans of $\mathbf{Z }$ and $(\mathbf{X }^T)^\dagger$ are equal. Since vector $(\mathbf{Z }\tilde{{\varvec{\beta }}} - \mathbf{y })$ is orthogonal to $\mathcal R (\mathbf{Z })$, it is also orthogonal to $\mathcal R ((\mathbf{X }^T)^\dagger )$, thus the second component on the RHS of (74) equates to 0, yielding

$$\begin{aligned} \mathbf{X }^T\mathbf{Z }\tilde{{\varvec{\beta }}} = \mathbf{X }^T\mathbf{y }. \end{aligned}$$

(75)

Comparing (75)–(71) proves $\tilde{{\varvec{\beta }}} = \breve{{\varvec{\beta }}}$, i.e., the mixed OLS-TLS estimate $\breve{{\varvec{\beta }}}$ coincides with the solution of the OLS (70).

To prove that Proposition 2 also holds for mixed OLS-TLS, it is sufficient to show that, given a non-minimal data subset $\nu$ of size $m+i \le n$,

$$\begin{aligned} |\mathbf{X }(\nu )^T \mathbf{X }(\nu )| \propto |\mathbf{Z }(\nu )^T \mathbf{Z }(\nu )|. \end{aligned}$$

(76)

To begin, let $\mathbf{X }= \mathbf{U }\mathbf{S }\mathbf{V }^T$ be the SVD of $\mathbf{X }$. Then $(\mathbf{X }^T)^\dagger = \mathbf{U }\mathbf{S }^{-1}\mathbf{V }^T$ is the SVD of $(\mathbf{X }^T)^\dagger$. Also, since $n > m$

$$\begin{aligned} \mathbf{X }= \mathbf{U }_m \mathbf{S }_m \mathbf{V }^T~~~~(\mathbf{X }^T)^\dagger = \mathbf{U }_m \mathbf{S }^{-1}_m \mathbf{V }^T. \end{aligned}$$

(77)

Then, from (72)

$$\begin{aligned} \mathbf{Z }&= \mathbf{U }_m ( \mathbf{S }_m \mathbf{V }^T - \sigma ^2_{m_2+1} \mathbf{S }_m^{-1}\mathbf{V }^T \mathbf L ) = \mathbf{U }_m \varvec{\varGamma } \end{aligned}$$

(78)

where we define the square matrix

$$\begin{aligned} \varvec{\varGamma } := ( \mathbf{S }_m \mathbf{V }^T - \sigma ^2_{m_2+1} \mathbf{S }_m^{-1}\mathbf{V }^T \mathbf L ). \end{aligned}$$

(79)

Also, observe that

$$\begin{aligned} \mathbf{Z }(\nu )&= \mathbf{U }_m(\nu ) \varvec{\varGamma }. \end{aligned}$$

(80)

In (66), the following determinant has been established

$$\begin{aligned} |\mathbf{X }(\nu )^T \mathbf{X }(\nu )|&= |\mathbf{U }_m(\nu )^T \mathbf{U }_m(\nu )| |\mathbf{S }_m \mathbf{V }^T|^2. \end{aligned}$$

(81)

The determinant $|\mathbf{Z }(\nu )^T \mathbf{Z }(\nu )|$ is then

$$\begin{aligned} |\mathbf{Z }(\nu )^T \mathbf{Z }(\nu )|&= | \mathbf{U }_m(\nu )^T \mathbf{U }_m(\nu ) | |\varvec{\varGamma }|^2 \end{aligned}$$

(82)

which implies $|\mathbf{X }(\nu )^T \mathbf{X }(\nu )| \propto |\mathbf{Z }(\nu )^T \mathbf{Z }(\nu )|$. Note that this result also holds for minimal subsets by setting $i = 0$.