Approximation properties of slice-matching operators

Abstract

Iterative slice-matching procedures are efficient schemes for transferring a source measure to a target measure, especially in high dimensions. These schemes have been successfully used in applications such as color transfer and shape retrieval, and are guaranteed to converge under regularity assumptions. In this paper, we explore approximation properties related to a single step of such iterative schemes by examining an associated slice-matching operator, depending on a source measure, a target measure, and slicing directions. In particular, we demonstrate an invariance property with respect to the source measure, an equivariance property with respect to the target measure, and Lipschitz continuity concerning the slicing directions. We furthermore establish error bounds corresponding to approximating the target measure by one step of the slice-matching scheme and characterize situations in which the slice-matching operator recovers the optimal transport map between two measures. We also investigate connections to affine registration problems with respect to (sliced) Wasserstein distances. These connections can be also be viewed as extensions to the invariance and equivariance properties of the slice-matching operator and illustrate the extent to which slice-matching schemes incorporate affine effects.

Appendices

Appendix A Proofs for Sect. 4

1.1 A.1 Key facts for proof of Remark 4

Proposition 16

Let ${\mathcal {D}}({\mathbb {R}}^n)$ be the set of differentiable vector fields from ${\mathbb {R}}^n$ to ${\mathbb {R}}^n$.

$$\begin{aligned} \Big (\bigcap _{P \in O(n)}{\mathfrak {S}}(P)\Big ) \cap {\mathcal {D}}({\mathbb {R}}^n)= \{x \mapsto a{{x}}+b: a>0 \text { and } b\in {\mathbb {R}}^n \}. \end{aligned}$$

Proof

For the proof, we need to show that a differentiable vector field $S \in \bigcap _{P \in O(n)}{\mathfrak {S}}(P)$ is an isotropic scaling with translation. Choose $P \in O(n)$ and write $S(x) = \sum _{i=1}^n f_i^P(x\cdot \theta _{i})\theta _{i}$ with $P=[\theta _1,\ldots ,\theta _n]$. Note that using the standard basis, we can also write $S(x)=\sum _{i=1}^n g_i(x_i)e_{i}$. Computing the Jacobian of S with respect to the two basis representations, we obtain

$$\begin{aligned} \begin{bmatrix} g_1'(x_1)&{}\\ {} &{}\ddots \\ {} &{} \quad &{}g_n'(x_n) \end{bmatrix} = P \begin{bmatrix} {f_1^P}^{'}({{x}}\cdot {{\theta }}_1)&{} &{}&{}\\ &{} {f_2^P}^{'}({{x}}\cdot {{\theta }}_2)&{}&{}\\ &{}&{} \ddots &{}\\ &{}&{}&{} \ {f_n^P}'({{x}}\cdot {{\theta }}_n) \end{bmatrix}P^t. \end{aligned}$$

Hence the two diagonal matrices above have the same diagonal entries, allowing for a possible reordering of the entries. Without loss of generality, we assume that $g_i^{\prime }(x_i)={f_i^P}^{'}({{x}}\cdot {{\theta }}_i), i= 1,...,n$ by possibly performing a column permutation of P and renaming $f_i^P$’s. Choosing an orthogonal matrix P such that one of its column $\theta _i$ with all entries being non-zero, one can immediately derive that the diagonal entries $g_i^{\prime }(x_i)$’s are the same for any fixed x. In summary,

$$\begin{aligned} g_1^{\prime }(x_1)= \cdots = g_n^{\prime }(x_n) = {f_1^P}^{\prime }({{x}}\cdot {{\theta }}_1) = \cdots = {f_n^P}^{\prime }({{x}}\cdot {{\theta }}_n) = a_{{x}}, \end{aligned}$$

where $a_{{x}}$ is a constant depending on ${{x}}= [x_1, \cdots ,x_n]^t\in {\mathbb {R}}^n$. Since the diagonal element $g_i^{\prime }(x_i)$ only depends on $x_i$, it follows that $a_{{{x}}}$ is a constant independent of ${{x}}$. Hence $S({{x}})=a{{x}}+b$ for some $a>0, b\in {\mathbb {R}}^n$. $\square$

Remark 11

In general, if $T\in \bigcap _{P \in O(n)}{\mathfrak {S}}(P)$ is differentiable on an open set $\Omega \subseteq {{\mathbb {R}}^n}$, the $T|_{\Omega }: \Omega \rightarrow {\mathbb {R}}^n$ is an isotropic scaling with translation. In particular, $\bigcap _{P \in O(n)}{\mathfrak {S}}(P)$ include some piecewise isotropic scalings with translations.

1.2 A.2 Proof of Proposition 10

We need the following proposition to derive the proof of Proposition 10:

Proposition 17

Consider two angles $\theta ,\nu \in S^{n-1}$, and assume that $T_{\sigma ^{\nu }}^{\mu ^{\nu }}$ is L-Lipschitz for all $\nu$, i.e. there exists $L>0$ such that $|T_{\sigma ^{\nu }}^{\mu ^{\nu }}(x)-T_{\sigma ^{\nu }}^{\mu ^{\nu }}(y)| \le L |x-y|$ for $x,y\in {\mathbb {R}}$ and $\nu \in S^{n-1}$, then

$$\begin{aligned} \Vert T_{{\sigma }^{\theta }}^{\mu ^{\theta }}\circ {\mathcal {P}}_{\theta } - T_{{\sigma }^{\nu }}^{\mu ^{\nu }}\circ {\mathcal {P}}_{\nu }\Vert _{\sigma } \le (2L+1)C\Vert \theta - \nu \Vert _2, \end{aligned}$$

where C is the max over the second moments of $\sigma$ resp. $\mu$.

Proof

$$\begin{aligned} \Vert T_{{\sigma }^{\theta }}^{\mu ^{\theta }}\circ {\mathcal {P}}_{\theta } - T_{{\sigma }^{\nu }}^{\mu ^{\nu }}\circ {\mathcal {P}}_{\nu }\Vert _{\sigma } \le \Vert T_{{\sigma }^{\theta }}^{\mu ^{\theta }}\circ {\mathcal {P}}_{\theta } - T_{{\sigma }^{\theta }}^{\mu ^{\nu }}\circ {\mathcal {P}}_{\theta }\Vert _{\sigma } + \Vert T_{{\sigma }^{\theta }}^{\mu ^{\nu }}\circ {\mathcal {P}}_{\theta } - T_{{\sigma }^{\nu }}^{\mu ^{\nu }}\circ {\mathcal {P}}_{\nu }\Vert _{\sigma } = (\diamond ). \end{aligned}$$

We bound these separately.

$$\begin{aligned} \Vert T_{{\sigma }^{\theta }}^{\mu ^{\theta }}\circ {\mathcal {P}}_{\theta } - T_{{\sigma }^{\theta }}^{\mu ^{\nu }}\circ {\mathcal {P}}_{\theta }\Vert _{\sigma }&= \Vert T_{{\sigma }^{\theta }}^{\mu ^{\theta }} - T_{{\sigma }^{\theta }}^{\mu ^{\nu }}\Vert _{\sigma ^{\theta }} = W_2\left( \mu ^{\theta }, \mu ^{\nu }\right) \\&\le \Vert {\mathcal {P}}_{\theta }-{\mathcal {P}}_{\nu }\Vert _{\mu } = \left( \int _{{\mathbb {R}}^n}|{\mathcal {P}}_{\theta }(x)-{\mathcal {P}}_{\nu }(x)|^2\, d\mu (x)\right) ^{1/2}\\&= \left( \int _{{\mathbb {R}}^n}|(\theta - \nu )\cdot x|^2\, d\mu (x)\right) ^{1/2}\\&\le \Vert \theta - \nu \Vert _2 \left( \int _{{\mathbb {R}}^n}\Vert x\Vert ^2\, d\mu (x)\right) ^{1/2}\\&\le C\Vert \theta - \nu \Vert _2, \end{aligned}$$

with C max of the second moments, which is bounded by assumption. Now for the second part, note that on ${\mathbb {R}}$ we have $T_{\sigma ^{\theta }}^{\mu ^{\nu }} = T_{\sigma ^{\nu }}^{\mu ^{\nu }} \circ T_{\sigma ^{\theta }}^{\sigma ^{\nu }}$

$$\begin{aligned}&\Vert T_{{\sigma }^{\theta }}^{\mu ^{\nu }}\circ {\mathcal {P}}_{\theta } - T_{{\sigma }^{\nu }}^{\mu ^{\nu }}\circ {\mathcal {P}}_{\nu }\Vert _{\sigma }\\&= \left( \int _{{\mathbb {R}}^n} |T_{\sigma ^{\nu }}^{\mu ^{\nu }}(T_{\sigma ^{\theta }}^{\sigma ^{\nu }}( {\mathcal {P}}_{\theta }(x))) - T_{{\sigma }^{\nu }}^{\mu ^{\nu }}({\mathcal {P}}_{\nu }(x))|^2\, d\sigma (x)\right) ^{1/2} = (\star ) \end{aligned}$$

Since $T_{\sigma ^{\nu }}^{\mu ^{\nu }}$ is L-Lipschitz, we get

$$\begin{aligned} (\star )&\le L\,\left( \int _{{\mathbb {R}}^n} | T^{\sigma ^{\nu }}_{\sigma ^{\theta }}({\mathcal {P}}_{\theta }(x)) - {\mathcal {P}}_{\nu }(x)|^2\, d\sigma (x)\right) ^{1/2} = L\Vert T^{\sigma ^{\nu }}_{\sigma ^{\theta }}\circ {\mathcal {P}}_{\theta } -{\mathcal {P}}_{\nu } \Vert _{\sigma } \\&\le L\left( \Vert T^{\sigma ^{\nu }}_{\sigma ^{\theta }}\circ {\mathcal {P}}_{\theta } - {\mathcal {P}}_{\theta }\Vert _{\sigma } + \Vert {\mathcal {P}}_{\theta }-{\mathcal {P}}_{\nu }\Vert _{\sigma }\right) \\&\le L \left( \Vert T^{\sigma ^{\nu }}_{\sigma ^{\theta }} - {\text {id}}\Vert _{\sigma ^{\theta }} + C\Vert \theta -\nu \Vert _2 \right) \\&= L\left( W_2(\sigma ^{\theta },\sigma ^{\nu })+C \Vert \theta - \nu \Vert _2 \right) \\&\le L\left( \Vert {\mathcal {P}}_{\theta }-{\mathcal {P}}_{\nu }\Vert _{\sigma }+C \Vert \theta - \nu \Vert _2 \right) \\&\le 2LC\Vert \theta - \nu \Vert _2 \end{aligned}$$

This implies

$$\begin{aligned} (\diamond ) \le (2L+1)C\Vert \theta - \nu \Vert _2. \end{aligned}$$

$\square$

Proof of Proposition 10

Based on (9), we let $T_{\sigma ,\mu ;P} = PD\circ P^t$ where $D(x)=[T_{\sigma ^{\theta _1}}^{\mu ^{\theta _1}}(x_1), T_{\sigma ^{\theta _2}}^{\mu ^{\theta _2}}(x_2), \cdots , T_{\sigma ^{\theta _n}}^{\mu ^{\theta _n}}(x_n)]^t$ for $x \in {\mathbb {R}}^n$ and $P = [\theta _1,\ldots ,\theta _n]$. Similarly, we let and $T_{\sigma ,\mu ;Q} = Q{\widetilde{D}}\circ Q^t$, with $Q = [\nu _1,\ldots ,\nu _n]$. We continue with deriving the bound:

$$\begin{aligned} \Vert T_{\sigma ,\mu ;P}-T_{\sigma ,\mu ;Q}\Vert _{\sigma }&= \Vert PDP^t-Q{\widetilde{D}}Q^t\Vert _{\sigma }\\&\le \Vert PDP^t - P{\widetilde{D}}Q^t\Vert _{\sigma } + \Vert P{\widetilde{D}}Q^t-Q{\widetilde{D}}Q^t\Vert _{\sigma } \\&= (1) + (2). \end{aligned}$$

We bound the two terms seperately. For (1), using Proposition 17, we get

$$\begin{aligned} \Vert PDP^t - P{\widetilde{D}}Q^t\Vert _{\sigma }^2&= \int _{{\mathbb {R}}^n} \Vert D(P^tx) - {\widetilde{D}}(Q^tx)\Vert _2^2\,d\sigma (x) \\&= \sum _{i=1}^n \int _{{\mathbb {R}}^n} |T_{\sigma ^{\theta _i}}^{\mu ^{\theta _i}}((P^tx)_i)-T_{\sigma ^{\nu _i}}^{\mu ^{\nu _i}}((Q^tx)_i)|^2\, d\sigma (x) \\&= \sum _{i=1}^n \Vert T_{\sigma ^{\theta _i}}^{\mu ^{\theta _i}}\circ {\mathcal {P}}_{\theta _i} - T_{\sigma ^{\nu _i}}^{\mu ^{\nu _i}}\circ {\mathcal {P}}_{\nu _i} \Vert _{\sigma }^2 \\&\le ((2L+1)C)^2 \sum _{i=1}^n \Vert \theta _i - \nu _i\Vert _2^2 \\&= ((2L+1)C)^2 \Vert P - Q\Vert _F^2. \end{aligned}$$

For (2) we get

$$\begin{aligned} \Vert P{\widetilde{D}}Q^t-Q{\widetilde{D}}Q^t\Vert _{\sigma }^2&= \int _{{\mathbb {R}}^n} \Vert (P-Q){\widetilde{D}}(Q^tx)\Vert _2^2 \, d\sigma (x) \\&\le \Vert P-Q\Vert _2^2\int _{{\mathbb {R}}^n} \Vert {\widetilde{D}}(Q^tx)\Vert _2^2 \, d\sigma (x) \\&\le \Vert P-Q\Vert _2^2 \, L^2 \int _{{\mathbb {R}}^n} \Vert Q^tx\Vert _2^2\, d\sigma (x) \le \Vert P-Q\Vert _2^2 \, L^2 C^2 \\&\le \Vert P-Q\Vert _F^2 \, L^2 C^2 \end{aligned}$$

Combining (1) and (2) gives the final bound. $\square$

Appendix B Proofs for Sect. 5

Proof of Proposition 13

Let $S^{\sigma ,\mu ,W_2}(x) = a^{W_2} x+ b^{W_2}$ and $S^{\sigma ,{{\mathcal {U}}}(\sigma ,\mu ,P), W_2}(x)= {\widetilde{a}}^{W_2}x+ {\widetilde{b}}^{W_2}$ be the critical functions for the associated minimization problem (22). By Proposition 18 and Corollary 20, we have

$$\begin{aligned} {\widetilde{a}}^{W_2} - a^{W_2}&=\frac{W_2^2(\sigma ,\mu )-\sum _{i=1}^nW^2_2(\sigma ^{\theta _i},\mu ^{\theta _i})}{2(M_2(\sigma )-\Vert E(\sigma )\Vert ^2)},\end{aligned}$$

(27)

$$\begin{aligned} {\widetilde{b}}^{W_2}-b^{W_2}&= -({\widetilde{a}}^{W_2} - a^{W_2})E(\sigma ), \end{aligned}$$

(28)

and the norm bound $\Vert S^{\sigma ,\mu ,W_2}-S^{\sigma ,{{\mathcal {U}}}(\sigma ,\mu ,P), W_2}\Vert _{\sigma }$ in (24) can be obtained via direct computation and the fact that the RHS is non-negative, see Lemma 24. It is left to show that these critical functions are indeed the minimizers by verifying

• 1. ${\widetilde{a}}^{W_2}\ge a^{W_2} >0$, see Lemmas 24, 26, and 27.

• 2. The Hessian associated H(a, b) with both the minimization problems are positive definite by a direct calculation and Lemma 26, where

  $$\begin{aligned} H(a,b)= 2 \begin{bmatrix} M_2(\sigma ) &{} (E(\sigma ))^t\\ E(\sigma ) &{} I_{n-1} \end{bmatrix}. \end{aligned}$$

  Here $I_{n-1}$ denotes the identity matrix of size $(n-1)\times (n-1)$.

The equality concerning the means follows from Corollary 19. $\square$

Proposition 18

Let $S^{\sigma ,\eta ,W_2}$ and $S^{\sigma ,\eta , SW_2}$ correspond to the critical points of the minimization problems in (22) and (30), respectively. Then the corresponding parameters satisfy

$$\begin{aligned} a^{W_2}&= \frac{\frac{1}{2}(M_2(\eta )+M_2(\sigma )-W_2^2(\sigma ,\eta ))-E(\sigma )\cdot E(\eta )}{M_2(\sigma )-\Vert E(\sigma )\Vert ^2},\\ b^{W_2}&= E(\eta )-a^{W_2}E(\sigma ),\\ a^{SW_2}&= \frac{\frac{1}{2}(M_2(\eta )+M_2(\sigma )-nSW_2^2(\sigma ,\eta ))-E(\sigma )\cdot E(\eta )}{M_2(\sigma )-\Vert E(\sigma )\Vert ^2},\\ b^{SW_2}&= E(\eta )-a^{SW_2}E(\sigma ), \end{aligned}$$

where $S^{\sigma ,\eta ,W_2}(x) = a^{W_2} x+ b^{W_2}$ and $S^{\sigma ,\eta ,SW_2}(x) = a^{SW_2} x+ b^{SW_2}$.

Proof

Given $\sigma \in {{\mathcal {W}}}_{2,ac}({\mathbb {R}}^n)$ and $\eta \in {{\mathcal {W}}}_2({\mathbb {R}}^n)$, let $M_2(\sigma )= \int \Vert x\Vert ^2d\sigma (x)$ (similarly define $M_2(\eta )$), $E(\sigma )= \int xd\sigma (x)$ (similarly define $E(\eta )$). For $S(x)=ax+b$, by the changes of variables formula and the fact that $T_{\sigma }^\eta = T_{S_{\sharp }\sigma }^{\eta }\circ S$, we have

$$\begin{aligned} W_2^2(S_{\sharp }\sigma ,\eta )&= \Vert T_{\sigma }^{\eta }-(ax+b)\Vert ^2_{\sigma } = M_2(\eta )+a^2M_2(\sigma )+2ab\cdot E(\sigma )\\ {}&-2a\int T_{\sigma }^{\eta }(x)\cdot x d\sigma (x)-E(\sigma )\cdot E(\eta )+\Vert b\Vert ^2-2E(\eta )\cdot b. \end{aligned}$$

Taking the partial derivatives gives

$$\begin{aligned} \frac{\partial }{\partial a}&= 2a M_2(\sigma )+2b\cdot E(\sigma )-2\int T_{\sigma }^{\eta }(x)\cdot x d\sigma (x), \\ \frac{\partial }{\partial b}&= 2b+2aE(\sigma )-2E(\eta ). \end{aligned}$$

Setting the above equations to zero and with the observation that $\int T_{\sigma }^{\eta }(x)\cdot x d\sigma (x) = \frac{1}{2}(M_2(\eta )+M_2(\sigma )-W_2^2(\sigma ,\eta ))$, we get the the desired formulas for $a^{W_2}$ and $b^{W_2}$. Similarly,

$$\begin{aligned} SW_2^2(S_{\sharp }\sigma ,\eta )&= \int _{S^{n-1}}W_2^2((S_{\sharp }\sigma )^{\theta }, \eta ^{\theta })du(\theta ) \\ {}&= \int _{S^{n-1}}\int _{{\mathbb {R}}} |T_{\sigma ^{\theta }}^{\eta ^\theta }(t)-(at+b\cdot \theta )|^2dt du(\theta )\\ {}&= \frac{1}{n}\Big (M_2(\eta )+a^2M_2(\sigma )+2ab\cdot E(\sigma )\\ {}&-2na\int _{S^{n-1}}\int _{{\mathbb {R}}} tT_{\sigma ^{\theta }}^{\eta ^\theta }(t)d\sigma ^{\theta }(t) du(\theta )-E(\sigma )\cdot E(\eta )+\Vert b\Vert ^2-2E(\eta )\cdot b\Big ). \end{aligned}$$

Taking the partial derivatives gives

$$\begin{aligned} \frac{\partial }{\partial a}&=\frac{1}{n} \Big (2a M_2(\sigma )+2b\cdot E(\sigma )-2n\int _{S^{n-1}}\int _{{\mathbb {R}}} tT_{\sigma ^{\theta }}^{\eta ^\theta }(t)dt du(\theta )\Big ), \\ \frac{\partial }{\partial b}&= \frac{1}{n} \Big (2b+2aE(\sigma )-2E(\eta )\Big ). \end{aligned}$$

Setting the above equations to zero and with the observation that $\int _{S^{n-1}}\int _{{\mathbb {R}}} tT_{\sigma ^{\theta }}^{\eta ^\theta }(t)d\sigma ^{\theta }(t) du(\theta ) = \frac{1}{2n}(M_2(\sigma )+M_2(\eta )-nSW_2^2(\sigma ,\eta ))$, we get the desired formulas for $a^{SW_2}$ and $b^{SW_2}$. We provide computational details in Appendix C. $\square$

Corollary 19

Given the same assumptions as in Proposition 18, for $D=W_2~\text {or}~ SW_2$

$$\begin{aligned} E({S^{\sigma ,\eta ,D}}_\sharp \sigma ) = E(\eta ). \end{aligned}$$

(29)

Proof

Upon direct calculation, we have $E({S^{\sigma ,\eta ,D}}_\sharp \sigma = a^DE(\sigma )+b^D$, where $a^D, b^D$ are as in Proposition 18. The conclusion can be derived from the expressions for $b^{W_2}$ and $b^{SW_2}$. $\square$

Corollary 20

Let $\eta = {{\mathcal {U}}}(\sigma ,\mu ,P)$ in Proposition 18. Then the parameters corresponding to $S^{\sigma ,{{\mathcal {U}}}(\sigma ,\mu ,P), W_2}$ and $S^{\sigma ,{{\mathcal {U}}}(\sigma ,\mu ,P), SW_2}$ satisfy

$$\begin{aligned} {\widetilde{a}}^{W_2}&= \frac{\frac{1}{2}(M_2(\mu )+M_2(\sigma )-\sum _{i=1}^nW_2^2(\sigma ^{\theta _i},\mu ^{\theta _i}))-E(\sigma )\cdot E(\mu )}{M_2(\sigma )-\Vert E(\sigma )\Vert ^2},\\ {\widetilde{b}}^{W_2}&= E(\mu )-{\widetilde{a}}^{W_2}E(\sigma ),\\ {\widetilde{a}}^{SW_2}&= \frac{\frac{1}{2}(M_2(\mu )+M_2(\sigma )-nSW_2^2(\sigma ,{{\mathcal {U}}}(\sigma ,\mu ,P)))-E(\sigma )\cdot E(\mu )}{M_2(\sigma )-\Vert E(\sigma )\Vert ^2},\\ {\widetilde{b}}^{SW_2}&= E(\mu )-a^{SW_2}E(\sigma ), \end{aligned}$$

where $S^{\sigma ,{{\mathcal {U}}}(\sigma ,\mu ,P), W_2}(x)= {\widetilde{a}}^{W_2}x+ {\widetilde{b}}^{W_2}$ and $S^{\sigma ,{{\mathcal {U}}}(\sigma ,\mu ,P), SW_2}(x)= {\widetilde{a}}^{SW_2}x+ {\widetilde{b}}^{SW_2}$.

Proof

The above formulas follows directly from Proposition 18, the fact that ${{\mathcal {U}}}(\sigma ,\mu ,P)$ and $\mu$ have the same mean (see (13)), and the formula (7) for $W_2^2(\sigma , {{\mathcal {U}}}(\sigma ,\mu ,P))$. $\square$

Proposition 21

Let

$$\begin{aligned} {{\mathcal {S}}}(P):= \{x\mapsto P\Lambda P^t x+b: \Lambda \text{ is } \text{ positive } \text{ and } \text{ diagonal }, b\in {\mathbb {R}}^n\}. \end{aligned}$$

Consider the minimization problem

$$\begin{aligned} S_P^{\sigma ,\eta }&:= \mathop {\mathrm {arg\,min}}\limits _{S_P\in {{\mathcal {S}}}(P)} \Vert {S_P}-T_\sigma ^\eta \Vert _{\sigma }. \end{aligned}$$

(30)

Let $S^{\sigma ,\mu }_{P}$ and $S^{\sigma ,{{\mathcal {U}}}(\sigma ,\mu ;P)}_{P}$ be the minimizers of (30) with $\eta =\mu$ and $\eta = {{\mathcal {U}}}(\sigma ,\mu ,P)$, respectively. We denote the diagonal entries of the corresponding $\Lambda$ by $a_i$ and ${\widetilde{a}}_i$, respectively. Similar notation holds for $b_i$ and ${\widetilde{b}}_i$. Then

$$\begin{aligned} {\widetilde{a}}_i - a_i&= \frac{\int |\theta _i\cdot (T_{\sigma }^{\mu }(x)-x)|^2d\sigma (x)- W_2^2(\sigma ^{\theta _i},\mu ^{\theta _i})}{2(M_2^{\sigma ^{\theta _i}}- (E^{\theta _i})^2)}\ge 0,\\ {\widetilde{b}}_i - b_i&= - \sum _{i=1}^n\theta _i E^{\sigma ^{\theta _i}} ( {\widetilde{a}}_i - a_i). \end{aligned}$$

Proof

The proof uses similar arguments in Proposition 18 and Corollary 20 except the partial derivatives are with respect to $a_i$ and $\widetilde{a_i}$ instead of a and ${\widetilde{a}}$. Note that following these arguments, we use the equations presented in Lemma 23. $\square$

Appendix C Other technical details

Lemma 22

Let $\sigma \in {{\mathcal {W}}}_{2,ac}({\mathbb {R}}^n)$ and $\eta , \mu \in {{\mathcal {W}}}_2({\mathbb {R}}^n)$. Then we get

$$\begin{aligned}&E({(T_{\sigma ,\mu ;P})_{\sharp }\sigma })=\int T_{\sigma ,\mu ;P}(x) d\sigma (x) = \int yd\mu (y)=E(\mu ),\\&\int T_{\sigma }^{\eta }(x)\cdot x d\sigma (x) = \frac{1}{2}(M_2(\eta )+M_2(\sigma )-W_2^2(\sigma ,\eta ))\\&M_2({(T_{\sigma ,\mu ;P})_{\sharp }\sigma })=\int \Vert T_{\sigma ,\mu ;P}(x)\Vert ^2 d\sigma (x) = M_2(\mu ) \end{aligned}$$

Proof

By the change of variables formula, we have

$$\begin{aligned} \int T_{\sigma ,\mu ;P}(x) d\sigma (x)&= \sum _{i=1}^n \theta _i\int T_{\sigma ^{\theta _i}}^{\mu ^{\theta _i}}(x\cdot \theta _i)d\sigma (x) = \sum _{i=1}^n \theta _i\int T_{\sigma ^{\theta _i}}^{\mu ^{\theta _i}}(t)d\sigma ^{\theta _i}(t)\\&= \sum _{i=1}^n \theta _i\int zd\mu ^{\theta _i}(z) = \sum _{i=1}^n \theta _i\int y\cdot \theta _i d\mu (y)\\&= \int y d\mu (y), \end{aligned}$$

$$\begin{aligned} \int T_{\sigma }^{\eta }(x)\cdot x d\sigma (x)&= \frac{1}{2}\Big (\int \Vert T_{\sigma }^{\eta }(x)\Vert ^2d\sigma (x)\\ {}&+\int \Vert x\Vert ^2d\sigma (x)- \int \Vert T_{\sigma }^{\eta }(x)-x\Vert ^2d\sigma (x)\Big )\\ {}&=\frac{1}{2}(M_2(\eta )+M_2(\sigma )-W_2^2(\sigma ,\eta )), \end{aligned}$$

$$\begin{aligned} \int \Vert T_{\sigma ,\mu ;P}(x)\Vert ^2 d\sigma (x)&=\int \sum _{i=1}^n |T_{\sigma ^{\theta _i}}^{\mu ^{\theta _i}}(x\cdot \theta )|^2d\sigma (x) = \sum _{i=1}^n \int |T_{\sigma ^{\theta _i}}^{\mu ^{\theta _i}}(t)|^2d\sigma ^{\theta _i}(t)\\&=\sum _{i=1}^n \int |w|^2d\mu ^{\theta _i}(w) = \sum _{i=1}^n\int |y\cdot \theta _i|^2d\mu (y)\\&= \int \Vert y\Vert ^2d\mu (y)= M_2(\mu ), \end{aligned}$$

where the last steps make use of the fact that $P=[\theta _1,\cdots ,\theta _n]$ is an orthogonal matrix. $\square$

Lemma 23

Let $\sigma \in {{\mathcal {W}}}_{2,ac}({\mathbb {R}}^n)$, $\eta \in {{\mathcal {W}}}_2({\mathbb {R}}^n)$, and $b\in {\mathbb {R}}^n$. Then

$$\begin{aligned}&\int _{S^{n-1}}\int _{{\mathbb {R}}} tT_{\sigma ^{\theta }}^{\eta ^\theta }(t)d\sigma ^{\theta }(t) du(\theta ) = \frac{M_2(\sigma )+M_2(\eta )-nSW_2^2(\sigma ,\eta )}{2n}\end{aligned}$$

(31)

$$\begin{aligned}&\int _{S^{n-1}}\int _{{\mathbb {R}}} t^2d\sigma ^{\theta }(t)du(\theta ) = \frac{M_2(\sigma ) }{n}\end{aligned}$$

(32)

$$\begin{aligned}&\int _{S^{n-1}}\int _{{\mathbb {R}}} |T_{\sigma ^{\theta }}^{\eta ^\theta }(t)|^2d\sigma ^{\theta }(t)du(\theta ) = \frac{M_2(\eta )}{n}\end{aligned}$$

(33)

$$\begin{aligned}&\int _{S^{n-1}}\int _{{\mathbb {R}}} (b\cdot \theta ) t d\sigma ^{\theta }(t)du(\theta ) = \frac{E(\sigma )\cdot b}{n}\end{aligned}$$

(34)

$$\begin{aligned}&\int _{S^{n-1}}\int _{{\mathbb {R}}} (b\cdot \theta ) T_{\sigma ^{\theta }}^{\eta ^\theta }(t)td\sigma ^{\theta }(t)du(\theta ) = \frac{E(\eta )\cdot b}{n} \end{aligned}$$

(35)

Proof

We note that (32) and (33) are analogous by the change of variables formula, so are (34) and (35). We will first show (32).

$$\begin{aligned} \int _{S^{n-1}}\int _{{\mathbb {R}}} t^2d\sigma ^{\theta }(t)du(\theta )&= \int _{S^{n-1}}\int _{{\mathbb {R}}^n}|x\cdot \theta |^2 d\sigma (x)du(\theta ) \\&\hspace{-.4cm}\overset{\text {Fubini}}{=} \int _{{\mathbb {R}}^n}\int _{S^{n-1}}|x\cdot \theta |^2 du(\theta )d\sigma (x)\\&= \int _{{\mathbb {R}}^n}\frac{\Vert x\Vert ^2}{2}d\sigma (x)\\&=\frac{M_2(\sigma )}{n}. \end{aligned}$$

For (34), we have

$$\begin{aligned}&\int _{S^{n-1}}\int _{{\mathbb {R}}} (b\cdot \theta ) t d\sigma ^{\theta }(t)du(\theta )\\&\quad = \int _{S^{n-1}}b\cdot \theta \int _{{\mathbb {R}}^n}x\cdot \theta d\sigma (x)du(\theta )\\&\quad = \int _{S^{n-1}}(b\cdot \theta )(E(\sigma )\cdot \theta ) du(\theta )\\&\quad = \int _{S^{n-1}}\frac{1}{2}\Big (|b\cdot \theta )|^2+|E(\sigma )\cdot \theta |^2-|(b-E(\sigma ))\cdot \theta |^2\Big )du(\theta )\\&\quad = \frac{1}{2n}\Big (\Vert b\Vert ^2+\Vert E(\sigma )\Vert ^2- \Vert b-E(\sigma )\Vert ^2\Big )\\&\quad = \frac{E(\sigma )\cdot b}{n}. \end{aligned}$$

With (32) and (33), we have (31):

$$\begin{aligned}&\int _{S^{n-1}}\int _{{\mathbb {R}}} tT_{\sigma ^{\theta }}^{\eta ^\theta }(t)d\sigma ^{\theta }(t) du(\theta ) \\ {}&=\frac{1}{2} \int _{S^{n-1}}\int _{{\mathbb {R}}}\Big (t^2+(T_{\sigma ^{\theta }}^{\eta ^\theta }(t))^2- (t-T_{\sigma ^{\theta }}^{\eta ^\theta }(t))^2\Big ) d\sigma ^{\theta }(t) du(\theta )\\&= \frac{1}{2n}\Big (M_2(\sigma )+M_2(\eta )- n\int _{S^{n-1}}W_2^2(\sigma ^{\theta },\eta ^{\theta })du(\theta )\Big )\\&= \frac{M_2(\sigma )+M_2(\eta )- nSW_2^2(\sigma ,\eta )}{2n}. \end{aligned}$$

$\square$

Lemma 24

Let $\sigma \in {{\mathcal {W}}}_{2,ac}({\mathbb {R}}^n)$ and $\mu \in {{\mathcal {W}}}_2({\mathbb {R}}^n)$ and $P = [\theta _1, \cdots , \theta _n]\in O(n)$. Then

$$\begin{aligned} W_2^2(\sigma ,\mu ) \ge \sum _{i=1}^nW_2^2(\sigma ^{\theta _i},\mu ^{\theta _i}). \end{aligned}$$

Proof

By [19, Proposition 5.1.3],

$$\begin{aligned} W_2^2(\sigma ^{\theta },\mu ^{\theta })\le \int |\theta \cdot x-\theta \cdot y|^2d\gamma ^*(x,y), \end{aligned}$$

where $\gamma ^*$ is the optimal transport plan between $\sigma$ and $\mu$. Then

$$\begin{aligned} \sum _{i=1}^n W_2^2(\sigma ^{\theta _i},\mu ^{\theta _i})&\le \int \sum _{i=1}^n |\theta _i\cdot (x-y)|^2d\gamma ^*(x,y)\\&= \int \Vert x-y\Vert ^2d\gamma ^*(x,y)\\&= W_2^2(\sigma ,\mu ). \end{aligned}$$

$\square$

Lemma 25

Let $h:{\mathbb {R}}^n\rightarrow {\mathbb {R}}^n$ and $\sigma ({\mathbb {R}}^n)=1$. Then

$$\begin{aligned} \int \Vert h(x)\Vert ^2d\sigma (x) \Vert \ge \Vert \int h(x)d\sigma (x)\Vert ^2, \end{aligned}$$

where equality holds if and only if $h(x)=v$ $\sigma$-a.e. for some $v\in {\mathbb {R}}^n$.

Proof

Let $h(x)= [h_1(x), \cdots , h_n(x)]^t$. By Hölder’s inequality,

$$\begin{aligned} \int |h_i(x)|d\sigma (x&)\le \left( \int |h_i(x)|^2d\sigma (x)\right) ^{1/2}\left( \int 1^2 d\sigma (x)\right) ^{1/2} \\&= \left( \int |h_i(x)|^2d\sigma (x)\right) ^{1/2}. \end{aligned}$$

Squaring the above inequality and summing over i gives the desired inequality. Observe that equality holds if and only if $h_i(x)=v_i$ for some constant $v_i\in {\mathbb {R}}$. $\square$

Lemma 26

Let $\sigma \in {{\mathcal {W}}}_{2,ac}({\mathbb {R}}^n)$, and $M_2(\sigma ), E(\sigma )$ be defined as in Proposition 18. Then

$$\begin{aligned} M_2(\sigma )- \Vert E(\sigma )\Vert ^2>0. \end{aligned}$$

Proof

Since x is not a constant vector $\sigma$-a.e. ($\sigma \in {{\mathcal {W}}}_{2,ac}({\mathbb {R}}^n)$), it follows from Lemma 25 with $h(x)=x$ that

$$\begin{aligned} \int \Vert x\Vert ^2 d\sigma (x) > \Vert \int x d\sigma (x)\Vert ^2. \end{aligned}$$

$\square$

Lemma 27

Let $\sigma \in {{\mathcal {W}}}_{2,ac}({\mathbb {R}}^n), \mu \in {{\mathcal {W}}}_2({\mathbb {R}}^n)$ and $\phi$ be a convex function such that $\triangledown \phi = T_{\sigma }^{\mu }$ given by Brenier’s theorem (see e.g., [37, Theorem 1.48]). If $\phi$ is differentiable at $E(\sigma )$, where $E(\sigma ) = \int xd\sigma (x)$, then

$$\begin{aligned} \int T_{\sigma }^{\mu }(x)\cdot x d\sigma (x)- \Big (\int xd\sigma (x)\Big )\cdot \Big (\int T_{\sigma }^{\mu }(x)d\sigma (x)\Big )\ge 0. \end{aligned}$$

(36)

Proof

Let $A = \{x\in {\mathbb {R}}^n: \phi \text{ is } \text{ differentiable } \text{ at } x\}$. Since $\phi$ is $\sigma$-a.e. differentiable, we have $\sigma (A)=1$. Then it follows from the convexity of $\phi$ that

$$\begin{aligned} (\triangledown \phi (x)-\triangledown \phi (E(\sigma )))\cdot (x-E(\sigma ))\ge 0, \quad \forall x\in A. \end{aligned}$$

(37)

Hence

$$\begin{aligned} \int _{A} (T_{\sigma }^{\mu }(x)-T_{\sigma }^{\mu }(E(\sigma )))\cdot (x-E(\sigma )) d\sigma (x)\ge 0, \end{aligned}$$

which is exactly the desired inequality (36) by a direct computation using $\sigma (A)=1$:

$$\begin{aligned}&-\int T_{\sigma }^{\mu }(E(\sigma ))\cdot x d\sigma (x)- \int T_{\sigma }^{\mu }(x)\cdot E(\sigma ) d\sigma (x) + T_{\sigma }^{\mu }(E(\sigma ))\cdot E(\sigma ) \\ {}&= -T_{\sigma }^{\mu }(E(\sigma )) \cdot E(\sigma ) - \Big (\int xd\sigma (x)\Big )\cdot \Big (\int T_{\sigma }^{\mu }(x)d\sigma (x)\Big )+ T_{\sigma }^{\mu }(E(\sigma )) \cdot E(\sigma )\\&= \Big (\int xd\sigma (x)\Big )\cdot \Big (\int T_{\sigma }^{\mu }(x)d\sigma (x)\Big ). \end{aligned}$$

$\square$

Remark 12

The same conclusion holds if the assumption were “$E(\sigma )$ lies in the support of $\sigma$" instead of $\phi$ being differentiable at $E(\sigma )$, which can be proved using the fact that the support of optimal transport plan is cyclically monotone.

Remark 13

Given the assumptions in:Lemma 27, one can show that the inequality is strict if in addition, there exists a ball B(x, r), where x lies in the support of $\sigma$, such that for any $\lambda \in (0,1)$ and $y\in B(x,r)$

$$\begin{aligned} \phi ((1-\lambda )y+\lambda E(\sigma )) < (1-\lambda )\phi (y)+\lambda \phi (E(\sigma )), \end{aligned}$$

which guarantees that the inequality (37) is strict for y in a set with positive measure. In particular, if furthermore $\phi$ in Lemma 27 is strictly convex, the strict inequality holds.

Proposition 28

Let $\sigma \in {{\mathcal {W}}}_{2,ac}({\mathbb {R}}^n)$ and $\mu = T^b_{\sharp }\sigma$ with $T^b(x)= x+b, b\ne 0\in {\mathbb {R}}^n$. Consider iteration $\sigma _{k+1}=(T_{\sigma _k,\mu ; \theta _k})_\sharp \sigma _k$, with $\sigma _0=\sigma$ and where $\theta _k$ is chosen i.i.d. according to the uniform measure on $S^{n-1}$. Then

$$\begin{aligned} \sigma _k \xrightarrow {a.s.} \mu \quad \text {in~} W_2. \end{aligned}$$

Proof

By a direct computation, $T_{\sigma _k}^{\mu }(x) =x+b_k$, where

$$\begin{aligned} b_{k+1}= b_k - \theta _k(\theta _k\cdot b_k). \end{aligned}$$

To show $\sigma _k\rightarrow \mu$ almost surely, it suffices to show that $b_k \rightarrow 0$ almost surely. By symmetry of $S^{n-1}$, we assume without of generality that $b_0 = [1,0,\cdots ,0]^t$. Note that $\Vert b_1\Vert ^2= 1-|\theta _0\cdot b_0|^2$. Consider the spherical coordinates for $S^{n-1}$ with $\phi _1,\ldots , \phi _{n-2}\in [0,\pi ]$ and $\phi _{n-1}\in [0,2\pi ]$:

$$\begin{aligned} \begin{aligned}&x_{1} = \cos (\varphi _{1}),\quad x_{2} = \sin (\varphi _{1})\cos (\varphi _{2}), \quad x_{3} = \sin (\varphi _{1})\sin (\varphi _{2})\cos (\varphi _{3}) \\&\hspace{6cm} \cdots \\&x_{n-1} = \sin (\varphi _{1})\cdots \sin (\varphi _{n-2})\cos (\varphi _{n-1}),\ x_{n} = \sin (\varphi _{1})\cdots \sin (\varphi _{n-2})\sin (\varphi _{n-1}). \\ \end{aligned} \end{aligned}$$

The corresponding Jacobian is $\sin ^{n-2}(\varphi _1)\sin ^{n-3}(\varphi _2)\cdots \sin \varphi _{n-2}$. A direct computation gives

$$\begin{aligned} {\mathbb {E}}[|\theta _0\cdot b_0|^2]&= \frac{\int _0^{\pi } \sin ^{n-2}(\varphi _1)\cos ^2(\varphi _1)d\varphi _1}{\int _0^{\pi }\sin ^{n-2}(\varphi _1)d\varphi _1}\\&= 1-\frac{\int _0^{\pi } \sin ^{n}(\varphi _1)d\varphi _1}{\int _0^{\pi }\sin ^{n-2}(\varphi _1)d\varphi _1} \\ {}&= \rho <1. \end{aligned}$$

Hence ${\mathbb {E}}[\Vert b_1\Vert ^2] = 1-\rho \in (0,1)$. By symmetry and induction, one can show that

$$\begin{aligned} {\mathbb {E}}[\Vert b_k\Vert ^2] = (1-\rho )^k \xrightarrow {k\rightarrow \infty } 0. \end{aligned}$$

Since $\Vert b_{k+1}\Vert \le \Vert b_k\Vert$, by the monotone convergence theorem, we have

$$\begin{aligned} {\mathbb {E}}[\Vert b_k\Vert ^2]\longrightarrow {\mathbb {E}}[\alpha _{\infty }^2], \end{aligned}$$

where $\alpha _{\infty } = \lim \alpha _k$ and $\alpha _k = \Vert b_k\Vert$, which implies $\alpha _{\infty }= 0$ almost surely and hence $b_k \rightarrow 0$ almost surely. $\square$

Lemma 29

Let $\sigma ,\mu \in {{\mathcal {W}}}_{2,ac}({\mathbb {R}}^n)$. Then $(T_{\sigma ,\mu ;P})_\sharp \sigma \in {{\mathcal {W}}}_{2,ac}({\mathbb {R}}^n)$, for any $P\in O(n)$.

Proof

Let $P = [\theta _1,\cdots , \theta _n]$. A direct computation shows

$$\begin{aligned} \triangledown T_{\sigma ,\mu ;P}(x) = P \begin{bmatrix} (T_{\sigma ^{\theta _2}}^{\mu ^{\theta _i}})^{\prime }({{x}}\cdot {{\theta }}_1)&{} &{}&{}\\ &{} (T_{\sigma ^{\theta _2}}^{\mu ^{\theta _i}})^{\prime }({{x}}\cdot {{\theta }}_2)&{}&{}\\ &{}&{} \ddots &{}\\ &{}&{}&{} (T_{\sigma ^{\theta _n}}^{\mu ^{\theta _i}})^{\prime }({{x}}\cdot {{\theta }}_n) \end{bmatrix}P^t. \end{aligned}$$

Following similar arguments as in [38, Proof of Lemma 1, p. 949], it suffices to show that there exists a set $\Sigma$ such that (i) $\sigma ({\mathbb {R}}^n\setminus \Sigma )=0$ (ii) $T_{\sigma ,\mu ;P}|_{\Sigma }$ is injective and $\triangledown T_{\sigma ,\mu ;P}$ is positive definite on $\Sigma$. To this end, it suffices to observe that $T_{\sigma ^{\theta _i}}^{\mu ^{\theta _i}}$ is injective and $(T_{\sigma ^{\theta _i}}^{\mu ^{\theta _i}})^{\prime }>0$ outside a set $U_i$ that is $\sigma ^{\theta _i}$-negligible, i.e., $\sigma ^{\theta _i}(U_i)=0$. Here we have used the fact that $T_{\sigma ^{\theta _i}}^{\mu ^{\theta _i}}$ exists and is unique given that $\sigma \in {\mathcal {P}}_{ac}({\mathbb {R}}^n)$ (and hence $\sigma ^{\theta _i}$ is absolutely continuous, see e.g., Box 2.4. in [37, p. 82]). The fact that $M_2((T_{\sigma ,\mu ;P})_\sharp \sigma )$ is finite follows from (14) and that $M_2(\mu )<\infty$. $\square$