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How to get the following plug-in empirical estimator?

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They introduce a new correlation. For $\pi\in \Pi(\mu,\nu)$ the set of coupling of two probability measures $\mu$ and $\nu$ on a Polish space $(X,d)$. The author introduces a plugin estimator.

Question: How to understand $\hat{\pi}$? For


For the estimator of the denominator, I use the same notation as the answer.

\begin{equation*} D(\hat{\pi})=\sum_{y,z}d(y,z)\hat{\pi}_2(\{y\})\hat{\pi}_2(\{z\}) =\sum_{y,z}d(y,z)\frac{1}{N^2}\sum 1[\phi(Y_n)\in\{y\}]\sum 1[\phi(Y_m)\in\{z\}] \end{equation*} Expand the product of these two summation, $$ =\frac{1}{N^2}\sum_{y,z}d(y,z)(\sum_{i=1}^N 1[\phi(Y_i)\in\{y\}]1[\phi(Y_i)\in\{z\}]+\sum_{i\neq j}1[\phi(Y_i)\in\{y\}]1[\phi(Y_j)\in\{z\}] ) $$

They introduce a new correlation. For $\pi\in \Pi(\mu,\nu)$ the set of coupling of two probability measures $\mu$ and $\nu$ on a Polish space $(X,d)$. The author introduces a plugin estimator.

How to understand $\hat{\pi}$? For the estimator of the denominator, I use the same notation as the answer.

\begin{equation*} D(\hat{\pi})=\sum_{y,z}d(y,z)\hat{\pi}_2(\{y\})\hat{\pi}_2(\{z\}) =\sum_{y,z}d(y,z)\frac{1}{N^2}\sum 1[\phi(Y_n)\in\{y\}]\sum 1[\phi(Y_m)\in\{z\}] \end{equation*} Expand the product of these two summation, $$ =\frac{1}{N^2}\sum_{y,z}d(y,z)(\sum_{i=1}^N 1[\phi(Y_i)\in\{y\}]1[\phi(Y_i)\in\{z\}]+\sum_{i\neq j}1[\phi(Y_i)\in\{y\}]1[\phi(Y_j)\in\{z\}] ) $$

They introduce a new correlation. For $\pi\in \Pi(\mu,\nu)$ the set of coupling of two probability measures $\mu$ and $\nu$ on a Polish space $(X,d)$. The author introduces a plugin estimator.

Question: How to understand $\hat{\pi}$?


For the estimator of the denominator, I use the same notation as the answer.

\begin{equation*} D(\hat{\pi})=\sum_{y,z}d(y,z)\hat{\pi}_2(\{y\})\hat{\pi}_2(\{z\}) =\sum_{y,z}d(y,z)\frac{1}{N^2}\sum 1[\phi(Y_n)\in\{y\}]\sum 1[\phi(Y_m)\in\{z\}] \end{equation*} Expand the product of these two summation, $$ =\frac{1}{N^2}\sum_{y,z}d(y,z)(\sum_{i=1}^N 1[\phi(Y_i)\in\{y\}]1[\phi(Y_i)\in\{z\}]+\sum_{i\neq j}1[\phi(Y_i)\in\{y\}]1[\phi(Y_j)\in\{z\}] ) $$

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In this paper https://arxiv.org/abs/2102.00356, theyThey introduce a new Wasserstein correlation for. For $\pi\in \Pi(\mu,\nu)$ the set of coupling of two probability measures $\mu$ and $\nu$ on a Polish space $(X,d)$ $$ W(\pi)=\frac{\int W_1(\pi_{x_1},\nu)\mu(dx_1)}{\int d(y,z)\nu(dy)\nu(dz)}, $$ where $\pi$ has a $\mu$-a.s. unique disintegration w.r.t. the first coordinate and $W_1$ is the 1-Wasserstein distance.

  The author introduceintroduces a plugin estimator. Given iid random samples $(X_1,Y_1),\dots, (X_N,Y_N)$. Let's partition the $[0,1]$ into a disjoint union of a finite number of intervals and let $\phi: [0,1]\mapsto [0,1]$ map each intervals to its center. Let $$ \hat{\pi}=\frac{1}{N}\sum_{i=1}^N \delta_{(\phi(X_i), \phi(Y_i))} $$ Define $\Phi:=\{(\phi)^{-1}(\{x\}): x\in \phi([0,1])\}$. (Then $[0,1]=\cup_{G\in \Phi} G$ disjoint.)

(1) How to understand $\hat{\pi}$?

(2) As on page 23, why do we have for any Borel set $G\subset [0,1]$ $$ W(\hat{\pi})=\frac{\sum_{G\in \phi} \frac{1}{N}|i\in\{1,\dots,N\} s.t. X_i\in G|\cdot W(\hat{\pi}_G, \hat{\pi}_2)}{\frac{1}{N^2}\sum_{n,m=1}^N|\phi(Y_n)-\phi(Y_m)|} $$ where $\pi_G$ is the conditional probability $$ \pi_G(\cdot)=\frac{1}{\pi_1(G)}\int_G \pi_{x_1}(\cdot)\pi_1(dx) $$

(3) What is $\hat{\pi}_G$? I am confused about this empirical measure.

ForFor the estimator of the denominator, I use the same notation as the answer.

\begin{equation*} D(\hat{\pi})=\sum_{y,z}d(y,z)\hat{\pi}_2(\{y\})\hat{\pi}_2(\{z\}) =\sum_{y,z}d(y,z)\frac{1}{N^2}\sum 1[\phi(Y_n)\in\{y\}]\sum 1[\phi(Y_m)\in\{z\}] \end{equation*} Expand the product of these two summation, $$ =\frac{1}{N^2}\sum_{y,z}d(y,z)(\sum_{i=1}^N 1[\phi(Y_i)\in\{y\}]1[\phi(Y_i)\in\{z\}]+\sum_{i\neq j}1[\phi(Y_i)\in\{y\}]1[\phi(Y_j)\in\{z\}] ) $$

I do not know how to go to the next step.

So for $G_1=\phi^{-1}(\{y\})$ and $G_2=\phi^{-1}(\{z\})$ $$ =\sum_{G_1\in\Phi,G_2\in\Phi}d(y,z)\frac{1}{N^2}\sum 1[Y_i\in G_1]\sum 1[Y_i\in G_2] $$ but I am not sure what is $d(y,z)=|y-z|$?

In this paper https://arxiv.org/abs/2102.00356, they introduce a new Wasserstein correlation for $\pi\in \Pi(\mu,\nu)$ the set of coupling of two probability measures $\mu$ and $\nu$ on a Polish space $(X,d)$ $$ W(\pi)=\frac{\int W_1(\pi_{x_1},\nu)\mu(dx_1)}{\int d(y,z)\nu(dy)\nu(dz)}, $$ where $\pi$ has a $\mu$-a.s. unique disintegration w.r.t. the first coordinate and $W_1$ is the 1-Wasserstein distance.

  The author introduce a plugin estimator. Given iid random samples $(X_1,Y_1),\dots, (X_N,Y_N)$. Let's partition the $[0,1]$ into a disjoint union of a finite number of intervals and let $\phi: [0,1]\mapsto [0,1]$ map each intervals to its center. Let $$ \hat{\pi}=\frac{1}{N}\sum_{i=1}^N \delta_{(\phi(X_i), \phi(Y_i))} $$ Define $\Phi:=\{(\phi)^{-1}(\{x\}): x\in \phi([0,1])\}$. (Then $[0,1]=\cup_{G\in \Phi} G$ disjoint.)

(1) How to understand $\hat{\pi}$?

(2) As on page 23, why do we have for any Borel set $G\subset [0,1]$ $$ W(\hat{\pi})=\frac{\sum_{G\in \phi} \frac{1}{N}|i\in\{1,\dots,N\} s.t. X_i\in G|\cdot W(\hat{\pi}_G, \hat{\pi}_2)}{\frac{1}{N^2}\sum_{n,m=1}^N|\phi(Y_n)-\phi(Y_m)|} $$ where $\pi_G$ is the conditional probability $$ \pi_G(\cdot)=\frac{1}{\pi_1(G)}\int_G \pi_{x_1}(\cdot)\pi_1(dx) $$

(3) What is $\hat{\pi}_G$? I am confused about this empirical measure.

For the estimator of the denominator, I use the same notation as the answer.

\begin{equation*} D(\hat{\pi})=\sum_{y,z}d(y,z)\hat{\pi}_2(\{y\})\hat{\pi}_2(\{z\}) =\sum_{y,z}d(y,z)\frac{1}{N^2}\sum 1[\phi(Y_n)\in\{y\}]\sum 1[\phi(Y_m)\in\{z\}] \end{equation*} Expand the product of these two summation, $$ =\frac{1}{N^2}\sum_{y,z}d(y,z)(\sum_{i=1}^N 1[\phi(Y_i)\in\{y\}]1[\phi(Y_i)\in\{z\}]+\sum_{i\neq j}1[\phi(Y_i)\in\{y\}]1[\phi(Y_j)\in\{z\}] ) $$

I do not know how to go to the next step.

So for $G_1=\phi^{-1}(\{y\})$ and $G_2=\phi^{-1}(\{z\})$ $$ =\sum_{G_1\in\Phi,G_2\in\Phi}d(y,z)\frac{1}{N^2}\sum 1[Y_i\in G_1]\sum 1[Y_i\in G_2] $$ but I am not sure what is $d(y,z)=|y-z|$?

They introduce a new correlation. For $\pi\in \Pi(\mu,\nu)$ the set of coupling of two probability measures $\mu$ and $\nu$ on a Polish space $(X,d)$. The author introduces a plugin estimator.

How to understand $\hat{\pi}$? For the estimator of the denominator, I use the same notation as the answer.

\begin{equation*} D(\hat{\pi})=\sum_{y,z}d(y,z)\hat{\pi}_2(\{y\})\hat{\pi}_2(\{z\}) =\sum_{y,z}d(y,z)\frac{1}{N^2}\sum 1[\phi(Y_n)\in\{y\}]\sum 1[\phi(Y_m)\in\{z\}] \end{equation*} Expand the product of these two summation, $$ =\frac{1}{N^2}\sum_{y,z}d(y,z)(\sum_{i=1}^N 1[\phi(Y_i)\in\{y\}]1[\phi(Y_i)\in\{z\}]+\sum_{i\neq j}1[\phi(Y_i)\in\{y\}]1[\phi(Y_j)\in\{z\}] ) $$

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