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 4 added 204 characters in body edited Dec 27 '17 at 14:46 Xi'an 63.5k88 gold badges103103 silver badges389389 bronze badges Pierre Jacob on Statisfaction explains how to implement a maximal coupling between two distributions $$p$$ and $$q$$. Which is essentially a sequence of two accept-reject steps. From there you can try to estimate the correlation $$\varrho$$ between both Gaussian variates. Here is for instance a comparison of empirical estimates of the correlations, depending on the values of $$\mu$$ and $$\sigma$$ (with the other Gaussian being standard): The correlation associated with a maximal coupling can be derived from \begin{align*}\mathbb{E}[XY]&=\mathbb{E}[XY\mathbb{I}_{X=Y}]+\mathbb{E}[XY\mathbb{I}_{X\ne Y}]\\ &=\int x^2 p(x)\wedge q(x)\text{d}x+ \int\int xy \{p(x)-p(x)\wedge q(x)\}\{q(y)-p(y)\wedge q(y)\}\text{d}x\text{d}y \end{align*} Note that, to answer the question, the correlation returned by maximal coupling is a function of the four parameters of the Gaussians, while the maximal correlation is always equal to one. Pierre Jacob on Statisfaction explains how to implement a maximal coupling between two distributions $$p$$ and $$q$$. Which is essentially a sequence of two accept-reject steps. From there you can try to estimate the correlation $$\varrho$$ between both Gaussian variates. Here is for instance a comparison of empirical estimates of the correlations, depending on the values of $$\mu$$ and $$\sigma$$ (with the other Gaussian being standard): The correlation associated with a maximal coupling can be derived from \begin{align*}\mathbb{E}[XY]&=\mathbb{E}[XY\mathbb{I}_{X=Y}]+\mathbb{E}[XY\mathbb{I}_{X\ne Y}]\\ &=\int x^2 p(x)\wedge q(x)\text{d}x+ \int\int xy \{p(x)-p(x)\wedge q(x)\}\{q(y)-p(y)\wedge q(y)\}\text{d}x\text{d}y \end{align*} Pierre Jacob on Statisfaction explains how to implement a maximal coupling between two distributions $$p$$ and $$q$$. Which is essentially a sequence of two accept-reject steps. From there you can try to estimate the correlation $$\varrho$$ between both Gaussian variates. Here is for instance a comparison of empirical estimates of the correlations, depending on the values of $$\mu$$ and $$\sigma$$ (with the other Gaussian being standard): The correlation associated with a maximal coupling can be derived from \begin{align*}\mathbb{E}[XY]&=\mathbb{E}[XY\mathbb{I}_{X=Y}]+\mathbb{E}[XY\mathbb{I}_{X\ne Y}]\\ &=\int x^2 p(x)\wedge q(x)\text{d}x+ \int\int xy \{p(x)-p(x)\wedge q(x)\}\{q(y)-p(y)\wedge q(y)\}\text{d}x\text{d}y \end{align*} Note that, to answer the question, the correlation returned by maximal coupling is a function of the four parameters of the Gaussians, while the maximal correlation is always equal to one. 3 added 300 characters in body edited Dec 26 '17 at 18:01 Xi'an 63.5k88 gold badges103103 silver badges389389 bronze badges Pierre Jacob on Statisfaction explains how to implement a maximal coupling between two distributions $$p$$ and $$q$$. Which is essentially a sequence of two accept-reject steps. From there you can try to estimate the correlation $$\varrho$$ between both Gaussian variates. Here is for instance a comparison of empirical estimates of the correlations, depending on the values of $$\mu$$ and $$\sigma$$ (with the other Gaussian being standard): The correlation associated with a maximal coupling can be derived from \begin{align*}\mathbb{E}[XY]&=\mathbb{E}[XY\mathbb{I}_{X=Y}]+\mathbb{E}[XY\mathbb{I}_{X\ne Y}]\\ &=\int x^2 p(x)\wedge q(x)\text{d}x+ \int\int xy \{p(x)-p(x)\wedge q(x)\}\{q(y)-p(y)\wedge q(y)\}\text{d}x\text{d}y \end{align*} Pierre Jacob on Statisfaction explains how to implement a maximal coupling between two distributions $$p$$ and $$q$$. Which is essentially a sequence of two accept-reject steps. From there you can try to estimate the correlation $$\varrho$$ between both Gaussian variates. Here is for instance a comparison of empirical estimates of the correlations, depending on the values of $$\mu$$ and $$\sigma$$ (with the other Gaussian being standard): Pierre Jacob on Statisfaction explains how to implement a maximal coupling between two distributions $$p$$ and $$q$$. Which is essentially a sequence of two accept-reject steps. From there you can try to estimate the correlation $$\varrho$$ between both Gaussian variates. Here is for instance a comparison of empirical estimates of the correlations, depending on the values of $$\mu$$ and $$\sigma$$ (with the other Gaussian being standard): The correlation associated with a maximal coupling can be derived from \begin{align*}\mathbb{E}[XY]&=\mathbb{E}[XY\mathbb{I}_{X=Y}]+\mathbb{E}[XY\mathbb{I}_{X\ne Y}]\\ &=\int x^2 p(x)\wedge q(x)\text{d}x+ \int\int xy \{p(x)-p(x)\wedge q(x)\}\{q(y)-p(y)\wedge q(y)\}\text{d}x\text{d}y \end{align*} 2 added 256 characters in body edited Dec 25 '17 at 11:55 Xi'an 63.5k88 gold badges103103 silver badges389389 bronze badges Pierre Jacob on Statisfaction explains how to implement a maximal coupling between two distributions $$p$$ and $$q$$. Which is essentially a sequence of two accept-reject steps. From there you can try to estimate the correlation $$\varrho$$ between both Gaussian variates. Here is for instance a comparison of empirical estimates of the correlations, depending on the values of $$\mu$$ and $$\sigma$$ (with the other Gaussian being standard): Pierre Jacob on Statisfaction explains how to implement a maximal coupling between two distributions $$p$$ and $$q$$. Which is essentially a sequence of two accept-reject steps. From there you can try to estimate the correlation $$\varrho$$ between both Gaussian variates. Pierre Jacob on Statisfaction explains how to implement a maximal coupling between two distributions $$p$$ and $$q$$. Which is essentially a sequence of two accept-reject steps. From there you can try to estimate the correlation $$\varrho$$ between both Gaussian variates. Here is for instance a comparison of empirical estimates of the correlations, depending on the values of $$\mu$$ and $$\sigma$$ (with the other Gaussian being standard): 1 answered Dec 25 '17 at 10:29 Xi'an 63.5k88 gold badges103103 silver badges389389 bronze badges