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My question comes form a potion of the long review paper, which is attached below enter image description here

In the set-up, $\sigma_1$ and $\sigma_2$ are possibly different, constant diffusion matrices. To my knowledge, if we take the ("cheap") synchronous coupling $B^1_t = B^2_t$, then we can estimate $\mathbb{E}[|X^1_t - X^2_t|^2]$ using for instance Ito isometry or BDG inequality. However, what will be the error estimate $\mathbb{E}[|X^1_t - X^2_t|^2]$ under this new ("the best") coupling? The reference paper FH16 published on The Annals of Probability is overwhelming long and it is also too technical for me. Thanks for any help!

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  • $\begingroup$ As the authors state, $W_2(\mathcal{L}(X_t^1),\mathcal{L}(X_t^2))^2$ is precisely a "dynamical version" (meaning that it involves time) of the standard 2-Wasserstein distance between two multivariate normal distributions: en.wikipedia.org/wiki/Wasserstein_metric#Normal_distributions with $\mu_1 = X^1_0$, $\mu_2=X^2_0$, $C_1 = t a_1$ and $C_2 = t a_2$. This result is straightforward to prove. $\endgroup$ Jul 15, 2021 at 15:01
  • $\begingroup$ @NawafBou-Rabee thanks! But what are $C_1$ and $C_2$? It seems that it appears out of the blue. I also have no clue as to why you mentioned that "This result is straightforward to prove" $\endgroup$
    – Fei Cao
    Jul 15, 2021 at 16:56
  • $\begingroup$ @NawafBou-Rabee thank you very much. But in my original post above, there are no constants such as $C_1$ and $C_2$, I am wondering where do these constants come from... $\endgroup$
    – Fei Cao
    Jul 15, 2021 at 19:25
  • $\begingroup$ @NawafBou-Rabee Thank you! But there is a mistake: $X_t^2 = X_0^2 + \sigma_2 U(a_1,a_2) B_t^1$ instead of $X_t^2 = X_0^2 + U(a_1,a_2) B_t^1$. Also, it is hard to "simplify" $\mathbb{E}[ | X_t^1 - X_t^2|^2 ]$ and compare with the corresponding bound obtained via the synchronous coupling... $\endgroup$
    – Fei Cao
    Jul 15, 2021 at 22:32

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Take $\sigma_1 = a_1^{1/2}$ and $\sigma_2 = a_2^{1/2}$. Consider the coupling $$ X^1_t = X_0^1 + a_1^{1/2} B^1_t \quad \text{and} \quad X^2_t = X_0^2 + a_2^{1/2} U(a_1, a_2) B_t^1 \;. $$ Here $U(a_1,a_2) = a_2^{-1/2} a_1^{-1/2} (a_1^{1/2} a_2 a_1^{1/2} )^{1/2}$ is an orthogonal matrix since $$ U(a_1,a_2) U(a_1, a_2)^T = a_2^{-1/2} a_1^{-1/2} (a_1^{1/2} a_2 a_1^{1/2} ) a_1^{-1/2} a_2^{-1/2} = I $$ where $I$ is the identity matrix. As a shorthand, let $A =a_1^{1/2} - a_2^{1/2} U(a_1, a_2) $, so that $$ X_t^1 - X_t^2 = X_0^1 - X_0^2 + A B_t^1 \;. $$

Then \begin{align} E[|X^1_t - X^2_t|^2] &= |X_0^1 - X_0^2|^2 + E[| A B^1_t|^2] = |X_0^1 - X_0^2|^2 + E[(A B^1_t)^T (A B^1_t)] \\ &= |X_0^1 - X_0^2|^2 + E[\operatorname{Trace}( ( A B^1_t)^T (A B^1_t)) ] \\ &= |X_0^1 - X_0^2|^2 + E[\operatorname{Trace}( A (B^1_t) (B^1_t)^T A^T ) ] \\ &= |X_0^1 - X_0^2|^2 + t \operatorname{Trace}( A A^T ) \end{align} where we used the cyclic property of the trace and the fact that $E[ (B^1_t) (B^1_t)^T]= t I$.

To finish, \begin{align} &\operatorname{Trace}( A A^T ) = \operatorname{Trace}( (a_1^{1/2} - a_2^{1/2} U(a_1, a_2)) (a_1^{1/2} - a_2^{1/2} U(a_1, a_2))^T ) \\ &= \operatorname{Trace}( a_1 - a_2^{1/2} U(a_1, a_2) a_1^{1/2} - a_1^{1/2} U(a_1, a_2)^T a_2^{1/2} + a_2^{1/2} U(a_1, a_2) U(a_1, a_2)^T a_2^{1/2} ) \\ &= \operatorname{Trace}( a_1 - 2 a_1^{1/2} a_2^{1/2} U(a_1, a_2) + a_2 ) \\ &= \operatorname{Trace}( a_1 - 2 (a_1^{1/2} a_2 a_1^{1/2} )^{1/2} + a_2 ) \end{align} where again we used the cyclic property of the trace and that $U(a_1,a_2)$ is an orthogonal matrix. Combining the above we obtain $$ E[|X^1_t - X^2_t|^2] = |X_0^1 - X_0^2|^2 + t \operatorname{Trace}( a_1 + a_2 - 2 (a_1^{1/2} a_2 a_1^{1/2} )^{1/2} ) \;, $$ as required. This is indeed a dynamical (i.e., time-dependent) version of the 2-Wasserstein distance between two multivariate normal distributions.

In contrast, for the synchronous coupling, $$ X^1_t = X_0^1 + a_1^{1/2} B^1_t \quad \text{and} \quad X^2_t = X_0^2 + a_2^{1/2} B_t^1 \;, $$ we obtain: $$ E[|X^1_t - X^2_t|^2] = |X_0^1 - X_0^2|^2 + t \operatorname{Trace}( a_1 + a_2 - 2 a_1^{1/4} a_2^{1/2} a_1^{1/4} ) \;. $$ As discussed further in Section 2 of the reference below, we note that the synchronous coupling is optimal with respect to the 2-Wasserstein distance when $a_1$ and $a_2$ commute, since in that case $U(a_1,a_2)=I$.

Givens, Clark R.; Shortt, Rae Michael, A class of Wasserstein metrics for probability distributions, Mich. Math. J. 31, 231-240 (1984). ZBL0582.60002.

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  • $\begingroup$ Thank you very much! Professor $\endgroup$
    – Fei Cao
    Jul 16, 2021 at 17:56

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