I am studying this book about Optimal Transport, and in Remark 2.19 it talks about translation in Variance, where it is stated that a nice property of Wasserstein Distances is the ability to factor out translations: $$\mathcal{W}_2 (T_{\tau\#}\alpha \; , \; T_{\tau'\#}\beta)^2= \mathcal{W}_2(\alpha,\beta)^2- 2\langle \tau -\tau' \; , \; \mathbf{m}_\alpha - \mathbf{m}_\beta \rangle + \|\tau -\tau'\|^2$$ where $\mathbf{m}_\alpha \triangleq \int_\mathcal{X}xd\alpha(x) \in \mathbb{R}^d$ is the mean of a measure $\alpha$.
The proceeded to write that: $$ \mathcal{W}_2(\alpha,\beta)^2=\mathcal{W}_2(\bar{\alpha},\bar{\beta})^2+\|\mathbf{m}_\alpha - \mathbf{m}_\beta \|^2$$
Where $\mathcal{W}_2(\bar{\alpha},\bar{\beta})^2=\mathcal{W}_2 (T_{\tau\#}\alpha \; , \; T_{\tau'\#}\beta)^2$.
Now, I can't wrap my head around how the term $\|\mathbf{m}_\alpha - \mathbf{m}_\beta \|^2$ came to be in the equation. I've tried to trace backwards, but I got stuck. Here is my attempt: \begin{align} 2 \langle \tau-\tau',\mathbf{m}_\alpha-\mathbf{m}_\beta\rangle - \|\tau -\tau'\|^2 &= \|\mathbf{m}_\alpha-\mathbf{m}_\beta\|^2 \\ 2 \langle \tau-\tau',\mathbf{m}_\alpha-\mathbf{m}_\beta\rangle &= \|\tau -\tau'\|^2 + \|\mathbf{m}_\alpha-\mathbf{m}_\beta\|^2 \\ \langle \tau-\tau',\mathbf{m}_\alpha-\mathbf{m}_\beta\rangle &= \frac{\|\tau -\tau'\|^2 + \|\mathbf{m}_\alpha-\mathbf{m}_\beta\|^2}{2} \end{align}
I haven't studied analysis or measure theory, but I am trying to understand as much as I could. I thought that this is an inner product and that $\tau-\tau'$ is the difference between translations, which could be viewed as some vector, and similarly for the difference between the means of the discrete measures which I assume that they yield the center of the discrete measures $\alpha$ and $\beta$.
I am pretty sure something is wrong in my understanding, or at least my basics. So, my questions are:
- where did I go wrong?
- is that an inner product?
- what is the definition of this $\langle \cdot , \cdot \rangle$ notation?
- and what is the definition of these norms $\|\cdot\|^2$ in the context of the Wasserstein distances?
Thanks
EDIT: I am using $\triangleq$ as the equal with a "def" on top.
EDIT: The purpose is self-study. I am not enrolled anywhere so it isn't a homework.
EDIT: From Remark 2.19 in the book: $(\bar{\alpha},\bar{\beta})$ are the "centered" zero mean measures $\bar{\alpha}=T_{\mathbf{m}_\alpha \#} \alpha$
Let ${\tau}=m_a,\tau'=m_b$, since
$2\langle \mathbf{m}_\alpha - \mathbf{m}_\beta \; , \; \mathbf{m}_\alpha - \mathbf{m}_\beta \rangle = 2\|\mathbf{m}_\alpha -\mathbf{m}_\beta'\|^2$,
you get from the first equation that
$\mathcal{W}_2 (T_{m_\alpha \#}\alpha \; , \; T_{m_\beta \#}\beta)^2= \mathcal{W}_2(\alpha,\beta)^2- 2\langle \mathbf{m}_\alpha - \mathbf{m}_\beta \; , \; \mathbf{m}_\alpha - \mathbf{m}_\beta \rangle + \|\mathbf{m}_\alpha -\mathbf{m}_\beta\|^2 $
$= \mathcal{W}_2(\alpha,\beta)^2- \|\mathbf{m}_\alpha -\mathbf{m}_\beta\|^2 $ .
Rearrange the items, you get the results that you want.