Point cloud Similarities

 Point to distribution:

$\overrightarrow{p}=(p_x,p_y)\Rightarrow P(x,y)=\frac{1}{2\pi {\sigma }_x{\sigma }_y}\exp \{-\frac{x^2}{2{\sigma }_x^2}-\frac{y^2}{2{\sigma }_y^2}\}$

$\overrightarrow{p}=\sum _{i=1}^D{p}^i{\hat{e}}_i\Rightarrow P(x^i)=(2\pi {)}^{-D/2}\left(\prod _{i=1}^D{\sigma }_{x^i}^{-1}\right)\exp \{-\sum _{i=1}^D\frac{{(x^i)}^2}{2{({\sigma }_{x^i})}^2}\}$

Correlation Function for between one point distributions:

$P(x^i;x_0^i):=P(x^i-x_0^i)$

$⟨P(x^i)P(x^i;x_0^i)⟩={(4\pi )}^{-D/2}\left(\prod _{i=1}^D{\sigma }_{x^i}^{-1}\right)\exp \{-\sum _{i=1}^D\frac{{(x_0^i)}^2}{4{({\sigma }_{x^i})}^2}\}$

For two point clouds $\{{\overrightarrow{p}}_i\}$ and $\{{\overrightarrow{q}}_i\}$,

$P(x^i;\{{\overrightarrow{p}}_j\})=\sum _{j}P(x^i;p_j^i)$

The correlation of two point clouds is

$C(\{{\overrightarrow{p}}_j\},\{{\overrightarrow{q}}_j\})=\frac{⟨P(x^i;\{{\overrightarrow{p}}_j\})P(x^i;\{{\overrightarrow{q}}_j\})⟩}{\sqrt{⟨P(x^i;\{{\overrightarrow{p}}_j\})P(x^i;\{{\overrightarrow{p}}_j\})⟩⟨P(x^i;\{{\overrightarrow{q}}_j\})P(x^i;\{{\overrightarrow{q}}_j\})⟩}}$

We want to write $⟨P(x^i;\{{\overrightarrow{p}}_j\})P(x^i;\{{\overrightarrow{q}}_j\})⟩$ with $P(x^i;\{{\overrightarrow{p}}_j\})=\sum _{j}P(x^i,p_j^i)$.

$C^{\prime }(\{{\overrightarrow{p}}_j\},\{{\overrightarrow{q}}_j\}):=⟨P(x^i;\{{\overrightarrow{p}}_j\})P(x^i;\{{\overrightarrow{q}}_j\})⟩$

$=⟨\sum _{j,k}P(x^i;p_j^i)P(x^i;q_k^i)⟩$

$\sim \sum _{j,k}\exp \{-\sum _{i=1}^D\frac{{(p_j^i-q_k^i)}^2}{4{({\sigma }_{x^i})}^2}\}$

To maximize $C^{\prime }(\{{\overrightarrow{p}}_j\},\{{\overrightarrow{q}}_j\})$ by shift of $q$, we need to $\frac{d}{d\mathrm{\Delta }{q}^i}{C}^{\prime }(\{{\overrightarrow{p}}_j\},\{{\overrightarrow{q}}_j+\mathrm{\Delta }\overrightarrow{q}\})$.

$$\frac{d}{d\mathrm{\Delta }{q}^i}{C}^{\prime }(\{{\overrightarrow{p}}_j\},\{{\overrightarrow{q}}_j+\mathrm{\Delta }\overrightarrow{q}\})\sim {\frac{d}{d\mathrm{\Delta }{q}^i}\sum _{j,k}\exp \{-\sum _{i=1}^D\frac{{(p_j^i-q_k^i-\mathrm{\Delta }{q}^i)}^2}{4{({\sigma }_{x^i})}^2}\}|}_{\mathrm{\Delta }q=0}$$

$$={-\frac{{\sigma }_{x^i}^2}{2}\sum _{j,k}(p_j^i-q_k^i-\mathrm{\Delta }{q}^i)\exp \{-\sum _{i=1}^D\frac{{(p_j^i-q_k^i-\mathrm{\Delta }{q}^i)}^2}{4{({\sigma }_{x^i})}^2}\}|}_{\mathrm{\Delta }q=0}$$

$$=-\frac{{\sigma }_{x^i}^2}{2}\sum _{j,k}(p_j^i-q_k^i)\exp \{-\sum _{i=1}^D\frac{{(p_j^i-q_k^i)}^2}{4{({\sigma }_{x^i})}^2}\}$$

In the same sense,

${\frac{d^2}{{(d\mathrm{\Delta }{q}^i)}^2}{C}^{\prime }(\{{\overrightarrow{p}}_j\},\{{\overrightarrow{q}}_j+\mathrm{\Delta }\overrightarrow{q}\})|}_{\mathrm{\Delta }q=0}$
$$\sim \sum _{j,k}[\frac{{\sigma }_{x^i}^4(p_j^i-q_k^i{)}^2}{4}+\frac{{\sigma }_{x^i}^2}{2}]\exp \{-\sum _{i=1}^D\frac{{(p_j^i-q_k^i)}^2}{4{({\sigma }_{x^i})}^2}\}$$

We can do Newton's method using these 1st and 2nd derivatives.