Point cloud Similarities

 Point to distribution:

\(\vec{p}=(p_x, p_y) \Rightarrow P(x,y)= \frac{1}{2\pi \sigma_x \sigma_y} \exp\left\{-\frac{x^2}{2\sigma_x^2}-\frac{y^2}{2\sigma_y^2}\right\}\)

\(\vec{p}=\sum_{i=1}^D p^i \hat{e}_i  \Rightarrow P(x^i)= (2\pi)^{-D/2}\left(\prod_{i=1}^D \sigma_{x^i}^{-1}\right) \exp\left\{-\sum_{i=1}^D \frac{{(x^i)}^2}{2{(\sigma_{x^i})}^2}\right\}\)

Correlation Function for between one point distributions:

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

\(\left<P(x^i)P(x^i; x_0^i)\right> = {(4\pi)}^{-D/2} \left(\prod_{i=1}^D \sigma_{x^i}^{-1}\right) \exp\left\{-\sum_{i=1}^D \frac{{(x_0^i)}^2}{4{(\sigma_{x^i})}^2}\right\} \)

For two point clouds \(\{\vec{p}_i\}\) and \(\{\vec{q}_i\}\),

\(P(x^i;\{\vec{p}_j\}) = \sum_j P(x^i; p_j^i)\)

The correlation of two point clouds is

\(C(\{\vec{p}_j\}, \{\vec{q}_j\}) = \frac{\left<P(x^i;\{\vec{p}_j\})P(x^i;\{\vec{q}_j\}) \right>}{\sqrt{\left<P(x^i;\{\vec{p}_j\}) P(x^i;\{\vec{p}_j\}) \right>\left<P(x^i;\{\vec{q}_j\}) P(x^i;\{\vec{q}_j\}) \right>}}\)

We want to write \(\left<P(x^i;\{\vec{p}_j\}) P(x^i;\{\vec{q}_j\}) \right>\) with \(P(x^i;\{\vec{p}_j\}) = \sum_j P(x^i, p_j^i)\).

\(C'(\{\vec{p}_j\}, \{\vec{q}_j\}):=\left<P(x^i;\{\vec{p}_j\}) P(x^i;\{\vec{q}_j\}) \right>\)

\(\qquad \qquad\qquad = \left< \sum_{j,k} P(x^i; p_j^i) P(x^i; q_k^i) \right>\)

\(\qquad \qquad\qquad \sim\sum_{j,k} \exp\left\{-\sum_{i=1}^D \frac{{(p_j^i - q_k^i)}^2}{4{(\sigma_{x^i})}^2}\right\} \)

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

\[\frac{d}{d\Delta q^i}C'(\{\vec{p}_j\}, \{\vec{q}_j + \Delta\vec{q}\}) \sim \left. \frac{d}{d\Delta q^i}\sum_{j,k} \exp\left\{-\sum_{i=1}^D \frac{{(p_j^i - q_k^i - \Delta q^i)}^2}{4{(\sigma_{x^i})}^2}\right\} \right|_{\Delta q = 0} \]

\[=\left. -\frac{\sigma_{x^i}^2}{2}\sum_{j,k}(p_j^i - q_k^i - \Delta q^i) \exp\left\{-\sum_{i=1}^D \frac{{(p_j^i - q_k^i - \Delta q^i)}^2}{4{(\sigma_{x^i})}^2}\right\}\right|_{\Delta q = 0} \]

\[= -\frac{\sigma_{x^i}^2}{2}\sum_{j,k}(p_j^i - q_k^i) \exp\left\{-\sum_{i=1}^D \frac{{(p_j^i - q_k^i )}^2}{4{(\sigma_{x^i})}^2}\right\} \]

In the same sense,

\(
\left.\frac{d^2}{{(d\Delta q^i)}^2}C'(\{\vec{p}_j\}, \{\vec{q}_j + \Delta\vec{q}\})\right|_{\Delta q = 0} \)
\[\sim
 \sum_{j,k}\left[\frac{\sigma_{x^i}^4 (p_j^i - q_k^i)^2}{4} + \frac{\sigma_{x^i}^2}{2}\right] \exp\left\{-\sum_{i=1}^D \frac{{(p_j^i - q_k^i )}^2}{4{(\sigma_{x^i})}^2}\right\}
\]

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