# Probability and Statistics in n-D, ProbMAN 2

Probabilistic Robotics Lecture 2

## Basics

\begin{aligned} f(\vec{x}) & \\ \int_{\mathbb{R}^n} f(\vec{x})d\vec{x} &= 1 \\ \vec\mu &= \int_{\mathbb{R}^n} \vec x\ \rho (\vec x)\ d \vec x \\ \Sigma &= \int_{\mathbb{R}^n} (\vec x - \vec \mu)(\vec x - \vec \mu)^T\ \rho (\vec x)\ d \vec x \\ \end{aligned}

### Expectation

\begin{aligned} \mu &= E[x] = \int_{\mathbb{R}^n} x\ \rho(x)\ dx \\ \sigma^2 &= E[(x - \mu)^2] = \int_{\mathbb{R}^n} (x - \mu)^2\ \rho(x)\ dx \\ E[f(x)] &= \int_{-\inf}^{\inf} f(x)\ \rho(x)\ dx \\ E[f(\vec x)] &= \int_{D \subset \mathbb{R}^n} f(\vec x)\rho(\vec x)\ d\vec x \\ \end{aligned}

## Multivariate Gaussian

\begin{aligned} \rho(x;\mu,\Sigma) &\dot= \frac{1}{(2\pi)^{\frac{n}{2}}|det\Sigma|^{\frac{1}{2}}}exp\{-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)\} \\ \int_{\mathbb{R}^n}& \rho(x; \mu, \Sigma)\ dx = 1 \\ \int_{\mathbb{R}^n}& x\ \rho(x; \mu, \Sigma)\ dx = \mu \\ \int_{\mathbb{R}^n}& (x - \mu)\ \rho(x; \mu, \Sigma)\ dx = \Sigma \\ \end{aligned}

## Crazy Math Tools

As Covariance matrix is symmetric matrix, it can be diagonalized by orthogonal matrix $Q$. \begin{aligned} \Sigma &= Q\Lambda Q^T \\ det|\Sigma| &= det|Q\Lambda Q^T| = det|Q|\ det|\lambda|\ det|Q^T| = det|QQ^T|\ det|\lambda| = det|\Lambda| \\ \end{aligned}

## Eigenvactors and Eigenvalues'

\begin{aligned} A\ \vec v &= \lambda\ \vec v \\ A^n\ \vec v &= \lambda^n\ \vec v \\ A &= Q\Lambda Q^T,\ Q\ is\ Eigen\ vectors,\ \Lambda\ is\ Eigen\ values \\ A^n &= Q\Lambda^n Q^T \\ \end{aligned}

## Homework 2: Entropy of Multivariate Gaussian

mu is not important as we can always add this x - mu to the original x. to make it simpler, we can assume mu = 0. \begin{aligned} S(f) &\dot= -\int_{x} f(x)logf(x) dx \\ &= - \int_{x} \frac{1}{(2\pi)^{n/2}|det\Sigma|^{1/2}}exp\{-\frac{1}{2}x^T\Sigma^{-1}x\}log\frac{1}{(2\pi)^{n/2}|det\Sigma|^{1/2}}exp\{-\frac{1}{2}x^T\Sigma^{-1}x\} dx \\ &= log(2\pi^{n/2}|det\Sigma|^{1/2}) \int_x \frac{-\frac{1}{2}x^T\Sigma^{-1}x}{2\pi^{n/2}|det\Sigma|^{1/2}} exp\{-\frac{1}{2}x^T\Sigma^{-1}x\} dx \\ &= log(2\pi^{n/2}|det\Sigma|^{1/2}) E[x^T\Sigma^{-1}x] \\ \\ \because \Sigma &= E[(x - \mu)(x - \mu)^T] = E[xx^T] \\ &= \int x x^T N(x; 0,\Sigma) dx \\ \\ \therefore E[x^T\Sigma^{-1}x] &= E[tr(x^T\Sigma^{-1}x)] \\ &= tr(\Sigma^{-1}E[xx^T]) \\ &= tr(\Sigma^{-1}\Sigma) \\ &= tr(I) \\ &= n \\ \\ \therefore S(f) &= log(2\pi^{n/2}|det\Sigma|^{1/2}) E[x^T\Sigma^{-1}x] \\ &= log(2\pi^{n/2}|det\Sigma|^{1/2}) n \\ \end{aligned}