Assignment 2

Task 1: Transformation of Gaussian random variables

Let $x\in {\mathbb{R}}^n$ be $\mathcal{N}(\mu ,\Sigma )$ , find the distribution

a) $z={\Sigma }^{-1/2}(x-\mu )$

Where ${\Sigma }^{1/2}({\Sigma }^{1/2}{)}^T=\Sigma$

We know that:

$$(x-\mu )\sim \mathcal{N}(0,\Sigma )$$

We also know from the linearity of the Gaussian (Theorem 3.2.2) that:

Linearity

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Where $F={\Sigma }^{-1/2}$, $A=\Sigma$, and $a=0$, which gives us

$$\begin{array}{rl}p(z)& =\mathcal{N}(z;0,{\Sigma }^{-1/2}\Sigma ({\Sigma }^{-1/2}{)}^T)\\ p(z)& =\mathcal{N}(z;0,{\mathbb{I}}_n)\end{array}$$

Which is the standard normal distribution.

b) $y_i=z_i^2$

Where $z_i$ is the $i$‘th variable in the vector $z$. We already know from example 2.10 that the product of Gaussians is a Chi Squared Distribution, but we will now prove it.

Given theorem 2.5.1

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Where $y_i=f(z_i)=z_i^2$ → $z_i=h(y_i)=\pm \sqrt{y_i}$. This means that $|\det (f_i^{-1}(y_i))|$ can be written as

$$\begin{array}{rl}|\det (F_i^{-1}(y_i))|& =|\det (\pm \sqrt{y_i})|\\ |\det (F_i^{-1}(y_i))|& =|\pm \frac{1}{2\sqrt{y_i}}|\\ |\det (F_i^{-1}(y_i))|& =\frac{1}{2\sqrt{y_i}}\end{array}$$

And we can write $g(f_i^{-1}(y_i))$ as

$$\begin{array}{rl}g(f_i^{-1}(y_i))& =g(\pm \sqrt{y_i})=\frac{1}{\sqrt{2\pi }}{e}^{(-y_i/2)}\end{array}$$

Combining this we get

$$\begin{array}{rl}h(y_i)& =\sum _{i}g(f_i^{-1}(y))|\det (F_i^{-1}(y))|\\ h(y_i)& =g(\sqrt{y_i})\cdot |\det (\sqrt{y_i})|+g(\sqrt{-y_i})\cdot |\det (-\sqrt{y_i})|\\ h(y_i)& =2\frac{1}{\sqrt{2\pi }}{e}^{-y_i/2}\frac{1}{2\sqrt{y_i}}\\ h(y_i)& =\frac{1}{\sqrt{2\pi y_i}}{e}^{-y_i/2}\end{array}$$

We see that $p(y_i)={\chi }_1^2(y_i)$, meaning that we have $1$ degree of freedom.

c) $y=(x-\mu {)}^T{\Sigma }^{-1}(x-\mu )=z^{T}z=\sum z_i^2=\sum y_i$

Of all these possible distributions, it is easiest to work with $y=\sum y_i$ if we use the Moment-Generating Function and multiplying.

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$M_{y_i}(s)$ is given by

$$\begin{array}{rl}{M}_{y_i}(s)& =\frac{1}{(1-2s{)}^{k/2}}{|}_{k=1}\\ M_{y_i}(s)& =\frac{1}{(1-2s{)}^{1/2}}\end{array}$$

Which then finally gives us

$$\begin{array}{rl}{M}_y(s)& =\frac{1}{(1-2s{)}^{n/2}}\end{array}$$

Which we know is a Gamma Distribution

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Where $\theta =2$ and $k=\frac{n}{2}$

Task 2: Sensor Fusion