114 lines
		
	
	
		
			2.4 KiB
		
	
	
	
		
			Plaintext
		
	
	
	
	
	
			
		
		
	
	
			114 lines
		
	
	
		
			2.4 KiB
		
	
	
	
		
			Plaintext
		
	
	
	
	
	
---
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title: Scratch Page
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toc: no
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format: markdown
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...
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# Math Stuff
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On observation of $\mathcal{D}$, the *likelihood* of hypothesis
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$\mathcal{R}_{\alpha}$ is
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$\mathit{P}(\mathcal{D}|\mathcal{R}_{\alpha})$.
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## Fingerprint Variance
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Additionally, we associate a collective **RSS Variance**
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$\sigma_{F_s}$ with each fingerprint, which is a weighed average of
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the RSS values of each of the vector elements using $C_i$ as the
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weight. It is calculated in this manner:
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$$
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\sigma_{F_s} = \frac{\sum_{i \in F_s} \sigma_i\, C_i}
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                    {\sum_{i \in F_s} C_i}
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$$
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# Bayesian Regression
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First, specify a set of probabilistic models of the data.
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Let a member of this set be denoted by $\mathcal{R}_\alpha$
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$\mathcal{R}_\alpha$ has a *prior* probability $P(\mathcal{H}_\alpha)$
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On observation of $\mathcal{D}$, the *likelihood* of hypothesis
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$\mathcal{R}_{\alpha}$ is
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$\mathit{P}(\mathcal{D}|\mathcal{R}_{\alpha})$.
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The *posterior* probability of $\mathcal{R}_{\alpha}$ is then given by
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$\mathit{P}(\mathcal{H}_{\alpha})\mathit{P}(\mathcal{D}|\mathcal{H}_{\alpha})$
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This follows from **Bayes' Theorem** which says
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$$
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P(A|B) = \frac{P(B | A)\, P(A)}{P(B)}
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$$
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# Matrix Stuff
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This is a column vector:
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$$\vec v = \left(\begin{matrix}
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1\\
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3\\
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7
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\end{matrix}\right)$$
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The *vector sum* of $\vec u$ and $\vec v$ is:
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$$\vec u + \vec v =
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\left(\begin{matrix}
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u_1\\
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\vdots\\
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u_n
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\end{matrix}\right) + \left(\begin{matrix}
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v_1\\
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\vdots\\
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v_n
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\end{matrix}\right) = \left(\begin{matrix}
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u_1 + v_1\\
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\vdots\\
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u_n + v_n
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\end{matrix}\right)
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$$
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The *scalar multiplication* of the real number $r$ and the vector $\vec v$ is:
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$$
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r \cdot \vec v = r \cdot \left(\begin{matrix}
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v_1\\ \vdots \\ v_n
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\end{matrix}\right) = \left(\begin{matrix}
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rv_1 \\ \vdots \\ rv_n
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\end{matrix}\right)
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$$
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This system:
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$$\begin{alignedat}{4}
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 x & {}-{} &  y & {}+{} &  z & = 1\\
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3x & {}+{} &    &       &  z & = 3\\
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5x & {}-{} & 2y & {}+{} & 3z & = 5
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\end{alignedat}$$
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reduces
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$$
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\left(\begin{array}{rrr|r}
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1&-1&1&1\\
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3&0&1&3\\
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5&-2&3&5\\
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\end{array}\right) \longrightarrow_{-5\rho_1+\rho_3}^{-3\rho_1+\rho_2}
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\left(\begin{array}{rrr|r}
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1&-1&1&1\\
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0&3&-2&0\\
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0&3&-2&0\\
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\end{array}\right) \longrightarrow^{-\rho_2+\rho_3}
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\left(\begin{array}{rrr|r}
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1&-1&1&1\\
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0&3&-2&0\\
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0&0&0&0\\
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\end{array}\right)
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$$
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to a one parameter solution set:
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$$
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\{ \begin{pmatrix}1\\ 0\\ 0\end{pmatrix} +
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   \begin{pmatrix}-1/3\\ 2/3\\ 1\end{pmatrix} z \mid z \in \Bbb{R} \}
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$$ |