updates to notes for PSIM and stochastic networks

Research/PSIM/notes.tex

 \documentclass{article}
 
-\usepackage{/home/mccoid/LaTeX/preamble}
+\usepackage{preamble}
+\usepackage{float}
+
+\floatstyle{boxed}
+\newfloat{algorithm}{t}{}
+\floatname{algorithm}{Algorithm}
 
 \title{Collocation matrices representing inverse operators - notes}
 \author{Conor McCoid}
@@ -57,14 +62,14 @@ These matrices will be denoted by:
 
 For the Wronskians of $E \setminus E_i$ we present the following lemma:
 
-\begin{lemma}[Wronskians of exponential functions]
+\begin{lem}[Wronskians of exponential functions]
 Let $\set{\lambda_k}_{k=1}^m \in \mathbb{R}$ and $\Lambda$ the matrix defined earlier for such a set, then
 \begin{equation*}
 \W{\set{e^{\lambda_k x}}_{k=1}^m}{x} = \abs{\Lambda}
 e^{x \sum_{k=1}^m \lambda_k} .
 \end{equation*}
 \label{lem:exp}
-\end{lemma}
+\end{lem}
 
 \begin{proof}
 The case of any two $\lambda_k$ being equal is trivially true as both sides are necessarily zero.
@@ -113,14 +118,14 @@ The special case is equivalent to $M=1$ and $\lambda_1 = 0$.
 As such, the set $E = \set{\Poly{j}}_{j=0}^{m-1}$.
 The following lemma presents the Wronskians for such polynomials:
 
-\begin{lemma}[Wronskians of polynomials]
+\begin{lem}[Wronskians of polynomials]
 \begin{align*}
 (i) && \Wpoly{k=0}{m} & = 1 \\
 (ii) && \Wpoly{k=0,k \neq j}{m} & = \Wpoly{k=1}{m-j} \\
 (iii) && \Wpoly{k=1}{m} & = \Poly{m}
 \end{align*}
 \label{lem:poly}
-\end{lemma}
+\end{lem}
 
 \begin{proof}
 \begin{description}
@@ -218,12 +223,12 @@ The case for one root with multiplicity $m$, represented by $M=1$, is a generali
 The set $E = E_1 = \set{\Poly{k} e^{\lambda_1 x}}_{k=1}^m$ which is the set $E$ from the previous case multiplied by $e^{\lambda_1 x}$.
 The following lemma then makes the generalization simple.
 
-\begin{lemma}
+\begin{lem}
 \begin{equation*}
 W(\{ f_k g \}_{k=1}^m ; x) = g^m W(\{ f_k \}_{k=1}^m ; x )
 \end{equation*}
 \label{lem:group}
-\end{lemma}
+\end{lem}
 
 \begin{proof}
 It is trivially true for $m=1$.
@@ -436,4 +441,78 @@ One may write out equation (\ref{eq:Wronskian system}) with this in mind:
 where $\tilde{I}_k $ indicates the selection of those rows corresponding to $\lambda_k$.
 Note that all $F_k^{-1}(x)$ are principal submatrices of $F_{k^*}^{-1}$, where $m_{k^*} \geq m_k$ for all $k=1,...,M$.
 
+\section{Other attempts}
+
+\newcommand{\Lcal}{\mathcal{L}}
+\newcommand{\ddx}{\frac{d}{dx}}
+
+Let $\hat{\Lcal}$ be the operator for the set $\set{\Poly{k} e^{\lambda_j x}}_{j \neq q}$ and $\tilde{\Lcal}$ the one for $\set{\Poly{k} e^{\lambda_q x}}_{k \neq n}$, then
+\begin{align*}
+\Lcal & = \hat{\Lcal} \tilde{\Lcal} \\
+& = \left [\left ( \ddx \right )^{m-l} +\sum_{j \neq q} \lambda_j \left ( \ddx \right )^{m-l-1} + \dots + \prod_{j \neq q} \lambda_j \right ] 
+\left [ \left ( \ddx \right )^l + r(x) \left ( \ddx \right )^{l-1} + \dots + s(x) \right ] \\
+& = \left ( \ddx \right )^m + \left ( \sum_{j \neq q} \lambda_j + r(x) \right ) \left ( \ddx \right )^{m-1} + \dots
+\end{align*}
+but then $\Lcal$ is the operator for 
+\begin{itemize}
+\item $\set{\Poly{k} e^{\lambda_j x} } \cup \set{\hat{\Lcal} \left ( \Poly{k} e^{\lambda_q x} \right )}_{k \neq n}$ or
+\item $\set{\tilde{\Lcal} \left ( \Poly{k} e^{\lambda_j x} \right )}_{k,j \neq q} \cup \set{\Poly{k} e^{\lambda_q x}}_{k \neq n}$.
+\end{itemize}
+
+\begin{align*}
+E & = \set{P_k(x)}_{k=1}^m \\
+& = \set{\Poly{j} e^{\lambda_i x}}_{j=0, i=1}^{m_i-1,M} \\
+& = \cup_{i=1}^M \set{\Poly{j} e^{\lambda_i x}}_{j=0}^{m_i-1} \\
+& = \cup_{i=1}^M E_i
+\end{align*}
+\begin{itemize}
+\item $E \to \Lcal$ such that $\forall f \in E$ $\Lcal f = 0$.
+\item $\cup_{i \neq k} E_i \to \tilde{\Lcal}$ in the same way: $\tilde{\Lcal} f = f^{(m-m_k)}(x)  + \sum_{i \neq k} \lambda_i^{m_i} f^{(m - m_k - 1)}(x) + \dots$.
+\item $E_k \setminus \set{\Poly{k} e^{\lambda_i x} } \to \hat{\Lcal}$, $\hat{\Lcal}f = f^{(m_k-1)}(x) + r(x) f^{(m_k-2)}(x) + \dots$.
+\item $E \setminus \set{\Poly{k} e^{\lambda_i x} } \to \bar{\Lcal}$, $\bar{\Lcal} f = f^{(m-1)}(x) + \left ( r(x) + \sum_{i \neq k} \lambda_i^{m_i} \right ) f^{(m-2)}(x) + \dots$.
+\end{itemize}
+Need $W(\hat{\Lcal} \left ( E_k \setminus \set{\Poly{k} e^{\lambda_i x}} ; x\right)$.
+
+\begin{align*}
+\hat{\Lcal} \left ( E_k \setminus \set{\Poly{n} e^{\lambda_k x}} \right ) & = \prod_{i \neq k} \left ( \ddx - \lambda_i \right )^{m_i} \Poly{j} e^{\lambda_k x}, \quad j \neq n \\
+& \neq \prod_{i \neq k} \left ( \ddx - \lambda_i \right )^{m_i - 1} \left ( \poly{j-1}+ \lambda_k \Poly{j} - \lambda_i \poly{j} \right ) e^{\lambda_k x}
+\end{align*}
+
+\begin{align*}
+\tilde{\lambda}_l = \begin{cases} \lambda_1 & 1 \leq l \leq m_1 \\ \lambda_i & \sum_{j < i, j \neq k} m_j < l \leq \sum_{j \leq i,j \neq k} m_j \end{cases}
+\end{align*}
+
+\begin{align*}
+\prod_{l=1}^{m-m_k} & \left ( \ddx - \tilde{\lambda}_l \right ) \Poly{j} e^{\lambda_k x} = \\
+ \prod_{l=2}^{m-m_k} & \left ( \ddx - \tilde{\lambda}_l \right ) \left ( \poly{j-1} + (\lambda_k - \tilde{\lambda}_l ) \Poly{j} \right ) e^{\lambda_k x} = \\
+\prod_{l=3}^{m-m_k} & \left ( \ddx - \tilde{\lambda}_l \right ) \left ( \poly{j-2} + (2 \lambda_k - \tilde{\lambda}_1 - \tilde{\lambda}_2 ) \poly{j-1} + (\lambda_k^2 - \lambda_k \tilde{\lambda}_1 - \tilde{\lambda}_2 \lambda_k + \tilde{\lambda}_1 \tilde{\lambda}_2 ) \Poly{j} \right ) e^{\lambda_k x}
+\end{align*}
+
+\begin{align*}
+\Lcal P_k(x) & = P_k^{(m)}(x) - \sum_{i=1}^{M} m_i \lambda_i P_k^{(m-1)}(x) + \dots = 0 \\
+\hat{\Lcal} P_k(x) & = P_k^{(m-1)}(x) + r(x) P_k^{(m-2)}(x) + \dots = 0 & \forall k \neq j \\
+\tilde{\Lcal} P_j(x) & = P_j'(x) + q(x) P_j(x) = 0 \\
+\bar{\Lcal} P_k(x) & = P_k^{(m-1)}(x) + p(x) P_k^{(m-2)}(x) + \dots & \text{such that } \Lcal P_k(x) = \bar{\Lcal} \tilde{\Lcal} P_k(x) \\
+\breve{\Lcal} P_k(x) & = P_k'(x) + s(x) P_k(x) & \text{such that } \Lcal P_k(x) = \breve{\Lcal} \hat{\Lcal} P_k(x) \\
+\implies -\sum_{i=1}^M m_i \lambda_i & = q(x) + p(x) = r(x) + s(x)
+\end{align*}
+
+$\Omega$ (in this document, pretty sure $\Lambda$ in the current version) has $\sum_{k=1}^M \left [ \frac{m_k (m_k+1)}{2} + (m-m_k)^2 \right ]$ elements.
+
+To make $\Omega$ make each piece separately.
+Separate into coefficients and roots.
+\begin{algorithm}
+$\lambda_k$, $m_k$
+
+$\Omega_{\lambda_k} = \text{spdiags} \left ( 0:-1:1-m, \lambda_k^{(0:m-1)}, m, m_k \right )$;
+
+$P_T$ (Pascal's triangle in lower triangular form, size $m \times m$)
+
+$\Omega_k = P_T(:,1:m_k) \otimes \Omega_{\lambda_k}$; ($\otimes$ is the Hadamard matrix product)
+
+$\Omega = \begin{bmatrix} \Omega_1 & \Omega_2 & \dots & \Omega_M \end{bmatrix}$;
+
+\caption{Algorithm to construct $\Omega$}
+\end{algorithm}
+
 \end{document}

Research/Stochastic networks/StochasticNetworks.tex

@@ -2,7 +2,8 @@
 
 \usepackage{preamble}
 \usepackage{tikz}
-\usetikzlibrary{positioning}
+\usetikzlibrary{positioning,calc}
+\usepackage{pgfplots}
 
 \begin{document}
 
@@ -53,42 +54,24 @@ Define $\dagger \bbp = \bbp \cup \bbr_- \cup \set{-\infty}$.
 $(\dagger \bbp, \lor)$ is a commutative group.
 \end{lemma}
 
-\subsection{The Probabilistic Semiring}
+\subsection{The Probabilistic Concentric Ring??}
 
 In addition to the OR operation, one can equip $\bbp$ with the AND operation.
 This operation is equivalent to standard multiplication and shall be referred to as such.
 
-\begin{lemma}
-$(\bbp, \lor, \cdot)$ is a commutative semiring.
-\end{lemma}
-
-\subsection{The Probabilistic Field}
+Unfortunately, the OR and AND operations do not distribute, meaning they do not interact in any meaningful way.
+Essentially, one can set up two groups that overlap in the region $\bbp$ and each operation considers consequences only on their respective group.
+We can call $(\bbp,\lor,\cdot)$ a concentric ring, though such an object has little use.
 
 The probadd inverses have already been included in $\dagger \bbp$.
 The multiplicative inverses that remain lie in $(1,\infty)$.
 Taken together, this forms the extended real numbers, $\bbr^*$.
 
-\begin{lemma}
-$(\bbr^*, \lor, \cdot)$ is a field.
-\end{lemma}
-
-\begin{proof}
-Not true, since the operations do not allow for distributivity.
-\end{proof}
-
-\subsection{The Probabilistic Inner Product Space}
-
-Since $(\bbr^*, \lor, \cdot)$ is a field, the space $({\bbr^*}^n, \bbr^*, \lor, \cdot)$ is a vector space. -not true, see proof of prev. lemma
-
-Define the inner product over this vector space as
+Define the inner product over this space as
 \begin{equation}
 \langle \vec{p}, \vec{q} \rangle = p_1 q_1 \lor p_2 q_2 \lor \dots \lor p_n q_n .
 \end{equation}
 
-\begin{lemma}
-$({\bbr^*}^n, \bbr^*, \lor, \cdot, \langle \cdot, \cdot \rangle)$ is an inner product space.
-\end{lemma}
-
 \subsection{Probabilistic Matrices}
 
 To represent the linear transformations of probabilistic vectors, one can construct matrices containing elements of $\bbr^*$.
@@ -98,6 +81,20 @@ This is made trivial by using the probabilistic inner product:
 A \vec{p} = \begin{bmatrix} \vec{a}_1^\top \\ \vdots \\ \vec{a}_n^\top \end{bmatrix} \vec{p} = \begin{bmatrix} \langle \vec{a}_1, \vec{p} \rangle \\ \vdots \\ \langle \vec{a}_n, \vec{p} \rangle \end{bmatrix} .
 \end{equation*}
 
+\begin{figure}
+\centering
+\begin{tikzpicture}
+\begin{axis}[samples=500, domain=-4:4, restrict y to domain=-4:4,
+xlabel={$p$},
+ylabel={$\dagger p = \frac{-p}{1-p}$}]
+\addplot[thick] plot (\x, {-\x/(1-\x)});
+\addplot[dotted] plot (\x, 1);
+\addplot[dotted] plot (1, \x);
+\end{axis}
+\end{tikzpicture}
+\caption{Visualization of the inverses $\dagger p$.}
+\end{figure}
+
 \section{Stochastic Networks}
 
 We define a stochastic network as a graph with vertices taking on a discrete set of $k$ values.
@@ -383,6 +380,23 @@ roundnode/.style={circle, fill=white, draw=black, very thick}]
 		The paths follow the relationship $\mathcal{D} \supset \mathcal{C} \subset \mathcal{A} \subset \mathcal{B}$.}
 \end{table}
 
+\subsection{Notes}
+
+If two paths cross and one arrives at the intersection before the other than the probability of infection is:
+\begin{equation*}
+p_1(t) + p_2(t) (1 - p_1(t)) = p_1(t) \lor p_2(t).
+\end{equation*}
+
+If two paths collide at the same time then the probability of infection is either
+$$1 - (1 - p_1(t))(1-p_2(t)) = p_1(t) \lor p_2(t)$$
+or
+$$p_1(t) (1-p_2(t)) + p_2(t) (1-p_1(t)) = \left ( p_1(t) \lor p_2(t) \right ) - \left (p_1(t) \land p_2(t) \right ).$$
+I think the first option makes more sense.
+
+The total probability for a node is then the probsum of all paths that run through it.
+
+Keywords: Bayesian networks, MCMC
+
 \appendix
 
 \section{Notes}