mathematics.rmd

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title: "Background Notes"
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---


# Seasonal Patterns

In pure mathematics, one of the angle sum identity theorems states that

\vspace{10pt}

> \begin{theorem}[Angle Sum Identity]
  \label{trigonometry}
  If $\alpha$ \& $\phi$ are each acute angles of two distinct right-angled triangles, and $\alpha$ \& $\phi$ are adjacent, then
  \begin{equation}
  A \: sin(\alpha + \phi) \; \equiv \; a \: sin \alpha \: cos \phi \; + \; b \: cos \alpha \: sin \phi  (\#eq:0001)
  \end{equation}
\end{theorem}

\vspace{10pt}

Hence, and for seasonal pattern modelling purposes, the expression

\begin{equation}
  A sin(2 \pi f t + \phi)  (\#eq:0002)
\end{equation}

is expressible as

\begin{equation}
  A \: sin(2 \pi f t + \phi) \; \equiv \; a \: sin(2 \pi f t) + b \: cos(2 \pi f t)  (\#eq:0003)
\end{equation}

if $A$ & $\phi$ values exist for the expression

\begin{equation}
  A sin(2{\pi}ft) \: cos{\phi} \; + \; A cos(2{\pi}ft) \: sin{\phi}  \; \equiv \;
a \: sin(2{\pi}ft) \; +  \; b \: cos(2{\pi}ft)  (\#eq:0004)
\end{equation}

\vspace{10pt}

**Proof**

By the coefficients of *\cref{eq:0004}*

\begin{align}
a & = A cos \phi (\#eq:0005) \\
b & = A sin \phi (\#eq:0006)
\end{align}

\vspace{10pt}

The quotient of *\cref{eq:0005}* & *\cref{eq:0006}* is

\begin{equation}
\frac{a}{b} = \frac{cos \phi}{sin \phi} = tan \phi (\#eq:0007)
\end{equation}

*\cref{eq:0007}* is in line with a right-angled triangle with acute angle $\phi$, i.e., *\cref{fig:0001}*.  By the Pythagoras theorem, the length of the hypotenuse side of the acute right-angled triangle is

\begin{equation}
\sqrt{a^{2} + b^{2}} (\#eq:0010)
\end{equation}

\vspace{10pt}

\begin{figure}
\begin{center}
\begin{tikzpicture}
\coordinate (A) at (3, 0);
\coordinate (B) at (6, 0);
\coordinate (C) at (6, 4);
\draw[orange, thick] (A) -- node[midway, below] {$a$}
                     (B) -- node[midway, right] {$b$}
                     (C) -- node[midway, sloped, above] {$\sqrt{a^2 + b^2}$} cycle;
\pic[draw = orange!50, angle radius = 7mm, font = \footnotesize, "$\phi$"] {angle = B--A--C};
\end{tikzpicture}
\end{center}
\caption{An acute right-angled triangle}
\label{fig:0001}
\end{figure}

\vspace{10pt}


*\cref{eq:0010}* is derivable from *\cref{eq:0005}* & *\cref{eq:0006}*, i.e.,

\begin{equation}
a^{2} + b^{2} = A^{2}cos^{2}\phi + A^{2}sin^{2}\phi = A^{2} (\#eq:0008)
\end{equation}

$\Rightarrow$

\begin{equation}
A =  \sqrt{a^{2} + b^{2}} (\#eq:0009)
\end{equation}

Therefore, $A$ exists; it is the length of the hypotenuse of the acute right-angled triangle (*\cref{fig:0001}*).  Altogether, $A$ & $\phi$ exists.


\clearpage

# Sum & Product Rules


\footnotesize
\begin{figure}
\begin{center}
\begin{tikzpicture}
\node (A) at (-0.01, 1.23) [lightgrey, left] {$\mbox{\scriptsize $y_{_{j}}$}$};
\node (B) at (1.75, -0.01) [lightgrey, below] {$\mbox{\scriptsize $x_{_{i}}$}$};
\node (C) at (1.75, 1.23) [lightgrey] {$\mbox{\scriptsize $n_{_{ij}}$}$};
\node (D) at (3.41, 1.23) [lightgrey, right, yshift = -1mm] { $\mbox{ \scriptsize $r_{j} = \sum\limits_{i} {n_{_{ij}}}$ }$ };
\node (E) at (1.75, 1.23) [lightgrey, rotate = 90, right, xshift = 10mm, yshift = -1mm] { $\mbox{ \scriptsize $c_{i} = \sum\limits_{j} {n_{_{ij}}}$ }$ };
\draw[step = 0.5cm, darkgrey, line width = 0.01mm] (0.01, 0.01) grid (3.4, 2.4);
\end{tikzpicture}
\end{center}
\caption{Events}
\label{fig:0002}
\end{figure}
\normalsize

\vspace{35pt}

## The marginal; sum rule

\begin{equation}
p(X = x_{i}) = \frac{c_{i}}{N}
\end{equation}

**but**

\begin{equation}
c_{i} = \sum\limits_{j} {n_{ij}}
\end{equation}

**and**

\begin{equation}
p(X = x_{i}, Y = y_{j}) = \frac{n_{ij}}{N}
\end{equation}

$\Rightarrow$

\begin{equation}
n_{ij} = N p(X = x_{i}, Y = y_{j})
\end{equation}

Hence

\begin{align}
p(X = x_{i}) & = \frac{c_{i}}{N}  \\
             & = \frac{1}{N} \sum\limits_{j} {n_{ij}} \\
             & = \frac{1}{N} \sum\limits_{j} {N p(X = x_{i}, Y = y_{j})} \\
             & = \sum\limits_{j} {p(X = x_{i}, Y = y_{j})}
\end{align}


\vspace{35pt}

## The joint; product rule

\begin{align}
p(X = x_{i}, Y = y_{j}) & = \frac{n_{ij}}{N} \\
                        & = \frac{n_{ij}}{c_{i}} \times  \frac{c_{i}}{N}
\end{align}

**but**

\begin{equation}
p(Y = y_{j} | X = x_{i}) = \frac{n_{ij}}{c_{i}}
\end{equation}

**and**

\begin{equation}
p(X = x_{i}) = \frac{c_{i}}{N}
\end{equation}


Hence

\begin{align}
p(X = x_{i}, Y = y_{j}) & = \frac{n_{ij}}{N} \\
                        & = \frac{n_{ij}}{c_{i}} \times  \frac{c_{i}}{N} \\
                        & = p(Y = y_{j} | X = x_{i}) p(X = x_{i})
\end{align}