\documentclass[methods-collated.tex]{subfiles}
\begin{document}
  \section{Statistics}

  \subsection*{Probability}

  \begin{align*}
    \Pr(A \cup B) &= \Pr(A) + \Pr(B) - \Pr(A \cap B) \\
    \Pr(A \cap B) &= \Pr(A|B) \times \Pr(B) \\
    \Pr(A|B) &= \frac{\Pr(A \cap B)}{\Pr(B)} \\
    \Pr(A) &= \Pr(A|B) \cdot \Pr(B) + \Pr(A|B^{\prime}) \cdot \Pr(B^{\prime})
  \end{align*}

  Mutually exclusive \(\implies \Pr(A \cup B) = 0\) \\

  Independent events:
  \begin{flalign*}
    \quad \Pr(A \cap B) &= \Pr(A) \times \Pr(B)& \\
    \Pr(A|B) &= \Pr(A) \\
    \Pr(B|A) &= \Pr(B)
  \end{flalign*}

  \subsection*{Combinatorics}

  \begin{itemize}
    \item Arrangements \({n \choose k} = \frac{n!}{(n-k)}\)
    \item \colorbox{important}{Combinations} \({n \choose k} = \frac{n!}{k!(n-k)!}\)
    \item Note \({n \choose k} = {n \choose k-1}\)
  \end{itemize}

  \subsection*{Distributions}

  \subsubsection*{Mean \(\mu\)}

  \textbf{Mean} \(\mu\) or \textbf{expected value} \(E(X)\)

  \begin{align*}
    E(X) &= \frac{\Sigma \left[ x \cdot f(x) \right]}{\Sigma f} \tag{\(f =\) absolute frequency} \\
    &= \sum_{i=1}^n \left[ x_i \cdot \Pr(X=x_i) \right] \tag{discrete}\\
    &= \int_\textbf{X} (x \cdot f(x)) \> dx
  \end{align*}

  \subsubsection*{Mode}

  Most popular value (has highest probability of all \(X\) values). Multiple modes can exist if \(>1 \> X\) value have equal-highest probability. Number must exist in distribution.

  \subsubsection*{Median}

  If \(m > 0.5\), then value of \(X\) that is reached is the median of \(X\). If \(m = 0.5 = 0.5\), then \(m\) is halfway between this value and the next. To find \(m\), add values of \(X\) from smallest to alrgest until the sum reaches 0.5.

  \[ m = X \> \text{such that} \> \int_{-\infty}^{m} f(x) dx = 0.5 \]

  \subsubsection*{Variance \(\sigma^2\)}

  \begin{align*}
    \operatorname{Var}(x) &= \sum_{i=1}^n p_i (x_i-\mu)^2 \\
    &= \sum (x-\mu)^2 \times \Pr(X=x) \\
    &= \sum x^2 \times p(x) - \mu^2 \\
    &= \operatorname{E}(X^2) - [\operatorname{E}(X)]^2
    &= E\left[(X-\mu)^2\right]
  \end{align*}

  \subsubsection*{Standard deviation \(\sigma\)}

  \begin{align*}
    \sigma &= \operatorname{sd}(X) \\
    &= \sqrt{\operatorname{Var}(X)}
  \end{align*}

  \subsection*{Binomial distributions}

  Conditions for a \textit{binomial distribution}:
  \begin{enumerate}
    \item Two possible outcomes: \textbf{success} or \textbf{failure}
    \item \(\Pr(\text{success})\) is constant across trials (also denoted \(p\))
    \item Finite number \(n\) of independent trials
  \end{enumerate}


  \subsubsection*{Properties of \(X \sim \operatorname{Bi}(n,p)\)}

  \begin{align*}
    \mu(X) &= np \\
    \operatorname{Var}(X) &= np(1-p) \\
    \sigma(X) &= \sqrt{np(1-p)} \\
    \Pr(X=x) &= {n \choose x} \cdot p^x \cdot (1-p)^{n-x}
  \end{align*}

  \begin{cas}
    Interactive \(\rightarrow\) Distribution \(\rightarrow\) \verb;binomialPdf; then input
    \begin{description}[nosep, style=multiline, labelindent=0.5cm, leftmargin=3cm, font=\normalfont]
      \item [x:] no. of successes
      \item [numtrial:] no. of trials
      \item [pos:] probability of success
    \end{description}
  \end{cas}

  \subsection*{Continuous random variables}

  A continuous random variable \(X\) has a pdf \(f\) such that:

  \begin{enumerate}
    \item \(f(x) \ge 0 \forall x \)
    \item \(\int^\infty_{-\infty} f(x) \> dx = 1\)
  \end{enumerate}

  \begin{align*}
    E(X) &= \int_\textbf{X} (x \cdot f(x)) \> dx \\
    \operatorname{Var}(X) &= E\left[(X-\mu)^2\right]
  \end{align*}

  \[ \Pr(X \le c) = \int^c_{-\infty} f(x) \> dx \]


  \subsection*{Two random variables \(X, Y\)}

  If \(X\) and \(Y\) are independent:
  \begin{align*}
    \operatorname{E}(aX+bY) & = a\operatorname{E}(X)+b\operatorname{E}(Y) \\
    \operatorname{Var}(aX \pm bY \pm c) &= a^2 \operatorname{Var}(X) + b^2 \operatorname{Var}(Y)
  \end{align*}

  \subsection*{Linear functions \(X \rightarrow aX+b\)}

  \begin{align*}
    \Pr(Y \le y) &= \Pr(aX+b \le y) \\
    &= \Pr\left(X \le \dfrac{y-b}{a}\right) \\
    &= \int^{\frac{y-b}{a}}_{-\infty} f(x) \> dx
  \end{align*}

  \begin{align*}
    \textbf{Mean:} && \operatorname{E}(aX+b) & = a\operatorname{E}(X)+b \\
    \textbf{Variance:} && \operatorname{Var}(aX+b) &= a^2 \operatorname{Var}(X) \\
  \end{align*}

  \subsection*{Expectation theorems}

  For some non-linear function \(g\), the expected value \(E(g(X))\) is not equal to \(g(E(X))\).

  \begin{align*}
    E(X^2) &= \operatorname{Var}(X) - \left[E(X)\right]^2 \\
    E(X^n) &= \Sigma x^n \cdot p(x) \tag{non-linear} \\
    &\ne [E(X)]^n \\
    E(aX \pm b) &= aE(X) \pm b \tag{linear} \\
    E(b) &= b \tag{\(\forall b \in \mathbb{R}\)}\\
    E(X+Y) &= E(X) + E(Y) \tag{two variables}
  \end{align*}

  \subsection*{Sample mean}

  Approximation of the \textbf{population mean} determined experimentally.

  \[ \overline{x} = \dfrac{\Sigma x}{n} \]

  where
  \begin{description}[nosep, labelindent=0.5cm]
    \item \(n\) is the size of the sample (number of sample points)
    \item \(x\) is the value of a sample point
  \end{description}

  \begin{cas}
    \begin{enumerate}[leftmargin=3mm]
      \item Spreadsheet
      \item In cell A1:\\ \path{mean(randNorm(sd, mean, sample size))}
      \item Edit \(\rightarrow\) Fill \(\rightarrow\) Fill Range
      \item Input range as A1:An where \(n\) is the number of samples
      \item Graph \(\rightarrow\) Histogram
    \end{enumerate}
  \end{cas}

  \subsubsection*{Sample size of \(n\)}

  \[ \overline{X} = \sum_{i=1}^n \frac{x_i}{n} = \dfrac{\sum x}{n} \]

  Sample mean is distributed with mean \(\mu\) and sd \(\frac{\sigma}{\sqrt{n}}\) (approaches these values for increasing sample size \(n\)).

  For a new distribution with mean of \(n\) trials, \(\operatorname{E}(X^\prime) = \operatorname{E}(X), \quad \operatorname{sd}(X^\prime) = \dfrac{\operatorname{sd}(X)}{\sqrt{n}}\)

  \begin{cas}

    \begin{itemize}
      \item Spreadsheet \(\rightarrow\) Catalog \(\rightarrow\) \verb;randNorm(sd, mean, n); where \verb;n; is the number of samples. Show histogram with Histogram key in top left
      \item To calculate parameters of a dataset: Calc \(\rightarrow\) One-variable
    \end{itemize}

  \end{cas}

  \subsection*{Normal distributions}


  \[ Z = \frac{X - \mu}{\sigma} \]

  Normal distributions must have area (total prob.) of 1 \(\implies \int^\infty_{-\infty} f(x) \> dx = 1\) \\
  \(\text{mean} = \text{mode} = \text{median}\)

  \begin{warning}
    Always express \(z\) as +ve. Express confidence \textit{interval} as ordered pair.
  \end{warning}

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    \begin{figure*}[hb]
      \centering
      \begin{tikzpicture}
        \begin{axis}[every axis plot post/.style={
            mark=none,domain=-3:3,samples=50,smooth}, 
          axis x line=bottom, 
          axis y line=left,
          enlargelimits=upper,
          x=\textwidth/10,
          ytick={0.55},
          yticklabels={\(\frac{1}{\sigma \sqrt{2\pi}}\)}, 
          xtick={-2,-1,0,1,2},
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          ylabel={\(\Pr(X=x)\)}]
          \addplot {gauss(0,0.75)};
          \fill[red!30] (-3,0)  -- plot[id=f3,domain=-3:3,samples=50] function {1/(0.75*sqrt(2*pi))*exp(-((x)^2)/(2*0.75^2))} -- (3,0) -- cycle;
          \fill[darkgray!30] (3,0)  -- plot[id=f3,domain=-3:3,samples=50] function {1/(0.75*sqrt(2*pi))*exp(-x*x*0.5/(0.75*0.75))} -- (3,0) -- cycle;
          \fill[lightgray!30] (-2,0)  -- plot[id=f3,domain=-2:2,samples=50] function {1/(0.75*sqrt(2*pi))*exp(-x*x*0.5/(0.75*0.75))} -- (2,0) -- cycle;
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          xlabel={\(Z\)},
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    \end{figure*}

  \subsection*{Confidence intervals}

  \begin{itemize}
    \item \textbf{Point estimate:} single-valued estimate of the population mean from the value of the sample mean \(\overline{x}\)
    \item \textbf{Interval estimate:} confidence interval for population mean \(\mu\)
    \item \(C\)\% confidence interval \(\implies\) \(C\)\% of samples will contain population mean \(\mu\)
  \end{itemize}

  \subsubsection*{95\% confidence interval}

  For 95\% c.i. of population mean \(\mu\):

  \[ x \in \left(\overline{x} \pm 1.96 \dfrac{\sigma}{\sqrt{n}} \right)\]

  where:
    \begin{description}[nosep, labelindent=0.5cm]
    \item \(\overline{x}\) is the sample mean
    \item \(\sigma\) is the population sd
    \item \(n\) is the sample size from which \(\overline{x}\) was calculated
  \end{description}

  \begin{cas}
    Menu \(\rightarrow\) Stats \(\rightarrow\) Calc \(\rightarrow\) Interval \\
    Set \textit{Type = One-Sample Z Int} \\ \-\hspace{1em} and select \textit{Variable}
  \end{cas}

  \subsection*{Margin of error}

  For 95\% confidence interval of \(\mu\):
  \begin{align*}
    M &= 1.96 \times \dfrac{\sigma}{\sqrt{n}} \\
    &= \dfrac{1}{2} \times \text{width of c.i.} \\
    \implies n &= \left( \dfrac{1.96 \sigma}{M} \right)^2
  \end{align*}

  Always round \(n\) up to a whole number of samples.

  \subsection*{General case}

  For \(C\)\% c.i. of population mean \(\mu\):

  \[ x \in \left( \overline{x} \pm k \dfrac{\sigma}{\sqrt{n}} \right) \]
  \hfill where \(k\) is such that \(\Pr(-k < Z < k) = \frac{C}{100}\)

  \begin{cas}
    Menu \(\rightarrow\) Stats \(\rightarrow\) Calc \(\rightarrow\) Interval \\
    Set \textit{Type = One-\colorbox{important}{Prop} Z Int} \\
    Input  x \(= \hat{p} * n\)
  \end{cas}

  \subsection*{Confidence interval for multiple trials}

  For a set of \(n\) confidence intervals (samples), there is \(0.95^n\) chance that all \(n\) intervals contain the population mean \(\mu\).

  \end{document}