From: Andrew Lorimer Date: Mon, 9 Sep 2019 09:51:42 +0000 (+1000) Subject: [methods] clean up statistics ref X-Git-Tag: yr12~36 X-Git-Url: https://git.lorimer.id.au/notes.git/diff_plain/65d62a134004e8996b150d20795dc93a3c206684?ds=sidebyside [methods] clean up statistics ref --- diff --git a/methods/statistics-ref.tex b/methods/statistics-ref.tex new file mode 100644 index 0000000..3461bce --- /dev/null +++ b/methods/statistics-ref.tex @@ -0,0 +1,335 @@ +\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} + + \pgfmathdeclarefunction{gauss}{2}{% + \pgfmathparse{1/(#2*sqrt(2*pi))*exp(-((x-#1)^2)/(2*#2^2))}% + } + \pgfkeys{/pgf/decoration/.cd, + distance/.initial=10pt + } + \pgfdeclaredecoration{add dim}{final}{ + \state{final}{% + \pgfmathsetmacro{\dist}{5pt*\pgfkeysvalueof{/pgf/decoration/distance}/abs(\pgfkeysvalueof{/pgf/decoration/distance})} + \pgfpathmoveto{\pgfpoint{0pt}{0pt}} + \pgfpathlineto{\pgfpoint{0pt}{2*\dist}} + \pgfpathmoveto{\pgfpoint{\pgfdecoratedpathlength}{0pt}} + \pgfpathlineto{\pgfpoint{(\pgfdecoratedpathlength}{2*\dist}} + \pgfsetarrowsstart{latex} + \pgfsetarrowsend{latex} + \pgfpathmoveto{\pgfpoint{0pt}{\dist}} + \pgfpathlineto{\pgfpoint{\pgfdecoratedpathlength}{\dist}} + \pgfusepath{stroke} + \pgfpathmoveto{\pgfpoint{0pt}{0pt}} + \pgfpathlineto{\pgfpoint{\pgfdecoratedpathlength}{0pt}} + }} + \tikzset{dim/.style args={#1,#2}{decoration={add dim,distance=#2}, + decorate, + postaction={decorate,decoration={text along path, + raise=#2, + text align={align=center}, + text={#1}}}} + } + \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}, + x tick label style = {font=\footnotesize}, + xticklabels={\((\mu-2\sigma)\), \((\mu-\sigma)\), \(\mu\), \((\mu+\sigma)\), \((\mu+2\sigma)\)}, + xlabel={\(x\)}, + every axis x label/.style={at={(current axis.right of origin)},anchor=north west}, + every axis y label/.style={at={(axis description cs:-0.02,0.2)}, anchor=south west, rotate=90}, + 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; + \fill[white!30] (-1,0) -- plot[id=f3,domain=-1:1,samples=50] function {1/(0.75*sqrt(2*pi))*exp(-x*x*0.5/(0.75*0.75))} -- (1,0) -- cycle; + \begin{scope}[<->] + \draw (-1,0.35) -- (1,0.35) node [midway, fill=white] {68.3\%}; + \draw (-2,0.25) -- (2,0.25) node [midway, fill=white] {95.5\%}; + \draw (-3,0.15) -- (3,0.15) node [midway, fill=white] {99.7\%}; + \end{scope} + \begin{scope}[-, dashed, gray] + \draw (-1,0) -- (-1, 0.35); + \draw (1,0) -- (1, 0.35); + \draw (-2,0) -- (-2, 0.25); + \draw (2,0) -- (2, 0.25); + \draw (-3,0) -- (-3, 0.15); + \draw (3,0) -- (3, 0.15); + \end{scope} + \end{axis} + \begin{axis}[every axis plot post/.append style={ + mark=none,domain=-3:3,samples=50,smooth}, + axis x line=bottom, + enlargelimits=upper, + x=\textwidth/10, + xtick={-2,-1,0,1,2}, + axis x line shift=30pt, + hide y axis, + x tick label style = {font=\footnotesize}, + xlabel={\(Z\)}, + every axis x label/.style={at={(axis description cs:1,-0.25)},anchor=south west}] + \addplot {gauss(0,0.75)}; + \end{axis} + \end{tikzpicture} + \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}