From: Andrew Lorimer Date: Sun, 4 Aug 2019 06:36:04 +0000 (+1000) Subject: [methods] binomial theorem X-Git-Tag: yr12~76 X-Git-Url: https://git.lorimer.id.au/notes.git/diff_plain/8809411f48d5ab2300a6c1ab3d94633711e6520d [methods] binomial theorem --- diff --git a/methods/statistics.pdf b/methods/statistics.pdf index 991c449..8ebb5fa 100644 Binary files a/methods/statistics.pdf and b/methods/statistics.pdf differ diff --git a/methods/statistics.tex b/methods/statistics.tex index ed51259..6bfbe48 100644 --- a/methods/statistics.tex +++ b/methods/statistics.tex @@ -1,5 +1,6 @@ \documentclass[a4paper]{article} \usepackage[a4paper,margin=2cm]{geometry} +\usepackage{array} \usepackage{amsmath} \usepackage{amssymb} \usepackage{tcolorbox} @@ -51,17 +52,27 @@ \begin{itemize} \item \textbf{Probability distribution graph} - a series of points on a cartesian axis representing results of outcomes. $\Pr(X=x)$ is on $y$-axis, $x$ is on $x$ axis. - \item \textbf{Mean $\mu$} - measure of central tendency. Also known as \textit{balance point} or \textit{expected value} of a distribution. Centre of a symmetrical distribution. - \item \textbf{Mode} - most popular value (has highest probability of \(X\) values). Multiple modes can exist if \(>1 \> X\) value have equal-highest probability. + \item \textbf{Mean $\mu$} or \textbf{expected value} \(E(X)\) - measure of central tendency. Also known as \textit{balance point}. Centre of a symmetrical distribution. + \begin{align*} + \overline{x} = \mu = E(X) &= \frac{\Sigma(xf)}{\Sigma(f)} \\ + &= \sum_{i=1}^n (x_i \cdot P(X=x_i)) \\ + &= \int_{-\infty}^{\infty} x\cdot f(x) \> dx \quad \text{(for pdf } f \text{)} + &= \sum_{-\infty}^{\infty} + \end{align*} + \item \textbf{Mode} - most popular value (has highest probability of \(X\) values). Multiple modes can exist if \(>1 \> X\) value have equal-highest probability. Number must exist in distribution. \item \textbf{Median \(m\)} - the value of \(x\) such that \(\Pr(X \le m) = \Pr(X \ge m) = 0.5\). 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. - \item \textbf{Variance $\sigma^2$} - measure of spread of data around the mean. Not the same magnitude as the original data. Represented by $\sigma^2=\operatorname{Var}(x) = \sum (x=\mu)^2 \times p(x) = \sum (x-\mu)^2 \times \Pr(X=x)$. Alternatively: $\sigma^2 = \operatorname{Var}(X) = \sum x^2 \times p(x) - \mu^2$ + \[ m = X \> \text{such that} \> \int_{-\infty}^{m} f(x) dx = 0.5 \] + \item \textbf{Variance $\sigma^2$} - measure of spread of data around the mean. Not the same magnitude as the original data. For distribution \(x_1 \mapsto p_1, x_2 \mapsto p_2, \dots, x_n \mapsto p_n\): + \begin{align*} + \sigma^2=\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 + \end{align*} \item \textbf{Standard deviation $\sigma$} - measure of spread in the original magnitude of the data. Found by taking square root of the variance: $\sigma =\operatorname{sd}(X)=\sqrt{\operatorname{Var}(X)}$ \end{itemize} \subsubsection{Expectation theorems} - \[ \overline{x} = \frac{\Sigma(xf)}{\Sigma(f)} = \Sigma (x p(x)) \tag{expected value} \] - \begin{align*} E(aX \pm b) &= aE(X) \pm b \\ E(z) &= z \\ @@ -71,7 +82,38 @@ \end{align*} + \section{Binomial Theorem} + + \begin{align*} + (x+y)^n &= {n \choose 0} x^n y^0 + {n \choose 1} x^{n-1}y^1 + {n \choose 2} x^{n-2}y^2 + \dots + {n \choose n-1}x^1 y^{n-1} + {n \choose n} x^0 y^n \\ + &= \sum_{k=0}^n {n \choose k} x^{n-k} y^k \\ + &= \sum_{k=0}^n {n \choose k} x^k y^{n-k} + \end{align*} + \begin{enumerate} + \item powers of \(x\) decrease \(n \rightarrow 0\) + \item powers of \(y\) increase \(0 \rightarrow n\) + \item coefficients are given by \(n\)th row of Pascal's Triangle where \(n=0\) has one term + \item Number of terms in \((x+a)^n\) expanded \& simplified is \(n+1\) + \end{enumerate} + + Combinations: \(^n\text{C}_r = {N\choose k}\) (binomial coefficient) + \begin{itemize} + \item Arrangements \({n \choose k} = \frac{n!}{(n-r)}\) + \item Combinations \({n \choose k} = \frac{n!}{r!(n-r)!}\) + \item Note \({n \choose k} = {n \choose k-1}\) + \end{itemize} + \subsubsection{Pascal's Triangle} + + \begin{tabular}{>{$}l<{$\hspace{12pt}}*{13}{c}} + n=\cr0&&&&&&&1&&&&&&\\ + 1&&&&&&1&&1&&&&&\\ + 2&&&&&1&&2&&1&&&&\\ + 3&&&&1&&3&&3&&1&&&\\ + 4&&&1&&4&&6&&4&&1&&\\ + 5&&1&&5&&10&&10&&5&&1&\\ + 6&1&&6&&15&&20&&15&&6&&1 + \end{tabular} \end{document}