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@bemasher
Created September 13, 2012 03:40
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\documentclass[twocolumn,10pt]{article}
\usepackage[top=0.5cm, bottom=0.5cm, left=0.5cm, right=0.5cm]{geometry}
\usepackage{amssymb, amsmath, amstext}
\usepackage{datetime}
\usepackage{listings}
\usepackage[pdftex]{graphicx}
\usepackage{subfigure}
\usepackage{siunitx}
\usepackage{enumitem}
\setlist{nosep}
\setlist{noitemsep}
\setlength{\parindent}{0.0cm}
\setlength{\parskip}{0.1cm}
\newcommand{\defhead}[1]{\textbf{\underline{#1}:} \\}
\newcommand{\vect}[1]{\hat{\boldsymbol{#1}}}
\renewcommand{\familydefault}{\sfdefault}
\renewcommand{\thefootnote}{\arabic{footnote}}
\everymath{\allowdisplaybreaks\displaystyle}
\makeatletter
\renewcommand\@makefnmark{\@textsuperscript{\normalfont(\@thefnmark)}}
\makeatother
\begin{document}
\textbf{Probability and Set Theory}:
\begin{eqnarray*}
P[\phi] &=& 0 \\
P[A^c] &=& 1 - P[A] \\
P[AB] &=& P[A] \cap P[B] \\
P[A\cup B] &=& P[A] + P[B] - P[A\cap B] \\
\text{If } A \subset B &\text{ then }& P[A] \leq P[B] \\
P[A|B] &=& \frac{P[AB]}{P[B]} = \frac{P[B|A]P[A]}{P[B]} \\
P[B] &=& P[B|A]P[A]+P[B|A^c]P[A^c] \\
\text{Indep. iff } P[AB] &=& P[A]P[B] \\
&=& P[A|B] = P[A] \land P[B|A] = P[B]
\end{eqnarray*}
\textbf{Sampling}
\begin{eqnarray*}
k \text{-permutations } (n)_k &=& \frac{n!}{(n-k)!} \\
n \text{ choose } k \; \binom{n}{k} &=& \binom{n}{n-k} = \frac{n!}{k!(n-k)!}
\end{eqnarray*}
\textbf{Bernoulli Trials\footnote{Where $p$ is probability of pass.}}:
\begin{eqnarray*}
P[\text{Exactly } k \text{ pass of }n] &=& \binom{n}{k}(1-p)^{n-k}p^k
\end{eqnarray*}
\textbf{Systems (Series)\footnotemark[\value{footnote}]}:
\begin{eqnarray*}
P[\text{Pass}] &=& p^n \\
P[\text{Fail}] &=& 1 - p^n
\end{eqnarray*}
\textbf{Systems (Parallel)\footnotemark[\value{footnote}]}:
\begin{eqnarray*}
P[\text{Pass}] &=& 1 - (1 - p)^n \\
P[\text{Fail}] &=& (1 - p)^n
\end{eqnarray*}
\textbf{Random Variable Distributions}:
\begin{itemize}[topsep=0pt]
\item Uniform - Equiprobable: $1/(l-k+1)$
\item Bernoulli - Only two outcomes: $\left\{\begin{matrix}
1-p & x=0 \\
p & x=1 \\
0 & else
\end{matrix}\right.$
\item Geometric - Pass after $n$ trials: $p(1-p)^{n-1}$
\item Binomial - Number of passes in $n$ trials: $\binom{n}{k}p^k(1-p)^{n-k}$
\item Pascal - Fails until pass on final in $n$ trials: $\binom{n-1}{k-1}p^k(1-p)^{n-k}$
\item Poisson - Average passes per time: $\left\{\begin{matrix}
\frac{\alpha^x e^{-\alpha}}{x!} & x \geq 0, \alpha=\lambda T \\
\lambda & \text{Pass per time} \\
\end{matrix}\right.$
\end{itemize}
\textbf{Operations on PMF}:
\begin{eqnarray*}
\text{CDF } F_X(x) &=& P[X\leq x] = \sum_{i=0}^{x} P[x_i] \\
\text{Mode } P_X(x_\text{mod}) &\geq& P_X(x) \forall x \\
\text{Median } P[X < x_\text{med}] &=& P[X > x_\text{med}] \\
\end{eqnarray*}
\newpage
\textbf{Expected}:
\begin{itemize}
\item Expected $E[X] = \mu_X = \sum_{x\in S_X}x P_X(x)$
\item Uniform $E[X] = (k+l)/2$
\item Binomial $E[X]=np$
\item Pascal $E[X]=k/p$
\end{itemize}
\textbf{Variance, Std Dev}:
\begin{itemize}
\item Variance - $\sigma_X^2 = E[(X-\mu_X)^2] = E[X^2]-\mu_X^2$
\item Std Dev - $\sqrt{\text{Var}[X]} = \sqrt{\sigma_X^2} = \sigma$
\item Nth Moment - $E[X^n]$
\item Nth Central Moment - $E[(X-\mu_X)^n]$
\item Thm. 2.14 - $\text{Var}[aX+b]=a^2\text{Var}[X]$
\end{itemize}
\textbf{Derived RV}:
\begin{itemize}
\item $E[Y] = \mu_Y = \sum_{x\in S_X}g(x)P_X(x)$
\item $E[aX+b] = aE[X]+b$
\end{itemize}
\textbf{Conditional RV}:
\begin{itemize}
\item $P_{X|B}(x) = P[X=x|B] = \left\{\begin{matrix}\frac{P_X(x)}{P[B]} & x \in B \\ 0 & \text{else}\end{matrix}\right.$
\end{itemize}
\textbf{Cont. Uniform RV}:
\begin{itemize}
\item $f_X(x) = \left\{\begin{matrix}\frac{1}{b-a} & a \leq x < b \\ 0 & \text{else}\end{matrix}\right.$
\item $F_X(x) = \left\{\begin{matrix}0 & x \leq a\\ \frac{x-a}{b-a} & a < x \leq b\\ 1 & x > b\end{matrix}\right.$
\item $E[X]=\frac{a+b}{2}$
\item Var$[X]=\frac{(b-a)^2}{12}$
\end{itemize}
\textbf{Cont. Exponential RV}:
\begin{itemize}
\item $f_X(x) = \left\{\begin{matrix}\lambda e^{-\lambda x} & x \geq 0\\ 0 & \text{else}\end{matrix}\right.$
\item $F_X(x) = \left\{\begin{matrix}1-e^{-\lambda x} & x \geq 0\\ 0 & \text{else}\end{matrix}\right.$
\item $E[X] = \frac{1}{\lambda}$
\item Var$[X]=\frac{1}{\lambda^2}$
\end{itemize}
\textbf{Cont. Gaussian RV}:
\begin{itemize}
\item $f_X(x) = \frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x-\mu)^2/2\sigma^2}$
\item $F_X(x) = \Phi\left(\frac{x-\mu}{\sigma}\right) = \frac{1}{2}+\frac{1}{2}\text{Erf}\left(\frac{x}{\sqrt{2}}\right)$
\item $\Phi(-z) = Q(z) = 1 - \Phi(z)$
\item $E[X] = \mu$
\item Var$[X] = \sigma^2$
\end{itemize}
\newpage
\textbf{Inequalities}:
\begin{itemize}
\item Markov: $P[X\geq c^2] \leq \frac{E[X]}{c^2}$ (for $X>0$)
\item Chebyshev: $P[|Y-\mu_Y|\geq c] \leq \frac{\text{Var}[Y]}{c^2}$ (for $c\geq 0$)
\end{itemize}
\textbf{Estimates}:
\begin{itemize}
\item Mean Square Error: $e=E[(\hat{R} - r)^2]$ (Var for unbiased)
\item Sample Mean: Var$[M_n(X)]=\frac{\text{Var}[X]}{n}$
\item Sample Var: $V_n(X)=\frac{1}{n}\sum_{i=1}^n(X_i-M_n(X))^2$
\item $E[V_n(X)] = \frac{n-1}{n} \text{Var}[X]$
\end{itemize}
\textbf{Confidence Intervals}
\begin{itemize}
\item $P[|M_n(X) - \mu_X| \geq c]\leq \frac{\text{Var}[X]}{nc^2}=\alpha$
\item $P[|M_n(X) - \mu_X| < c] \geq 1 - \frac{\text{Var}[X]}{nc^2}= 1 - \alpha$
\end{itemize}
\textbf{Gaussian Confidence Interval}:
\begin{itemize}
\item $M_n(X) - c \leq \mu \leq M_n(X) + c$
\item $\alpha/2=Q\left(\frac{c\sqrt{x}}{\sigma}\right)=1-\Phi\left(\frac{c\sqrt{n}}{\sigma}\right)$
\item $1-\alpha=1-2Q\left(\frac{c\sqrt{n}}{\sigma_X}\right)$
\end{itemize}
\textbf{Significance Testing}:
\begin{itemize}
\item Type I Error: False Rejection
\item Type II Error: False Acceptance
\end{itemize}
\end{document}
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