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@@ -11,10 +11,16 @@ tex_template = """\\documentclass[12pt]{article} \\usepackage{amsmath} \\usepackage{amsfonts} +\\newtheorem{definition}{Definition} +\\newtheorem{lemma}{Lemma} + {% if p is not none %} \\title{ {{title}} } {% endif %} +\\setlength{\parindent}{0cm} +\\setlength{\parskip}{0.3em} + \\begin{document} {% if title is not none %} \maketitle diff --git a/sem6/prob/m4/notes.tex b/sem6/prob/m4/notes.tex new file mode 100644 index 0000000..4a1aea0 --- /dev/null +++ b/sem6/prob/m4/notes.tex @@ -0,0 +1,145 @@ +\title{Special Probability Distributions} + +TODO Look at discrete distributions. + +\section{Moment Generations} +\emph{These are left over from the last lecture.} + +Two random variables with the same $E[X]$ and $Var(X)$ and variance are not the same. + +Instead one can calculate the expantance of a higher order of random variable. + +\begin{definition} + The $n$'th moment of a r.v. is defined as: \[ + E[X^n] = \sum_{i} x_{i}^{n} p(x_i) + .\] +\end{definition} + +% TODO Describe how to extract a moment +\begin{definition} + The \emph{Moment Generation Function} for a r.v. variable is defined as: \[ + \varphi(t) = E[e^{t X}] = \sum_{i} e^{t x_i} p(x_i) +.\] + The continues version is given by: \[ + \varphi(t) = E[e^{t X}] = \int_{-\infty}^{\infty} e^{tx} f(x) \,dx + .\] +\end{definition} + +From this function one can generate all moments of the random variable X. +The variance can be calculated from the first two moments. + +\begin{lemma} + If two r.v. have the same moments they can be said to be the same. +\end{lemma} + +\section{Discrete Distributions} + +These are all covered nicely in the book, in section 3.1.5. + +\subsection{Bernoulli} + +\begin{definition} + If a random variable is \emph{Bernoulli} with probability $p$, its PMF is: \[ + P_X(x) = \left\{ + \begin{array}{ll} + p & \mathrm{for} \: x = 1 \\ + 1 - p & \mathrm{for} \: x = 0 \\ + 0 & \mathrm{otherwise} \\ + \end{array} + \right. + .\] +\end{definition} + +The Bernoulli random variable can also be called the \emph{Indicator} random variable. +Because either event $A$ occurs or not. + +\subsection{Geometric} + +Is a series of independent Bernoulli tails, such as the number of coin tosses before a heads occurs. + +\begin{definition} + If X is \emph{geometric} with parameter $p$ its PMF is: \[ + P_X(k) = \left\{ + \begin{array}{ll} + p(1-p)^{k-1} & \mathrm{for} \: k = 1,2,3,... \\ + 0 & \mathrm{otherwise} + \end{array} + \right. + .\] + where $0 < p < 1$. +\end{definition} + +\subsection{Binomial} + +Suppose a coin toss with $P(H) = p$. +If the coin is tossed $n$ times $X$ defines the number of heads that are observed. + +\begin{definition} + If $X \sim Binomial(n,p)$, X is said to be \emph{binomial} and its PMF is: \[ + P_X(k) = \left\{ + \begin{array}{ll} + \binom{n}{k} p^k (1 - p)^{n-k} & \mathrm{for} \: k = 0,1,2,...,n \\ + 0 & \mathrm{otherwise} + \end{array} + \right. + .\] + where $0 < p < 1$. +\end{definition} + +\subsection{Pascal} + +Is also called \emph{Negative binomial} and describes the number of trails before $m$ successes. + +\begin{definition} + If $X \sim Pascal(m,p)$ its PMF is: \[ + P_X(k) = \left\{ + \begin{array}{ll} + \binom{k-1}{m-1} p^m (1-p)^{k-m} & \mathrm{for} \: k=m,m+1,m+2,... \\ + 0 & \mathrm{otherwise} + \end{array} + \right. + .\] + where $0 < p < 1$. +\end{definition} + +\subsection{Hyper geometric} + +Suppose that a bag contains $b$ blue and $r$ red marbles, and $k \leq b + r$ marbles are chosen. +Then $X$ is the number of chosen blue marbles. + +\begin{definition} + If $X \sim Hypergeometric(b,r,k)$ its PMF is: \[ + P_X(k) = \left\{ + \begin{array}{ll} + \frac{\binom{b}{x} \binom{r}{k-x}}{\binom{b+r}{k}} & \mathrm{for} \: x \in R_X \\ + 0 & \mathrm{otherwise} + \end{array} + \right. + .\] + where $R_X = \{\max(0, k-r), \max(0,k-r)+1,...,\min(k,b)\}$. +\end{definition} + +\subsection{Poisson} + +Can be used very well to model random variables in nature. + +\begin{definition} + A random variable with values 0,2,3,... can be said to be Poisson with parameter $\lambda > 0$, with PMF: \[ + P(X = i) = e^{-\lambda} \frac{\lambda^i}{i!} + .\] +\end{definition} + +The expected value is: $ + E[X] = \lambda +$ + +And the variance is: $ + Var(X) = \lambda +$ + +The Poisson distribution can be used to approximate binomial distribution. + +\begin{lemma} + Two independent Poisson r.v. added together give a poisson distribution with $\lambda = \lambda_1 + \lambda_2$. +\end{lemma} + |