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Random Vectors

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Vectors  ::  Covariance  ::  Functions  ::  Application  ::  Markov Model

Random Vectors

Random vectors are an extension of the bivariate random variables.

$n$ r.v.s $X_1, X_2, \ldots, X_n$ define a measurable mapping from an underlying sample space $(\Omega, \Fc)$ to $(\Rbb^n,\Bc^n)$ , where $\Bc^n$ is the smallest $\sigma$ -field containing all sets of the form

\begin{displaymath}\{(x_1,x_2,\ldots,x_n): a_1 < x_1 \leq b_1, a_2 < x_2 \leq b_2,
\cdots, a_n < x_n \leq b_n\}.

The {\bf joint distribution} of $X_1, \ldots, X_n$ is
\end{displaymath}This probability is denoted as $P_\Xbf(B)$.

The {\bf joint cumulative distribution function} c.d.f. if
... a_n) = F_\Xbf(\abf),
\qquad \abf \in \Rbb^n.

The {\bf joint probability mass function} (p.m.f.) is
\begin{...}p_\Xbf(\abf) = P(X_1=a_1,\ldots, X_n = a_n)

The {\bf joint probability density function} (p.d.f.) $f_\Xb...
..._2 \cdots dx_n
\end{displaymath}for a continuous random vector.
Fact: $X_1, X_2, \ldots, X_n$ are independent if $F_\Xbf$ or $p_\Xbf$ or $f_\Xbf$ factor into products of marginals.

Suppose $g:(\Rbb^n,\Bc^n) \rightarrow (\Rbb^n,\Bc^n)$ is measurable. Then $g(X_1,\ldots, X_n)$ is a random variable. Law of unconscious statistician:

\begin{displaymath}E[g(X_1,\ldots, X_n)] =
\int \cdots \int g(x_...
x_{i_2}, \ldots, x_{i_n}) & \text{discrete}.

Copyright 2008, Todd Moon. Cite/attribute Resource . admin. (2006, May 31). Random Vectors. Retrieved January 07, 2011, from Free Online Course Materials — USU OpenCourseWare Web site: This work is licensed under a Creative Commons License Creative Commons License