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| Title of object: |
random vector |
| Canonical Name: |
RandomVector |
| Type: |
Definition |
| Created on: |
2004-06-30 12:25:09 |
| Modified on: |
2005-06-30 21:30:46 |
| Classification: |
msc:62H99, msc:15A52 |
| Defines: |
random matrix, distribution of a random vector, distribution of a random matrix |
Preamble:
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Content:
A \emph{random vector} is a finite-dimensional formal (column) vector of random variables. So if $X_1,X_2,\ldots,X_n$ are random variables, $\textbf{X}=(X_1,X_2,\ldots,X_n)^T$ is a random vector. One can similarly define a
\emph{random matrix} to be a formal matrix whose entries are all random variables.
Let $\textbf{X}=(X_1,X_2,\ldots,X_n)^T$ be a random vector. If E($X_i$) exists ($<\infty$) for each $i$, then the expectation of $\textbf{X}$, called the
\emph{mean vector} and denoted by $\mathbf{E}(\textbf{X})$, is defined to be:
$$\mathbf{E}(\textbf{X}):=(\mbox{E}(X_1),\mbox{E}(X_2),\ldots,\mbox{E}(X_n))^T.$$ One obvious consequence is that $\mathbf{E}(\textbf{X})^T=\mathbf{E}(\textbf{X}^T)$. The expectation of a random matrix is similarly defined.
Again, let $\textbf{X}=(X_1,X_2,\ldots,X_n)^T$ be a random vector. If $\boldsymbol{\mu}$=$\mathbf{E}(\textbf{X})$ is defined and E($X_iX_j$) are defined for all $1\leq i,j \leq n$, then the variance of $\textbf{X}$, denoted by $\textbf{Var}(\textbf{X})$, is defined to be: $$\textbf{Var}(\textbf{X}):=\mathbf{E}((\textbf{X}-\boldsymbol{\mu})(\textbf{X}-\boldsymbol{\mu})^T).$$ It is not hard to see that $\textbf{Var}(\textbf{X})$ is an $n\times n$ symmetric matrix and it is equal to the covariance matrix of the $X_i$'s.
The \emph{distribution of a random vector} is just the joint distribution of its coordinates. Similary, the \emph{distribution of a random matrix} is the joint distribution of its matric components.
\textbf{\PMlinkescapetext{Properties}:}
\begin{enumerate}
\item If \textbf{X} is an $n$-dimensional random vector with \textbf{A} a $m\times n$ constant matrix and $\boldsymbol{\alpha}$ an $m$-dimensional constant vector, then $$\mathbf{E}(\mathbf{AX}+\boldsymbol{\alpha})=\mathbf{AE}(\mathbf{X})+\boldsymbol{\alpha}.$$
\item Same set up as above. Then $$\mathbf{Var}(\mathbf{AX}+\boldsymbol{\alpha})=\mathbf{AVar}(\mathbf{X})\mathbf{A}^T.$$ If the ${X_i}$'s are \emph{iid} (independent identically distributed), with variance $\boldsymbol{\sigma}^2$, then $$\mathbf{Var}(\mathbf{AX}+\boldsymbol{\alpha})=\boldsymbol{\sigma}^2\mathbf{AA}^T.$$
\item Let $\mathbf{X}$ be an $n$-dimensional random vector with $\boldsymbol{\mu}=\mathbf{E(X)}$,
$\boldsymbol{\Sigma}=\mathbf{Var(X)}$. $\mathbf{A}$ is an $n\times n$ constant matrix. Then
$$\mathbf{E}(\mathbf{X}^T\mathbf{AX})=\operatorname{tr}(\mathbf{A}\boldsymbol{\Sigma})+
\boldsymbol{\mu}^T\mathbf{A}\boldsymbol{\mu}.$$
\end{enumerate} |
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