

\subsection{\bf 1.6\hspace{1em} Means--Moments--Correlations}
\subsection{\bf for Multidimensional Distributions I}

\begin{slide}
$\bullet$ Given $n$-dimensional distribution $p(x_1, \ldots x_n)$:

\begin{itemize}
\item The mean now becomes an $n$-dimensional vector
\[
\left\langle x_{k} \right\rangle = \int\limits \otimes\,
dx_i \hspace{2em} x_k\, p(x_1 \ldots x_n)
\]

\item The variance becomes an $n\times n$ moment matrix
{\bf M}

\[
\mbox{\bf M}_{\,kl} = \int\limits \otimes\, dx_i \ \ \left\lbrack
\mbox{\bf x}_{\,k} - \left\langle \mbox{\bf x}_{\,k} \right\rangle
\right\rbrack \left\lbrack \mbox{\bf x}_{\,l} - \left\langle \mbox{\bf
x}_{\,l} \right\rangle \right\rbrack p\left(x_1 \ldots x_n \right)
\]

\item Note for we can show---analogously to Equation~(9)---that
\begin{equation}
M_{kl} = \left\langle x_k x_l \right\rangle - \left\langle x_k \right\rangle
\left\langle x_l \right\rangle
\end{equation}

\item Also note that {\bf M} is symmetric and positive semi-definite.
[This follows because $y_k\, M_{kl}\, y_l$---any $y$---clearly has a
$\geq 0$ integrand].

\item Now the diagonal terms of {\bf M} are called variances which are
$\geq 0$:

\item Their square roots are standard deviations again.

\item Further, when $k\neq l$, we put
${\bf \rho\,}_{kl} = M_{kl}/\sqrt{M_{kk}\, M_{ll}}$,
which is $\leq 1$ and is called the correlation coefficient.
\end{itemize}
\end{slide}

