CentralMoments.tex

\ifx\wholebook\relax\else
\documentclass[twoside]{book}
\usepackage[active]{srcltx}
\usepackage[LY1]{fontenc}
\input{DhbDefinitions}
\begin{document}
\fi

\chapter{Accurate accumulation of expectation values}
\section{Accurate accumulation of central moments}
\label{sec:centralmoments} This section shows the detailed
derivation of equation \ref{eq:deltamoment} of section
\ref{sec:robustmoment}. The aim of this demonstration is to
expressed the central moment of order $k$ estimated over a sample
of $n+1$ measurements as a function of the central moments of
order lower or equal to $k$ estimated over a sample of $n$
measurements. The estimator of the central moment of order $k$ is
defined by:
\begin{equation}
\label{eq:momstart}
 \left\langle\left(x-\bar{x}\right)^k\right\rangle_{n+1}
   =\displaystyle{1\over n+1}\sum_{i=1}^{n+1} \left(x_i-\left\langle
   x\right\rangle_{n+1}\right)^k.
\end{equation}
We shall now concentrate on changing the sum of equation
\ref{eq:momstart} in such way as to bring quantities which are
already computed. The sum of \ref{eq:momstart} is equal to
\begin{equation}
  \begin{array}{ll}
 S&=\left(n+1\right)\left\langle\left(x-\bar{x}\right)^k\right\rangle_{n+1}\\
   &=\displaystyle\sum_{i=1}^{n+1} \left(x_i-\left\langle x\right\rangle_{n+1}\right)^k\\
   &=\displaystyle\sum_{i=1}^{n+1} \left(x_i-\left\langle x\right\rangle_n+
   \left\langle x\right\rangle_n-\left\langle
   x\right\rangle_{n+1}\right)^k\\
   &=\displaystyle\sum_{i=1}^{n+1} \left[\left(x_i-\left\langle x\right\rangle_n\right)
   +\Delta_{n+1}\right]^k,
  \end{array}
\end{equation}
where we have introduced the correction defined in equation
\ref{eq:deltaverage}. We can now transform the expression inside
the sum using the binomial expansion:
\begin{equation}
\label{eq:deltamomentsstep}
  \begin{array}{ll}
 S&=\displaystyle\sum_{i=1}^{n+1} \sum_{l=0}^k \left({l\atop
   k}\right)
   \left(x_i-\left\langle x\right\rangle_n\right)^l
   \Delta_{n+1}^{k-l}\\
   &=\displaystyle\sum_{l=0}^k \left({l\atop
   k}\right)\sum_{i=1}^{n+1}
   \left(x_i-\left\langle x\right\rangle_n\right)^l
   \Delta_{n+1}^{k-l}\\
  \end{array}
\end{equation}
In the second part of equation \ref{eq:deltamomentsstep} the two
sums have been permuted. Like in the case of the average, we now
make the last term of the inner sum explicit. The remaining sum
can then be expressed as a function of the estimators of the
central moments over $n$ measurements. The term containing the
$(n+1)\th$ measurement can be rewritten as a function of the
correction defined in equation \ref{eq:deltaverage}. We have:
\begin{equation}
\label{eq:deltamomentsstep1}
  \begin{array}{ll}
 S&=\displaystyle\sum_{l=0}^k \left({l\atop
   k}\right)\left[\left(x_{n+1}-\left\langle
   x\right\rangle_n\right)^l \Delta_{n+1}^{k-l}+
   \Delta_{n+1}^{k-l}\sum_{i=1}^{n+1} \left(x_i-\left\langle x\right\rangle_n\right)^l
   \right]\\
   &=\displaystyle\sum_{l=0}^k \left({l\atop
   k}\right)\left[\left(n+1\right)^l\left({x_{n+1}-\left\langle
   x\right\rangle_n\over n+1}\right)^l \Delta_{n+1}^{k-l}+
   n\left\langle \left(x-\bar{x}\right)^l\right\rangle_n\Delta_{n+1}^{k-l}
   \right]\\
   &=\displaystyle\sum_{l=0}^k \left({l\atop
   k}\right)\left[\left(-n-1\right)^l \Delta_{n+1}^k+
   n\left\langle \left(x-\bar{x}\right)^l\right\rangle_n\Delta_{n+1}^{k-l}
   \right]\\
   &=\displaystyle\sum_{l=0}^k \left({l\atop
   k}\right)\left(-n-1\right)^l \Delta_{n+1}^k+\sum_{l=0}^k \left({l\atop
   k}\right)
   n\left\langle \left(x-\bar{x}\right)^l\right\rangle_n\Delta_{n+1}^{k-l}
   .\\
  \end{array}
\end{equation}
In the last line of equation \ref{eq:deltamomentsstep1} the first
term contains the binomial expansion of the following expression
\begin{equation}
  \sum_{l=0}^k \left({l\atop k}\right)\left(-n-1\right)^l
  = \left[1+\left(-n-1\right)\right]^k=\left(-n\right)^k.
\end{equation}
Thus,we have:
\begin{equation}
  \begin{array}{ll}
 S&=\displaystyle\left(-n\Delta_{n+1}\right)^k+n\sum_{l=0}^k \left({l\atop
   k}\right)\left\langle \left(x-\bar{x}\right)^l\right\rangle_n\Delta_{n+1}^{k-l}
   .\\
  \end{array}
\end{equation}
In this last equation, the first term of the sum is just
$\Delta_{n+1}^k$ and the second term of the sum vanishes by
definition of the average $\bar{x}$. This gives us the final
expression to compute the estimator of the central moment computed
over $n+1$ measurements as a function of the estimator of the
central moment computed over $n$ measurements
\begin{equation}
  \left\langle\left(x-\bar{x}\right)^k\right\rangle_{n+1}
  ={n\over n+1}\left\{
  \left[ 1 - \left(-n\right)^{k-1}\right]\Delta_{n+1}^k
  +\sum_{l=2}^k \left({l\atop k}\right)
  \left\langle\left(x-\bar{x}\right)^l\right\rangle_n
  \Delta_{n+1}^{k-l}
  \right\}.
\end{equation}
{\sl Quod erat demonstrandum$\ldots$}

\section{Accurate accumulation of the covariance}
\label{sec:robustcov} This section shows the detailed derivation
of equation \ref{eq:robustcov}. To simplify notation, the
components $x_i$ and $x_j$ have been renamed $x$ and $y$
respectively.

The estimator of the covariance of two random variables $x$ and
$y$ over $n$ measurements is defined by:
\begin{equation}
 {\cov}_n\left(x,y\right)=\left\langle \left(x_i-\left\langle
   x\right\rangle_n\right)\left(y_i-\left\langle
   y\right\rangle_n\right)
 \right\rangle_n
   ={1\over n}\sum_{i=1}^n \left(x_i-\left\langle
   x\right\rangle_n\right)\left(y_i-\left\langle
   y\right\rangle_n\right).
\end{equation}
The estimator of the covariance of $x$ and $y$ over $n+1$
measurements is then given by:
\begin{equation}
 {\cov}_{n+1}\left(x,y\right)={1\over n+1}\sum_{i=1}^{n+1} \left(x_i-\left\langle
   x\right\rangle_{n+1}\right)\left(y_i-\left\langle
   y\right\rangle_{n+1}\right).
\end{equation}
The sum in the equation above can then be expressed as:
\begin{equation}
  \begin{array}{ll}
 C_{n+1}&= \left(n+1\right)\left\langle \left(x_i-\left\langle
   x\right\rangle_{n+1}\right)\left(y_i-\left\langle
   y\right\rangle_{n+1}\right)
 \right\rangle_{n+1}\\
 &=\displaystyle\sum_{i=1}^{n+1} \left(x_i-\left\langle
   x\right\rangle_{n+1}\right)\left(y_i-\left\langle
   y\right\rangle_{n+1}\right)\\
 &=\displaystyle\sum_{i=1}^{n+1} \left(x_i-\left\langle
   x\right\rangle_n + \left\langle x\right\rangle_n-\left\langle
   x\right\rangle_{n+1}\right)\left(y_i-\left\langle
   y\right\rangle_n + \left\langle y\right\rangle_n-\left\langle
   y\right\rangle_{n+1}\right)\\
 &=\displaystyle\sum_{i=1}^{n+1} \left(x_i-\left\langle
   x\right\rangle_n + \Delta_{x,n+1}\right)\left(y_i-\left\langle
   y\right\rangle_n + \Delta_{y,n+1}\right),\\
  \end{array}
\end{equation}
where we have introduce the corrections to the estimation of the
expectation value of $x$ and $y$ as follow:
\begin{equation}
 \left\{
 \begin{array}{lcl}
   \Delta_{x,n+1}&=&\left\langle x\right\rangle_n-\left\langle
   x\right\rangle_{n+1}\\[1.5ex]
   &=&{\displaystyle\left\langle x\right\rangle_n-x_{n+1}\over\displaystyle
   n+1},\\[4ex]
   \Delta_{y,n+1}&=&\left\langle y\right\rangle_n-\left\langle
   y\right\rangle_{n+1}\\[1.5ex]
   &=&{\displaystyle\left\langle y\right\rangle_n-y_{n+1}\over\displaystyle
   n+1}.\\
 \end{array}
 \right.
\end{equation}
Thus, we have:
\begin{equation}
  \begin{array}{ll}
 C_{n+1}&=\displaystyle\sum_{i=1}^{n+1}\left[\left(x_i-\left\langle
   x\right\rangle_n\right)\left(y_i-\left\langle
   y\right\rangle_n\right)+\Delta_{y,n+1}\left(x_i-\left\langle
   x\right\rangle_n\right)\right.\\
   &\qquad\qquad\qquad\left.+\Delta_{x,n+1}\left(y_i-\left\langle
   y\right\rangle_n\right)+\Delta_{x,n+1}\Delta_{y,n+1}\right]\\
   &=\displaystyle\sum_{i=1}^{n+1}\left(x_i-\left\langle
   x\right\rangle_n\right)\left(y_i-\left\langle
   y\right\rangle_n\right)+\left(n+1\right)\Delta_{x,n+1}\Delta_{y,n+1}\\
   &\qquad\qquad\qquad+\Delta_{y,n+1}\left(x_{n+1}-\left\langle
   x\right\rangle_n\right)+\Delta_{x,n+1}\left(y_{n+1}-\left\langle
   x\right\rangle_n\right).\\
 \end{array}
\end{equation}
The last line is obtained from the definition of the expectation
values $\left\langle x\right\rangle_n$ and $\left\langle
y\right\rangle_n$.Using the definitions of $\Delta_{x,n+1}$ and
$\Delta_{y,n+1}$ we have:
\begin{equation}
  \begin{array}{ll}
 C_{n+1}&=\displaystyle\sum_{i=1}^{n+1}\left(x_i-\left\langle
   x\right\rangle_n\right)\left(y_i-\left\langle
   y\right\rangle_n\right)-\left(n+1\right)\Delta_{x,n+1}\Delta_{y,n+1}\\
   &=\displaystyle\sum_{i=1}^{n}\left(x_i-\left\langle
   x\right\rangle_n\right)\left(y_i-\left\langle
   y\right\rangle_n\right)-\left(n+1\right)\Delta_{x,n+1}\Delta_{y,n+1}\\
   &\qquad\qquad\qquad+\left(x_{n+1}-\left\langle
   x\right\rangle_n\right)\left(y_{n+1}-\left\langle
   y\right\rangle_n\right)\\[2ex]
   &=n\ {\cov}_n\left(x,y\right)+n\left(n+1\right)\Delta_{x,n+1}\Delta_{y,n+1}.\\
  \end{array}
\end{equation}
Now, we obtain the expression for the estimator of the covariance
over $n+1$ measurements as a function of the estimator of the
covariance over $n$ measurements:
\begin{equation}
 \ {\cov}_{n+1}\left(x,y\right)={n\over
 n+1}{\cov}_n\left(x,y\right)+n\Delta_{x,n+1}\Delta_{y,n+1}.
\end{equation}
Note that this equation yields equation \ref{eq:accumvariance} if
one put $y=x$.

\ifx\wholebook\relax\else\end{document}\fi