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diff --git a/manual/chapters/appendices.tex b/manual/chapters/appendices.tex new file mode 100644 index 0000000..d310873 --- /dev/null +++ b/manual/chapters/appendices.tex @@ -0,0 +1,154 @@ +\chapter*{Errors} + +\section*{Propagation of Random Errors} +Suppose one measures basically the same quantity twice. This might be the +number of $\gamma$-rays detected in 10 minutes with a scintillation detector. +Let $n_1$ be the number detected the first time and $n_2$ the number the +second time. Assume that the average number for many such measurements is +$\overline{n}$. We may then consider a variety of averages denoted by $<>$: +\begin{eqnarray*} +\overline{n}&=&<n>\\ +\overline{n_1}&=&<n>=\overline{n}\\ +<n_1-\overline{n}>&=&0\\ +\overline{n_2}&=&<n>\\ +\sigma_n&=&\sqrt{<(n-\overline{n})^2>} +\end{eqnarray*} + +The root-mean-square(rms) deviation from the mean, ( $\sigma$) is what is +often called the +error in a measurement. +We now determine the ``variance'' ($\sigma^2$) expected for various combinations of +measurements. One only needs to take the square root of $\sigma^2$ to obtain +the error. +\begin{eqnarray*} +\sigma^2&=&<(n_1-\overline{n}+n_2-\overline{n})^2>\\ +&=&<(n_1-\overline{n})^2+(n_2-\overline{n})^2+2(n_1-\overline{n})(n_2-\overline{n})>\\ +&=&<(n_1-\overline{n})^2>+<(n_2-\overline{n})^2>+2<(n_1-\overline{n})(n_2-\overline{n})>\\ +&=&<(n_1-\overline{n})^2>+<(n_2-\overline{n})^2>+2<(n_1-\overline{n})><(n_2-\overline{n})>\\ +\sigma^2&=&\sigma_1^2+\sigma_2^2+0 +\end{eqnarray*} +The average value of the last term is zero since the two measurements are +independent and one can take the averages of each part separately. + +With this result it is easy to get the variance in a linear combination of +$n_1$ and $n_2$. If + +\begin{displaymath} +f=a\cdot n_1 +b\cdot n_2 +\end{displaymath} + +then: +\begin{displaymath} +\sigma_f^2=a^2\sigma_1^2+b^2\sigma_2^2 +\end{displaymath} + +If the errors are small and $f$ is a function of $n_1$ and $n_2$: $f(n_1,n_2)$ +then: +\begin{equation}\label{ssgen} +\sigma_f^2=\left(\frac{\partial f}{\partial n_1}\right)^2\sigma_1^2+\left(\frac{\partial f}{\partial n_2}\right)^2\sigma_2^2 +\end{equation} +It should be clear that one can extend Eq. \ref{ssgen} to arbitrary numbers of +parameters. + +As an example of this latest form suppose $f=n_1\cdot n_2$ then: +\begin{displaymath} +\sigma_f^2=n_2^2\sigma_1^2+n_1^2\sigma_2^2 +\end{displaymath} +or +\begin{displaymath} +\frac{\sigma_f^2}{f^2}=\frac{\sigma_1^2}{n_1^2}+\frac{\sigma_2^2}{n_2^2} +\end{displaymath} + +Thus in this case the fractionial variances add. + +Note: the $\sigma_m$ the error in the mean of $n$ measurements of the +same thing is: $\sigma_m=\sigma /\sqrt{n}$. +\subsection*{Probability Distribution Functions} +\subsubsection*{Binomial} +If the probability of {\it success} in a trial is $p$ then +the probability of $n$ {\it successes} in $N$ trials is: +\begin{displaymath} +P(n)=\frac{N!}{(N-n)!n!}p^n(1-p)^{N-n} +\end{displaymath} +This distribution has a mean $\mu=Np$ and variance $\sigma^2=Np(1-p)$. +This is the starting point for figuring the odds in card games, for example. +\subsubsection*{Poisson} +The probability of $n$ events is: +\begin{displaymath} +P(n)=\frac{e^{-\mu}\mu^n}{n!} +\end{displaymath} +where is the $\mu$ is the mean value and the variance, $\sigma^2=\mu$. +This is the distribution one gets, e.g., with the number of radioactive +decays detected in a fixed finite amount of time. It can be derived from +the binomial distribution in an appropriate limit. +\subsubsection*{Normal or Gaussian Distribution} +This is the first continuous probability distribution. +\begin{displaymath} +P(x)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{\frac{-(x-\mu)^2}{2\sigma^2}} +\end{displaymath} +This function, as you might guess, has mean $\mu$ and variance $\sigma^2$. +If one makes averages of almost anything one finds that the result is +almost always well described by a Normal distribution. Both the binomial and +Poisson distributions approach this distribution in appropriate limits as +does the $\chi^2$ described below. +\subsubsection*{Chi-square distribution: $\chi^2$} +This probability density function (pdf) has the parameter: $N_f$, the number of +degrees of freedom. It is: +\begin{displaymath} +P(x)=\frac{\frac{1}{2}\left(\frac{x}{2}\right)^{(N_f/2)-1}}{\Gamma\left( +\frac{N_f}{2}\right)} +\end{displaymath} +The mean of this pdf is: $\mu=N_f$ and the variance is: $\sigma^2=2N_f$. +The pdf is of considerable use in physics. It is used extensively in the +fitting of histogrammed data. +\newpage + +\appendix{Linear Least Squares} + + +Consider a set of experimental results measured as a function of some +parameter $x$, i.e., $E(x_i)$. Suppose that these results are expected to +be represented by a theoretical function $T(x_i)$ and that $T(x_i)$ is +in turn linearly expandable in terms of independent functions $f_j(x_i)$: +\begin{displaymath} +T(x_i)=\sum_ja_jf_j(x_i) +\end{displaymath} +Suppose now one wants to find the coefficients $a_j$ by minimizing $\chi^2$, +the sum of differences between the experimental results and the theoretical +function, squared, i.e., minimize: +\begin{displaymath} +\chi^2=\sum_i\left(\sum_ja_jf_j(x_i)-E(x_i)\right)^2 +\end{displaymath} +This is found by finding: +\begin{displaymath} +0=\frac{\partial}{\partial a_k}\chi^2= +2\cdot \sum_i\left(\sum_ja_jf_j(x_i)-E(x_i)\right)\cdot f_k(x_i) +\end{displaymath} +This may be rewritten as: +\begin{equation}\label{meq} +\sum_i\left(\sum_ja_jf_j(x_i)f_k(x_i)\right)=\sum_iE(x_i)f_k(x_i) +\end{equation} +The rest is algebra. The formal solution, which can in fact be +easily implemented, is to first define: +\begin{eqnarray} +M_{j,k}&=&\sum_if_j(x_i)f_k(x_i\\ +V_k(i)&=&\sum_iE(x_i)f_k(x_i) +\end{eqnarray} +So that Eq. \ref{meq}. becomes: +\begin{displaymath} +\sum a_jM_{j,k}=V_k +\end{displaymath} +The $a_j$ may then be found by finding the inverse of $M_{j,k }$ +\begin{figure}: +\begin{displaymath} +a_j=\sum_kV_k\cdot M^{-1}_{k,j} +\end{displaymath} +Question: How does this procedure change if: +\begin{displaymath} +\chi^2=\sum_i\frac{(T(x_i)-E(x_i))^2}{\sigma(x_i)^2} +\end{displaymath} +where $\sigma(x_i)$ is the error in the measurement of $E(x_i)$? + +\centerline{\epsfig{width=\linewidth,angle=-90, file=datafg.eps}} +\caption{\label{lsqf} Data Fit to a Straight Line.} +\end{figure} |