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author | Eugeniy Mikhailov <evgmik@gmail.com> | 2013-08-30 17:38:34 -0400 |
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committer | Eugeniy Mikhailov <evgmik@gmail.com> | 2013-08-30 17:42:57 -0400 |
commit | b3c921f6e78472fcbd69ae248ca636686cd5d4cc (patch) | |
tree | 047333fc81ffd6c08e54e1593eb86f13307e5a4d /manual/chapters/appendices.tex | |
parent | 8a985f9892c45ceda9de2fccfb42d4bf5122b0a9 (diff) | |
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diff --git a/manual/chapters/appendices.tex b/manual/chapters/appendices.tex deleted file mode 100644 index d310873..0000000 --- a/manual/chapters/appendices.tex +++ /dev/null @@ -1,154 +0,0 @@ -\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} |