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+\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}