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diff --git a/Master/texmf-dist/doc/latex/exam-n/sample/numerical3.tex b/Master/texmf-dist/doc/latex/exam-n/sample/numerical3.tex new file mode 100644 index 00000000000..554de298e9d --- /dev/null +++ b/Master/texmf-dist/doc/latex/exam-n/sample/numerical3.tex @@ -0,0 +1,42 @@ +\documentclass[compose]{exam-n} +\begin{document} + +\begin{question}{30} \comment{by Graham Woan} +Distinguish between frequentist and Bayesian definitions of +probability, and explain carefully how parameter estimation is +performed in each regime.\partmarks{10} + +A square ccd with $M\times M$ pixels takes a dark frame for +calibration purposes, registering a small number of electrons in +each pixel from thermal noise. The probability of there being $n_i$ +electrons in the $i$th pixel follows a Poisson distribution, i.e. +\begin{equation*} + P(n_i|\lambda) = \exp(-\lambda)\lambda^{n_i}/n_i!, +\end{equation*} +where $\lambda$ is the same constant for all pixels. Show that the +expectation value of is $\langle n_i \rangle = \lambda$. +\partmarks{5} [You may assume the relation $\sum_0^\infty \frac{x^n}{n!}=\exp(x)$.] + +Show similarly that +\begin{equation*} + \langle n_i(n_i-1) \rangle = \lambda^2. +\end{equation*} +and hence, or otherwise, that the variance of $n_i$ is also +$\lambda$. +\partmarks{5} + +The pixels values are summed in columns. Show that these sums, +$S_j$, will be drawn from a parent probability distribution that is +approximately +\begin{equation*} p(S_j|\lambda)=\frac{1}{\sqrt{2\pi +M\lambda}}\exp\left[-\frac{(S_j-M\lambda)^2}{2M\lambda}\right], +\end{equation*} +clearly stating any theorems you use. +\partmarks{5} + +Given the set of $M$ values $\{S_j\}$, and interpreting the above +as a Bayesian likelihood, express the posterior probability for +$\lambda$, justifying any assumptions you make. +\partmarks{5} +\end{question} +\end{document} |