... ... @@ -3,7 +3,7 @@ \usepackage{csquotes} \usepackage{hyperref} \usepackage{graphicx} \usepackage{graphicx, amsmath} \graphicspath{{img/}} \usepackage[backend=biber, sorting=none]{biblatex} ... ... @@ -17,9 +17,54 @@ \begin{document} \maketitle \begin{abstract} Abstract \end{abstract} \section*{The need for a guide} \section*{Mathematical description of what NIFTy needs} For first order minimization: \begin{itemize} \item $\mathcal H (d|s) = - \log \mathcal P (d|s) + \text{constant in }s$ \item $\mathcal H' = \frac{\partial}{\partial s} \mathcal H (d|s)$ \end{itemize} For second order minimization additionally: \begin{itemize} \item $\langle \mathcal H' \mathcal H'^\dagger \rangle_{\mathcal P (d|s)}$. \end{itemize} Note, that for Gaussian, Poissonian and Bernoulli likelihoods this term doesn't need to be calculated and implemented because NIFTy does it automatically. \section*{Becoming more specific} If the likelihood is Gaussian ($\mathcal H(d|s) \propto (d-R(s))^\dagger N^{-1} (d-R(s))$) or Poissonian ($\mathcal H(d|s) \propto - \log (R(s))^\dagger d+\sum_i R(s)_i$ ), NIFTy needs: \begin{itemize} \item $R$. \item $R'^\dagger := (\frac{\partial}{\partial s} R(s))^\dagger$. \item $R' = \frac{\partial}{\partial s} R(s)$. (only for 2nd order minimization) \end{itemize} \section*{Example $\gamma$-ray imaging} The information a $\gamma$-ray astronomer would provide to the algorithm (in the simplest case): \begin{itemize} \item Data has Poissonian statistics. \item Two functions: \verb|R(s)| and \verb|R_adjoint(d)| where $R$ applies a convolution with a Gaussian kernel of given size to \verb|s| and picks out the positions to which there exists a data point in \verb|d|. \verb|R_adjoint(d)| implements $R^\dagger$. \end{itemize} Why is this already sufficient? \begin{itemize} \item The Hamiltonian $\mathcal H$ is given by: $\mathcal H (d|s) = - \log (R(s))^\dagger d+\sum_i R(s)_i$. Implementing $R$ and stating that the data is Poissonian determines this form. \item Since $R$ is a composition of a convolution and a sampling both of which is a linear operation, $R$ itself is a linear operator.\footnote{I.e. $R(\alpha s_1+s_2) = \alpha R(s_1) + R(s_2)$.} Thus, $R' = R$ and $R'^\dagger = R^\dagger$. All in all, we need an implementation for $R$ and $R^\dagger$. \end{itemize} % \section*{Bibliography test} % RESOLVE was first presented in \cite{Resolve2016}. ... ...