Commit 8cb33343 authored by Philipp Arras's avatar Philipp Arras
Browse files

Add first text

parent 9a0bb68d
Pipeline #35431 passed with stage
in 1 minute and 53 seconds
......@@ -3,7 +3,7 @@
\usepackage{graphicx, amsmath}
\usepackage[backend=biber, sorting=none]{biblatex}
......@@ -17,9 +17,54 @@
\section*{The need for a guide}
\section*{Mathematical description of what NIFTy needs}
For first order minimization:
\item $\mathcal H (d|s) = - \log \mathcal P (d|s) + \text{constant in }s$
\item $\mathcal H' = \frac{\partial}{\partial s} \mathcal H (d|s)$
For second order minimization additionally:
\item $\langle \mathcal H' \mathcal H'^\dagger \rangle_{\mathcal P (d|s)}$.
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:
\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)
\section*{Example $\gamma$-ray imaging}
The information a $\gamma$-ray astronomer would provide to the algorithm (in the
simplest case):
\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 $R^\dagger$.
Why is this already sufficient?
\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 $R^\dagger$.
% \section*{Bibliography test}
% RESOLVE was first presented in \cite{Resolve2016}.
Supports Markdown
0% or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment