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\begin{document}
\large\sffamily

%\begin{document}

  \postertitle{0.96\textwidth}
              {shield.eps}
              {Bayes linear kinematics in the design of experiments}
              {Kevin J.\ Wilson and Malcolm Farrow }
              {School of Mathematics and Statistics, University of Newcastle
	            upon Tyne, U.K.}
              {E-mail: k.j.wilson@ncl.ac.uk, malcolm.farrow@ncl.ac.uk}
              {whiteblue}
              {dblue}

  \begin{multicols}{3}
    \heading{0.82\columnwidth}{whiteblue}{black}{1: The Design of Experiments}

\begin{itemize}
\item View the \textcolor{blue}{design} of an experiment  as a \textcolor{blue}{decision problem}.
\item Trade offs between the \textcolor{blue}{costs} of  the experiment and the \textcolor{blue}{benefits} from it. 
\item \textcolor{blue}{Benefits} in terms of beliefs about future unknowns: either increase in information or some terminal payoff. 
%The \textcolor{blue}{optimal design} of an experiment is the best possible choice of design, found in a decision theoretic way.
\end{itemize}

\vspace{-2ex}
\begin{figure}
\begin{center}

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\caption{\textcolor{mygreen}{The process of Bayesian experimental design}}
\label{pretty}
\end{center}
\end{figure}

\begin{itemize}
\item Specify a \textcolor{blue}{utility function} $U$ which expresses the decision maker's preferences over uncertain outcomes.
\item The solution to the experimental design problem is the design which maximises the \textcolor{blue}{expected utility}.
\end{itemize}

\begin{tabular}{ll}
Initial decision (design choice): & \textcolor{mygreen}{$d_{1}$} \\ 
Terminal decision: & \textcolor{mygreen}{$d_{2}$} \\ 
Unknown parameters: & \textcolor{blue}{$\theta$} \\ 
Observed data: & \textcolor{blue}{$x$} \\
Utility function: & $U(\textcolor{mygreen}{d_{1}},\textcolor{mygreen}{d_{2}},\textcolor{blue}{\theta},\textcolor{blue}{x})$ 
\end{tabular}

The expected utility of the best decision is given by
\begin{displaymath}
U(d_{1})=\int_{\mathcal{X}} \max_{d_{2}\in D_{2}} \int_{\Theta}U(d_{1},d_{2},\theta,x)f(\theta|x,d_{1})f(x|d_{1})d\theta dx
\end{displaymath} 
The optimal design $d_{1}^{*}$ maximises this:
\begin{equation}
\label{Exp.des}
U(d_{1}^{*})=\max_{d_{1}\in D_{1}}\int_{\mathcal{X}}\max_{d_{2}\in D_{2}}\int_{\Theta}U(d_{1},d_{2},\theta,x)f(\theta|x,d_{1})f(x|d_{1})d\theta dx
\end{equation}

\begin{itemize}
\item Full Bayesian computation can be difficult.
\item Except in special cases neither the maximisation nor the integration can be solved analytically and approximation or simulation based methods are needed.
\end{itemize}

\heading{0.82\columnwidth}{whiteblue}{black}{2: Bayes linear kinematics}

\begin{itemize}
\item \textcolor{blue}{Bayes linear methods} specify only first and second order moments for unknown quantities (say $\bm\alpha$ and $\bm\beta$).
\item  Bayes linear adjusted expectation and variance for $\bm\beta$ given $\bm\alpha$:
\begin{eqnarray*}
\E_{1}(\bm\beta ;\bm\alpha) & = & \E_{0}(\bm\beta)+\Cov_{0}(\bm\beta,\bm\alpha)\Var_{0}^{-1}(\bm\alpha)\left[\bm\alpha-\E_{0}(\bm\alpha)\right] \\
\Var_{1}(\bm\beta ; \bm\alpha) & = & \Var_{0}(\bm\beta)-\Cov_{0}(\bm\beta,\bm\alpha)\Var_{0}^{-1}(\bm\alpha)\Cov_{0}(\bm\alpha,\bm\beta)
\end{eqnarray*}
\end{itemize}

\begin{itemize}
\item \textcolor{blue}{Bayes linear kinematics} gives us a way of propagating changes in belief about some quantities through to other quantities in a Bayes linear structure.
\item The Bayes linear kinematic adjusted expectation and variance of $\bm\beta$ given updated beliefs about $\bm\alpha$ of $\E_{1}(\bm\alpha)$ and $\Var_{1}(\bm\alpha)$ are 
\begin{eqnarray*}
\E_{1}(\bm\beta ; \bm\alpha) & = & \E_{0}(\bm\beta)+\Cov_{0}(\bm\beta,\bm\alpha)\Var_{0}^{-1}(\bm\alpha)\left[\E_{1}(\bm\alpha)-\E_{0}(\bm\alpha)\right]
\end{eqnarray*}
\begin{multline*}
\Var_{1}(\bm\beta ; \bm\alpha) = \Var_{0}^{*}(\bm\alpha)+\Cov_{0}(\bm\beta,\bm\alpha)\Var_{0}^{-1}(\bm\alpha) \\  \Var_{1}(\bm\alpha)\Var_{0}^{-1}(\bm\alpha)\Cov_{0}(\bm\alpha,\bm\beta)
\end{multline*}
where $\Var_{0}^{*}(\bm\alpha)$ is the Bayes linear adjusted variance.
\item Successive Bayes linear kinematic updates are \textcolor{blue}{not necessarily commutative}.
\item However, \cite{Gol04} give conditions under which the requirement of commutativity leads to a unique Bayes linear kinematic update.
\item If a single update is made for each of scalars $U_{1},\ldots,U_{p}$ then a \textcolor{blue}{sufficient condition} for a unique commutative update is
\begin{displaymath}
\Var_{0}^{-1}(U_{k})\Var_{1}(U_{k})<1,
\end{displaymath}
for each $U_{k}$.
\end{itemize}

\heading{0.82\columnwidth}{whiteblue}{black}{3: Usability experiment}

Before software goes on sale or a website goes live \textcolor{blue}{usability} \textcolor{blue}{testing} is undertaken.
This is done by taking a sample of ``users'' (assumed exchangeable) and asking each to perform a number of tasks. 
The successes and failures at these tasks  are recorded and the software can be \textcolor{blue}{launched} as it is or \textcolor{blue}{rewritten} (i.e.\ modified). 
\\
\\
To design such an experiment for a single task the number of users must be chosen and the decision whether to launch or rewrite the product made. 

\begin{figure}
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    \uput[u](42,56){$n$}
    \cnodeput(40,45){O}{\textcolor{blue}{outcome}}
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  \end{pspicture}
  \caption{\textcolor{mygreen}{Decision tree for usability test}}
  \label{dectree}
\end{figure}

\begin{itemize}
\item If $n$ users are asked to perform a task then the number who successfully complete that task, $X$, follows a binomial distribution
\begin{displaymath}
X\mid\theta_{l}\sim\textrm{bin}(n,\theta_{l}),
\end{displaymath}
where $\theta_{l}$ is the probability a user successfully completes the task after launch. We are also interested in the probability a user would successfully complete the task after rewrite, $\theta_{r}$. 
\item The two probabilities can be given beta distributions
\begin{eqnarray*}
\theta_{l}\sim\textrm{beta}(a_{l},b_{l}),& \theta_{r} \mid a_{r},b_{r} \sim\textrm{beta}(a_{r},b_{r}).
\end{eqnarray*}
\end{itemize}
\begin{itemize}
\item $\theta_{l}$ and $\theta_{r}$ will not in general be independent in our prior beliefs. If we transform them
\begin{eqnarray*}
\eta_{l}=\log\left(\dfrac{\theta_{l}}{1-\theta_{l}}\right), & \eta_{r}=\log\left(\dfrac{\theta_{r}}{1-\theta_{r}}\right),
\end{eqnarray*}
we can then link $\eta_{l}$ and $\eta_{r}$ in a Bayes linear structure. 
\item Updating of $\theta_{l}$ is conjugate. Then use the Bayes linear kinematic updating equations to update $\theta_{r}$.
%\item The utility function takes the form
%\begin{displaymath}
%U(s_{n},t_{k},Y,C_{k})=p_{1}U_{f}(C_{k})+p_{2}U_{s}(Y)+p_{3}U_{f}(C_{k})U_{s}(Y),
%\end{displaymath}
%where $s_{n}$ and $T_{k}$ are the initial and terminal decisions, benefit is measured in terms of $Y$, the number of future customers who successfully complete the task and $C_{k}$ is the financial cost given terminal decision $t_{k}$, $k=l,r$.
\item The sample size problem can be solved using (\ref{Exp.des}).
\end{itemize}

\begin{figure}
\begin{center}
\includegraphics{exp.utility.users.ps}
\label{exp.util}
\caption{\textcolor{mygreen}{This gives optimal sample size as $n=11$. `Difference' is the difference between the expected utilities of launch and rewrite}}
\end{center}
\end{figure}

 \heading{0.82\columnwidth}{whiteblue}{black}{5: References}

    \renewcommand\refname{}
    \vspace{-7ex}
     \begin{normalsize}
    \begin{thebibliography}{99}
      \bibitem{Chu09} Chukwu, L.O., Samuel, O.B., and Olaogun, M.O., (2009) Combined Effects of Binary Mixtures of Commonly Used Agrochemicals: Patterns of Toxicity in Fish. \emph{Research Journal of Agriculture and Biological Sciences} {\bf 5}, 883-891.
      \bibitem{Gol04} Goldstein, M., and Shaw, S., (2004) Bayes linear kinematics and Bayes linear Bayes graphical models. \emph{Biometrika} {\bf 9}, 425-446.
    \end{thebibliography}
    \end{normalsize}
    
\heading{0.82\columnwidth}{whiteblue}{black}{4: Bioassay}

\cite{Chu09} investigated the effect of \textcolor{blue}{5 different doses} of a fertiliser, NPK 20:10:10, on a certain fish, \emph{Oreochromis niloticus}. 
A number of fish were exposed to each dose. The \textcolor{blue}{response} was death within 4 days. 
Interest lies in the answers to two questions;
\begin{enumerate}
\item How many fish should be included in the experiment?
\item How many fish should be given each dose?
\end{enumerate}

\begin{itemize}
 \item Again binomial counts $X_{1},\ldots,X_{5}$ for the number of deaths of fish under dose $i=1,\ldots,5$.
 \item This time we don't specify a link function but simply say that
 \begin{displaymath}
 \eta_{i}=g(\theta_{i})
 \end{displaymath}
 with \textcolor{blue}{pseudo expectation} and \textcolor{blue}{pseudo variance}
 \begin{eqnarray*}
 \hat{\E}_{0}(\eta_{i})=g_{1}\left(\dfrac{a_{i}}{a_{i}+b_{i}}\right), & \hat{\Var}_{0}(\eta_{i})=g_{2}\left(\dfrac{1}{a_{i}+b_{i}}\right),
 \end{eqnarray*}
 where $g_{1}$ and $g_{2}$ are suitable monotonic functions.
 \item We take $g_{1}(x)=\log(x/1-x)$ (so the expectation of $\eta_{i}$ is unrestricted) and $g_{2}(x)=x$.
 \item Variance will clearly always decrease upon observation of data and so using the BLK sufficient condition a unique commutative solution always exists.
\end{itemize}

\begin{figure}
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  \caption{\textcolor{mygreen}{Decomposition of the utility function for a bioassay experiment}}
  \label{bioassay utility}
\end{figure}

\begin{itemize}
 \item Assume mutual utility independence at each node.
 \item Overall utility $U$:
 \begin{displaymath}
 U=p_{1}U_{B}+p_{2}U_{C}+p_{3}U_{B}U_{C}
 \end{displaymath}
 \item Cost utility $U_{C}$:
 \begin{displaymath}
 U_{C}=q_{1}U_{E}+q_{2}U_{F}+q_{3}U_{E}U_{F}
 \end{displaymath}
 \item We use an information gain benefit utility which can be calculated using
 \begin{displaymath}
 U(\bm\eta)=1-\frac{1}{5}\{\Var_{0}^{-1}(\bm\eta)\Var_{5}(\bm\eta ;\bm n)\},
 \end{displaymath}
 where $\Var_{5}(\bm\eta ;\bm n)$ is the BLK adjusted variance having chosen sample sizes of $\bm n=(n_{1},\ldots,n_{5})^{'}$ at the doses.
 \item Crucially this does not depend on how many fish die at each dose and so the experimental design problem can be solved without having knowledge of the full joint distribution of $\bm X$.
\end{itemize}
   
\begin{itemize}
\item \textcolor{blue}{Additive node}: $(p_{1},p_{2},p_{3})=(0.8,0.2,0)$ gives optimal allocation $\bm n^{*}=(21,8,4,3,5)$ i.e., sample size of $N=41$.
\item \textcolor{blue}{Binary node}: $(p_{1},p_{2},p_{3})=(0.6,0.2,0.2)$ gives $\bm n^{*}=(19,6,3,2,3)$ i.e., sample size of $N=33$.
\item Take terminal decision: `report adjusted expectation of $\bm\theta$'.
\item \textcolor{blue}{Additive node}: $\bm x=(3,5,3,3,5)$ gives $\E_{5}(\bm\theta ; \bm x)=(0.23,0.52,0.68,0.76,0.85)$.
\item \textcolor{blue}{Binary node}: $\bm x=(3,3,2,1,3)$ gives $\E_{5}(\bm\theta ; \bm x)=(0.20,0.45,0.60,0.68,0.78)$.
\item Compare with \textcolor{dred}{$\E_{0}(\bm\theta)=(0.24,0.48,0.61,0.68,0.77)$}.
\end{itemize}   
   
  \end{multicols}

%\hrule

\end{document}