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\section{Absolute Formulation}  \subsection{Rate}  Next we define an absolute rate which is the mixed formulation.  \begin{equation}  k_{abs}\equiv \frac{d}{dt}[\ln S_i(t)]  \end{equation}  \subsection{Fisher Information}  Likewise we define the ``mixed formulation'' which is truly the absolute formulation.  \begin{equation}  I_a=\sum_i \gamma_i k_{abs}=\sum_i\gamma_i\frac{d}{dt}[\ln S_i(t)]  \end{equation}  With these definitions in place the picture would be completed with one final formulation the all S formulation $k_S$ and an associated $I_s$.  \subsection{GFE to FE}  For the absolute formulation the GFE is defined as follows  \begin{equation}  GFE_a=\frac{d}{dt}\sum [\gamma_i(t)k_a(t)]  \end{equation}  We also know that $k_a\equiv \frac{d}{dt}[\ln S_i]$ and that $\gamma_i=\frac{S_i}{\sum[S_i]}$. With these definition we can determine the time derivative of the average rate as follows.   \begin{equation}  \frac{d}{dt}\sum [\gamma_i(t)k_a(t)]=\sum[\frac{d}{dt}[\gamma_i(t)k_a(t)]]  \end{equation}  Now applying the product rule and distributing the sum we find the following.   \begin{equation}  GFE_a=\sum[\gamma_i\frac{d}{dt}[k_a]]+\sum[k_a\frac{d}{dt}[\frac{S_i}{\sum[S_i]}]]  \end{equation}  The second term can be evaluated and simplified using the product rule for the derivative as follows.  \begin{equation}  GFE_a=\sum[\gamma_i\frac{d}{dt}[k_a]]+\sum[k_a(S_i\frac{d}{dt}[(\Sigma[S_i])^{-1}]+(\Sigma[S_i])^{-1}\frac{d}{dt}[S_i])]  \end{equation}  Simplifying further we find  \begin{equation}  GFE_a=\sum[\gamma_i\frac{d}{dt}[k_a]]+\sum[k_a(\frac{-S_i}{(\Sigma S_i)^2}\frac{d}{dt}[(\Sigma S_i)]+(\Sigma S_i)^{-1}\frac{d}{dt}[S_i])]  \end{equation}  Now substituting in the definition of the $\gamma_i$ and distributing the sum we find the following.   \begin{equation}  GFE_a=\sum[\gamma_i\frac{d}{dt}[k_a]]+\sum[-k_a\gamma_i \frac{d}{dt}[\ln (\Sigma S_i)]]+\sum[\frac{k_a}{\Sigma S_i}\frac{d}{dt}[S_i]]  \end{equation}  For simplicity lets call these three terms A, B, and C respectively. Term A is already simplified however lets analyze terms B and C individually.  \begin{equation}  B: \sum[-k_a\gamma\frac{d}{dt}[\ln \Sigma S_i(t)]]  \end{equation}  This sum can really be broken up the derivative term is already being summed therefore we can write the following.  \begin{equation}  -(\sum[k_a\gamma_i])(\frac{d}{dt}[\ln \Sigma S_i])=-(\sum[k_a\gamma_i])(\frac{1}{\Sigma S_i}\Sigma\frac{d}{dt}[ S_i])  \end{equation}  From here we remember that $\frac{d}{dt}[S_i]=k_aS_i$ and we find a nice simplification.  \begin{equation}  -(\sum[k_a\gamma_i])(\sum[k_a\gamma_i])=-(\sum k_a\gamma_i)^2=(\overline{k})^2  \end{equation}  With this form we can see the GFE forming, we now need to analyze term C.  \begin{equation}  C:\sum[\frac{k_a}{\Sigma S_i}\frac{d}{dt}[S_i]]=\sum[\frac{k_a S_i}{\Sigma S_i S_i}\frac{d}{dt}[S_i]]=\sum[k_a\gamma_i\frac{d}{dt}[\ln S_i]]=\sum[k_a^2\gamma_i]  \end{equation}  This last term can easily be identified as $=I_a=\overline{k_a^2}$ which completes our GFE for the absolute formulation.   \begin{equation}  GFE_a=\sum[\gamma_i\frac{d}{dt}[k_a]]-(\sum k_a\gamma_i)^2+\sum[k_a^2\gamma_i]=\sum[\gamma_i\frac{d}{dt}[k_a]]-(\overline{k})^2+\overline{k_a^2}  \end{equation}  \subsection{Inequality}