Notes added from 041720

main.tex

@@ -8,6 +8,7 @@
 \usepackage{amssymb}
 %\let\proof\relax
 %
+
 \let\endproof\relax
 \usepackage{amsthm}
 \usepackage{mathtools}

sop.tex

@@ -80,7 +80,8 @@ Notice we have two cases of $\cvar{j, j'}$, the first is when $j = j'$, i.e. $(\
 &= \sum_{\substack{\wElem_1,\cdots,\wElem_\prodsize,\\\wElem'_1,\cdots,\wElem'_\prodsize\\\in W}}\prod_{i = 1}^{\prodsize}v_i(\wElem_i)v_i(\wElem'_i)\left(\ex{\prod_{i = 1}^{\prodsize}s(\wElem_i)\conj{s(\wElem'_i)}\ind{h(\wElem_i) = j}\ind{h(\wElem'_i) = j'}} - \ex{\prod_{i = 1}^{\prodsize}s(\wElem_i)\ind{h(\wElem_i) = j}}\cdot\ex{\prod_{i = 1}^{\prodsize}\conj{s(\wElem'_i)}\ind{h(\wElem_i') = j'}} \right).\label{eq:var-lambda-j-j'}
 \end{align}
 \AH{How can I present the derivation of the bounds below in a \textit{better} way?}
-
+\AH{Have subsections for $j = j'$ and $j \neq j'$}
+\AH{A better argument might be: Let us assume that there $i \neq i'$, $w_i \neq w'_i$ which is $0$ in expectation.  Don't forget parameters for $\term_1$ and change this notation.  The cardinal rule is: if you don't need it, then don't use it (notation).  Define $\term_1$ once and then use later.}
 Equation ~\eqref{eq:var-lambda-j-j'} for $j \neq j'$ bounds to the rightmost sum of \cref{eq:sigsq}.  For $\term_1^{\cvar{j, j'}} = \ex{\prod_{i = 1}^{\prodsize}s(\wElem_i)s(\wElem'_i)\ind{h(\wElem_i) = j}\ind{h(\wElem'_i) = j'}}$, because hash function $h$ cannot bucket the same world to two different buckets, the only instance $\term_1^{\cvar{j, j'}} = 1$ occurs when there is no overlap between the $\wElem_i$ and $\wElem'_i$ variables.  Given the condition of no overlap, $\term_1^{\cvar{j, j'}} = 1$ only with the further condition that $\forall i \in [\prodsize], \wElem_i = \wElem, \wElem'_i = \wElem', \wElem \neq \wElem'$.  Notice, however, given the conditions, the product of the remaining expectations will cancel this out.  Looking at the remaining two expectations $\term_2^{\cvar{j, j'}} = \ex{\prod_{i = 1}^{\prodsize}\sine(\wElem_i) \ind{\hfunc(\wElem_i) = j}} \cdot \ex{\prod_{i = 1}^{\prodsize}\conj{\sine(\wElem'_i)} \ind{\hfunc(\wElem'_i) = j'}}$, that $\term_2^{\cvar{j, j'}} = 1$ only when $\forall i \in [\prodsize], \wElem_i = \wElem, \wElem'_i = \wElem'$.  Taken together, the constraints leave us with only one possible case for $\term_1^{\cvar{j, j'}} - \term_2^{\cvar{j, j'}} \neq 0$, when all variables are the same world.  Thus, 
 \begin{align}
 &\sum_{j \neq j'}\cvar{j, j'} = - \frac{1}{B^2}\sum_{\wElem \in W}\prod_{i = 1}^{\prodsize}v_i^2(\wElem)\label{eq:cvar-bound}.
@@ -126,7 +127,7 @@ We next describe the nonzero terms of \cref{eq:sig-j-last}.
 \begin{Definition}
 Define and then fix the total ordering of the $|\wSet|$ worlds in $\wSet$ to follow the total order of the natural numbers. Further we require that $\forall i, j \in [\dist], i < j \implies \dw_i < \dw_j, i.e. \wElem_1 \prec\ldots\prec\wElem_\prodsize$.
 %Given a fixed order $\wSet_{\order}: \left(\wSet, \wSet\right)\mapsto \mathbb{B}$ of possible worlds, define the lexographical order of distinct worlds $\wSet_\dist$ to be the ordering which complies to the identity mapping of elements in $[\prodsize]$ to elements in $[\dist]$ up to $\dist$, such that .  In other worlds, $\forall \wElem, \wElem' \in \wSet_\dist, \dw < \wElem' \leftrightarrow \wSet_{\order}\left(\wElem, \wElem'\right) = T$.
-\end{Definition}
+\end{Definition}\AH{You only need to define order on the elements of $\wSet$, as we haven't defined $\dist$ yet.  Introduce $\prec$ to denote ordering}
 
 To help describe all possible world value matchings we introduce functions $f$ and $f'$.
 \begin{Definition}
@@ -148,7 +149,7 @@ We rewrite equation \eqref{eq:sig-j-last} in terms of $\dist$ distinct worlds, w
 \label{eq:sig-j-distinct}
 \end{equation}
 
-Thje functions $f, f'$ are used to produce the mapping $\left(\wElem_1,\ldots, \wElem_{\prodsize}\right) \mapsto \left(\dw_1,\ldots, \dw_{\prodsize}\right)$.  Observe that the cartesian product of world values assigned to $\wElem_1,\ldots,\wElem_\prodsize$ throughout the summation can be rearranged into groups of world variables with distinct world values, for each distinct element $\dist$ in the set $[\prodsize]$.  For each $\dist \in [\prodsize]$, all possible combinations of $\dist$ world values is equivalently modeled by taking the set of surjective functions $f:[\prodsize]\mapsto [\dist]$ and summing over all of their respctive mappings$\left(\wElem_1,\ldots, \wElem_{\prodsize}\right) \mapsto \left(\dw_1,\ldots, \dw_{\prodsize}\right)$.\AR{Again total ordering is on worlds in $W$-- $\dw_{f(1)}\prec\cdots\prec\dw_{f(m)}$ does not make sense since some of these world values could be the same.}  
+Thje functions $f, f'$ are used to produce the mapping $\left(\wElem_1,\ldots, \wElem_{\prodsize}\right) \mapsto \left(\dw_{f(1)},\ldots, \dw_{f(\prodsize)}\right)$.  Observe that the cartesian product of world values assigned to $\wElem_1,\ldots,\wElem_\prodsize$ throughout the summation can be rearranged into groups of world variables with distinct world values, for each distinct element $\dist$ in the set $[\prodsize]$.  For each $\dist \in [\prodsize]$, all possible combinations of $\dist$ world values is equivalently modeled by taking the set of surjective functions $f:[\prodsize]\mapsto [\dist]$ and summing over all of their respctive mappings$\left(\wElem_1,\ldots, \wElem_{\prodsize}\right) \mapsto \left(\dw_1,\ldots, \dw_{\prodsize}\right)$.\AR{Again total ordering is on worlds in $W$-- $\dw_{f(1)}\prec\cdots\prec\dw_{f(m)}$ does not make sense since some of these world values could be the same.}  
 \AH{The point I am trying to make here, is that we are avoiding double counting such that the values assigned to $\dw_1 \prec \cdots \prec \dw_\dist$ will always satisfy $\dw_i < \dw_j$ if $i < j$.}
 For any $\dist$, the set of all non-symmetric surjective mappings $f$ constitute all unique mappings, while additionally including all symmetrical counterparts allows for double counting.  This double counting is mitigated by imposing the requirement that $\dw_1 \prec \cdots \prec \dw_\dist$ based on the fixed order of the world values, which then \AR{I do not see what the ``symmetrical counterparts" comment adds here. Just remove it}.  
 \AH{Symmetrical counterparts  of $f, f'$ constitute double counting without the ordering imposed on $\dw_i$.}