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sop.tex

@@ -80,8 +80,8 @@ Notice we have two cases of $\cvar{j, j'}$, the first is when $j = j'$, i.e. $(\
 \subsection{$\cvar{j, j'}~|~j \neq j'$}
 For notational convenience set 
 \begin{align*}
-\term_1\left(\wElem_1,\ldots, \wElem_{\prodsize}\right) = &\ex{\prod_{i = 1}^{\prodsize}\sine(\wElem_i)\conj{\sine(\wElem'_i)}\ind{\hfunc(\wElem_i) = j} \ind{\hfunc(\wElem'_i) = j'}}\\
-\term_2\left(\wElem_1,\ldots, \wElem_{\prodsize}\right) = &\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'}} 
+\term_1\left(\wElem_1,\ldots, \wElem_{\prodsize}, \wElem'_1,\ldots, \wElem'_{\prodsize}\right) = &\ex{\prod_{i = 1}^{\prodsize}\sine(\wElem_i)\conj{\sine(\wElem'_i)}\ind{\hfunc(\wElem_i) = j} \ind{\hfunc(\wElem'_i) = j'}}\\
+\term_2\left(\wElem_1,\ldots, \wElem_{\prodsize}, \wElem'_1,\ldots, \wElem'_{\prodsize}\right) = &\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'}} 
 \end{align*}
 Focusing on $\term_1$, observe that $\term_1 = 1$ if and only if all the $\wElem_i$'s are equal, all the $\wElem'_i$'s are equal, and the two groups of variables do not equal each other, 
 \begin{equation*}
@@ -142,9 +142,9 @@ Taking a look at the leftmost term of \cref{eq:sigsq}, we establish bounds the v
 \end{align}
 
 Before proceeding, we introduce some notation  and terminology that will aid in communicating the bounds we are about to establish.  We refer to the leftmost expectation of \cref{eq:sig-j-last} in the following way:
-
-\[\term_1\left(\wElem_1,\ldots,\wElem_\prodsize, \wElem_1',\ldots, \wElem_\prodsize'\right) = \ex{\prod_{i = 1}^\prodsize s(w_i)\overline{s(w'_i)}\ind{h(w_i) = j}\ind{h(w'_i) = j}}.%\text{, and}
-\]
+\begin{equation}
+\term_1\left(\wElem_1,\ldots,\wElem_\prodsize, \wElem_1',\ldots, \wElem_\prodsize'\right) = \ex{\prod_{i = 1}^\prodsize s(w_i)\overline{s(w'_i)}\ind{h(w_i) = j}\ind{h(w'_i) = j}}\label{eq:term-1}.%\text{, and}
+\end{equation}
 %\[\term_2\left(\wElem_1,\ldots,\wElem_\prodsize, \wElem_1',\ldots, \wElem_\prodsize'\right)  = \ex{\prod_{i = 1}^ks(w_i)\ind{h(w_i) = j}}\cdot \ex{\prod_{i = 1}^\prodsize\overline{s(w'_i)}\ind{h(w'_i) = j}}. \]
 
 We will use the vocabulary 'term' to denote $\prod_{i = 1}^{\prodsize}\vect_i(\wElem_i)\vect_i(\wElem_i') \cdot\term_1\left(\wElem_1,\ldots,\wElem_\prodsize\right)$ given a specific set of world values.  %To say that a term survives \AR{You should not care about whether the $T_1$ term survives or not. See the above comment on why.} the expectation is to mean that $\term_1 - \term_2 \neq 0$.  Note, that the only terms that survive the expectation above are mappings of $w_i = w'_j = w$ for $i, j \in [\prodsize]$, such that each $w_i$ has a match, i.e., no $w_i$ or $w'_j$ stands alone without a matching world in its complimentary set.  In other words, the set of values in $\wElem_1,\ldots,\wElem_k$ has a bijective mapping to the set of values in $\wElem'_1,\ldots,\wElem'_k$.
@@ -206,7 +206,7 @@ Next, we argue that $S_1 \subseteq S_2$.  Pick an arbitrary tuple $\left(\wElem_
 \begin{Lemma}
 Given a distinct $\dw_1 \prec \cdots \prec \dw_{\dist}$ over every distinct $\surj:[\prodsize]\mapsto [\dist]$, the tuples $\left(\dw_{\surj(1)},\ldots, \dw_{\surj(\prodsize)}\right)$ are distinct.
 \end{Lemma}
-\AH{Needs a formal proof.  Working on that.}
+
 \begin{proof}
 For a given $\prodsize, \dist$, every function $\surj$ in the set of surjective functions ($\surjSet$), applies its mapping to the set of distinct value combinations that respect the total order ($\mtupSet$).  In other words, an alternative view is to take the cartesian product of the world values that can be mapped to the $\dist$ $\dw_i$ variables respecting $\prec$, i.e.,
 \[\mtupSet = \left\{\left(\dw_1,\ldots, \dw_{\dist}\right) ~|~ \dw_1,\ldots, \dw_{\dist} \in \wSet, \dw_1 \prec \cdots \prec \dw_{\dist}\right\},\]
@@ -221,13 +221,13 @@ For the second case, assume that $\disttup = \disttup'$, which then requires tha
 \end{proof}
 %, while additionally including all symmetrical counterparts allows for double counting.  This double counting is mitigated by the fact 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}.  
 %yields exactly the world value combinations containing $\dist$ distinct values which appear in the cartesian product of the sum. 
-\AR{Overall comments: (1) The main thing missing if explicitly stating that $(w_1,\dots,w_k)\mapsto (\dw_{\surj(1)},\ldots,\dw_{\surj(\dist)})$. (2) After stating the map you should argue in words why all distinct tuples with $m$ distinct world values are covered.}  
+%\AR{Overall comments: (1) The main thing missing if explicitly stating that $(w_1,\dots,w_k)\mapsto (\dw_{\surj(1)},\ldots,\dw_{\surj(\dist)})$. (2) After stating the map you should argue in words why all distinct tuples with $m$ distinct world values are covered.}  
 
 
 \begin{Definition}
 Functions $\surj:[\prodsize]\mapsto [\dist], \surj':[\prodsize]\mapsto [\dist']$ are said to be matching, denoted $\match{\surj}{\surj'}$, if and only if 
 	\begin{enumerate}
-		\item $\dist = \dist'$
+		%\item $\dist = \dist'$
 		\item $\forall i \in [\dist], |\surj^{-1}(i)| = |\surj'^{-1}(i)|$, i.e., the cardinality of variables mapped to $\dw_i$ equals the cardinality of variables mapped to $\dw_i'$, for all $i \in [\dist]$.
 	\end{enumerate}
 \end{Definition}
@@ -235,9 +235,9 @@ Functions $\surj:[\prodsize]\mapsto [\dist], \surj':[\prodsize]\mapsto [\dist']$
 \AH{handle $\term_1$ first, then use the lemma to tackle eq 95; use lemma that was stated in terms of $\term_1$, and create another lemma for 95}
 \AH{explicity note the independence of h and s, to justify pushing expectations through products}
 \begin{Lemma}\label{lem:sig-j-survive}
-When $\surj, \surj'$are matching, where $\forall j \in[\dist], \dw_{_j} = \dw_{'_j}$, \cref{eq:sig-j-distinct} is exactly 
+When $\surj, \surj'$are matching, where for every $j \in[\dist], \dw_{j} = \dw'_{j}$, %\cref{eq:sig-j-distinct} is exactly 
 \[
-\sum_{\substack{\dw_{_1}, \ldots,\dw_{_\dist},\\\dw_{'_1},\ldots,\dw_{'_{\dist'}}\\ \in W}}\prod_{i = 1}^{\prodsize}\vect_i(\dw_{_{\surj(i)}})\vect_i(\dw_{'_{\surj'(i)}})
+\term_1(\dw_{\surj(1)},\dots, \dw_{\surj(\prodsize)}, \dw_{\surj'(1)},\dots, \dw_{\surj'(\prodsize')}) = \sum_{\substack{\dw_{1} \prec \cdots \prec \dw_{_\dist}\\ \in \wSet}}\prod_{i = 1}^{\prodsize}\vect_i(\dw_{\surj(i)})\vect_i(\dw_{\surj'(i)})
 \]
 and $0$ otherwise.
 \end{Lemma}
@@ -262,19 +262,20 @@ and $0$ otherwise.
 %In the above, since we have more than pairwise independence for $\wElem \neq \wElem'$, we can push the expectation into the product.  Then by \cref{lem:exp-sine} we get 0 for both expectations.\newline
 
 
-\AR{First some typos/things that are incorrect below-- note this is \textbf{not} an exhaustive list. (1) In the proof below the $w_i$ and $w'_i$ should be $\tilde{w}_i$ and $\tilde{w'}_i$ respectively. (2) The expression for $T_1$ below is incorrect since it seems to assume that all the pre-image sizes are $1$-- the expression for $T_2$ is fine except the $j_i$ terms are not defined. However, ``taking out" one term for $\tilde{w'}_{m'}$ for $T_2$ is incorrect since e.g. we could have the pre-image of $m'$ have size $>1$. (3) The proof below never explicitly argues why the condition $\dw_{_j} = \dw_{'_j}$ is needed.}
+\AR{First some typos/things that are incorrect below-- note this is \textbf{not} an exhaustive list. (1) In the proof below the $w_i$ and $w'_i$ should be $\dw_i$ and $\dw_i$ respectively. (2) The expression for $T_1$ below is incorrect since it seems to assume that all the pre-image sizes are $1$-- the expression for $T_2$ is fine except the $j_i$ terms are not defined. However, ``taking out" one term for $\dw_{m'}$ for $T_2$ is incorrect since e.g. we could have the pre-image of $m'$ have size $>1$. (3) The proof below never explicitly argues why the condition $\dw_{_j} = \dw_{'_j}$ is needed.}
 \AR{Here is how I recommend that you re-write the proof. First as mentioned earlier, you should only consider the $T_1$ terms (as you account for the $T_2$ terms later on. Second you should first start off by re-stating the $T_1$ term like so. Consider the ``generic term"--
-\[T_1(\tilde{w}_{\surj(1)},\dots, \tilde{w}_{\surj(m)}, \tilde{w'}_{\surj'(1)},\dots, \tilde{w'}_{\surj'(m')}).\]
+\[T_1(\dw_{\surj(1)},\dots, \dw_{\surj(m)}, \dw_{\surj'(1)},\dots, \dw_{\surj'(m')}).\]
 Then re-write the what the above term is based on the exact definition (BTW I'm dropping the $\mathbf{E}$ terms for convenience but they should be all there below.) In particular, the above term by definition is exactly
-\[\prod_{i=1}^k s(\tilde{w}_{\surj(i)})\cdot \overline{s(\tilde{w'}_{\surj'(i)})}.\]
+\[\prod_{i=1}^k s(\dw_{\surj(i)})\cdot \overline{s(\dw_{\surj'(i)})}.\]
 Now re-write the above in terms of ``powers" of distinct worlds:
-\[ (\prod_{i=1}^m s(\tilde{w}_{i})^{|\surj^{-1}(i)|})\cdot \overline{(\prod_{j=1}^m s(\tilde{w'}_j)^{|\surj^{-1}(j)|})}\]
+\[ (\prod_{i=1}^m s(\dw_{i})^{|\surj^{-1}(i)|})\cdot \overline{(\prod_{j=1}^m s(\dw_j)^{|\surj^{-1}(j)|})}\]
 Now once you have the above expression, then it will be much easier to argue why if any of the matching conditions are not satisfied then the expression is $0$. I also believe that working with the above expression will also make it more ``obvious" as to why the different conditions are required. Currently the arguments below do not explicitly bring this out...
 }
+\textit{*In the subsequent proofs, at most we assume 2k wise independence for both $\hfunc$ and $\sine$, but we really would like less*}.
 \begin{proof}
 
-Consider the "generic term"--
-\[T_1(\tilde{w}_{\surj(1)},\dots, \tilde{w}_{\surj(\prodsize)}, \tilde{w'}_{\surj'(1)},\dots, \tilde{w'}_{\surj'(\prodsize')}).\]
+Note \cref{eq:term-1}, and consider the "generic term"--
+\[\term_1(\dw_{\surj(1)},\dots, \dw_{\surj(\prodsize)}, \dw_{\surj'(1)},\dots, \dw_{\surj'(\prodsize')}).\]
 
 Let's rewrite the term based on its exact definition:
 \begin{align*}
@@ -286,9 +287,9 @@ Further see how the requirement that $\dw_i = \dw'_i$ gives us the precise combi
 
 To prove that \cref{lem:sig-j-survive} is true, consider what the expectation looks like when $\surj, \surj'$ are not matching.  The first condition for $\surj, \surj'$ to be matching is violated when $\dist \neq \dist'$.  
 \AH{The following observation isn't necessary to complete the proof.  I'll just leave it for now in case it may be valuable down the road.}
-\AH{Don't use $\exists, \forall$ in prose; mathematcial notation shouldn't be used to replace the prose, only to enhance the prose.}
-Observe that $\forall \dist \in [\prodsize], \sum_{i = 1}^{\dist}|\surj^{-1}(i)| = \prodsize$ and that this fact implies for $\dist, \dist' \in [\prodsize] ~|~\dist \neq \dist', \exists i \in [m] ~|~\forall j \in [m], \surj^{-1}(i)| \neq |\surj'^{-1}(j)|$, meaning that if we have that $\dist \neq \dist'$, then the second matching condition is also violated.  
-\AH{Moving on...}Note that $\dw_1\ldots\dw_\dist, \dw_1'\ldots\dw_{\dist'}'$ are distinct world values such that $\forall i \neq j \in [\dist], \dw_i = \dw_i' \neq \dw_j = \dw_j'$.  To make things easier, assume that $\dist < \dist'$.  The opposite case of $\dist > \dist'$ has a symmetrical proof.  Fixing variables $\dw_1\ldots\dw_\dist, \dw_1'\ldots\dw_\dist$, we have at least one $\dw_i$ without a conjugate, and at least one extra distinct value, $\dw_{\dist'}'$, for which no $\dw_{\dist'}$ exists.  This (these) distinct term(s) cancel(s) out all the other values in the expectations.
+
+Observe that for all $\dist \in [\prodsize], \sum_{i = 1}^{\dist}|\surj^{-1}(i)| = \prodsize$ and that this fact implies for $\dist, \dist' \in [\prodsize]$ such that $\dist \neq \dist'$, there exists $i \in [m]$ such that $|\surj^{-1}(i)| \neq |\surj'^{-1}(i)|$, meaning that if we have that $\dist \neq \dist'$, then the second matching condition is also violated.  
+\AH{Moving on...}Note that $\dw_1\ldots\dw_\dist, \dw_1'\ldots\dw_{\dist'}'$ are distinct world values such that $\forall i \neq j \in [\dist], \dw_i = \dw_i' \neq \dw_j = \dw_j'$.  To make things easier, assume that $\dist < \dist'$.  The opposite case of $\dist > \dist'$ has a symmetrical proof.  Fixing variables $\dw_1\ldots\dw_\dist, \dw_1'\ldots\dw'_{\dist'}$, we have at least one $\dw_i$ without a conjugate, and at least one extra distinct value, $\dw_{\dist'}'$, for which no $\dw_{\dist'}$ exists.  This (these) distinct term(s) cancel(s) out all the other values in the expectations.
 
 
 \begin{align}
@@ -297,8 +298,7 @@ Observe that $\forall \dist \in [\prodsize], \sum_{i = 1}^{\dist}|\surj^{-1}(i)|
 = &\ex{\left(\prod_{i = 1}^{\dist}\sine(\dw_{i})^{|\surj^{-1}(i)|}\right) \cdot \left(\prod_{j = 1}^{\dist}\conj{\sine(\dw'_{j})}^{|\surj^{-1}(j)|}\right)} \cdot \prod_{l = \dist + 1}^{\dist'} \ex{\conj{\sine(\dw_l)}^{|\surj^{-1}(l)|}} = 0.\label{eq:lem-fmatch-pt1}
 \end{align}
 In \cref{eq:lem-fmatch-pt1} the expectation can be pushed through the last product group since we know that all the operands are distinct from any others appearing in the overall product.  Then, by \cref{lem:exp-sine} we get $0$ for that rightmost term, and this cancels out the rest of the terms in the overall product.
-
-\textit{Here at most we assume 2k wise independence, but we really would like less}. 
+ 
 \AH{Can we bring it down to k-wise, since we have $<$ k terms which we are pushing the expectation through?}
 To complete the proof, we now approach the case where $\dist = \dist'$, but there is a $\dw_i, \dw_i'$ with an unequal number of mappings.  
 
@@ -317,16 +317,16 @@ Let $n = \{i ~|~ |\surj^{-1}(i)| \neq |\surj'^{-1}(i)|\}$.  Further, let $\dist_
 
 \begin{align}
 \term_1 = &\ex{\left(\prod_{i \in [\dist_*]} \sine(\dw_i)^{|\surj^{-1}(i)|} \conj{\sine(\dw'_i)}^{|\surj'^{-1}(i)|}\right) 
-	\left(\prod_{j \in [n]}\sine(\dw_i)^{|\surj^{*-1}|} \conj{\sine(\dw'_i)}^{|\surj^{*-1}|}\right) 
+	\left(\prod_{j \in [n]}\sine(\dw_i)^{|\surj^{*-1}(j)|} \conj{\sine(\dw'_i)}^{|\surj^{*-1}(j)|}\right) 
 		\left(\prod_{\substack{i' \in [n],\\\surj^{-1}(i') > \surj'^{-1}(i')}} \sine(\dw_{i'})^{|\surj^{-1}(i')| - |\surj'^{-1}(i')|} \prod_{\substack{j' \in n ~|~\\ \surj'^{-1}('j) > \surj^{-1}(j')}} \conj{\sine(\dw'_{j'})}^{|\surj'^{-1}(j')| - |\surj^{-1}(j')|}\right)} \label{eq:lem-match-pt2-line1}\\
 = &\ex{\left(\prod_{i \in [\dist_*]} \sine(\dw_i)^{|\surj^{-1}(i)|} \conj{\sine(\dw'_i)}^{|\surj'^{-1}(i)|}\right) 
-	\left(\prod_{j \in [n]}\sine(\dw_i)^{|\surj^{*-1}|} \conj{\sine(\dw'_i)}^{|\surj^{*-1}|}\right)} \cdot
+	\left(\prod_{j \in [n]}\sine(\dw_i)^{|\surj^{*-1}(j)|} \conj{\sine(\dw'_i)}^{|\surj^{*-1}(j)|}\right)} \cdot
 		\ex{\left(\prod_{\substack{i' \in n ~|~\\\surj^{-1}(i') > \surj'^{-1}(i')}} \sine(\dw_{i'})^{|\surj^{-1}(i')| - |\surj'^{-1}(i')|} \prod_{\substack{j' \in n ~|~\\ \surj'^{-1}('j) > \surj^{-1}(j')}} \conj{\sine(\dw'_{j'})}^{|\surj'^{-1}(j')| - |\surj^{-1}(j')|}\right)}\nonumber\\
 = &\ex{\left(\prod_{i \in [\dist_*]} \sine(\dw_i)^{|\surj^{-1}(i)|} \conj{\sine(\dw'_i)}^{|\surj'^{-1}(i)|}\right) 
-	\left(\prod_{j \in [n]}\sine(\dw_i)^{|\surj^{*-1}|} \conj{\sine(\dw'_i)}^{|\surj^{*-1}|}\right)} \cdot
+	\left(\prod_{j \in [n]}\sine(\dw_i)^{|\surj^{*-1}(j)|} \conj{\sine(\dw'_i)}^{|\surj^{*-1}(j)|}\right)} \cdot
 		\ex{\prod_{\substack{i' \in n ~|~\\\surj^{-1}(i') > \surj'^{-1}(i')}} \sine(\dw_{i'})^{|\surj^{-1}(i')| - |\surj'^{-1}(i')|}} \cdot \ex{\prod_{\substack{j' \in n ~|~\\ \surj'^{-1}('j) > \surj^{-1}(j')}} \conj{\sine(\dw'_{j'})}^{|\surj'^{-1}(j')| - |\surj^{-1}(j')|}}\nonumber\\ 
 = &\ex{\left(\prod_{i \in [\dist_*]} \sine(\dw_i)^{|\surj^{-1}(i)|} \conj{\sine(\dw'_i)}^{|\surj'^{-1}(i)|}\right) 
-	\left(\prod_{j \in [n]}\sine(\dw_i)^{|\surj^{*-1}|} \conj{\sine(\dw'_i)}^{|\surj^{*-1}|}\right)} \cdot
+	\left(\prod_{j \in [n]}\sine(\dw_i)^{|\surj^{*-1}(j)|} \conj{\sine(\dw'_i)}^{|\surj^{*-1}(j)|}\right)} \cdot
 		\prod_{\substack{i' \in n ~|~\\\surj^{-1}(i') > \surj'^{-1}(i')}} \ex{\sine(\dw_{i'})^{|\surj^{-1}(i')| - |\surj'^{-1}(i')|}} \cdot \prod_{\substack{j' \in n ~|~\\ \surj'^{-1}('j) > \surj^{-1}(j')}} \ex{\conj{\sine(\dw'_{j'})}^{|\surj'^{-1}(j')| - |\surj^{-1}(j')|}}\label{eq:lem-match-pt2-last}\\
 & = 0.\nonumber
 \end{align}