Done with pass on Sec 11 (till Sec 1.1)

sop.tex

@@ -13,6 +13,7 @@ For notational convenience define
 &\term_j = \sum_{\wElem \in \wSet_j} \prod_{i = 1}^{\prodsize}\vect_i(\wElem)
 \end{align*}
 Let us show first that the expectation of the estimate does in fact yield the value we are estimating, $\term_j$.
+\AR{You should convert the above statement into a formal lemma. Otherwise it is weird to see a proof without any formal statement of what it is proving.}
 \begin{proof}
 
 \begin{align*}
@@ -21,7 +22,7 @@ Let us show first that the expectation of the estimate does in fact yield the va
 = &\ex{\sum_{\substack{\wElem_1,\ldots, \wElem_{\prodsize}\\ \in \wSet_j}} \prod_{i = 1}^{\prodsize}\vect_i(\wElem_i)\prod_{i = 1}^{\prodsize}\sine(\wElem_i)}\\
 = &\sum_{\substack{\wElem_1,\ldots, \wElem_{\prodsize}\\ \in \wSet_j}} \prod_{i = 1}^{\prodsize}\vect_i(\wElem_i)\ex{\prod_{i = 1}^{\prodsize}\sine(\wElem_i)}
 \end{align*}
-Fix the variables $\wElem_1,\ldots, \wElem_{\prodsize}$. Define $\dist$ to be the number of distinct worlds in $\wElem_1,\ldots, \wElem_{\prodsize}$ and $e_l$ to be the number of repitions for the $l_{th}$ distinct world value.  For $\term_1^{\est_j} = \ex{\prod_{i = 1}^{\prodsize} \sine(\wElem_i)}$, we get
+Fix the variables $\wElem_1,\ldots, \wElem_{\prodsize}$. Define $\dist$ to be the number of distinct worlds in $\wElem_1,\ldots, \wElem_{\prodsize}$ and $e_l$ to be the number of repitions for the $l_{th}$ \AR{General typesetting comments. (1) You shoud laway use $\ell$ instead of $l$. (2) Typeset $l_{th}$ as $\ell^{\text{th}}$-- note that ``th" is in superscript and not in math mode.} distinct world value.  For $\term_1^{\est_j} = \ex{\prod_{i = 1}^{\prodsize} \sine(\wElem_i)}$, \AR{Why are you defining the new notation $\term_1^{\est_j}$. You should always be wary of introducing new notation since it makes things hard to read.} we get
 \begin{align*}
 \term_1^{\est_j} = &\ex{\prod_{i = 1}^{\prodsize}\sine(\wElem_i)}\\
 = &\ex{\prod_{l = 1}^{\dist} \sine(\wElem_l)^{e_l}}\\
@@ -30,6 +31,7 @@ Fix the variables $\wElem_1,\ldots, \wElem_{\prodsize}$. Define $\dist$ to be th
 		1	& \dist = 1.
 	\end{cases}	
 \end{align*}
+\AR{Why is the last equality true? You need to justify it by explicitly showing how you are using Lemma~\ref{lem:exp-sine} to prove it.}
 Notice, that the above leaves us with the condition that $\forall i, j \in [\prodsize], \wElem_i = \wElem_j$,
 \begin{align*}
 = &\sum_{\wElem \in \wSet_j}\prod_{i = 1}^{\prodsize} \vect_i(w) \cdot \term_1^{\est_j} = \term_j.
@@ -46,10 +48,11 @@ Therefore, substituting in the definition of variance for complex numbers,
 \begin{align}
 \sigsq &= \ex{\sum_j \est_j \cdot \conj{\sum_{j'} \est_j'}} - \ex{\sum_j \est_j}\cdot\ex{\conj{\sum_{j'} \est_{j'}}}\nonumber\\
 &= \ex{\sum_j \est_j \cdot \sum_{j'} \conj{\est_j'}} - \ex{\sum_j \est_j}\cdot\ex{\sum_{j'} \conj{\est_{j'}}}\nonumber\\
-&= \sum_{j, j'}\left(\ex{\est_j \cdot \overline{\est_j'}} - \ex{\est_j}\ex{\overline{\est_{j'}}} = \cvar{j, j'}\right)\nonumber\\
+&= \sum_{j, j'}\left(\ex{\est_j \cdot \overline{\est_j'}} - \ex{\est_j}\ex{\overline{\est_{j'}}} = \cvar{j, j'}\right)\nonumber\\ 
 &= \sum_j\ex{\est_j \cdot \overline{\est_j'}} - \ex{\est_j}\ex{\overline{\est_j}} + \sum_{j \neq j'}\cvar{j, j'}\nonumber\\
 &= \sum_j \sigsq_j + \sum_{j \neq j'}\cvar{j, j'} \label{eq:sigsq-jneqj}
 \end{align}
+\AR{The above is a terrible way to define $\lambda(j,j')$. Pretty much any reader will miss the fact that you defined  it here. Define $\lambda(j,j')$ ideally outside the align statement especially since this definition will be used later on as well.}
 Notice that assuming independence of  $\sigsq_j ~\forall j \in \sketchCols$, we can push the variance through the sum and obtain the result
 \begin{align*}
 &\sigsq - \sum_j \sigsq_j = \cvar{j, j'}\\
@@ -57,6 +60,7 @@ Notice that assuming independence of  $\sigsq_j ~\forall j \in \sketchCols$, we
 \end{align*}
 Recall that we started this section out by seeking to prove \cref{eq:var-to-prove}.  Should this be true, the use of $\leq$ in the above implication results from the fact that $\sigsq \leq \sum_j \sigsq_j \implies \cvar{j, j'} \leq 0$.
 \AH{I'm really not so sure about the above results.  This was from a conversation we had months ago, but we're basing an implication on something we haven't proved.  That doesn't seem right to me.}
+\AR{Yeah, the para above does not make sense.}
 
 One can see that \cref{eq:sigsq-jneqj} is composed of two addends.  We now bound each of them separately.
 \subsection{Bounding $\sum_{j \neq j'}\cvar{j, j'}$}