From 2ff236556d6f104908faec5f4fea6210c32e9cb4 Mon Sep 17 00:00:00 2001
From: Atri Rudra <atri@buffalo.edu>
Date: Thu, 9 Jul 2020 15:59:57 -0400
Subject: [PATCH] Minor corrects+ new comments

---
 ra-to-poly.tex | 6 +++---
 1 file changed, 3 insertions(+), 3 deletions(-)

diff --git a/ra-to-poly.tex b/ra-to-poly.tex
index 2216575..9531ae6 100644
--- a/ra-to-poly.tex
+++ b/ra-to-poly.tex
@@ -63,17 +63,17 @@ In the general case, the binary value of $\vct{w}$ uniquely identifies a potenti
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 
 
-Assume a domain of $\{0, 1\}$ for each $X_i \in \vct{X}$.  The polynomial $\llbracket\poly(\rel)\rrbracket(\tup)$ can be viewed as a function $\{0, 1\}^M \mapsto \mathbb{N}$.  Since, from this point on, our discussion will involve one polynomial for an arbirtrary $\tup$, we thus abuse notation by using $\poly(\vct{X})$ to be the annotated polynomial $\llbracket\poly(\rel)\rrbracket(\tup)$.  
+Assume a domain of $\{0, 1\}$ for each $X_i \in \vct{X}$.  The polynomial $\llbracket\poly(\rel)\rrbracket(\tup)$ can be viewed as a function $\{0, 1\}^M \mapsto \mathbb{N}$.\AR{Technically, the function view is for {\em evaluations} of the polynomial and the equivalence only holds for the polynomial mod $X_i^2-X_i$ for all $i$.}  Since, from this point on, our discussion will involve one polynomial for an arbirtrary $\tup$, we thus abuse notation by using $\poly(\vct{X})$ to be the annotated polynomial $\llbracket\poly(\rel)\rrbracket(\tup)$.  
 
 One of the aggregates we desire to compute over the annotated polynomial is the expectation, denoted,
 
 \AH{With our notation, I no longer think that $\vct{w} \sim \pd$ is necessary footer for $\expct$.  We can probably just have $\expct\limits_{\vct{w}}$ instead.  Do you agree?}
 \AR{No. How would you state Lemma 4 without explicitly using $P$ in the definition of expectation?}
 
-\[\expct_{\vct{w} \sim \pd}\pbox{\poly(\rw)} = \sum\limits_{\wVec \in \{0, 1\}^\numTup} \poly(\wVec)\cdot \pd[\rw = \vct{w}].\]
+\[\expct_{\vct{\rw} \sim \pd}\pbox{\poly(\rw)} = \sum\limits_{\wVec \in \{0, 1\}^\numTup} \poly(\wVec)\cdot \pd[\rw = \vct{w}].\]
 
 The $\ti$ model has features that we can exploit.  Since the powerset of $[\numTup]$ is exactly $\wSet$, the bit-string world value $\vct{w}$ can be used as indexing to determine which tuples are present in the $\vct{w}$ world.  Given an $\numTup$-sized vector $\vct{p}$, where the $i^{th}$ element, $\prob_i$ is the probability of the $i^{th}$ tuple, denote the vector $\vct{p}$ according to the probability distributation $\pd$ as $\pd^{(\vct{p})}$.  We can then write an equivalent expectation for $\ti$ model,
 
-\[\expct_{\wVec\sim \pd^{(\vct{p})}}\pbox{\poly(\wVec)} = \sum\limits_{\wVec \in \{0, 1\}^\numTup} \poly(\wVec)\prod_{\substack{i \in [\numTup]\\ s.t. \wElem_i = 1}}\prob_i \prod_{\substack{i \in [\numTup]\\s.t. w_i = 0}}\left(1 - \prob_i\right).\]
+\[\expct_{\rw\sim \pd^{(\vct{p})}}\pbox{\poly(\rw)} = \sum\limits_{\wVec \in \{0, 1\}^\numTup} \poly(\wVec)\prod_{\substack{i \in [\numTup]\\ s.t. \wElem_i = 1}}\prob_i \prod_{\substack{i \in [\numTup]\\s.t. w_i = 0}}\left(1 - \prob_i\right).\]