Notes on potential bugs.

sop.tex

@@ -181,6 +181,7 @@ Functions \surj, \surj' are the set of surjective mappings from $\prodsize$ to $
 The functions $\surj, \surj'$ are used to produce the mappings $w_i \mapsto \dMap{w_{\surj(i)}}$.  In particular, $\surj$ and $\surj'$ are machinery for mapping $\prodsize$ $\wElem$-world variables to $\dist$ distinct values.  
 
 We rewrite equation \eqref{eq:sig-j-last} in terms of $\dist$ distinct worlds, with $\surj, \surj'$ mappings.
+\AH{A bug here?  I think it is necessary to include $\term_2$ for the rewriting to be equivalent.}
 \begin{equation}
 \sum_{\dist  = 1}^{\prodsize}\sum_{\dist'  = 1}^{\prodsize}\sum_{\surj, \surj'}\sum_{\substack{\dw_1 \prec \cdots \prec \dw_\dist,\\ \dw'_{1} \prec \cdots \prec \dw'_{\dist'}\\ \in W}}\prod_{i = 1}^{\prodsize}\vect_i(\dw_{\surj(i)})\vect_i(\dw'_{\surj'(i)})\cdot \term_1\left(\dw_{\surj(1)},\ldots,\dw_{\surj(\prodsize)}, \dw'_{\surj'(1)},\ldots, \dw'_{\surj'(\prodsize)}\right)
 \label{eq:sig-j-distinct}
@@ -232,9 +233,6 @@ Functions $\surj:[\prodsize]\mapsto [\dist], \surj':[\prodsize]\mapsto [\dist']$
 	\end{enumerate}
 \end{Definition}
 
-\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{\^---next on the agenda!}
-
 \begin{Lemma}\label{lem:sig-j-survive}
 When $\surj, \surj'$are matching, where for every $i \in[\dist], \dw_{i} = \dw'_{i}$, %\cref{eq:sig-j-distinct} is exactly 
 \[
@@ -263,18 +261,19 @@ 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 $\dw_i$ 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"--
-\[T_1(\dw_{\surj(1)},\dots, \dw_{\surj(m)}, \dw_{\surj'(1)},\dots, \dw_{\surj'(m')}).\]
-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
-\[\prod_{i=1}^k s(\dw_{\surj(i)})\cdot \overline{s(\dw_{\surj'(i)})}.\]
-Now re-write the above in terms of ``powers" of distinct worlds:
-\[ (\prod_{i=1}^m s(\dw_{i})^{|\surj^{-1}(i)|})\cdot \overline{(\prod_{j=1}^m s(\dw_j)^{|\surj^{-1}(j)|})}\]
-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...
-}
+%\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"--
+%\[T_1(\dw_{\surj(1)},\dots, \dw_{\surj(m)}, \dw_{\surj'(1)},\dots, \dw_{\surj'(m')}).\]
+%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
+%\[\prod_{i=1}^k s(\dw_{\surj(i)})\cdot \overline{s(\dw_{\surj'(i)})}.\]
+%Now re-write the above in terms of ``powers" of distinct worlds:
+%\[ (\prod_{i=1}^m s(\dw_{i})^{|\surj^{-1}(i)|})\cdot \overline{(\prod_{j=1}^m s(\dw_j)^{|\surj^{-1}(j)|})}\]
+%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}
+\AH{This generic term was introduced earlier in \cref{eq:sig-j-distinct}}
 
+\begin{proof}
 Note \cref{eq:term-1}, and consider the "generic term"--
 \[\term_1(\dw_{\surj(1)},\dots, \dw_{\surj(\prodsize)}, \dw_{\surj'(1)},\dots, \dw_{\surj'(\prodsize)}).\]
 
@@ -385,6 +384,8 @@ Using the above definitions, we can now present the variance bounds for $\sigsq_
 When $\match{\surj}{\surj'}$, with $\dw_i = \dw'_i$ for all $i \in [\dist]$, \cref{eq:sig-j-distinct} = 
 \[\sum_{\dist  = 2}^{\prodsize}\frac{1}{\sketchCols^{\dist}}\sum_{\substack{\surj, \surj'\\\match{\surj}{\surj'}}}\sum_{\substack{\dw_{_1}, \ldots,\dw_{_\dist}\\ \in W}}\prod_{i = 1}^{\prodsize}\vect_i(\dw_{\surj(i)})\vect_i(\dw_{\surj'(i)}).\]
 \end{Lemma}
+\AH{A bug below?  When we have that for all $i \in [\prodsize]$ $\dw_i = \dw_{i'}$, doesn't $\term_2 = \frac{1}{\sketchCols^2}$ while $\term_1 = \frac{1}{\sketchCols}$, yielding a difference of $\frac{B - 1}{\sketchCols^2}$?}
+\AH{Should we mention this weird case for $\dist = \dist' = 1$, where for all $i \in [\prodsize]$ $\dw_i = \wElem$ and for all $i' \in [\prodsize]$ $\dw'_{i'} = \wElem'$ \textit{but} $\wElem \neq \wElem'$?  This case makes $\term_1 - \term_2 = 0$, and doesn't change the results.}
 
 By the fact that the expectations cancel when $\forall i, i', j, j'\in [\prodsize], \wElem_i = \wElem_j = \wElem, \wElem_{i'}' = \wElem_{j'}' = \wElem'$, for both $\wElem = \wElem'$ and $\wElem \neq \wElem'$, we can rid ourselves of $\term_2$, (\cref{eq:term-2}), the case when there exists only one distinct world value.  This is precisely why we have not needed to account for the last two expectations in \cref{eq:sig-j-last}.  We then need to sum up all the $\dist$ distinct world value possibilities for $\dist \in [2, \prodsize]$.  Starting with \cref{eq:sig-j-distinct},