ODIn/paper-BagRelationalPDBsAreHard
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Aaron Huber 2019-09-03 15:16:29 -04:00
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combining.tex
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@ -79,10 +79,11 @@ Note that with an odd number of sketches being multiplied, such as 3, we get the
For $\est{3}$, multiplying an even number of sketches yields
\begin{align*}
&\expect{\sum_{j \in \sketchCols}\sCom{1}{j} \cdot \sCom{2}{j}}\\
=&\expect{\sum_{j \in \sketchCols}\left(\sum_{\substack{\wVec \in \pw \st\\\hashP{\wVec} = j}}\gVP{1}{\wVec}\polP{\wVec}\cdot \sum_{\substack{\wVecPrime \in \pw \st\\\hashP{\wVecPrime} = j}}\gVP{2}{\wVecPrime}\polP{\wVecPrime}\right)}\\
=&\expect{\sum_{j \in \sketchCols}\sum_{\substack{\wVec, \wVecPrime \in \pw \st\\\hashP{\wVec} = j\\\wVec = \wVecPrime}}\gVP{1}{\wVec}\gVP{2}{\wVec}\polP{\wVec}\polP{\wVec}\sum_{\substack{\wVec, \wVecPrime \in \pw \st\\\hashP{\wVec} = j\\\wVec \neq \wVecPrime}}\gVP{1}{\wVec}\gVP{2}{\wVecPrime}\polP{\wVec}\polP{\wVecPrime}}\\
=&\expect{\sum_{\wVec \in \pw}\gVP{1}{\wVec}\gVP{2}{\wVec}}\\
=&\gVP{1}{\wVec}\gVP{2}{\wVec}
=&\expect{\sum_{j \in \sketchCols}\left(\sum_{\substack{\wOne \in \pw \st\\\hashP{\wOne} = j}}\gVP{1}{\wOne}\polP{\wOne}\cdot \sum_{\substack{\wTwo \in \pw \st\\\hashP{\wTwo} = j}}\gVP{2}{\wTwo}\polP{\wTwo}\right)}\\
=&\mathbb{E}\big[\sum_{j \in \sketchCols}\sum_{\substack{\wOne, \wTwo \in \pw \st\\\hashP{\wOne} = j\\\wOne = \wTwo}}\gVP{1}{\wOne}\gVP{2}{\wOne}\polP{\wOne}\polP{\wOne} +\\
&\qquad \gVP{1}{\wOne}\polP{\wOne}\sum_{\substack{\wOne, \wTwo \in \pw \st\\\hashP{\wOne} = j\\\wOne \neq \wTwo}}\gVP{2}{\wTwo}\polP{\wTwo}\big]\\
=&\expect{\sum_{\wOne \in \pw}\gVP{1}{\wOne}\gVP{2}{\wOne}}\\
=&\sum_{\wOne \in \pw}\gVP{1}{\wOne}\gVP{2}{\wOne}
\end{align*}
Following the reversal of the pattern of $\est{2}$, an odd number of sketches would produce an expectation of $0$, since each product in the sum has an operand whose expectation evaluates to $0$, as seen in the following,
\begin{align*}
@ -112,7 +113,9 @@ The case for an odd number of sketches can be reduced to the even case by includ
\wTwo \neq \wOne}}\gVP{2}{\wTwo}\gVP{3}{\wTwo}.
\end{align*}
One potential work around would be to store additional sketches with independent $\pol$ functions. For $\est{2}$, this would result in
\begin{align*}
\end{align*}
For the case of multiplication, when assumming independent variables, it is a known result that
\[
\varParam{X \cdot Y} = \expect{X^2}\expect{Y^2} - (\expect{X})^2 (\expect{Y})^2.