ODIn/paper-BagRelationalPDBsAreHard
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New Chebyshev bounds

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Aaron Huber 2019-08-16 11:32:16 -04:00
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analysis.tex
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@ -327,9 +327,41 @@ Substituting $\Delta = k\sigma \rightarrow k = \frac{\Delta}{\sigma} \rightarrow
\begin{equation*}
Pr\left[~|X - \mu|~> \Delta~\right] < \frac{\sigma^2}{\Delta^2}
\end{equation*}
\AH{This next bit needs to be redone.}
For the case when $\Delta = \mu\epsilon$, taking both Chebyshev bounds, setting them equal to each other, simplifying and solving for $\sketchCols$ results in
\begin{align*}
\frac{\sigma^2}{\Delta^2} &= \frac{1}{3}\\
\frac{\norm{\genV}_2^2 \cdot \left(|\pw|\right) + \norm{\genV}_1^2}{\sketchCols \norm{\genV}_1^2 \cdot \epsilon^2} &= \frac{1}{3}\\
\frac{3\norm{\genV}_2^2\left(|\pw|\right) + \norm{\genV}_1^2}{\norm{\genV}_1^2 \cdot \epsilon^2} &= \sketchCols
\end{align*}
A brief digression is desirable for the purpose of simply the above bounds. Recall the Cauchy Schwarts inequality which states:
\[\sum_i a_i \cdot b_i \leq \norm{a}_2 \cdot \norm{b}_2\].
The L1 norm can be expanded to the following expression,
\[\norm{\genV}_1 = \sum_{\wVec \in \pw} 1 \cdot \genVParam{\wVec}.\]
Notice that the constant term can be viewed as a vector of $1$'s with size $n$ (the size of $\genV$). Calling this vector $x$ and taking the L2 norm gives\begin{align}
\norm{x} &= \sqrt{1_1^2 + 1_2^2 + \cdots + 1_n^2}\nonumber\\
&= \sqrt{n * 1} \nonumber\\
&= \sqrt{n}\nonumber\\
&= \sqrt{|\pw|}\label{eq:w-card}
\end{align}
By \eqref{eq:w-card} and Cauchy Swarts, we then have
\[
\norm{\genV}_1 \leq \sqrt{|\pw|} \cdot \norm{\genV}_2,
\]
which squared yields
\[
\norm{\genV}_1^2 \leq |\pw| \cdot \norm{\genV}_2^2.
\]
Substituting the Cauchy Schwarts bounds into the Chebyshev calculations gives
\begin{align}
&\sketchCols \leq \frac{3\norm{\genV}_2^2\left(|\pw|\right) + \norm{\genV}_2^2\left(|\pw|\right)}{\norm{\genV}_2\sqrt{|\pw|}}\nonumber\\
&\sketchCols \leq \frac{4\norm{\genV}_2^2\left(|\pw|\right)}{\norm{\genV}_2\sqrt{|\pw|}}\nonumber\\
&\sketchCols \leq 4\norm{\genV}_2\sqrt{|\pw|}
\end{align}
\AH{\textbf{BEGIN}: Old Bound calculations}
\begin{align*}
\frac{\sigma^2}{\Delta^2} &= \frac{1}{3}\\
\frac{ 2^{2N}\big(\frac{2\prob}{\sketchCols}\big)}{\mu^2\epsilon^2} &= \frac{1}{3}\\
@ -348,6 +380,7 @@ Setting $\Delta = \epsilon\numWorlds$ gives
\end{align*}
Other cases for $\Delta$ can be solved similarly.
\AH{\textbf{END}: Old Bound calculations}