Norm upper bounds

master
Aaron Huber 2019-08-20 12:01:56 -04:00
parent e24db7dc23
commit 689ced732f
3 changed files with 68 additions and 15 deletions

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@ -337,6 +337,7 @@ Equation \eqref{eq:spaceTwo} has each of the $|\pw|$ worlds times all the rest o
\eqref{eq:spaceOne} and \eqref{eq:spaceTwo} together form
\AH{We cannot use L0 'norm' here because \eqref{eq:spaceOne} relies on the cardinality of worlds independent of whether world existence.}
\begin{align}
&\norm{\genV}^2_2\left(\frac{|\pw|}{\sketchCols}- 1\right) + \frac{\norm{\genV}_1^2 - \norm{\genV}_2^2}{\sketchCols}\\
& < \frac{\norm{\genV}_2^2\left(|\pw|\right) + \norm{\genV}_1^2}{\sketchCols} \label{eq:variance}
@ -365,38 +366,88 @@ Pr\left[~|X - \mu|~> \Delta~\right] < \frac{\sigma^2}{\Delta^2}
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*}
\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*}
\frac{3\norm{\genV}_2^2\left(|\pw|\right) + \norm{\genV}_1^2}{\norm{\genV}_1^2 \cdot \epsilon^2} &= \sketchCols\label{eq:bucket-bounds-no-sub}
\end{align}
A brief digression is desirable for the purpose of simplifying 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}
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
\SR{Simplify further with L0 'norm', although that makes the simplification more difficult.}
\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
\[
\norm{\genV}_1 \leq \sqrt{|\pw|} \cdot \norm{\genV}_2,
\]
\begin{equation}
\norm{\genV}_1 \leq \sqrt{|\pw|} \cdot \norm{\genV}_2\label{eq:norm1-cauchy},
\end{equation}
which squared yields
\begin{equation}
\norm{\genV}_1^2 \leq |\pw| \cdot \norm{\genV}_2^2\label{eq:norm1-sq-cauchy}.
\end{equation}
Note that \eqref{eq:norm1-sq-cauchy} can be further tightened to
\begin{equation}
\norm{\genV}_1^2 \leq \norm{\genV}_0 \cdot \norm{\genV}_2^2
\end{equation}
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|}\label{eq:b-cauchy}
&\sketchCols \leq \frac{3\norm{\genV}_2^2\left(|\pw|\right) + \norm{\genV}_2^2\left(|\pw|\right)}{\norm{\genV}_2\sqrt{|\pw|}\cdot \epsilon^2}\nonumber\\
&\sketchCols \leq \frac{4\norm{\genV}_2^2\left(|\pw|\right)}{\norm{\genV}_2\sqrt{|\pw|}\cdot \epsilon^2}\nonumber\\
&\sketchCols \leq \frac{4\norm{\genV}_2\sqrt{|\pw|}}{\epsilon^2}\label{eq:b-cauchy}
\end{align}
\AH{\textbf{BEGIN}: Old Bound calculations}
\AH{Justify this.}
\begin{Justification}
\hfill
\begin{itemize}
\item stuff goes here.
\end{itemize}
\end{Justification}
To further tighten the bounds calculations above, we can bound the square of the L2 norm.
\begin{align}
\norm{\genV}_2^2 &= \sum_{i = 1}^{n}|\genV|^2 \\
&\leq \sum_{i = 1}^{n}\left(max_{i}|\genV_i|\right)\left(\genV_i\right)\\
&\leq \sum_{i = 1}^{n}\norm{\genV}_\infty |\genV_i|\\
&\leq \norm{\genV}_\infty \sum_{i = 1}^{n}|\genV_i|\\
&\leq \norm{\genV}_\infty \cdot \norm{\genV}_1 \label{eq:l2-bounds}
\end{align}
\AH{Justify this.}
\begin{Justification}
\hfill
\begin{itemize}
\item stuff goes here.
\end{itemize}
\end{Justification}
Going back to equation \eqref{eq:bucket-bounds-no-sub} and substituting in the above bounds obtains the following.
\begin{align}
\sketchCols &= \frac{3\norm{\genV}_2^2\left(|\pw|\right) + \norm{\genV}_1^2}{\norm{\genV}_1^2 \cdot \epsilon^2} \\
&\leq \frac{3\norm{\genV}_\infty\norm{\genV}_1 \left(|\pw|\right) + \norm{\genV}_1^2}{\norm{\genV}_1^2\cdot \epsilon^2}\label{eq:sub-bounds1}\\
&\leq \frac{3\norm{\genV}_\infty\sqrt{\norm{\genV}_0}\norm{\genV}_2\left(|\pw|\right) + \norm{\genV}_0\norm{\genV}_2^2}{\norm{\genV}_0\norm{\genV}_2^2 \cdot \epsilon^2}\label{eq:sub-bounds2}\\
&\leq \frac{3\norm{\genV}_\infty \sqrt{\norm{\genV}_0} \sqrt{\norm{\genV}_\infty\norm{\genV}_1}\left(|\pw|\right) + \norm{\genV}_0\norm{\genV}_\infty\norm{\genV}_1}{\norm{\genV}_0\norm{\genV}_\infty\norm{\genV}_1\epsilon^2}\label{eq:sub-bounds3}\\
&\leq \frac{\norm{\genV}_\infty \sqrt{\norm{\genV}_0\norm{\genV}_1}\left(3\sqrt{\norm{\genV}_\infty}\left(|\pw|\right) + \sqrt{\norm{\genV}_0\norm{\genV}_1}\right)}{\norm{\genV}_0\norm{\genV}_\infty\norm{\genV}_1\epsilon^2}\label{eq:sub-bounds4}\\
&\leq \frac{3\sqrt{\norm{\genV}_\infty}\left(|\pw|\right) + \sqrt{\norm{\genV}_0\norm{\genV}_1}}{\sqrt{\norm{\genV}_0\norm{\genV}_1} \epsilon^2} \label{eq:sub-bounds5}\\
&\leq \frac{3\sqrt{\norm{\genV}_\infty}\left(|\pw|\right)}{\sqrt{\norm{\genV}_0\norm{\genV}_1} \epsilon^2} + \frac{1}{\epsilon^2}\label{eq:sub-bounds-final}
\end{align}
\begin{Justification}
\hfill
\begin{itemize}
\item \eqref{eq:sub-bounds1} results from substituting \eqref{eq:l2-bounds} for the L2 norm.
\item \eqref{eq:sub-bounds2} is obtained from substituting \eqref{eq:norm1-cauchy} for the L1 norm and \eqref{eq:norm1-sq-cauchy} for the L1 norm squared terms in both the numerator and denominator.
\item \eqref{eq:sub-bounds3} is the result of further substituting \eqref{eq:l2-bounds} for the newly introduced L2 norm terms in the numerator.
\item \eqref{eq:sub-bounds4} is the result of factoring out common terms in the numerator.
\item \eqref{eq:sub-bounds5} is the result of cancelling out common terms in the numerator and denominator.
\item \eqref{eq:sub-bounds-final} is simply a rearrangement of the two numerator terms, for the purpose of making things simpler.
\end{itemize}
\end{Justification}
\startOld{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}\\
@ -415,7 +466,7 @@ Setting $\Delta = \epsilon\numWorlds$ gives
\end{align*}
Other cases for $\Delta$ can be solved similarly.
\AH{\textbf{END}: Old Bound calculations}
\finOld

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@ -2,7 +2,7 @@
\section{Instantiation}
\label{sec:instantiation}
\AH{This section has been started, but needs to be completed.}
\subsection{TIDB}
Consider the case of a TIDB with $\numTup$ tuples, with $\prob = \frac{1}{2}$ for given tuple $t$. Because TIDB has the property of set semantics, the vector $\genV$ can then be defined as a binary bit vector $\{0, 1\}^\numTup$, whose value represents a possible world, and, where each index represents a specific tuple $t$ id. Under these semantics, with $w_t$ representing the index mapped to a a tuple $t$'s identity, $\genV$ can alternatively be viewed as a function

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@ -18,7 +18,6 @@
\newcommand{\buck}{\textbf{j}}
\newcommand{\jVec}{\textbf{j}}
\newcommand{\lenB}{b}
%\newcommand{\log}{log}
\newcommand{\hVec}{\textbf{h}_{i,k}}
\newcommand{\matrixH}{H}
\newcommand{\hVecMatrix}{\begin{pmatrix*}[l]
@ -134,7 +133,10 @@
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