Proof for \tilde{Q}(p,...p)

poly-form.tex

@@ -8,12 +8,16 @@ Let $\vect_1,\ldots, \vect_\numTup$ be vectors annotating $\numTup$ tuples in a
 \end{equation}
 Here we define vector indexing by the $\numTup$-bit binary tuple $\wVec = (\wbit_1,\ldots,\wbit_\numTup)$ such that the possible world $\wVec$ is identified by its bit vector binary value.
 
-Futher we define the polynomial $\poly(\vect_1,\ldots,\vect_\numTup)$ as an arbitrary polynomial defined over the input vectors, whose addition and multiplication operations are defined as the traditional point-wise vector addition and multiplication.  We overload notation and denote the $i^{th}$ world of $\poly$ as $\poly[\wVec]$, where $\poly$ can be viewed as the output annotation vector, and the L-1 norm can be represented as $\poly = \sum\limits_{\wVec \in \wSet} \poly[\wVec]$.
+Futher we define the polynomial $\poly(\vect_1,\ldots,\vect_\numTup)$ as an arbitrary polynomial defined over the input vectors, whose addition and multiplication operations are defined as the traditional point-wise vector addition and multiplication.  We overload notation and denote the $i^{th}$ world of $\poly$ as $\poly[\wVec]$, where $\poly$ can be viewed as the output annotation vector, and the L-1 norm can be represented as 
 
-By noting that \cref{eq:vec-def} is equivalent to $\vect_i[(\wbit_1,\ldots, \wbit_\numTup)] = \wbit_i$, we can further reformulate the problem by saying that $\poly[\wVec] = \poly(\wbit_1,\ldots, \wbit_\numTup)$, where each element of the RHS input is a bit element of the $\wVec$ bit vector, and we can thus replace each variable of $\poly$ with its corresponding input bit, and solve for the particular world $\wVec$.  We can then further rewrite our L-1 norm output as $\sum\limits_{(\wbit_1,\ldots, \wbit_\numTup) \in \{0, 1\}^\numTup} \poly(\wbit_1,\ldots, \wbit_\numTup)$.
+\[\norm{\poly}_1 = \sum\limits_{\wVec \in \wSet} \poly[\wVec].\]
+
+By noting that \cref{eq:vec-def} is equivalent to $\vect_i[(\wbit_1,\ldots, \wbit_\numTup)] = \wbit_i$, we can further reformulate the problem by saying that $\poly[\wVec] = \poly(\wbit_1,\ldots, \wbit_\numTup)$, where each element of the RHS input is a bit element of the $\wVec$ bit vector, and we can thus replace each variable of $\poly$ with its corresponding input bit, and solve for the particular world $\wVec$.  We can then further rewrite our L-1 norm output as 
+
+\[\norm{\poly}_1 = \sum\limits_{(\wbit_1,\ldots, \wbit_\numTup) \in \{0, 1\}^\numTup} \poly(\wbit_1,\ldots, \wbit_\numTup).\]
 
 Further, define $\rpoly(\wbit_1,\ldots, \wbit_\numTup)$ as the reduced version of $\poly(\wbit_1,\ldots, \wbit_\numTup)$, formally defined as
-\[\rpoly(\wbit_1,\ldots, \wbit_\numTup) = \poly(\wbit_1,\ldots, \wbit_\numTup) \mod \wbit_1^2-\wbit\cdots\mod \wbit_\numTup^2 - \wbit_\numTup.\]  Intuitively, $\rpoly(\wbit_1,\ldots, \wbit_\numTup)$ is the expanded sum of products form of $\poly(\wbit_1,\ldots, \wbit_\numTup)$ such that if any $\wbit_j$ term  has an exponent $e > 1$, it is reduced to $1$, i.e. $\wbit_j^e\mapsto \wbit_j$ for any $e > 1$.  This usefulness of this reduction will be seen shortly.
+\[\rpoly(\wbit_1,\ldots, \wbit_\numTup) = \poly(\wbit_1,\ldots, \wbit_\numTup) \mod \wbit_1^2-\wbit\cdots\mod \wbit_\numTup^2 - \wbit_\numTup.\]  Intuitively, $\rpoly(\wbit_1,\ldots, \wbit_\numTup)$ is the expanded sum of products form of $\poly(\wbit_1,\ldots, \wbit_\numTup)$ such that if any $\wbit_j$ term  has an exponent $e > 1$, it is reduced to $1$, i.e. $\wbit_j^e\mapsto \wbit_j$ for any $e > 1$.  The usefulness of this reduction will be seen shortly.
 
 First, note the following fact:
 \[\text{For all } (\wbit_1,\ldots, \wbit_\numTup) \in \{0, 1\}^\numTup, \poly(\wbit_1,\ldots, \wbit_\numTup) = \rpoly(\wbit_1,\ldots, \wbit_\numTup).\]
@@ -22,10 +26,11 @@ First, note the following fact:
 For all $b \in \{0, 1\}$ and all $e \geq 1$, $b^e = 1$.\qed
 \end{proof}
 
-As stated previously, the aggregate output we seek to compute is the L1-norm of the output vector $\poly$, which we'll denote as $\out$.  Assuming each tuple has a probability $p = 0.5$, we note that
-\begin{Property}
+Assuming each tuple has a probability $\prob = 0.5$, we note that
+\begin{Property}\label{prop:l1-rpoly-numTup}
+The L-1 norm of Q is equal to $\rpoly$ times the number of possible worlds, $|\wSet| = 2^\numTup$.
 \begin{equation*}
-\out = \sum_{(\wbit_1,\ldots, \wbit_\numTup) \in \{0, 1\}^\numTup} \poly(\wbit_1,\ldots, \wbit_\numTup) = 2^\numTup \cdot \rpoly(\frac{1}{2},\ldots, \frac{1}{2}).
+\norm{\poly}_1 = \sum_{(\wbit_1,\ldots, \wbit_\numTup) \in \{0, 1\}^\numTup} \poly(\wbit_1,\ldots, \wbit_\numTup) = 2^\numTup \cdot \rpoly(\frac{1}{2},\ldots, \frac{1}{2}).
 \end{equation*}
 \end{Property}
 
@@ -38,4 +43,28 @@ Note that for any single monomial, this is indeed the case since 1) each tuple a
 The final result follows by noting that $\rpoly$ is a sum of monomials, and we can, by rearranging the sum, equivlently push the sum into the monomials.\qed
 \end{proof}
 
+\begin{Property}
+For the case of general $\prob$, where each tuple in the TIDB is present with probability $\prob$, the expectation of polynomial $\poly$ is equal to $\rpoly(\prob,\ldots, \prob).$
+\end{Property}
+
+\begin{proof}
+Note that $\poly$ has an equivalent sum of products form such that $\poly$ is the sum of monomials.  By linearity of expectation, the expectation of $\poly$ is equivalent to expectation of each monomial in $\poly$.  
+
+Note further, that for the general monomial, the only operation is product.  In the binary (TIDB) setting, if any of the variables in the monomial are zero, then the whole monomial becomes zero.  Note that this case contributes nothing to the expectation.  Notice also, that if the variables are all one, then their product is one, and the product of the identity element with the product of probabilities is the product of probabilities.  This is the only condition which contributes to the expectation.  It is therefore necessary to know which worlds the variables in the monomial are all equal to one, and the expectation then is the sum of the probabilities of each world for which the monomial's variables are equal to one.  This sum of probabilities is also known as the marginal probability, and this sum is always equal the overall probability of the variables in the monomial.  It then stands that the general monomial's expectation is equal to the product of its variable's overall probabilities.  The sum of all such expecations is exactly the definition of $\rpoly(\prob,\ldots, \prob)$.  Let $M$ be a monomial, $t$ be the number of monomials in $\poly$, and $x_1(\cdots x_v)$ represent the product variable(s) in $M_i$.  Then
+\begin{equation*}
+\ex{M_1 + \cdots + M_t} = \ex{M_1} +\cdots+\ex{M_t}
+\end{equation*}
+For any $M_i$,
+\begin{align*}
+&\ex{M_i} = \ex{x_1(\cdots x_v)} = \sum_{(\wElem_1,\ldots, \wElem_N) \in \{0, 1\}^\numTup} x_1(\cdots x_v) \cdot p_1\cdots p_v\\
+&=\sum_{\substack{(\wElem_1,\ldots, \wElem_\numTup) \in \{0, 1\}\\\
+				s.t. \forall i' \in \{j | x_J \in M_i\}\\
+				\wElem_{i'} = 1}} x_1(\cdots x_v) \prod_{i' \in \{j | x_j \in M_i\}} p_i \prod_{\substack{i'' \not\in \{j | x_j \in M_i\}\\ \wElem_{i''} = 1}}p_{i''}\prod_{\substack{i'' \not\in \{j | x_j \in M_i\}\\ \wElem_{i'''} = 0}} 1 - p_{i'''}\\
+&=\prod_{i' \in \{j | x_j \in M_i\}}\prob_{i'}\\ 	
+&\implies \ex{M_1} +\cdots+\ex{M_t} = \prod_{i_1 \in \{j | x_j \in M_1\}}\prob_{i_1} +\cdots+\prod_{i_v \in \{j | x_j \in M_v\}} \prob_{i_v}\\
+&=\rpoly(\prob_1,\ldots, \prob_\numTup).\qed
+\end{align*}
+I just realized, that I could have saved a lot of time by noting that for the case of TIDB, all monomial variables in $M_i$ are independent, and then using linearity of expectation to conclude the proof.  
+\end{proof}
+