Merge branch 'master' of https://git.overleaf.com/61d6263016ff472ac9308dea

introduction.tex

@@ -172,7 +172,8 @@ The lineage polynomial for $\query_1^2$ is $\poly_1^2\inparen{A, B, C, E, U, Y,
 $$
 =A^2U^2B^2 + B^2Y^2E^2 + B^2Z^2C^2 + 2AUB^2YE + 2AUB^2ZC + 2B^2YEZC.
 $$
-To compute $\expct\pbox{\poly_1^2}$ we can use linearity of expectation and push the expectation through each summand.  To keep things simple, let us focus on the monomial $\monomial{1}(A,U,B) = A^2U^2B^2$ as the procedure is the same for all other monomials of $\poly_1^2$.  Let $\randWorld_X$ be the random variable corresponding to a variable $X$. Because the distinct variables in the product are independent, we can push expectation through them yielding $\expct\pbox{\randWorld_A^2\randWorld_U^2\randWorld_B^2}=\expct\pbox{\randWorld_A^2}\expct\pbox{\randWorld_U^2}\expct\pbox{\randWorld_B^2}$.  Since $\randWorld_A, \randWorld_B\in \inset{0, 1}$ we can simplify to $\expct\pbox{\randWorld_A}\expct\pbox{\randWorld_U^2}\expct\pbox{\randWorld_B}$ by the fact that for any $W\in \inset{0, 1}$, $W^2 = W$.  Observe that if $W_U\in\inset{0, 1}$, then we further would have $\expct\pbox{\randWorld_A}\expct\pbox{\randWorld_U}\expct\pbox{\randWorld_B} = \prob_A\cdot\prob_U\cdot\prob_B = \rmonomial{1}\inparen{\prob_A, \prob_U, \prob_B}$ (denoting $\probOf\pbox{\randWorld_X = 1} = \prob_X$).  However, in this example, we get stuck with $\expct\pbox{\randWorld_U^2}$, since $\randWorld_U\in\inset{0, 1, 2}$ and for $\randWorld_U \gets 2$, $\randWorld_U^2 \neq \randWorld_U$.
+To compute $\expct\pbox{\poly_1^2}$ we can use linearity of expectation and push the expectation through each summand.  To keep things simple, let us focus on the monomial $\monomial{1}(A,U,B) = A^2U^2B^2$ as the procedure is the same for all other monomials of $\poly_1^2$.  Let $\randWorld_X$ be the random variable corresponding to a variable $X$. Because the distinct variables in the product are independent, we can push expectation through them yielding $\expct\pbox{\randWorld_A^2\randWorld_U^2\randWorld_B^2}=\expct\pbox{\randWorld_A^2}\expct\pbox{\randWorld_U^2}\expct\pbox{\randWorld_B^2}$.  Since $\randWorld_A, \randWorld_B\in \inset{0, 1}$ we can simplify to $\expct\pbox{\randWorld_A}\expct\pbox{\randWorld_U^2}\expct\pbox{\randWorld_B}$ by the fact that for any $W\in \inset{0, 1}$, $W^2 = W$.  Observe that if $W_U\in\inset{0, 1}$, then we further would have $\expct\pbox{\randWorld_A}\expct\pbox{\randWorld_U}\expct\pbox{\randWorld_B} = \prob_A\cdot\prob_U\cdot\prob_B$ % = \rmonomial{1}\inparen{\prob_A, \prob_U, \prob_B}$ 
+(denoting $\probOf\pbox{\randWorld_X = 1} = \prob_X$).  However, in this example, we get stuck with $\expct\pbox{\randWorld_U^2}$, since $\randWorld_U\in\inset{0, 1, 2}$ and for $\randWorld_U \gets 2$, $\randWorld_U^2 \neq \randWorld_U$.
 
 The simple insight to get around this issue to note that the random variables $\randWorld_U$ and $\randWorld_{U_1}+2\randWorld_{U_2}$ have exactly the same distribution, where $\randWorld_{U_1},\randWorld_{U_2}\in\inset{0,1}$ and $\probOf\pbox{\randWorld_{U_j} = 1} = \probOf\pbox{\randWorld_{U} = j}$. Thus, the idea is to replace the variable $U$ by $U_1+2U_2$ (where $U_j$ corresponds to the event that $U$ has multiplicity $j$) yielding% to obtain the following polynomial:
 %