Dealing with sub- and supersets.

Francisco Coelho
1 parent 64dd7c8f
Showing 3 changed files with 51 additions and 8 deletions Show diff stats
code/asp/disj.lp
text/pre-paper.pdf
text/pre-paper.tex
 % prob(a) = 0.3
 a ; -a.
-b ; c :- a.
 \ No newline at end of file
+b ; c :- a.
+
+d ; -d.
+b :- c, d.
 \ No newline at end of file
 \documentclass[a4paper, 12pt]{article}
  
 \usepackage[x11colors]{xcolor}
+%
+\usepackage{tikz}
+\usetikzlibrary{calc}
+%
 \usepackage{hyperref}
 \hypersetup{
     colorlinks=true,
     linkcolor=blue,
 }
-
+%
 \usepackage{commath}
 \usepackage{amsthm}
 \newtheorem{assumption}{Assumption}
@@ -155,17 +159,39 @@ Equation \ref{eq:world.fold.superset} results from conditional independence of t
     Stable models are conditionally independent, given their total choices.
 \end{assumption}
  
-Consider the stable models $ab, ac$ from the example above. They result from the clause $b \vee c \leftarrow a$ and the total choice $a$. These formulas alone impose no relation between $b$ and $c$ (given $a$), so none should be assumed. Dependence relations are discussed in Subsection (\ref{subsec:dependence}).
+Consider the stable models $ab, ac$ from the example above. They result from the clause $b \vee c \leftarrow a$ and the total choice $a$. These formulas alone impose no relation between $b$ and $c$ (given $a$), so none should be assumed. Dependence relations are further discussed in Subsection (\ref{subsec:dependence}).
  
 % SUBSET
+\hrule
  
+\bigskip
 I'm not sure about what to say here.\marginpar{todo}
  
+My first guess was
 \begin{equation*}
     \pr{W  = w \given C = c} = \sum_{s \supset w}\pr{S = s \given C = c}.
 \end{equation*}
  
-But! $\pr{W = w \given C = c}$ already separates $\pr{W}$ into \textbf{disjoint}
+$\pr{W = w \given C = c}$ already separates $\pr{W}$ into \textbf{disjoint} events!
+
+Also, I am assuming that stable models are independent. 
+
+This would entail $p(w) = p(s_1) + p(s_2) - p(s_1)p(s_2)$ \textit{if I'm bound to set inclusion}. But I'm not. I'm defining a relation
+
+Also, if I set $p(w) =  p(s_1) + p(s_2)$ and respect the laws of probability, this entails $p(s_1)p(s_2) = 0$.
+
+So, maybe what I want is (1) to define the cover $\hat{w} = \cup_{s \supset w} s$
+
+\begin{equation*}
+    \pr{W  = w \given C = c} = \sum_{s \supset w}\pr{S = s \given C = c} - \pr{W = \hat{w} \given C = c}.
+\end{equation*}
+
+But this doesn't works, because we'd get $\pr{W = a \given C = a} < 1$.
+%
+
+%
+\bigskip
+\hrule
  
 % INDEPENDENCE
  
@@ -188,9 +214,9 @@ $
 $. 
  
  
-The interesting case is the subtree of the total choice $ac$. Notice that no stable model $s$ contains $adc$ because (1)  $adb$ is a stable model and (2) no stable model contains $adc$, because if $adc \subset s$ then $b \in s$ and $adb \subset s$.
+The interesting case is the subtree of the total choice $ad$. Notice that no stable model $s$ contains $adc$ because (1)  $adb$ is a stable model and (2) if $adc \subset s$ then $b \in s$ so $adb \subset s$. 
  
-Following the case of equations (\ref{eq:world.fold.stablemodel}) and (\ref{eq:world.fold.independent}) this sets 
+Following equations (\ref{eq:world.fold.stablemodel}) and (\ref{eq:world.fold.independent}) this sets 
 \begin{equation*}
     \begin{cases}
         \pr{W = adc \given C = ad} = 0,\cr
@@ -200,7 +226,7 @@ Following the case of equations (\ref{eq:world.fold.stablemodel}) and (\ref{eq:w
 which concentrates all probability mass from the total choice $ad$ in the $adb$ branch, including the node $W = adbc$. This leads to the following cases:
 $$
 \begin{array}{l|r}
-    x & \pr{x \given C = ad}\
+    x & \pr{W = x \given C = ad}\
     \hline
     ad & 1 \\
     adb & 1\\
@@ -208,8 +234,22 @@ $$
     adbc & 1    
 \end{array}
 $$
-so, for $C = ad$, $\pr{b} = 2/3, \pr{c} = 1/3, \pr{b,c} = 1/3 \not= \pr{b}\pr{c}$.
+so, for $C = ad$, 
+$$
+\begin{aligned}
+    \pr{W = b} &= \frac{2}{4} \cr
+    \pr{W = c} &= \frac{1}{4} \cr
+    \pr{W = bc} &= \frac{1}{4} \cr
+                &\not= \pr{W = b}\pr{W = c}
+\end{aligned}
+$$
+\textit{i.e.} the events $W = b$ and $W = c$ are dependent and that dependence results directly from the segment $0.2::d, b \leftarrow c \wedge d$ in the specification.
+
  
+%
+
+%
+\hrule
 \begin{quotation}\note{Todo}
  
     Prove the four world cases (done), support the product (done) and sum (tbd) options, with the independence assumptions.
1	1	% prob(a) = 0.3
2	2	a ; -a.
3		-b ; c :- a.
4	3	\ No newline at end of file
	4	+b ; c :- a.
	5	+
	6	+d ; -d.
	7	+b :- c, d.
5	8	\ No newline at end of file
...	...
1	1	\documentclass[a4paper, 12pt]{article}
2	2
3	3	\usepackage[x11colors]{xcolor}
	4	+%
	5	+\usepackage{tikz}
	6	+\usetikzlibrary{calc}
	7	+%
4	8	\usepackage{hyperref}
5	9	\hypersetup{
6	10	colorlinks=true,
7	11	linkcolor=blue,
8	12	}
9		-
	13	+%
10	14	\usepackage{commath}
11	15	\usepackage{amsthm}
12	16	\newtheorem{assumption}{Assumption}
...	...	@@ -155,17 +159,39 @@ Equation \ref{eq:world.fold.superset} results from conditional independence of t
155	159	Stable models are conditionally independent, given their total choices.
156	160	\end{assumption}
157	161
158		-Consider the stable models $ab, ac$ from the example above. They result from the clause $b \vee c \leftarrow a$ and the total choice $a$. These formulas alone impose no relation between $b$ and $c$ (given $a$), so none should be assumed. Dependence relations are discussed in Subsection (\ref{subsec:dependence}).
	162	+Consider the stable models $ab, ac$ from the example above. They result from the clause $b \vee c \leftarrow a$ and the total choice $a$. These formulas alone impose no relation between $b$ and $c$ (given $a$), so none should be assumed. Dependence relations are further discussed in Subsection (\ref{subsec:dependence}).
159	163
160	164	% SUBSET
	165	+\hrule
161	166
	167	+\bigskip
162	168	I'm not sure about what to say here.\marginpar{todo}
163	169
	170	+My first guess was
164	171	\begin{equation*}
165	172	\pr{W = w \given C = c} = \sum_{s \supset w}\pr{S = s \given C = c}.
166	173	\end{equation*}
167	174
168		-But! $\pr{W = w \given C = c}$ already separates $\pr{W}$ into \textbf{disjoint}
	175	+$\pr{W = w \given C = c}$ already separates $\pr{W}$ into \textbf{disjoint} events!
	176	+
	177	+Also, I am assuming that stable models are independent.
	178	+
	179	+This would entail $p(w) = p(s_1) + p(s_2) - p(s_1)p(s_2)$ \textit{if I'm bound to set inclusion}. But I'm not. I'm defining a relation
	180	+
	181	+Also, if I set $p(w) = p(s_1) + p(s_2)$ and respect the laws of probability, this entails $p(s_1)p(s_2) = 0$.
	182	+
	183	+So, maybe what I want is (1) to define the cover $\hat{w} = \cup_{s \supset w} s$
	184	+
	185	+\begin{equation*}
	186	+ \pr{W = w \given C = c} = \sum_{s \supset w}\pr{S = s \given C = c} - \pr{W = \hat{w} \given C = c}.
	187	+\end{equation*}
	188	+
	189	+But this doesn't works, because we'd get $\pr{W = a \given C = a} < 1$.
	190	+%
	191	+
	192	+%
	193	+\bigskip
	194	+\hrule
169	195
170	196	% INDEPENDENCE
171	197
...	...	@@ -188,9 +214,9 @@ $
188	214	$.
189	215
190	216
191		-The interesting case is the subtree of the total choice $ac$. Notice that no stable model $s$ contains $adc$ because (1) $adb$ is a stable model and (2) no stable model contains $adc$, because if $adc \subset s$ then $b \in s$ and $adb \subset s$.
	217	+The interesting case is the subtree of the total choice $ad$. Notice that no stable model $s$ contains $adc$ because (1) $adb$ is a stable model and (2) if $adc \subset s$ then $b \in s$ so $adb \subset s$.
192	218
193		-Following the case of equations (\ref{eq:world.fold.stablemodel}) and (\ref{eq:world.fold.independent}) this sets
	219	+Following equations (\ref{eq:world.fold.stablemodel}) and (\ref{eq:world.fold.independent}) this sets
194	220	\begin{equation*}
195	221	\begin{cases}
196	222	\pr{W = adc \given C = ad} = 0,\cr
...	...	@@ -200,7 +226,7 @@ Following the case of equations (\ref{eq:world.fold.stablemodel}) and (\ref{eq:w
200	226	which concentrates all probability mass from the total choice $ad$ in the $adb$ branch, including the node $W = adbc$. This leads to the following cases:
201	227	$$
202	228	\begin{array}{l\|r}
203		- x & \pr{x \given C = ad}\
	229	+ x & \pr{W = x \given C = ad}\
204	230	\hline
205	231	ad & 1 \\
206	232	adb & 1\\
...	...	@@ -208,8 +234,22 @@ $$
208	234	adbc & 1
209	235	\end{array}
210	236	$$
211		-so, for $C = ad$, $\pr{b} = 2/3, \pr{c} = 1/3, \pr{b,c} = 1/3 \not= \pr{b}\pr{c}$.
	237	+so, for $C = ad$,
	238	+$$
	239	+\begin{aligned}
	240	+ \pr{W = b} &= \frac{2}{4} \cr
	241	+ \pr{W = c} &= \frac{1}{4} \cr
	242	+ \pr{W = bc} &= \frac{1}{4} \cr
	243	+ &\not= \pr{W = b}\pr{W = c}
	244	+\end{aligned}
	245	+$$
	246	+\textit{i.e.} the events $W = b$ and $W = c$ are dependent and that dependence results directly from the segment $0.2::d, b \leftarrow c \wedge d$ in the specification.
	247	+
212	248
	249	+%
	250	+
	251	+%
	252	+\hrule
213	253	\begin{quotation}\note{Todo}
214	254
215	255	Prove the four world cases (done), support the product (done) and sum (tbd) options, with the independence assumptions.
...	...