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+# Probabilistic Answer Set Programming
+
+## Non-stratified programs
+
+> Minimal example of **non-stratified program**.
+
+The following annotated LP, with clauses $c_1, c_2, c_3$ respectively, is non-stratified (because has a cycle with negated arcs) but no head is disjunctive:
+```prolog
+0.3::a.                 % c1
+s :- not w, not a.      % c2
+w :- not s.             % c3
+```
+
+This program has three stable models:
+$$
+\begin{aligned}
+m_1 &= \set{ a, w } \cr
+m_2 &= \set{ \neg a, s } \cr
+m_3 &= \set{ \neg a, w }
+\end{aligned}
+$$
+
+The probabilistic clause `0.3::a.` defines a **total choice**
+$$
+\Theta = \set{
+    \theta_1 = \set{ a }, 
+    \theta_2 = \set{ \neg a }
+}
+$$
+such that
+$$
+\begin{aligned}
+P(\Theta = \set{ a }) &= 0.3\cr
+P(\Theta = \set{ \neg a }) &= 0.7 \cr
+\end{aligned}
+$$
+
+> While it is natural to extend $P( m_1 ) = 0.3$ from $P(\theta_1) = 0.3$ there is no clear way to assign $P(m_2), P(m_3)$ since both models result from the total choice $\theta_2$.
+
+
+
+Under the **CWA**, $\sim\!\!q \models \neg q$, so $c_2, c_3$ induce probabilities:
+
+$$
+\begin{aligned}
+p_a &= P(a | \Theta) \cr
+p_s &= P(s | \Theta) &= (1 - p_w)(1 - p_a) \cr
+p_w &= P(w | \Theta) &= (1 - p_w)
+\end{aligned}
+$$
+from which results
+$$
+\begin{equation}
+p_s = p_s(1 - p_a).
+\end{equation}
+$$
+
+So, if $\Theta = \theta_1 = \set{ a }$ (one stable model):
+
+- We have $p_a = P(a | \Theta = \set{ a }) = 1$.
+- Equation (1) becomes $p_s = 0$.
+- From $p_w = 1 - p_s$ we get $P(w | \Theta) = 1$.
+
+and if $\Theta = \theta_2 = \set{ \neg a }$ (two stable models):
+
+- We have $p_a = P(a | \Theta = \set{ \neg a }) = 0$.
+- Equation (1) becomes $p_s = p_s$; Since we know nothing about $p_s$, we let $p_s =  \alpha \in \left[0, 1\right]$.
+- We still have the relation $p_w = 1 - p_s$ so $p_w = 1 - \alpha$.
+
+We can now define the **marginals** for $s, w$:
+$$
+\begin{aligned}
+P(s) &=\sum_\theta P(s|\theta)P(\theta)= 0.7\alpha            \cr
+P(w) &=\sum_\theta P(s|\theta)P(\theta)= 0.3 + 0.7(1 - \alpha) \cr 
+\alpha &\in\left[ 0, 1 \right]
+\end{aligned}
+$$
+
+> The parameter $\alpha$ not only **expresses insufficient information** to sharply define $p_s$ but also **relates** $p_s$ and $p_w$.
+
+## Disjunctive heads
+
+> Minimal example of **disjunctive heads** program.
+
+Consider this LP
+
+```prolog
+0.3::a.
+b ; c :- a.
+```
+
+with three stable models:
+$$
+\begin{aligned}
+m_1 &= \set{ \neg a } \cr
+m_2 &= \set{ a, b } \cr
+m_3 &= \set{ a, c }
+\end{aligned}
+$$
+
+Again, $P(m_1) = 0.3$ is quite natural but there are no clear assignments for $P(m_2), P(m_3)$.
+
+The total choices here are
+$$
+\Theta = \set{
+    \theta_1 = \set{ a }
+    \theta_2 = \set{ \neg a }
+}
+$$
+such that
+$$
+\begin{aligned}
+P(\Theta = \set{ a }) &= 0.3\cr
+P(\Theta = \set{ \neg a }) &= 0.7 \cr
+\end{aligned}
+$$
+and the LP induces
+$$
+P(b \vee c | \Theta) = P(a | \Theta).
+$$
+
+Since the disjunctive expands as
+$$
+\begin{equation}
+P(b \vee c | \Theta) = P(b | \Theta) + P( c | \Theta)  - P(b \wedge c | \Theta)
+\end{equation}
+$$
+and we know that $P(b \vee c | \Theta) = P(a | \Theta)$ we need two independent parameters, for example
+$$
+\begin{aligned}
+P(b | \Theta) &= \beta \cr
+P(c | \Theta) &= \gamma \cr
+\end{aligned}
+$$
+where
+$$
+\begin{aligned}
+    \alpha & \in \left[0, 0.3\right] \cr
+    \beta & \in \left[0, \alpha\right]
+\end{aligned}
+$$
+
+This example also calls for reconsidering the CWA since it entails that **we should assume that $b$ and $c$ are conditionally independent given $a$.**
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+# Probabilistic ASP
+
+## Weighted Approach
+
+1. **Total Choices.** $N(C = x) = \prod_{a \in x} w_a \prod_{\neg a \in x} (1 - w_a)$.
+2. **Stable Models.** $N(S = x | C = c) = \alpha_{x,c}$,
+    where the set of parameters $\alpha_{x,c}$ is such that:
+    $$
+    \begin{cases}
+        \alpha_{x,c} \geq 0, & \forall c, x\cr
+        \alpha_{x,c} = 0, & \forall x \not\supseteq c \cr
+        \sum_{x} \alpha_{x,c} = 1, & \forall c.
+    \end{cases}
+    $$
+3. **Worlds.** $N(W = x)$
+   1. If $x$ is a _total choice_: $
+    N(W = x) = \prod_{a \in x} w_a \prod_{\neg a \in x} (1 - w_a).
+    $$
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