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--- public:t-622-arti-13-1:lab_5_-_bayesian_networks [2013/03/15 10:06] – [Material] stephan
+++ public:t-622-arti-13-1:lab_5_-_bayesian_networks [2024/04/29 13:33] (current) – external edit 127.0.0.1
@@ Line 11: / Line 11: @@
 The following laws of probabilities might be helpful:
-<latex>
+  * <latex>$P(\neg A) = 1 - P(A)$</latex>
-	\begin{itemize}
+  * Bayes Law: <latex>$P(A \land B) = P(A|B) * P(B)$</latex>
-		\item $P(A \land B) = P(A|B) * P(B)$
+  * Marginalization/Summing out: <latex>$P(A) = \sum_b P(A \land b) = P(A \land B) + P(A \land \neg B)$</latex>
-		\item $P(A \land B) = P(B|A) * P(A)$
+  * Independence 1: <latex>$P(A|B,C) = P(A|C)$</latex>, if <latex>$A$</latex> and <latex>$B$</latex> are independent given <latex>$C$</latex>.
-		\item $P(\neg A) = 1 - P(A)$
+  * Independence 2: <latex>$P(A \land B|C) = P(A|C) * P(B|C)$</latex>, if <latex>$A$</latex> and <latex>$B$</latex> are independent given <latex>$C$</latex>. (This follows from Independence 1 and Bayes Law.)
-		\item $P(A) = \sum_b P(A \land b) = P(A \land A) + P(A \land \neg B)$
-		\item $P(A \land B|C) = P(A|C) * P(B|C)$, if $A$ and $B$ are independent given $C$.
+All those laws allow arbitrary additional conditions as long as all probabilities have the condition:
-	\end{itemize}
+  * Bayes Law: <latex>$P(A \land B|C) = P(A|B,C) * P(B|C)$</latex>
-</latex>
+  * Marginalization/Summing out: <latex>$P(A|C) = \sum_b P(A \land b|C) = P(A \land B|C) + P(A \land \neg B|C)$</latex>
 ==== Problem 1: Smelly Doors ====