Riemann surface
matlab

Exercise 2. Having the network/graph shown in figure below, decide on the validity of following statements:

a) $P_1, P_5 \Perp P_6 \mid P_8$,
b) $P_2 \pi P_6 \mid \theta$,
2
c) $P_1 \Perp P_2 \mid P_8$,
d) $P_1 \Perp P_2, P_5 \mid P_4$,
e) Markov equivalence class that contains the shown graph contains exactly three directed graphs.

Solution:
a) FALSE, the path through $P_3, P_4$ and $P_7$ is opened, neither the nodes $P_1$ and $P_6$ nor $P_5$ and $P_6$ are d-separated,
b) FALSE, the path is blocked, namely the node $P_7$,
c) FALSE, unobserved linear $P_3$ is opened, converging $P_4$ is opened due to $P_8$, the path is opened,
d) FALSE, information flows through unobserved linear $P_3$,
e) TRUE, $P_1 \rightarrow P_3$ direction can be changed (second graph) then $P_3 \rightarrow P_5$ can also be changed (third graph).

Exercise 3. Let us have an arbitrary set of (conditional) independence relationships among $N$ variables that is associated with a joint probability distribution.
a) Can we always find a directed acyclic graph that perfectly maps this set (perfectly maps $=$ preserves all the (conditional) independence relationships, it neither removes nor adds any)?
b) Can we always find an undirected graph that perfectly maps this set?

Solution:
a) No, we cannot. An example is ${A \Perp C|B \cup D, B \Perp D| A \cup C}$ in a four-variable problem. This pair of conditional independence relationships leads to a cyclic graph or converging connection which introduces additional independence relationships. In practice if the perfect map does not exist, we rather search for a graph that encodes only valid (conditional) independence relationships and it is minimal in such sense that removal of any of its edges would introduce an invalid (conditional) independence relationship.
b) No, we cannot. An example is ${A \Perp C \mid \emptyset}$ in a three-variable problem (the complementary set of dependence relationships is ${A \pi B|\emptyset, A \pi B| C, B \pi C|\emptyset, B \pi C| A, A \pi C \mid B}$ ). It follows that $A$ and $B$ must be directly connected as there is no other way to meet both $A \pi B \mid \emptyset$ and $A \pi B \mid C$. The same holds for $B$ and $C$. Knowing $A \Perp C \mid \emptyset$, there can be no edge between $A$ and $C$. Consequently, it necessarily holds $A \Perp C \mid B$ which contradicts the given set of independence relationships (the graph encodes an independence relationship that does not hold).

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