In addition to the material in recent lectures, consider a brief tutorial on the topic of influence in Bayes nets.

Consider the following directed graphical model (i.e., Bayes Net) over five binary random variables and their associated conditional probability tables (CPTs). Note that no probabilities are actually given, since you do not need to do any arithmetic in the following sub-problems.

1. (4 points) Assess the truth of the following statements of conditional independence in the model above. We say “given $X$” when we mean that the value of $X$ is known. If the value of a particular variable is not given (i.e., known / on the left side of the “|” (given) bar), assume its value is not known. Answer “True” or “False” for each question:

- $Y \perp Z|X$
- $V \perp X|Y$
- $V \perp Z|Y$
- $V \perp W|X$

**Note**: Assume that the (unspecified) CPTs in the model contain values such that if influence is possible between a pair of nodes – given values of other variables in the model – using the heuristics we have discussed, then conditional independence will not hold. Under that assumption, the structure of the graph will be sufficient to tell you that a given conditional independence statement is false.

2. (3 points) Factor the joint distribution (over arbitrary values $v, w, x, y$ and $z$ of the random variables in the model) $P(v, w, x, y, z)$ according to the independence assumptions represented in the model.

3. (3 points) In terms of the given entries in the CPTs only, write an expression for the probability $P(V = 1,W = 0,X = 1,Y = 0,Z = 1)$.

4. (5 points each) In terms of the given entries in the CPTs only, write the expressions for the following events and simplify them algebraically. Here the meaning of algebraic simplification includes at least factoring out common terms from sums and simplifying sums. In the case of an unspecified value of a random variable (e.g., $P(V = v)$), there is no need to break down further in terms of complements.

- $P(V=1,W=0,Y=1,Z=1)$
- $P(W=0,X=0,Y=1)$
- $P(Y = 1)$

5. (5 points) In terms of the given entries in the CPTs only, write an expression for the conditional probability $P(W = 1|V = 1,X = 0,Y = 1,Z = 1)$, and simplify it algebraically. Here the meaning of algebraic simplification includes at least factoring out common terms from sums, simplifying sums, and canceling common terms that occur in both numerator and denominator. In the case of an unspecified value of a random variable (e.g., $P(V = v)$), there is no need to break down further in terms of complements.

Consider the following directed graphical model over nine binary random variables. Note that no probabilities are actually given, since you do not need to do any arithmetic in the following sub-problems.

Note: $Y_{-1}$ and $Y_0$ always have value 1, meaning “<START>”.

6. (16 points) Assess the truth of the following statements of conditional independence in the model above. We say “given $X$” when we mean that the value of $X$ is known. If the value of a particular variable is not given (i.e., known / on the left side of the “|” (given) bar), assume its value is not known. Answer “True” or “False” for each question:

- $C \perp Y_0|Y_2$
- $C \perp X_1|Y_1$
- $C \perp X_2|Y_1$
- $Y_0 \perp X_1|Y_1$
- $Y_0 \perp X_2|C$
- $Y_0 \perp X_2|Y_2$
- $Y_1 \perp Y_2|C$
- $Y_1 \perp Y_3|C$

**Note**: Assume that the (unspecified) CPTs in the model contain values such that if influence is possible between a pair of nodes – given values of other variables in the model – using the heuristics we have discussed, then conditional independence will not hold. Under that assumption, the structure of the graph will be sufficient to tell you that a given conditional independence statement is false.

7. (4 points) Factor the joint distribution (over arbitrary values $c, x_1, x_2, x_3, y_{-1}, y_0, y_1, y_2$ and $y_3$ of the random variables in the model) $P(c,x_1,x_2,x_3,y_{-1},y_0,y_1,y_2,y_3)$ according to the independence assumptions represented in the model.

8. (10 points each) In terms of the given entries in the CPTs only, write the following expressions. Remember that the random variables are binary-valued.

- $P(C=1,X_1 =0,X_2 =0,X_3 =0,Y_{-1}=1,Y_0=1,Y_1 =1,Y_2 =1,Y_3 =0)$
- $P(C=0,X_1 =1,X_2 =0,X_3 =1,Y_{-1}=1,Y_0=1,Y_1 =1,Y_2 =0,Y_3 =1)$

9. (10 points each) In terms of the given entries in the CPTs or complements of those entries, write the following expressions and simplify them algebraically. (See above for further guidance on necessary simplifications.) Regarding complements: for this problem there is no need to convert an expression for $P(A=0|\ldots)$ into $1-P(A=1|\ldots)$.

- $P(C=1,X_1 =0,X_2 =0,X_3 =0)$
- $P(C=0)$

10. (10 points) In terms of the given entries in the CPTs or complements of those entries, write an expression for the conditional probability $P(Y_1 = 0,Y_2 = 1,Y_3 = 1|C = 1,X_1 = 1,X_2 = 0,X_3 = 1)$, and simplify it algebraically. (See above for further guidance on necessary simplifications.) Regarding complements: for this problem there is no need to convert an expression for $P(A=0|\ldots)$ into $1-P(A=1|\ldots)$.