PostBayesian Network / Assignment

Bayesian-As-1

2025-10-11

BN-Assignment-1

Question 1.

Given $X$ , $Y$ and $Z$ are 3 sets of random variables. Prove if

P (X, Y ∣ Z) = P (X ∣ Z) P (Y ∣ Z)

then it follows that

P (X ∣ Y, Z) = P (X ∣ Z)

Since $P (X, Y ∣ Z) = P (X ∣ Z) P (Y ∣ Z)$

P (X ∣ Z) = \frac{P (X, Y ∣ Z)}{P (Y ∣ Z)}

\frac{P (X, Y ∣ Z)}{P (Y ∣ Z)} = \frac{\frac{P (X, Y ， Z)}{P (Z)}}{\frac{P (Y, Z)}{P (Z)}} = \frac{P (X, Y, Z)}{P (Y, Z)} = P (X, Y | Z)

Question 2.

Consider the following Bayesian networks with the following nodes

R (Rain): it is raining or not, with binary values $r$ : it is raining and $r^{c}$ : it is not raining. $v a l (R) = {r, r^{c}}$
L (Leaves): there are juicy leaves or not, $v a l (L) = {l, l^{c}}$
Q (Quokkas Happiness): the quokkas are happy or unhappy, $v a l (Q) = {q, q^{c}}$
T (Tourists): there are lots of tourists or not many, $v a l (T) = {t, t^{c}}$
S (Selfie Quokka): people are taking lots of quokka selfies, or not, $v a l (S) = {s, s^{c}}$

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According to the given Bayesian network, factorize the joint distribution $P (R, L, Q, T, S)$ Solution:

P (R, L, Q, T, S) = P (R) P (L | R) P (Q | L, R) P (T | Q, L, R) P (S | T, Q, L, R)

= P (R) P (L | R) P (Q | L) P (T | Q, R) P (S | T, Q)

Specify all the conditional independencies encoded in this Bayesian network. Solution:
- $(Q ⊥ R | L) \in I (P)$
- $(T ⊥ L | R, Q) \in I (P)$
- $(S ⊥ R, L | T, Q) \in I (P)$
What is the probability that there are lots of tourists? Solution:

P (t) = \sum_{Q \in q, q^{c}, R \in r, r^{c}} P (t ∣ Q, R) P (Q, R)

\begin{aligned} P (t) = & P (T = t ∣ Q = q, R = r) P (Q = q, R = r) \\ + P (T = t ∣ Q = q, R = r^{c}) P (Q = q, R = r^{c}) \\ + P (T = t ∣ Q = q^{c}, R = r) P (Q = q^{c}, R = r) \\ + P (T = t ∣ Q = q^{c}, R = r^{c}) P (Q = q^{c}, R = r^{c}) \end{aligned}

P (Q, R) = \sum_{L \in l, l^{c}} P (Q, L, R) = \sum_{L \in l, l^{c}} P (Q ∣ L) P (L ∣ R) P (R)

P (t) = \sum_{Q \in q, q^{c}, R \in r, r^{c}} [P (t ∣ Q, R) \times (\sum_{L \in l, l^{c}} P (Q ∣ L) P (L ∣ R) P (R))]

\begin{aligned} P (t) & = P (r) [P (t ∣ q, r) \sum_{l \in {l, l^{c}}} P (q ∣ l) P (l ∣ r) + P (t ∣ q^{c}, r) \sum_{l \in {l, l^{c}}} P (q^{c} ∣ l) P (l ∣ r)] \\ + P (r^{c}) [P (t ∣ q, r^{c}) \sum_{l \in {l, l^{c}}} P (q ∣ l) P (l ∣ r^{c}) + P (t ∣ q^{c}, r^{c}) \sum_{l \in {l, l^{c}}} P (q^{c} ∣ l) P (l ∣ r^{c})] \end{aligned}

= 0.2 \times (0.504 + 0.098 + 0.016 + 0.012) + 0.8 \times (0.243 + 0.009 + 0.063 + 0.441) = 0.7308

What is the probability that the quokkas are happy, given there are lots of quokka selfies being taken and it is raining? Solution:

P (Q = q | S = s, R = r) = \frac{P (Q = q, S = s, R = r)}{P (S = s, R = r)}

P (Q = q, S = s, R = r) = \sum_{L \in {l, l^{c}}} \sum_{T \in {t, t^{c}}} P (r) P (L | r) P (q | L) P (T | q, r) P (s | q, T)

= \sum_{T \in {t, t^{c}}} P (r) P (l | r) P (q | l) P (T | q, r) P (s | q, T) + \sum_{T \in {t, t^{c}}} P (r) P (l^{c} | r) P (q | l^{c}) P (T | q, r) P (s | q, T)

\begin{aligned} = & P (r) P (l | r) P (q | l) P (t | q, r) P (s | q, t) \\ + P (r) P (l | r) P (q | l) P (t^{c} | q, r) P (s | q, t^{c}) \\ + P (r) P (l^{c} | r) P (q | l^{c}) P (t | q, r) P (s | q, t) \\ + P (r) P (l^{c} | r) P (q | l^{c}) P (t^{c} | q, r) P (s | q, t^{c}) \end{aligned}

\begin{aligned} = & 0.2 \times 0.8 \times 0.9 \times 0.7 \times 0.9 \\ + 0.2 \times 0.8 \times 0.9 \times 0.3 \times 0.6 \\ + 0.2 \times 0.2 \times 0.7 \times 0.7 \times 0.9 \\ + 0.2 \times 0.2 \times 0.7 \times 0.3 \times 0.6 \end{aligned}

= 0.09072 + 0.02592 + 0.01764 + 0.00504 = 0.13932

P (S = s, R = r) = P (Q = q, S = s, R = r) + P (Q = q^{c}, S = s, R = r)

= 0.13932 + P (Q = q^{c}, S = s, R = r)

= 0.13932 + \sum_{L \in {l, l^{c}}} \sum_{T \in {t, t^{c}}} P (r) P (L | r) P (q^{c} | L) P (T | q^{c}, r) P (s | q^{c}, T)

= 0.13932 + 0.00448 = 0.14380

P (Q = q | S = s, R = r) = \frac{P (Q = q, S = s, R = r)}{P (S = s, R = r)} = \frac{0.13932}{0.14380} \approx 0.9688