SOLUTION: Consider the following Bayesian graph: B←A→C The probabilities and conditional probabilities are given below: A P(A) ¦ A B P(B︳A) ¦ A B

Question 1163566: Consider the following Bayesian graph:
B←A→C
The probabilities and conditional probabilities are given below:
A P(A) ¦ A B P(B︳A) ¦ A B P(C︳A)
f 0.2 ¦ f f 0.8 ¦ f f 0.3
t 0.8 ¦ f t 0.2 ¦ f t 0.7
¦ t f 0.1 ¦ t f 0.2
¦ t t 0.9 ¦ t t 0.8
a) Prove that B and C are conditionally independent given A
b) Prove that B and C are not unconditionally dependent.

Answer by CPhill(2264) (Show Source): You can put this solution on YOUR website!
To address these proofs, we first note that the graph $B \leftarrow A \rightarrow C$ is a **diverging connection**. In Bayesian networks, the joint probability distribution for this graph is factorized as:
$$P(A, B, C) = P(A)P(B|A)P(C|A)$$
### a) Prove that $B$ and $C$ are conditionally independent given $A$
By definition, $B$ and $C$ are conditionally independent given $A$ if:
$$P(B, C | A) = P(B|A)P(C|A)$$
**Proof:**
Using the definition of conditional probability:
$$P(B, C | A) = \frac{P(A, B, C)}{P(A)}$$
Substitute the factorization of the joint distribution from the graph structure:
$$P(B, C | A) = \frac{P(A)P(B|A)P(C|A)}{P(A)}$$
The $P(A)$ terms cancel out, leaving:
$$P(B, C | A) = P(B|A)P(C|A)$$
This equality holds for all values of $A, B,$ and $C$. Thus, **$B$ and $C$ are conditionally independent given $A$**.
---
### b) Prove that $B$ and $C$ are not unconditionally independent
*Note: The prompt asks to prove they are "not unconditionally dependent," but the standard property of a diverging graph is that they are **not unconditionally independent** (meaning they are dependent). I will prove they are dependent by showing $P(B, C) \neq P(B)P(C)$.*
**1. Calculate $P(B=t)$:**
Using the law of total probability:
$$P(B=t) = P(B=t|A=t)P(A=t) + P(B=t|A=f)P(A=f)$$
$$P(B=t) = (0.9)(0.8) + (0.2)(0.2) = 0.72 + 0.04 = \mathbf{0.76}$$
**2. Calculate $P(C=t)$:**
$$P(C=t) = P(C=t|A=t)P(A=t) + P(C=t|A=f)P(A=f)$$
$$P(C=t) = (0.8)(0.8) + (0.7)(0.2) = 0.64 + 0.14 = \mathbf{0.78}$$
**3. Calculate $P(B=t, C=t)$:**
$$P(B=t, C=t) = \sum_{a \in \{t,f\}} P(B=t|A=a)P(C=t|A=a)P(A=a)$$
$$P(B=t, C=t) = (0.9 \times 0.8 \times 0.8) + (0.2 \times 0.7 \times 0.2)$$
$$P(B=t, C=t) = 0.576 + 0.028 = \mathbf{0.604}$$
**4. Compare $P(B, C)$ to $P(B)P(C)$:**
$$P(B=t)P(C=t) = 0.76 \times 0.78 = \mathbf{0.5928}$$
Since $0.604 \neq 0.5928$, the joint probability is not equal to the product of the marginal probabilities. Therefore, **$B$ and $C$ are unconditionally dependent**. Knowing the state of $B$ provides information about the likely state of $A$, which in turn provides information about the likely state of $C$.
How would the relationship between B and C change if the arrow between A and B was reversed?
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