Question 1163566
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 $A$**.

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### 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?