从给定恒等式推导事件B与C独立性的逆问题求解

Alright, let's break down this proof step by step. We'll start by restating the given information clearly, then use basic conditional probability definitions to manipulate the identity until we reach the independence condition.

已知条件 #

We're given three key pieces of information:

1. The conditional probability identity holds:

   $P(A|B) = P(A|B \cap C)P(C) + P(A|B \cap C^C)P(CC)$

2. $P(A|B \cap C) \neq P(A|B)$ (this critical constraint eliminates a trivial case)
3. $P(A) > 0$ (ensures we don't deal with zero-probability events that would break conditional probability definitions)

Our goal is to prove that B and C are independent, meaning $P(B \cap C) = P(B)P(C)$.

推导过程 #

First, recall the core definition of conditional probability: for any events X and Y where $P(Y) > 0$, $P(X|Y) = \frac{P(X \cap Y)}{P(Y)}$. We'll rewrite every conditional probability in the identity using this rule.

Step 1: Expand the identity using conditional probability definitions #

• Left-hand side (LHS): $P(A|B) = \frac{P(A \cap B)}{P(B)}$
• Right-hand side (RHS):
  $P(A|B \cap C)P(C) = \frac{P(A \cap B \cap C)}{P(B \cap C)} \cdot P(C)$
  $P(A|B \cap C^C)P(CC) = \frac{P(A \cap B \cap C^C)}{P(B \cap C^C)} \cdot (1 - P(C))$

Substitute these into the original identity:
$$
\frac{P(A \cap B)}{P(B)} = \frac{P(A \cap B \cap C) \cdot P(C)}{P(B \cap C)} + \frac{P(A \cap B \cap C^C) \cdot (1 - P(C))}{P(B \cap C^C)}
$$

Step 2: Simplify using set theory for probabilities #

Note two key set-based probability rules:

1. $P(A \cap B) = P(A \cap B \cap C) + P(A \cap B \cap C^C)$ (since $C$ and $C^C$ are mutually exclusive and cover the entire sample space)
2. $P(B \cap C^C) = P(B) - P(B \cap C)$ (the portion of B that doesn't overlap with C)

Let's use shorthand to simplify the algebra:

• $X = P(A \cap B)$
• $X_1 = P(A \cap B \cap C)$ (so $P(A \cap B \cap C^C) = X - X_1$)
• $Z = P(B)$
• $Y = P(B \cap C)$ (so $P(B \cap C^C) = Z - Y$)

Substituting these into the equation gives:
$$
\frac{X}{Z} = \frac{X_1 \cdot P(C)}{Y} + \frac{(X - X_1) \cdot (1 - P(C))}{Z - Y}
$$

Step 3: Rearrange and factor the equation #

Multiply both sides by $Z \cdot Y \cdot (Z - Y)$ to eliminate denominators (all terms are positive because $P(A) > 0$ implies $X > 0$, and $P(B) > 0$ implies $Z > 0$, $Y > 0$, $Z-Y > 0$):
$$
X \cdot Y \cdot (Z - Y) = X_1 \cdot P(C) \cdot Z \cdot (Z - Y) + (X - X_1) \cdot (1 - P(C)) \cdot Z \cdot Y
$$

Expand and collect like terms:

• Expand the left-hand side: $X Y Z - X Y^2$
• Expand and combine right-hand side terms:
  • Terms with $X_1$: $X_1 Z (Y - P(C) Z)$
  • Terms with $X$: $X Y (Z P(C) - Y)$

Putting it all together, the equation simplifies to:
$$
(Z P(C) - Y)(X Y - X_1 Z) = 0
$$

Step 4: Use the given constraint to eliminate the trivial case #

We know $P(A|B \cap C) \neq P(A|B)$. Translating this to our shorthand:
$$
\frac{X_1}{Y} \neq \frac{X}{Z} \implies X_1 Z \neq X Y \implies X Y - X_1 Z \neq 0
$$

Since the second factor is non-zero, the first factor must equal zero:
$$
Z P(C) - Y = 0 \implies P(B) P(C) = P(B \cap C)
$$

结论 #

This is exactly the definition of independence for events B and C. We've successfully proven the result using basic conditional probability rules and algebraic manipulation.

内容的提问来源于stack exchange，提问作者Samayita