经济学论文证明中条件联合概率表达式等价性的验证及文献参考请求

Hey there! Let’s work through your probability equivalence question step by step, and I’ll point you to some solid textbook resources that cover this material.

First, let’s clarify the core of your question: you’re asking if this expression
$$\frac{p(A & B | C)}{p(A & B)}$$
is equivalent to
$$\frac{p(A|C)p(B|A,C)}{p(A)p(B|C)} $$

Let’s break down the math using basic probability rules (chain rule and definition of conditional probability):

1. Analyzing the left-hand side (LHS) #

Using the definition of conditional probability, we can rewrite the numerator of the LHS:
$$p(A & B | C) = \frac{p(A & B & C)}{p(C)}$$
The denominator of the LHS is just the joint probability, which we can expand via the chain rule:
$$p(A & B) = p(A)p(B|A)$$

Putting these together, the LHS becomes:
$$\frac{\frac{p(A & B & C)}{p(C)}}{p(A)p(B|A)} = \frac{p(A & B | C)}{p(A & B)}$$

2. Analyzing the right-hand side (RHS) #

Let’s expand each term in the RHS using the chain rule and conditional probability definition:

• The numerator: $p(A|C)p(B|A,C) = \frac{p(A & C)}{p(C)} \cdot \frac{p(A & B & C)}{p(A & C)} = \frac{p(A & B & C)}{p(C)}$
• The denominator: $p(A)p(B|C) = p(A) \cdot \frac{p(B & C)}{p(C)}$

Putting these together, the RHS simplifies to:
$$\frac{\frac{p(A & B & C)}{p(C)}}{p(A) \cdot \frac{p(B & C)}{p(C)}} = \frac{p(A & B & C)}{p(A) \cdot p(B & C)}$$

3. Comparing LHS and RHS #

For these two expressions to be equal, we’d need:
$$\frac{p(A & B & C)}{p(C) \cdot p(A & B)} = \frac{p(A & B & C)}{p(A) \cdot p(B & C)}$$
Canceling the common numerator (assuming $p(A & B & C) \neq 0$), this reduces to:
$$p(A) \cdot p(B & C) = p(C) \cdot p(A & B)$$
Which can be rearranged to:
$$p(A | B) = p(A | B & C)$$
This is the definition of conditional independence between $A$ and $C$ given $B$. So your proposed equivalence only holds in cases where this conditional independence is true—it’s not a universal identity.

4. The corrected universal equivalence #

If you intended the denominator of the RHS to be $p(A)p(B|A)$ (instead of $p(A)p(B|C)$), then the equivalence holds universally. Let’s confirm that:

• RHS becomes $\frac{p(A|C)p(B|A,C)}{p(A)p(B|A)} = \frac{\frac{p(A & B & C)}{p(C)}}{p(A & B)} = \frac{p(A & B | C)}{p(A & B)}$ which is exactly the LHS.

Textbook references #

For a thorough treatment of these probability rules, here are some standard resources used in economics and econometrics:

• Probability and Statistics for Economists by Bruce E. Hansen: This textbook has a clear, rigorous chapter on conditional probability, joint distributions, and the chain rule—perfect for thesis-level work.
• Econometric Analysis by William H. Greene: The introductory chapters on probability theory cover these identities in the context of econometric applications, which might align well with your thesis.
• A First Course in Probability by Sheldon Ross: A classic introductory probability text that breaks down these rules with intuitive examples, great for reinforcing the basics.

Hope this clears things up! Let me know if you need further clarification on any step.

备注：内容来源于stack exchange，提问作者JohnBlack