Is P(A,B|C) = P(A|C) P(B|C), if P(A,B) = P(A)P(B)?

In summary, the conversation discusses the relationship between two events, A and B, and how they are separable with a conditional added. The speaker is looking for a way to formally prove this relationship using Bayes' rules. However, it is shown that this relationship does not hold true in all cases, as demonstrated with the example of two fair coin flips. The conversation concludes with an additional example that further proves the lack of dependence between A and B when a condition is added.
  • #1
natski
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As stated in my subject line, I know that P(A|B) = P(A) and P(B|A) = P(B), i.e. A and B are separable as P(A,B) = P(A) P(B). I strongly suspect that this holds with a conditional added, but I can't find a way to formally prove it... can anyone prove this in a couple of lines via Bayes' rules? This is not a homework question, but part of my research and I can't find the answer anywhere.

Thanks to anyone who can help in advanced!
natski
 
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  • #2
No, this isn't true. Consider two fair coins flipped independently, let A be the event that the first coin comes up heads, B the event that the second coin comes up heads, and C be the event that at least one of the coins comes up heads. Then P(A) = P(B) = 1/2, P(A,B) = P(A)P(B) = 1/4, but P(A|C) = P(B|C) = 2/3 and P(A,B|C) = 1/3 [itex]\neq[/itex] P(A|C)P(B|C) = 4/9
 
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  • #3
Citan Uzuki said:
No, this isn't true. Consider two fair coins flipped independently, let A be the event that the first coin comes up heads, B the event that the second coin comes up heads, and C be the event that at least one of the coins comes up heads. Then P(A) = P(B) = 1/2, P(A,B) = P(A)P(B) = 1/4, but P(A|C) = P(B|C) = 2/3 and P(A,B|C) = 1/3 [itex]\neq[/itex] P(A|C)P(B|C) = 4/9
Even more obvious is C= exactly one coin is a head. Then the condition C forces a complete dependence between A and B.
 

FAQ: Is P(A,B|C) = P(A|C) P(B|C), if P(A,B) = P(A)P(B)?

1. What does P(A,B|C) represent in this equation?

P(A,B|C) represents the conditional probability of both event A and event B occurring, given that event C has already occurred.

2. Can you explain the meaning of P(A|C) and P(B|C) in this equation?

P(A|C) represents the conditional probability of event A occurring, given that event C has already occurred. Similarly, P(B|C) represents the conditional probability of event B occurring, given that event C has already occurred.

3. Why is P(A,B)=P(A)P(B) necessary for this equation to hold true?

P(A,B)=P(A)P(B) is necessary for this equation to hold true because it represents the product rule of probability, which states that the probability of two independent events both occurring is equal to the product of their individual probabilities.

4. Is this equation always true for any values of P(A), P(B), and P(C)?

Yes, this equation is always true for any values of P(A), P(B), and P(C) as long as P(A,B)=P(A)P(B). This is known as the multiplication rule of probability.

5. Can you provide an example to illustrate this equation?

Suppose we have a deck of cards with 52 cards, where 26 are red and 26 are black. Let A represent the event of drawing a red card, B represent the event of drawing a black card, and C represent the event of drawing a card. The probability of event A occurring is P(A)=26/52=1/2. Similarly, the probability of event B occurring is P(B)=26/52=1/2. The probability of event C occurring is P(C)=1. If we draw two cards from the deck without replacement, the probability of drawing a red card and a black card is P(A,B)=(26/52)(26/51)=1/4. Using the equation P(A,B|C) = P(A|C) P(B|C), we can calculate the conditional probabilities as P(A|C)=1/2 and P(B|C)=1/2. Plugging these values into the equation, we get 1/4 = (1/2)(1/2), which shows that the equation holds true for this example.

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