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An association rule has the form $A\Rightarrow C$, where $A$ is the antecedent and $C$ is the consequent. Suppose you have a database of one million transactions.

The questions are:

  1. If an association rule has 90% support, how many transactions contain all the items in $A$?

  2. For the same rule, how many transactions contain all the items in $C$?

  3. For the same rule, how many transactions contain all the items in both $A$ and $C$?

I am studying association rule learning and I couldn't answer the question. I think the question doesn't provide enough data to solve the questions. Because as far as I know: $$ \text{Support}(A \Rightarrow C) = P(A \cup C) = P(A) + P(C) - P(A \cap C) $$ Therefore to find $A$ I need to know $P(C)$ and $(A \cap C)$. And same for the others.

So how can I find the answer to my questions?

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    $\begingroup$ My apologies, @yns, I couldn't figure out how $\ge$ made any sense; I just wasn't thinking. Let me know if anything else is off. Note that there is information about the markup / formatting options CV supports here. Notably, CV supports $\LaTeX$ via mathjax; documentation can be found here. $\endgroup$ Jun 24, 2013 at 3:32
  • $\begingroup$ @gung thanks for the information about formatting and also for the editing. $\endgroup$
    – yns
    Jun 24, 2013 at 4:22

2 Answers 2

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Warning: I don't know much about association rule mining.

That being said from the definitions you've given you can't know the answers to questions 1, 2, or 3 from the given information. To see this simply note that you can have $$ P(A \cup C) = 0.9 $$ for all $P(A), P(C), P(A \cap C) \in [0,1]$ such that $$ \begin{align*} P(A) + P(C) - P(A \cap C) &= 0.9 &\text{with}\\ P(A), P(C) &\ge P(A \cap C), \end{align*} $$ for which you can find infinitely many solutions.

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So, are you taking the test tomorrow?

The answer is: 1. Don't know 2. Don't know 3. 90%

Given that in this case U is not the same as in set theory. Here it means cases where both X and Y occur.

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    $\begingroup$ Hi Juan, welcome to the site! As this question has the "self-study" tag, it would be helpful for the OP if you'd edit your answer and provided more details and information about the solutions. Otherwise, I don't see how your answer has any pedagogical value. $\endgroup$ Jun 24, 2013 at 9:26
  • $\begingroup$ Cool, didn't know that, I'll change it. Thanks $\endgroup$ Jun 24, 2013 at 11:08
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    $\begingroup$ If $X \cup Y$ means both $X$ and $Y$ occur, then what does $X \cap Y$ mean? $\endgroup$
    – alto
    Jun 24, 2013 at 11:49
  • $\begingroup$ what this means is, support for Y when X happened, so they both have to occur at the same time, right? $\endgroup$ Jun 25, 2013 at 8:43
  • $\begingroup$ I mean, from set theory, that sounds like union to you? Or like intersection? $\endgroup$ Jun 25, 2013 at 8:44

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