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I'm looking at a question for association rule mining and this comes up:

  • What is the confidence for $\emptyset \implies A $?
  • What is the confidence for $ A \implies \emptyset $?

Given: $$\newcommand{\Conf}{{\rm Conf}}\newcommand{\Support}{{\rm Support}}\Conf(A \implies B) = \frac {\Support(A \cup B)} {\Support(A)}$$

Therefore, $$\Conf(\emptyset \implies A) = \frac {\Support(A \cup \emptyset)} {\Support(\emptyset)} = \frac {\Support(A)} {\Support(\emptyset)}$$

Similarly,

$$\Conf(A\implies \emptyset) = \frac {\Support(A \cup \emptyset)} {\Support(A)} = \frac {\Support(A)} {\Support(A)} = 100\%$$

However, it doesn't sound right to me.

I would really appreciate any pointers.

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  • $\begingroup$ Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. $\endgroup$ Jun 7, 2017 at 14:28
  • $\begingroup$ it is. but from a past exam. we have a few different answers but none makes much sense to all of us. I was hoping someone could give us a hint $\endgroup$
    – Thang Do
    Jun 7, 2017 at 14:34

1 Answer 1

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Why do you say it doesn't sound right to you?

Your current answer is perfectly correct : $$Conf(\emptyset \Rightarrow A) = \frac{Support(A)}{Support(\emptyset)} = Support(A)$$ and $$Conf(A \Rightarrow \emptyset) = 1$$.

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  • $\begingroup$ why would $\Support(\emptyset) = 1$? that's one of the things I don't understand $\endgroup$
    – Thang Do
    Jun 7, 2017 at 14:46
  • $\begingroup$ That is because $\emptyset$ is included in any itemset : it is thus present in all the transactions $\endgroup$ Jun 7, 2017 at 15:22

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