# What is the confidence for rule for $\emptyset \implies A$?

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.

• Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. – gung - Reinstate Monica Jun 7 '17 at 14:28
• 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 – Thang Do Jun 7 '17 at 14:34

## 1 Answer

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 $$Conf(A \Rightarrow \emptyset) = 1$$.

• why would $\Support(\emptyset) = 1$? that's one of the things I don't understand – Thang Do Jun 7 '17 at 14:46
• That is because $\emptyset$ is included in any itemset : it is thus present in all the transactions – Julie SOULAS Jun 7 '17 at 15:22