Given a set of rule such as:
$A \rightarrow B$; $B \rightarrow C$; that satisfy minimum confidence in the context of apriori algorithm meaning: $$\text{Conf}(A \rightarrow B) \geq \text{min. confidence}$$ and $$\text{Conf}(B \rightarrow C) \geq \text{min. confidence}$$

Is it possible to state that $\text{Conf}(A \rightarrow C) \geq \text{min. confidence}$?

I believe it's not possible to assert the later given that the relation between $\text{Support}(A \rightarrow B)$ and $\text{Support}(B \rightarrow C)$ doesn't give any information about the $\text{Support}(A \rightarrow C)$

Does this make sense?

Note: it's just a book exercise I'm trying to solve(not homework!) Here is the chapter link, it's in page 405 exercise 3.


$\newcommand{\Conf}{{\rm Conf}} \newcommand{\confidence}{{\rm confidence}} \newcommand{\Support}{{\rm Support}}$You cannot directly state that $\Conf(A \rightarrow C) \geq \min \confidence$. Because it depends on the support of $A$, $B$ and $C$. Let's say: \begin{align} \Support(A,B) &= 60\%, &\Support(A) &= 90\% \\ \Support(A,C) &= 20\%, &\Support(B) &= 70\% \\ \Support(B,C) &= 50\%, &\Support(C) &= 60\% \end{align} Let $\min \confidence = 50\%$, therefore: \begin{align} \Conf(A \rightarrow B) &= 66\% > \min \confidence \\ \Conf(B \rightarrow C) &= 71\% > \min \confidence \\ &{\rm But,} \\ \Conf(A \rightarrow C) &= 22\% < \min \confidence \\ \end{align}

  • $\begingroup$ The previous answer has a mistake: with the support of a and c, the support of their union can't hold, because it is too low, it has to be at least 0.5. It can be fixed by taking support(a) = 0.9, support(b) = 0.4, support(c) = 0.3, support(a,b) = 0.3, support(b,c) = 0.3, support(a,c) = 0.2. and min = 0.3. Check by yourself, and make a table to fit these conditions. $\endgroup$ – asaf Oct 31 '18 at 18:24

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