Assume we have a rule {X} -> {Y} which means customers buy item X, they also buy item Y in a supermarket. I know the LIFT is used to measure the independence of X and Y which is equal $P(XY)/P(X)P(Y)$, the higher the better for the rules

Lets consider 2 cases:

1) We have the following probabilities for a rule: $P(X)=0.52, P(Y)=0.77, P(XY)=0.5, P(X|Y)=0.65$ and $P(Y|X)=0.98$

The LIFT would be : $$\frac{0.5}{0.52(0.77)}=1.25$$

2) We have the following probabilities for another rule {A}->{B}: $P(A)=0.21, P(B)=0.4, P(AB)=0.18, P(A|B)=0.45$ and $P(B|A)=0.85$

The LIFT would be : $$\frac{0.18}{0.21(0.4)}=2.14$$

A and B should be less dependent to each other compare to X and Y since they have less conditional probabilities ($P(A|B)<P(X|Y)$ and $P(B|A)<P(Y|X)$), but they still have higher lift.

Can it be said that lift does not linearly correlated with dependence and thus is not a good measure?

  • $\begingroup$ In what sense do $A$ and $B$ 'have less conditional probabilities'? $\endgroup$ – Juho Kokkala Aug 19 '16 at 15:15
  • $\begingroup$ Because $P(A|B)<P(X|Y)$ and $P(B|A)<P(Y|X)$? $\endgroup$ – BigData Aug 19 '16 at 15:42

Random variables W and Z being independent implies:


If we look at the probabilities you have there, the X and Y should have lower lift as the conditional probabilities are close to the unconditional ones, while for A and B, the conditional probabilities are farther away. This is part of what lift measures. The high conditional probability does not imply they are correlated.

So rewriting LIFT: $ \frac{P(Y|X)}{P(Y)}$ a weighted measure of conditional probability.

  • $\begingroup$ So A and B are more dependent to each other? However their associations are weaker. So we cannot use lift to measure how strong the association between the items on the the LHS and RHS of a rule? $\endgroup$ – BigData Aug 19 '16 at 15:45
  • $\begingroup$ "their associations are weaker" - How are you figuring this? The conditional probabilities do not measure the ''strength'' of the association. They measure how the events overlap, but that can happen with independent events. $\endgroup$ – VCG Aug 19 '16 at 15:45
  • $\begingroup$ $P(A|B)=0.45 < P(X|Y)=0.65$ means more likely people buy X when they have already bought Y compare to people buy A when they have already bought B. $P(B|A)=0.85<P(Y|X)=0.98$ means similar. So X and Y should have stronger association? $\endgroup$ – BigData Aug 19 '16 at 15:53
  • $\begingroup$ Is it because P(X) and P(Y) are higher than P(A) and P(B), so that their conditional probability will be higher as well? Thus conditional probability should not be a measure of association? $\endgroup$ – BigData Aug 19 '16 at 16:03
  • $\begingroup$ So P(Y)=.77, so it is already likely. So conditioning on X does not affect the probability of Y as much as it does for A and B. $\endgroup$ – VCG Aug 19 '16 at 16:03

In you example, A and B had lower probabilities than X and Y. Therefore the denominator of lift is small and the lift may easily get high. In an extreme case, if C and D occurred on just one row both, but together, the lift would be maximal possible, N (the data size). So the pattern would appear very strong but it could well be just noise. However, if you measured the strength with leverage (P(CD)-P(C)P(D)), you would get quite a different result, leverage<1/N (when maximum leverage is 0.25).

In general, lift measures dependence between two variable-value combinations (events), while leverage measures dependence between the corresponding binary variables. It depends on your modelling purposes which one you want to measure. Leverage tends to be more robust in the sense that the patterns with highest leverage tend to have approximately the same leverage in future data, while lift values may vary a lot (a pattern with highest possible lift may actually express negative dependence in future data). On the other hand, leverage favours patterns where marginal probabilities P(C) and P(D) are near 0.5 and you may miss interesting but less frequent patterns.

If you use lift to measure the strength, I suggest to evaluate also statistical significance of the pattern. Fisher's exact test is reliable even for infrequent patterns.


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