Dependence Factor as a Rule Evaluation Measure

Abstract

Certainty factor and lift are known evaluation measures of association rules. Nevertheless, they do not guarantee accurate evaluation of the strength of dependence between rule’s constituents. In particular, even if there is a strongest possible positive or negative dependence between rule’s constituents X and Y, these measures may reach values quite close to the values indicating independence of X and Y. Recently, we have proposed a new measure called a dependence factor to overcome this drawback. Unlike in the case of the certainty factor, when defining the dependence factor, we took into account the fact that for a given rule $X \rightarrow Y$, the minimal conditional probability of the occurrence of Y given X may be greater than 0, while its maximal possible value may less than 1. In this paper, we first recall definitions and properties of all the three measures. Then, we examine the dependence factor from the point of view of an interestingness measure as well as we examine the relationship among the dependence factor for X and Y with those for $\bar{X}$ and Y, X and $\bar{Y}$, as well as $\bar{X}$ and $\bar{Y}$, respectively. As a result, we obtain a number of new properties of the dependence factor.

Appendix

Proof of Lemma 2

In the proof, we will use the following equations:

• $P(\bar{X}) = 1 - P(X)$,         $P(\bar{Y}) = 1 - P(Y)$,

• $P(\bar{X}Y)=P(Y)-P(XY),\quad P(X\bar{Y}) = P(X)-P(XY),$

• $P(\bar{X}\bar{Y}) = P(\bar{X}) - P(\bar{X}Y) = 1 - P(X)-P(Y)+P({ XY})$.

Ad (a)

Case $P(\bar{X}\bar{Y}) >P(\bar{X}) \times P(\bar{Y})$:

This case is equivalent to the case when $P({ XY}) >P(X)\times P(Y)$ (by Lemma 1a). Then:

$df(\bar{X} \rightarrow \bar{Y}) = \text { /* by Proposition 3a */}$

$$\begin{aligned}&=\frac{P(\bar{X}\bar{Y})-P(\bar{X})\times P(\bar{Y})}{{max}\,{\_}{P(\bar{X}\bar{Y}|_{P(\bar{X}),P(\bar{Y})})}-P(\bar{X})\times P(\bar{Y})} = \text { /* by Proposition 6a */}\\&= \frac{(1 - P(X)-P(Y)+P({XY}))-(1 - P(X))\times (1 - P(Y))}{(1-\max \{ P({X}),P({Y})\}) - (1-P({X})) \times (1-P({Y}))}\\&=\frac{P({X}{Y})-P({X})\times P({Y})}{\min \{ P({X}),P({Y})\} -P({X})\times P({Y})} = \text { /* by Theorem 3 */ }\\&=df(X \rightarrow Y). \end{aligned}$$

Case $P(\bar{X}\bar{Y}) = P(\bar{X})\times P(\bar{Y})$:

This case is equivalent to the case when $P(XY) = P(X) \times P(Y)$ (by Lemma 1b). Then:

${ df}(\bar{X} \rightarrow \bar{Y} ) = \text { /* by Proposition 3a */}$

         $\begin{aligned}&= 0 = \text { /* by Proposition 3a */}\\&= { df}(X \rightarrow Y). \end{aligned}$

Case $P(\bar{X}\bar{Y}) < P(\bar{X}) \times P(\bar{Y})$ and $P(\bar{X}) + P(\bar{Y}) \le 1$:

This case is equivalent to the case when $P({ XY}) < P(X) \times P(Y)$ (by Lemma 1c) and $P(X)+P(Y)\ge 1$ (by Proposition 5c). Then:

${ df}(\bar{X} \rightarrow \bar{Y}) = \text { /* by Proposition 3a */}$

$$\begin{aligned}&-\frac{P(\bar{X})\times P(\bar{Y})-P(\bar{X}\bar{Y})}{P(\bar{X})\times P(\bar{Y})-\textit{min}\,{\_}{P(\bar{X}\bar{Y}|_{P(\bar{X}),P(\bar{Y})})}} = \text { /* by Proposition 6b */}\\&=- \frac{(1 - P(X))\times (1 - P(Y))-1 (- P(X)-P(Y)+P({XY}))}{(1-P({X})) \times (1-P({Y}))-\text {max}\{0,1- P({X}),P({Y})\} }\\&=-\frac{P({X})\times P({Y})-P({X}{Y})}{(1-P({X})-P({Y})+P({X}) \times P({Y}))-(0)}\\&=-\frac{P({X})\times P({Y})-P({X}{Y})}{(P({X})\times P({Y})- (P({X}) + P({Y})-1)}\\&=-\frac{P({X})\times P({Y})-P({X}{Y})}{P({X})\times P({Y})- \max \{0,P({X}) + P({Y})-1\}} = \text { /* by Theorem 3 */ }\\&= df(X\rightarrow Y). \end{aligned}$$

Case $P(\bar{X}\bar{Y}) < P(\bar{X}) \times P(\bar{Y})\,\mathrm{and}\,P(\bar{X}) + P(\bar{Y}) > 1$:

This case is equivalent to the case when $P(XY) < P(X) \times P(Y)$ (by Lemma 1c) and $P(X)+P(Y) <$ 1 (by Proposition 5c). Then:

$df(\bar{X}\rightarrow \bar{Y}) = \text { /* by Proposition 3a */}$

$$\begin{aligned}&= -\frac{P(\bar{X})\times P(\bar{Y})-P(\bar{X}\bar{Y})}{P(\bar{X})\times P(\bar{Y})-\textit{min}\,{\_}{P(\bar{X}\bar{Y}|_{P(\bar{X}),P(\bar{Y})})}} \text { /* by Proposition 6b */} \\&=- \frac{(1 - P(X))\times (1 - P(Y))- (1- P(X)-P(Y)+P({XY}))}{(1-P({X})) \times (1-P({Y}))-\text {max}\{0,1- P({X}),P({Y})\} }\\&=-\frac{P({X})\times P({Y})-P({X}{Y})}{(1-P({X})-P({Y})+P({X}) \times P({Y}))-(1-P(X)-P(Y))}\\&=-\frac{P({X})\times P({Y})-P({X}{Y})}{(P({X})\times P({Y})-0}\\&=-\frac{P({X})\times P({Y})-P({X}{Y})}{(P({X})\times P({Y})- \max \{0,P({X}) + P({Y})-1\}} = \text { /* by Theorem 3 */} \\&= df(X\rightarrow Y). \end{aligned}$$

Ad (b)

The proof is analogous to the proof of Lemma 1a.

Ad (c)

Case $P(X\bar{Y})>P(X) \times P(\bar{Y})$ and $P(X)\le P(\bar{Y})$:

This case is equivalent to the case when $P(XY)< P(X)\times P(Y)$ (by Lemma 1c) and $P(X) \le 1 - P(Y)$. Then:

$df(X \rightarrow \bar{Y}) = \text { /* by Proposition 3a */}$

$$\begin{aligned}&=\frac{P(X\bar{Y})-P(X)\times P(\bar{Y})}{max \,{\_}P(X\bar{Y}|_{P(X), P(\bar{Y})})-P(X)\times P(\bar{Y})} = \text { /* by Proposition 6c */} \\&= \frac{(P(X)-P(XY))-P(X)\times (1-P(Y))}{\min \{P(X), 1-P(Y)\}-P(X)\times (1-P(Y))}\\&= \frac{P(X)\times P(Y)-P(XY)}{P(X)\times P(Y)-0}\\&= \frac{P(X)\times P(Y)-P(XY)}{P(X)\times P(Y)-\max \{ 0, P(X)+P(Y)-1\}} = \text { /* by Theorem 3 */}\\&= - df(X \rightarrow Y). \end{aligned}$$

Case $P(X\bar{Y}) > P(X) \times P(\bar{Y})\,\mathrm{and}\,P(X) > P(\bar{Y})$.

This case is equivalent to the case when $P(XY)< P(X) \times P(Y)$ (by Lemma 1c) and $P(X) > 1 - P(Y)$. Then:

$df(X \rightarrow \bar{Y}) = \text { /* by Proposition 3a */}$

$$\begin{aligned}&=\frac{P(X\bar{Y})-P(X)\times P(\bar{Y})}{max {\_}P(X\bar{Y}|_{P(X), P(\bar{Y})})-P(X)\times P(\bar{Y})} = \text { /* by Proposition 6c */}\\&= \frac{(P(X)-P(XY))-P(X)\times (1-P(Y))}{\min \{P(X), 1-P(Y)\}-P(X)\times (1-P(Y))}\\&= \frac{P(X)\times P(Y)-P(XY)}{(1- P(Y))-P(X)\times (1-P(Y))}\\&= \frac{P(X)\times P(Y)-P(XY)}{P(X)\times P(Y)-\max \{ 0, P(X)+P(Y)-1\}} = \text { /* by Theorem 3 */}\\&= - df(X \rightarrow Y). \end{aligned}$$

Case $P(X\bar{Y})=P(X) \times P(\bar{Y})$:

This case is equivalent to the case when $P(XY)=P(X)\times P(Y)$ (by Lemma 1b). Then:

$df(\bar{X}\rightarrow \bar{Y}) = \text { /* by Proposition 3a */}$

         $\begin{aligned}&= 0 = \text { /* by Proposition 3a */} \\&= -df(X\rightarrow Y). \end{aligned}$

Case $P(X\bar{Y}) < P(X)\times P(\bar{Y})$ and $P(X)+ P(\bar{Y}) \le 1$.

This case is equivalent to the case when $P(XY) > P(X) \times P(Y)$ (by Lemma 1a) and $P(X)\le P(Y)$. Then:

$df(X \rightarrow \bar{Y}) = \text { /* by Proposition 3a */}$

$$\begin{aligned}&= -\frac{P(X)\times P(\bar{Y})-P(X\bar{Y})}{P(X)\times P(\bar{Y})- min \, {\_}P(X\bar{Y}|_{P(X),P(\bar{Y})})}= \text { /* by Proposition 6d */} \\&=- \frac{P(X)\times (1-P(Y))-(P(X)-P(XY))}{P(X)\times (1-P(Y))-\max \{0, P(X)-P(Y)\}}\\&= - \frac{P(XY)-P(X)\times P(Y)}{(P(X)-P(X)\times P(Y))-(0)}\\&= - \frac{P(XY)-P(X)\times P(Y)}{\min \{P(X),P(Y)\}-P(X)\times P(Y)}= \text { /* by Theorem 3 */}\\&= - df(X\rightarrow Y). \end{aligned}$$

Case $P(X\bar{Y}) < P(X) \times P(\bar{Y})$ and $P(X)+P(\bar{Y}) > 1$.

This case is equivalent to the case when $P(XY) > P(X) \times P(Y)$ (by Lemma 1a) and $P(X)>P(Y)$. Then:

$df(X \rightarrow \bar{Y}) = \text { /* by Proposition 3a */}$

$$\begin{aligned}&= -\frac{P(X)\times P(\bar{Y})-P(X\bar{Y})}{P(X)\times P(\bar{Y})- min\, {\_}P(X\bar{Y}|_{P(X),P(\bar{Y})})}= \text { /* by Proposition 6d */}\\&=- \frac{P(X)\times (1-P(Y))-(P(X)-P(XY))}{P(X)\times (1-P(Y))-\max \{0, P(X)-P(Y)\}}\\&= - \frac{P(XY)-P(X)\times P(Y)}{P(X)\times (1-P(Y))-(P(X)-P(Y))}\\&= - \frac{P(XY)-P(X)\times P(Y)}{P(Y) -P(X)\times P(Y)}\\&= - \frac{P(XY)-P(X)\times P(Y)}{\min \{P(X),P(Y)\}-P(X)\times P(Y)}= \text { /* by Theorem 3 */}\\&= - df(X\rightarrow Y). \square \end{aligned}$$