Numerical Problem

Question:

Consider an itemset I = X ∪Y = X′ ∪Y′ and let X′ ⊆ X. Show that confidence of the association rule

X ⇒ Y is greater than or equal to that of X′ ⇒ Y ′.

Answer:

Given:    I=X∪Y=X′∪Y′    and  X′⊆X

Proof:     Since,  I=X∪Y=X′∪Y′,  we know: Support(X∪Y) = Support(X′∪Y′)

This is because $X\mathrm{<span\; class="mspace"></span><span\; class="mbin">\cup Y\; and </span>}$$X' \cup Y'$ represent the same set of items, so their support must be equal.

Given $X' \subseteq X$, it follows that:  Support(X′) ≥ Support(X).

This is because $X'$ is a subset of $X$, meaning every transaction that contains $X$ will also contain $X'$.

The number of transactions containing X' is at least equal to those containing X.

Comparing Confidence: ​

1. Confidence of $𝑋\Rightarrow 𝑌$: Confidence(X⇒Y) = Support(X∪Y) / Support(X)
2. Confidence of ${𝑋}^{\mathrm{\prime }}\Rightarrow {𝑌}^{\mathrm{\prime }}$: Confidence(X′⇒Y′) = Support(X′∪Y′) / Support(X′)

Since, $\text{Support}(X \cup Y) = \text{Support}(X' \cup Y')$ and $\text{Support}(X) \leq \text{Support}(X')$ we have:

Confidence(X⇒Y)= Support(X∪Y)/Support(X) ​ ≥  Support(X∪Y)​/Support(X′) = Confidence(X′⇒Y′)

The confidence of the association rule $X\Rightarrow \mathrm{Y\; is\; greater\; than\; or\; equal\; to\; that\; of }$$X' \Rightarrow Y'$ when $X' \subseteq X$.