1 Inclusion-Exclusion Definition Boolean polynomials. (The definition is recursive. Boolean polynomial is what we can build with iteration of the procedure below.) • Variable names like A1;A2; ::: are Boolean polynomials. • If B1 and B2 are Boolean polynomials, then so are (B1) \ (B2), (B1) [ (B2), B1. For example, A1, A1 \ A2,(A1 \ A2) [ A1, etc. are all Boolean polynomials. Given sets (or events in a probability space) S1;S2; :::, it makes sense to evaluate the Boolean polynomial by substituting Ai Si, and completing the operations we obtain a set (event) as value of the polynomial with this substitution. We say that a Boolean polynomial is n-variable, if it uses n distinct variable names. A variable name is a 1-variable Boolean polynomial. Definition Indicator function Given a probability space (Ω; A; p) (for events) or a universe U (for sets), the indicator function of an A 2 A or A ⊆ U is a function χA :Ω ! R (or a function χA : U ! R) such that ( 1; if a 2 A; χA(a) = 0; if a2 = A: (The definition for sets can be considered as special case of the definition for probability spaces, if we associate with a universe U the probability space (Ω; A; p) with Ω = U, A = 2U and for A 2 A and p(A) = jAj=jΩj.) Lemma 1 For any n-variable Boolean polynomial B(A1; :::; An) there exists an an n-variable real function fB : n R ! R, such that for any substitution events S1; :::; Sn 2 A from any probability space into the variables A1; :::; An, the following Ω ! R functions are equal: χB(S1;:::;Sn) = fB(χS1 ; :::; χSn ): Proof. As Boolean polynomials were defined recursively, we have to prove the statement by induction that mimicks the recursive definition. If A is a variable name, then we define fA : R ! R by fA(x) = x. Clearly for any S 2 A and any ! 2 !, χS(!) = (x ◦ χs)(!) as required. If B(A1; :::; An) was defined as B = (B1) \ (B2) from already defined Boolean polynomials, then we must have that the variable sets of B1 and B2 are subsets of the variable set A1; :::; An, and some real functions fB1 and fB2 satisfy the requirements of the Lemma for them. Rewrite fB1 and fB2 in a form with n variables such that their variables correspond in order to the variables A1; :::; An, possibly not using at all some variables. Define fB(x1; :::; xn) := fB1 (x1; :::; xn)fB2 (x1; :::; xn). It clearly suffices. Similarly, if B = B1, set fB = 1 − fB1 ; and if B = (B1) [ (B2), set fB(x1; :::; xn) := fB1 (x1; :::; xn) + fB2 (x1; :::; xn) − fB1 (x1; :::; xn)fB2 (x1; :::; xn), after rewriting fB1 and fB2 as in the case of intersection. Let us consider the following two statements. For a fixed (Ω; A; p) probability space, Boolean polynomials Bj(A1; :::; An) and constants cj 2 R (j = 1; 2; :::; m) and events S1; :::; Sn 2 A, Pm • (a) 8! 2 Ω j=1 cjχBj (S1;:::;Sn)(!) ≥ 0 Pm • (b) j=1 cjp(Bj(S1; :::; Sn)) ≥ 0: It is clear the (a) implies (b): compute the expectation of both sides of (a), use that expectation of a non-negative random variable is non-negative, and use the linearity of expectation on the left-hand side. Theorem 2 Let us be given an (Ω; A; p) probability space, Boolean polynomials Bj(A1; :::; An) and constants cj 2 R (j = 1; 2; :::; m). The following statements are equivalent: Pm • (i) 8! 2 Ω and substitution Ai Si (Si 2 A); we have j=1 cjχBj (S1;:::;Sn)(!) ≥ 0. Pm • (ii) for all substitution Ai Si (Si 2 A); we have j=1 cjp(Bj(S1; :::; Sn)) ≥ 0: 1 Pm • (iii) for all substitution Ai Si = ; or Ai Si = Ω; we have j=1 cjp(Bj(S1; :::; Sn)) ≥ 0: Proof. It is obvious that (i) implies (ii) and that (ii) implies (iii). We show (iii) implies (i) by proving the Pm contrapostive. Assume that for some events S1; :::; Sn and an ! 2 Ω, we have j=1 cjχBj (S1;:::;Sn)(!) < 0. Define the following new events: ( ∗ Ω; if ! 2 Si; Si = ;; if !2 = Si: Next, using the lemma for the first equation, ∗ ∗ χ (!) = f (χ (!); :::; χ (!)) = f (χ ∗ (!); :::; χ ∗ (!)) = f (χ ∗ (! ); :::; χ ∗ (! )); Bj (S1;:::;Sn) Bj S1 Sn Bj S1 Sn Bj S1 Sn and the last equation holds for every !∗ 2 Ω. Observe that ∗ ∗ p(B (S ; :::; S )) = E(χ ∗ ∗ ) = E(f (χ ∗ ; :::; χ ∗ )) = χ (!): j 1 n Bj (S1 ;:::;Sn) Bj S1 Sn Bj (S1;:::;Sn) Repeating this for j = 1; 2; :::; m, we obtain the expected conclusion: m m X ∗ ∗ X cjp(Bj(S1 ; :::; Sn)) = cjχBj (S1;:::;Sn)(!) < 0: j=1 j=1 As every equality a = b can be written as simultaneous inequalities a ≤ b and b ≤ a, we obtain: Corollary 3 The Theorem applies if ≥ 0 is changed to = 0 in all three clauses. One easily obtains the inclusion-exclusion formula as an application for the corollary. First, for events Si (i 2 f1; 2; :::; ng) and I ⊆ f1; 2; :::; ng, define SI := \i2I Si. The inclusion-exclusion formula says: n n [ X X i−1 p Si = (−1) p(SI ): (1.1) inclexcl i=1 i=1 jIj=i I⊆{1;2;:::;ng The proof, according the the theorem, only requires checking the identity for events where for every i, Si = ; or Si = Ω. If every Si = ;, both sides in (1.1) are equal to 0. So assume that we have ` ≥ events = Ω, the other n − ` ` ` ` `−1` events are = ;. Now the lefthand side in (1.1) is = 1, while the righthand side is 1 − 2 + 3 − ::: + (−1) ` . These are equal as the expansion of (1 − 1)` = 0 according to binomial theorem shows. Theorem 4 (Bonferroni Inequalities) 2t n 2t+1 X X i−1 [ X X i−1 (−1) p(SI ) ≤ p Si ≤ (−1) p(SI ): i=1 jIj=i i=1 i=1 jIj=i I⊆{1;2;:::;ng I⊆{1;2;:::;ng Proof. The result is trivial when all are Si = ;. Otherwise, assume that exactly ` ≥ 1 of them are = Ω. With this substitution, the theorem boils down to 2t 2t+1 X ` X ` (−1)i−1 ≤ 1 ≤ (−1)i−1 : i i i=1 i=1 Pk i` Pk i`−1 `−1 k`−1 This follows from the identity i=0(−1) i = 1 + i=1(−1) i−1 + i = (−1) k : 2.