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Home , Complement (set theory), Power set, Probability space

OF OCT 1, 2018

Sets and operations on sets. Appendix B Review Venn diagrams and operations: (i) A ∪ B, (ii) intersection A ∩ B (also written as AB), (iii) Ac, (iv) difference A\B.

Problem. Let A, B, C be of Ω. Represent the following in set notation. (1) Set of elements that are in each of the three subsets = A ∩ B ∩ C. (2) Set of elements that are in A but neither in B nor in C = A ∩ Bc ∩ Cc. (3) Set of elements that are in at least one of the sets A or B = A ∪ B. (4) Set of elements that re in both A and B, but not in C = A ∩ B ∩ Cc. (5) Set of elements that are in A, but not in B or C or both = A\(B ∪ C).

Axiomatic definition of a probability . Let Ω (pronounced ‘omega’) be a set called sample . Elements of Ω are called outcomes while subsets of Ω are called events. Let F be the (i.e., set of all subsets of Ω) of Ω. Then a probability, or , is a from F to [0, 1] satisfying the following conditions: • P (Ω) = 1. P (∅) = 0. • Countable additivity: If A1,A2,... is a sequence of pairwise disjoint (i.e. Ai ∩ Aj = ∅ for all i 6= j) events then ∞ ∞ X P (∪i=1Ai) = P (Ai). i=1 In other words, with every event E ⊆ Ω, a probability measure associates a number 0 ≤ P (E) ≤ 1 which is the probability that we observe the event E happening. The triplet (Ω, F,P ) is called a . Remark 1. More generally F is taken to be a σ-field (pronounced ‘sigma-field’), a concept from measure theory. But you don’t need to know that right now. Some consequences of the definition.

• Finite additivity: If A1,...,Ak is a finite pairwise disjoint collection of events, then k  P ∪i=1Ai = P (A1) + ... + P (Ak).

This follows from countable additivity by taking Ak+1 = Ak+2 = ... = ∅. • Complements: If A is an event, its complement Ac is the event Ac = Ω\A. Clearly A and Ac are disjoint. Thus, by finite additivity, P (A) + P (Ac) = P (A ∩ Ac) = P (Ω) = 1. Thus P (Ac) = 1 − P (A).

Equally likely outcomes. Suppose the Ω is finite and has N ele- ments {ω1, . . . , ωN }. Say that all outcomes are equally likely if P (ωi) := P ({ωi}) = 1/N for all i. In that case, it is very easy to find the probability measure. If E is 1 2 an event, then E is a of Ω. If #E is the number of elements in E then, by finite additivity, #E P (E) = . N Check that obviously P (Ω) = 1 and P (∅) = 0. Example 1. Toss a coin twice. Then Ω = {HH,TT,HT,TH}, and 1 P {HH} = P {TT } = P {TH} = P {HT } = . 4 For example, the event A = {HH,TT }, i.e., the event that both tosses are the same, has probability P (A) = 1/2, under the assumption of equally likely outcomes. Example 2. Roll a die twice. Here Ω is the set of 36 possible outcomes {(1, 1), (1, 2), (1, 3),... (6, 5), (6, 6)}. Let E be the event that the sum of the two rolls is 5. Then E consists of 4 sample points: (1, 4), (2, 3), (3, 2), (4, 1). Hence P (E) = 4/36 = 1/9, under the assumption of equally likely outcomes. Example 3. Sampling with replacement. Imagine an urn with N ≥ 2 balls la- beled by {1, 2,...,N}. Pick one ball without looking, note its label, and put it back. Repeat this a total of k times. The k labels picked form a k- (a1, a2, . . . , ak). This is one random . The sample space Ω is the set of all k- with entries in {1, 2,...,N}. Clearly Ω is finite and has N k elements. We say that the sampling is random, or, more correctly uniformly at random, if every outcome in Ω is equally likely. Hence the probability of every tuple is exactly 1/N k, under the assumption of equally likely outcomes. What is the probability that every label sampled is either 1 or 2? Here the event k E := {(ω1, . . . , ωk): ωi is 1 or 2 for each i}. Clearly there are #E = 2 . Hence P (E) = 2k/N k = (2/N)k, under the assumption of equally likely outcomes. Example 4. Ordered sampling without replacement. In sampling without replacement, we still have an urn with N ≥ 2 balls labeled by {1, 2,...,N}. We pick balls successively and note its label. However, we do not put it back. Hence Ω is the set of arrangements of k labels taken from {1, 2,...,N}. Thus #Ω = (n)k. A random ordered sample without replacement means each of these (n)k outcomes are equally likely. What is the probability that every label sampled is either 1 or 2? It is zero if k > 2, since after the first two labels are 1 or 2, the rest cannot be either. Example 5. (Unordered) sampling without replacement. Take an urn with N ≥ 2 balls labeled by {1, 2,...,N}. By a random sample without replace- n ment we mean that pick any subset of k balls equally likely. Thus #Ω = k .