Why Do We Need Sigma-Algebras of Events?

In principle we could define a probability measure on an algebra Ж of sub­sets of the sample space rather than on a a-algebra. We only need to change condition (1.10) as follows: For disjoint sets Aj є Ж such that Uj= Aj є Ж, P(UjO0=1 Aj) = Yj=1 P(Aj). By letting all but a finite number of these sets be equal to the empty set, this condition then reads as follows: For disjoint sets Aj є Ж, j = 1, 2,…,n < to, P(Un=1 Aj) = =1 P(Aj). However, if

we confined a probability measure to an algebra, all kinds of useful results would no longer apply. One of these results is the so-called strong law of large numbers (see Chapter 6).

As an example, consider the following game. Toss a fair coin infinitely many times and assume that after each tossing you will get one dollar if the outcome is heads and nothing if the outcome is tails. The sample space ^ in this case can be expressed in terms of the winnings, that is, each element ш of ^ takes the form of a string of infinitely many zeros and ones, for example, ш = (1, 1, 0,1, 0, 1…). Now consider the event: “After n tosses the winning is kdollars.” This event corresponds to the set Ak, n of elements ш of ^ for which the sum of the first n elements in the string involved is equal to k. For example, the set A12 consists of all ш of the type (1, 0,…) and (0, 1,…). As in the example in Section 1.2.3, it can be shown that

P (Ak, n) = (k) (1/2 )n for k = 0, 1, 2,…,n,

P(Ak, n) = 0 for k > n or k < 0.

Next, for q = 1,2,…, consider the events after n tosses the average winning k/n is contained in the interval [0.5 – 1/q, 0.5 + 1/q]. These events corre­spond to the sets Bq, n = uk”=2n/2-qnl/q)]+1 Ak, n, where [x] denotes the smallest integer > x. Then the set ПТО=И Bq, m corresponds to the following event:

From the nth tossing onwards the average winning will stay in the interval [0.5 – 1/q, 0.5 + 1 /q]; the set u£Lj n“=n Bq, m corresponds to the event there exists an n (possibly depending on a>) such that from the nth tossing onwards the average winning will stay in the interval [0.5 – 1/q, 0.5 + 1/q]. Finally, the set n°=1 U^L1 n“=n Bq, m corresponds to the event the average winning converges to 1/2 as n converges to infinity. Now the strong law of large numbers states that the latter event has probability 1: P [n^=1 U^L1 n“=n Bq, m] = 1. However, this probability is only defined if n^=1 U^=1 n“=„ Bq, m є &. To guarantee this, we need to require that & be a a-algebra.

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