Category Introduction to the Mathematical and Statistical Foundations of Econometrics

Bayes’ Rule

Let A and B be sets in &. Because the sets A and A form a partition of the sample space ^, we have B = (B П A) U (B П A); hence,

P(B) = P(B П A) + P(B П A) = P(B|A)P(A) + P(B |A)P(A).

Moreover,

P(AB)= P(A П B) P(B|A)P(A)

( 1 ) P (B) P (B) ■

Combining these two results now yields Bayes’ rule:

Подпись: P (A| B)P (B | A) P (A)

P(B | A)P(A) + P(B| A)P(A) ■

Thus, Bayes’ rule enables us to compute the conditional probability P(A |B) if P(A) and the conditional probabilities P(B | A) and P(B | A) are given.

More generally, if Aj, j = 1, 2,…,n (< to) is a partition of the sample space ^ (i. e., the Aj’s are disjoint sets in Ж such that ^ = Un=j Aj), then

Подпись: P(Ai |B)P (B | Ai) P (Ai)

TTj=1 P (B| Aj) P (Aj) •

Bayes’ rule plays an important role in a special branch of statistics (and econometrics) called Bayesian ...

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Quality Control in Practice

The problem in applying this result in quality control is that K is unknown. Therefore, in practice the following decision rule as to whether K < R or not is followed. Given a particular number r < n, to be determined at the end of this subsection, assume that the set of N bulbs meets the minimum quality requirement K < R if the number k of defective bulbs in the sample is less than or equal to r. Then the set A(r) = {0, 1,…, r} corresponds to the assumption that the set of N bulbs meets the minimum quality requirement K < R, hereafter indicated by “accept,” with probability

r

P(A(r)) = £ P({k}) = pr(n, K), (1.12)

k=0

say, whereas its complement A(r) = {r + 1n} corresponds to the assump­tion that this set of N bulbs does not meet this quality requirement, hereafter indicated by “reject...

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Sampling with Replacement

As a third example, consider the quality control example in the previous section except that now the light bulbs are sampled with replacement: After a bulb is tested, it is put back in the stock of N bulbs even if the bulb involved proves to be defective. The rationale for this behavior may be that the customers will at most accept a fraction R/N of defective bulbs and thus will not complain as long as the actual fraction K/N of defective bulbs does not exceed R /N. In other words, why not sell defective light bulbs if doing so is acceptable to the customers?

The sample space ^ and the a-algebra & are the same as in the case of sampling without replacement, but the probability measure P is different. Con­sider again a sample Sj of size n containing k defective light bulbs...

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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...

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