A Course in Probability Theory (3rd Edition) by Kai Lai Chung

By Kai Lai Chung

Because the e-book of the 1st variation of this vintage textbook over thirty years in the past, tens of hundreds of thousands of scholars have used A direction in likelihood Theory. New during this variation is an creation to degree conception that expands the marketplace, as this remedy is extra in step with present classes.

While there are a number of books on chance, Chung's e-book is taken into account a vintage, unique paintings in likelihood concept because of its elite point of sophistication.

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DEFINITION. v. X is called discrete (or countably valued ) iff there is a countable set B ² R1 such that P X 2 B D 1. f. is. v. need not have a range that is discrete in the sense of Euclidean topology, even apart from a set of probability zero. v. f. in Example 2 of Sec. 1. The following terminology and notation will be used throughout the book for an arbitrary set , not necessarily the sample space. DEFINITION. For each 1 ² 8ω 2 , the function 11 Ð defined as follows: : 11 ω D if ω 2 1, if ω 2 n1, 1, 0, is called the indicator (function) of 1.

The axioms of finite additivity and of continuity together are equivalent to the axiom of countable additivity. PROOF. Let En #. We have the obvious identity: 1 1 En D Ek nEkC1 [ kDn Ek . kD1 If En # ∅, the last term is the empty set. 1 P En D 0. Hence (1) is true. 1 D 0. Ek kDnC1 Now if finite additivity is assumed, we have 1 1 n Ek P DP kD1 Ek CP kD1 1 n D Ek kDnC1 P Ek C P kD1 Ek . kDnC1 This shows that the infinite series 1 kD1 P Ek converges as it is bounded by the first member above. Letting n !

Precisely: 9 F D fE ² :E1F 2 N for some F 2 F g. It is easy to verify, using Exercise 1 of Sec. F. Clearly it contains F . For each E 2 F , we put P E DP F , where F is any set that satisfies the condition indicated in (7). To show that this definition does not depend on the choice of such an F, suppose that E 1 F1 2 N , E 1 F2 2 N . Then by Exercise 2 of Sec. 1, E 1 F1 1 E 1 F2 D F1 1 F2 1 E 1 E D F1 1 F2 . Hence F1 1 F2 2 N and so P F1 1 F2 D 0. This implies P F1 D P F2 , as was to be shown. We leave it as an exercise to show that P is a measure on F .

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