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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**Extra resources for A Course in Probability Theory (3rd Edition)**

**Sample text**

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 .