Chung K.L. a Course in Probability Theory (3ed., AP, ) - Free ebook download as PDF File .pdf), Text File .txt) or read book online for free. A first course in probability / Sheldon Ross. — 8th ed. p. cm. Includes A class in probability theory consists of 6 men and 4 women. An examination is given. This book contains about exercises consisting mostly of special cases and examples, second thoughts and alternative arguments, natural extensions, and.

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A Course on Elementary. Probability Theory. Statistics and Probability African Society. (SPAS) Books Series. Saint-Louis, Calgary, Alberta. A COURSE IN PROBABILITY THEORY THIRD EDITION KalLalChung Stanford University f\.CADEl\lIC PRESS --" po: il1r({"Jurt SCIence and Technology. A Course in Probability Theory, Third Edition. Home · A Course in Size Report. DOWNLOAD PDF Probability Theory: A Comprehensive Course. Read more.

X determines p, and therefore F, the converse is obviously false. A family of r. First of all, F is increasing by property viii of the measure. A brief description will now be given of the nine chapters, with some suggestions for reading and instruction. Regarding the function w as the "composIte mappmg": By "countable" we mean always "finite possibly empty or countably infinite".

They are among the most useful tools of measure theory, and serve to extend certain relations which are easily verified for a special class of sets or functions to a larger class.

Many versions of such theorems are known; see Exercise 10, 11, and 12 below. The best way to define the symmetric difference is through indicators of sets as follows: All properties of t; follow easily from this definition, some of which are rather tedious to verify otherwise. As examples: If Q has exactly 11 points, then J has 21Z members. The B.

If Q is countable, then J is generated by the singletons, and conversely. All countable subsets of Q and their complements form a B. The intersection of any collection of B. The union of a countable collection of B. Fa, ex EA. Prove that if each? Let ,'7" be a B. Consider the class of all sets with the asserted property and show that it is a B.

Let Q be a class of subsets of Q having the closure property iii ; let. Then sf contains the B. This is Dynkin's form of a monotone class theorem which is expedient for certain applications. The proof proceeds as in Theorem 2.

Each positive Borel measurable function is the limit of an increasing sequence of simple finitely-valued functions. An or a separable metric space containing all the open sets and closed sets. Prove that f: Show that the minimal such class is a field. A probability measure g'J. The abbreviation "p. These axioms imply the following consequences, where all sets are members of: The following proposition 1 Ell ,J.. It is a particular case of the monotone property x above, which may be deduced from it or proved in the same way as indicated below.

The axioms of finite additivity and of continuity together are equivalent to the axiom of countable additivity. Let Ell ,J We have the obvious identity: Hence 1 is true. For a later application Theorem 3. Then ii holds whenever Uk Ek E If. The triple Q, 2'ft, 9 is called a probability space triple ; Q alone is called the sample space, and co is then a sample point.

C Q, then the trace of the B. It is easy to see that this is a B. Example 1. Let Q be a countable set: Clearly axioms i , ii , and iii are satisfied. Conversely, let any such is?

The entire first volume of Feller's well-known book [13] with all its rich content is based on just such spaces. It is easily seen that? IJ and m may be defined as before. Example 3. The Euclidean B. A set in. Jel will be called a linear Borel set when there is no danger of ambiguity. However, the Borel-Lebesgue measure m on 0? Jeo, En t 0? For any countably infinite set Q, the collection of its finite subsets and their complements forms a field: Let Q be the space of natural numbers.

Hence ' is not a field. In the preceding example show that for each real number a in [0, I] there is an E in such that: Give an example of E that is not in r'-. Prove the nonexistence of a p. Hence criticize a sentence such as; "Choose an integer at random". Prove that the trace of a B.

Prove that the trace of Q, ;: T is a probability space, if? Now let t T be such that t Such a set is called thick in Q,. This procedure is called the adjunction of t T, fP. But it is not generated by all the singletons of ,gzl nor by any finite collection of subsets of. For the last two kinds of sets see, e.

T be the B. Determine the most general p. Instead of requiring that the E j 's be pairwise disjoint, we may make the broader assumption that each of them intersects only a finite number in the collection. Carry through the rest of the problem. The question of probability measures on:: There is in fact a one-toone correspondence between the set functions on the one hand, and the point functions on the other.

Both points of view are useful in probability theory. We establish first the easier half of this correspondence. Each p. Furthermore, let D be any dense subset of fRI, then the correspondence is already determined by that in 4 restricted to xED, or by any of the four relations in 5 when a and b are both restricted to D.

Let us write "Ix E fRI: We shall show that F is a d. First of all, F is increasing by property viii of the measure. The relations in 5 follow easily from the following complement to 4: Since Ix" t , x , we have by ix: To prove the last sentence in the theorem we show first that 4 restricted to XED implies 4 unrestricted.

For this purpose we note that M , x] , as well as F x , is right continuous as a function of x, as shown in 6. Hence the two members of the equation in 4 , being both right continuous functions of x and coinciding on a dense set, must coincide everywhere.

Now 2. Hence 4 follows. Incidentally, the correspondence 4 "justifies" our previous assumption that F be right continuous, but what if we have assumed it to be left continuous? Now we proceed to the second-half of the correspondence. Each d. F determines a p. This is the classical theory of Lebesgue-Stieltjes measure; see, e. However, we shall sketch the basic ideas as an important review. The d. Such a function is seen to be countably additive on its domain of definition.

What does this mean" Now we proceed to extend its domain of definition while preserving this additivity. If S is a countable union of such intervals which are disjoint: Next, we notice that any open interval a, b is in the extended domain why? Now it is well known that any open set U in!? But even with the class of open and closed sets we are still far from the B. Although it has been shown to be possible to proceed this way by transfinite induction, this is a rather difficult task.

There is a more efficient way to reach the goal via the notions of outer and inner measures as follows. For any subset S of. C closed, CCS u. It is clear that u. Equality does not in general hold, but when it does, we call S "measurable" with respect to F.

In this case the common value wiII be denoted by fL S. This new definition requires us at once to check that it agrees with the old one for all the sets for which fL has already been defined. The next task is to prove that: Details of these proofs are to be found in the references given above.

To finish: II with this property. It may be larger than:?: II, indeed it is see below , but this causes no harm, for the restriction of fL to gjl is a p. Let us mention that the introduction of both the outer and inner measures is useful for approximations. There is an alternative way of defining measurability through the use of the outer measure alone and based on Caratheodory's criterion. It should also be remarked that the construction described above for: In the general case of an "algebraic measure space", in which there is no topological structure, the role of the open sets is taken by an arbitrary field Sii, and a measure given on:: Ji may be extended to the minimal B.

In the case of: I, such an 0? Io of sets, each of which is the union of a finite number of intervals of the form a, b], , b], or a, 00 , where a E: Indeed the definition of the 2. For another case where such a construction is required see Sec. There is one more question: It is important to realize that this question is not answered by the preceding theorem. It is also worthwhile to remark that any p. Hence we should phrase the question more precisely by considering only p.

Let I-'- and v be two measures defined on the same B. We give the proof only in the case where I-'- and v are both finite, leaving the rest as an exercise. Q by hypothesis. J 27, which proves the theorem. Let I-'- and v be a-finite measures on.: In order to apply the theorem, we must verify that any of the hypotheses implies that JL and v agree on a field that generates:? Let us take intervals of the first kind and consider the field J30 defined above. If JL and v agree on such intervals, they must agree on by countable additivity.

This finishes the proof. Returning to Theorems 2. Given the p. JL on gel, there is a unique d. F satisfying 4. Conversely, given the d. We shall simply call JL the p. Instead of? We can either proceed analogously or reduce it to the case just discussed, as follows. The most interesting case is when F is the "uniform distribution" on W: The corresponding measure m on 93 is the usual Borel measure on [0, I], while its extension on L as described in Theorem 2.

It is well known that j is actually larger than M; indeed 1, m is the completion of.: The probability space n, ,:: Jf, g'l can be completed according to the next theorem. Let us call a set in ,J! A property that holds except on a null set is said to hold almost everywhere a.

Let e i. It is easy to verify, using Exercise 1 of Sec.

For each E E. P 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 t;.

F1 t;. F2 t;. Hence FJ t;. This implies? We leave it as an exercise to show that? Hence any subset of E also belongs to v'Y and so to? This proves that Q,: What is the advantage of completion? We need the measurability of the exact exceptional set to facilitate certain dispositions, such as defining or redefining a function on it; see Exercise 25 below.

The number of atoms of any o-finite measure is countable. This is the case if and only if F is discrete. This is the case if and only if F is singular. One half is proved by using Theorems 1. Translate Theorem 1. Translate the construction of a singular continuous d. What if the latter has positive measure? Describe a probability scheme to realize this. Show by a trivial example that Theorem 2.

Show that Theorem 2. Show that the: Let st, 27,. Prove that there exists a minimal B. Y2, where. Suppose that F has all the defining properties of a d. What modification is necessary in Theorem 2. For an arbitrary measure? Prove that for a measure f. Prove that if the p. Prove first that there exists E with "arbitrarily small" probability. A quick proof then follows from Zorn's lemma by considering a maximal collection of disjoint sets, the sum of whose probabilities does not exceed ct.

But an elementary proof without using any maximality principle is also possible. A point x is said to be in the support of a measure f.

The set of all such points is called the support of u, Prove that the support is a closed set whose complement is the maximal open set on which f. Show that Let f be measurable with respect to g;, and Z be contained in a null set. Show that the conclusion may be false otherwise. A real, extended-valued random variable is a function X whose domain is a set. A complex-valued random variable is a function on a set.

This definition in its generality is necessary for logical reasons in many applications, but for a discussion of basic properties we may suppose. This restricted meaning 3. Condition 1 then states that X-I carries members of 9'31 onto members of Such a function is said to be measurable twith respect to S'T. Thus, an r. The next proposition, a standard exercise on inverse mapping, is essential.

Theorem 3.

The preceding condition may be written as 3 'v'x: X-1 CC-oo, x] E Y. From Theorem 3. This B. Thus X is an r. This proves the "if" part of the theorem; the "only if' part is trivial. The next theorem relates the p. Each r. I, , J-L by means of the following correspondence: If the Bn's are disjoint sets in: Hence n n Finally X-I?!? Thus J-L is a p. It is the smallest Borel subfield of?

Thus 4 is a convenient way of representing the measure: This J-L is called the "probability distribution measure" or p. F according to Theorem 2. While the r.

X determines jL and therefore F, the converse is obviously false. A family of r. Let n, J be a discrete sample space see Example I of Sec. Every numerically valued function is an r. CW, 03, m. In this case an r. The two r.

The definition of a Borel measurable function is not affected, since no measure is involved; so any such function is an r. As in Example 2, there exists an r. We proceed to produce new r.

The quickest proof is as follows. Regarding the function f X of w as the "composite mapping": We must now discuss the notion of a random vector. This is just a vector each of whose components is an r.

It is sufficient to consider the case of two dimensions, since there is no essential difference in higher dimensions apart from complication in notation. A fortiori, it is also generated by product sets of the fonn where HI and H2 belong to. A function from 9 2 into 9'l'1 is called a Borel measurable function of two variables iff r: Jill C d Written out, this says that for each I-dimensional Borel set E, viz. Now let X and Y be two r. The random vector X, Y induces a probability v on qp as follows: This v is called the 2-dimensional, probability distribution or simply the p.

Let us also define, in imitation of X-1, the inverse mapping X, y -1 by the following formula: This mapping has properties analogous to those of X-1 given in Theorem 3. We can now easily generalize Theorem 3.

The last inclusion says the inverse mapping X, y -1 carries each 2- dimensional Borel set into a set in. This is proved as follows.

Now the collection of sets A in! It follows from what has just been shown that this B. J6 hence it must also contain: Hence each set in: Here are some important special cases of Theorems 3. Throughout the book we shall use the notation for numbers as well as functions: Generalization to a finite number of r. Passing to an infinite sequence, let us state the following theorem, although its analogue in real function theory should be well known to the reader.

To see, for example, that SUPjXj is an r. Here already we see the necessity of the general definition of an r. It is easy to see that X is discrete if and only if its d. Perhaps it is worthwhile to point out that a discrete r. Consider, for example, an r. The following terminology and notation will be used throughout the book for an arbitrary set Q, not necessarily the sample space. For each.

C Q, the function It. We have then More generally, let b, be arbitrary real numbers, then the function tp defined below: Each discrete T. X belongs to a certain partition. If j ranges over a finite index set, the partition is called finite and the r. Prove Theorem 3. For the "direct mapping" X, which of these properties of X-I holds? If two r. Given any p. Let e be uniformly distributed on [0,1]. For each d. Then G e has the d. Suppose X has the continuous d. F, then F X has the uniform distribution on [0,1].

What if F is not continuous? Is the range of an r. The sum, difference, product, or quotient denominator nonvanishing of the two discrete r. If Q is discrete countable , then every r. Conversely, every r. Use Exercise 23 of Sec.

If f is Borel measurable, and X and Y are identically distributed, then so are f X and f y. Is this B unique? Can there be a set A Generalize the assertion in Exercise 11 to a finite set of r. The reader is supposed to have some acquaintance with this, at least in the particular case 11,:: The r. For each positive discrete r. It is trivial that if X belongs to different partitions, the corresponding values given by 1 agree.

For each m, let Xm denote the r. Consequently there is monotone convergence: It should be shown that when X is discrete, this new definition agrees with the previous one. We say X has a finite or infinite expectation or expected value according as d: In the expected case we shall say that the expectation of X does not exist. The expectation, when it exists, is also denoted by in X w.

More generally, for each A in. This classical notation is really an anachronism, originated in the days when a point function was more popular than a set function. Here m is atomless, so the notation is adequate and there is no need to distinguish between the different kinds of intervals.

The general integral has the familiar properties of the Lebesgue integral on [0,1]. We list a few below for ready reference, some being easy consequences of others. As a general notation, the left member of 4 will be abbreviated to III X d? In the following, X, Y are r.

If the An's are disjoint, then iv Positivity. If Xl I X2 d9. On A, then a9 A IX" I If then L: If X; Now the partial sums of the series on the left may be rearranged Abel's method of partial summationl to yield, for N J IXI Jl IXI P lXI Let X Then the set function v defined on Y; as follows: P, is a probability measure on 7. Y on A with fA Y dPf! Deduce from Fatou's lemma: JA " Show this is false if the condition involving Y is omitted.

Given the r. X E see the end of Sec. Hence there exists a sequence of simple r. IXI for all m. Y ; call the resulting metric space M sr,: Prove that for each integrable r. X the mapping of M S? A2 are all continuous. If see Sec. Deduce Exercise 2 above as a special case. There is a basic relation between the abstract integral with respect to. We give the version in one dimension first. Let X on n,.

Then we have 11 in! X Ii dx provided that either side exists. Let B E. They are equal by the definition of J. This proves the theorem for j We shall need the generalization of the preceding theorem in several dimensions. No change is necessary except for notation, which we will give in two dimensions. Instead of the v in 5 of Sec.

L2 dx, dy. A Theorem 3. Let X, Y on st,: J2, J. L 2 and let j be a Borel measurable function of two variables. As a consequence of Theorem 3.

Lx and Fx denote, respectively, the p. L2 be as in Theorem 3. This result is a case of the linearity of 0 given but not proved here; the proof above reduces this property in the general case of Q,: Such a reduction is frequently useful when there are technical difficulties in the abstract treatment.

We end this section with a discussion of "moments". L and F are, respectively, the p. The moments about the mean are called central moments. That of order 2 is particularly important and is called the variance, var X ; its positive square root the standard deviation.

We note the inequality J"2 X It is a special case of the next inequality, of which we will sketch a proof. Jensen's inequality. Convexity means: We shall prove 22 for a simple r. Let then X take the value Yj with probability Aj, 1: Then we have by definition 1: AjCP Yj.

Finally, we prove a famous inequality that is almost trivial but very useful. Chebyshev inequality. We have by the mean value theorem: Another proof of O and Y For any d. Thus if X is a positive r.

Use Exercise 17 to express the mean of the maximum. They are said to be pairwise independent iff every two of them are independent. Note that 1 implies that the r. On the other hand, 1 is implied by the apparently weaker hypothesis: In terms of the p.

JL" induced by the random vector X , Jen, Y3" , and the p. Finally, we may introduce the n-dimensional distribution function corresponding to JL ", which is defined by the left side of 2 or in alternative notation: F Xl, Xj, By Theorem 3.

The proof of the next theorem is similar and is left as an exercise. Let We give two proofs in detail of this important result to illustrate the methods.

First proof Suppose first that the two r. Since X and Yare independent, we have for every j and k: Thus 5 is true in this case. Now let X and Y be arbitrary positive r. Then, according to the discussion at the beginning of Sec. Furthermore, for each m, X", and Y", are independent.

Note that for the independence of discrete r. Finally, it is clear that XmYIIl is increasing with m and o: For the general case, we use 2 and 3 of Sec. This again can be seen directly or as a consequence of Theorem 3. Hence we have, under our finiteness hypothesis: The first proof is completed.

Second proof. Consider the random vector X, Y and let the p. Then we have by Theorem 3. X t Y , finishing the proof! Observe that we are using here a very simple form of Fubini's theorem see below. Indeed, the second proof appears to be so much shorter only because we are relying on the theory of "product measure" f. L2 on CA2, jj2. This is another illustration of the method of reduction mentioned in connection with the proof of 17 in Sec.

A rigorous proof of this fact may be supplied by Theorem 3. Do independent random variables exist? Here we can take the cue from the intuitive background of probability theory which not only has given rise historically to this branch of mathematical discipline, but remains a source of inspiration, inculcating a way of thinking peculiar to the discipline.

It may be said that no one could have learned the subject properly without acquiring some feeling for the intuitive content of the concept of stochastic independence, and through it, certain degrees of dependence. Briefly then: If an unbiased coin is tossed and the two possible outcomes are recorded as 0 and 1, this is an r. Repeated tossing will produce a sequence of outcomes. If now a die is cast, the outcome may be similarly represented by an r.

Next we may draw a card from a pack or a ball from an urn, or take a measurement of a physical quantity sampled from a given population, or make an observation of some fortuitous natural phenomenon, the outcomes in the last two cases being r. Now it is very easy to conceive of undertaking these various trials under conditions such that their respective outcomes do not appreciably affect each other; indeed it would take more imagination to conceive the opposite!

In this circumstance, idealized, the trials are carried out "independently of one another" and the corresponding r. We have thus "constructed" sets of independent r. Can such a construction be made rigorous? We begin by an easy special case. The product B. Since Qn is also a countable set, we may define a p.

This p. It is trivial to verify that this is indeed a p. Furthermore, it has the following product property, extending its definition 7: Let 'll" be the n-dimensional cube immaterial whether it is closed or not: J and m" the usual Borel field and measure, is a probability space. The p. The reader may recall the term "independent variables" used in calculus, particularly for integration in several variables.

The two usages have some accidental rapport. The point of Example 2 is that there is a ready-made product measure there. Xi" it is possible to construct such a one based on given p. It remains to extend this definition to all of 28n, or, more logically speaking, to prove that there exists a p. The situation is somewhat more complicated than in Example 1, just as Example 3 in Sec. Indeed, the required construction is exactly that of the corresponding Lebesgue-Stieltjes measure in n dimensions.

This will be subsumed in the next theorem. Assuming that it has been accomplished, then sets of n independent r. Can we construct r. The simplest case will now be described and we shall return to it in the next chapter. This expansion is unique except when x is of the form ml'I"; the set of such x is countable and so of probability zero, hence whatever we decide to do with them will be immaterial for our purposes.

For the sake of definiteness, let us agree that only expansions with infinitely many digits "1" are used. Hence they are r. Then the set n x: It is clear that this set is just an interval of length 1 12n, hence of probability 1 12n. On the other hand for each i. This example seems extremely special, but easy extensions are at hand see Exercises 13, 14, and 15 below.

We are now ready to state and prove the fundamental existence theorem of product measures. Let a finite or infinite sequence of p. There exists a probability space Q,? P and a sequence of independent r. Without loss of generality we may suppose that the given sequence is infinite. Why" For each n , let Qn,: XII with fLn as its p. Indeed this is possible if we take Qn, 3;;,. Exercise 3 of Sec. Let the collection of subsets of Q, each of which is the union of a finite number of disjoint finiteproduct sets, be 3t.

It is easy to see that the collection giQ is closed with respect to complementation and pairwise intersection, hence it is a field. We shall take the? T in the theorem to be the B. This;j6 is called the product B. We define a set function 9 on. Markov Property.

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