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Summer 2009

Stochastic Summary by Mapun

Experiment, Sample Spaces, and Events Experiment:

() E E5 5 Outcome:

xi 1 E5

HHTHT Event:

C = Sample space:

sample space discrete (nite) (innite) continuous continuum sample space S

: 1. Sample Space outcome 1 = H 2 = T 2. Sample Space outcome 1 = 1 2 = 2 3 = 3 4 = 4 5 = 5 6 = 6 3. 1 = (1,1) 2 = (1,2) . . . 36 = (6,6)

A = { 4} B = {} C = {}

Sample Space Sample Space S1 S2 4. Sample Space n,n = 1,2,3, ...

1 = (T) 2 = (H,T) 3 = (H,H,T) ...

Countably Innite Finite 5. (semi-closed) [0,1) Sample Space x 0

Stochastic Midterm Summary by Mapun

1

1 = (T), 2 = (H, T), 3 = (H, H, T), . . . .

12Summer 2009

Stochastic Summary by Mapun

Chapter 2 Introduction to Probability of Note that in this case, n can extend to innity. This is another example Theory a combined sample space resulting from conducting independent but identical experiments. In this example, the sample space is countably innite, while depending sample spaces were nite. to various events, the previouson the situation. Before we do that, however, we start with three axioms that any method for assigning probabilities must satisfy:

EXAMPLE any event A, Pr(A) consider a random number generaAXIOM 2.1: For 2.5: As a last example, 0 (a negative probability does not make tor sense). that selects a number in an arbitrary manner from the semi-closed interval [0, 1). The sample space consists of all real numbers, x, for which x < 1 x This is an example of an experiment with a continuous sampleSample 0 < 1. is the sample space for a given experiment, Pr(S) = 1 (probabilities AXIOM 2.2: If SSample Space Continuous Space normalized so that the maximuma continuous space as well, such as can dene events on value is unity). are space. We A = {x < 1/2}, AXIOM 2.3a:If A < 1/4}, then Pr(A B) = Pr(A) + Pr(B). B = {|x 1/2| B = , C = {x = 1/2}.

As the word axiom implies, these statements are taken to be self-evident and require Other examples of experiments with really more of a self-imposed convenno proof. In fact,of Probability axioms arecontinuous sample spaces include 2.2 Axioms the rst two tion. the measurement of the for probabilities to be negative, or we or the have We could have allowed voltage of thermal noise in a resistor could measurement of (x, y, z) position of an oxygen molecule in the Axioms of Probability the probability to be something other than one. However, normalized the maximum atmosphere. Examples 2.1 to 2.4 illustrate discrete sample spaces. this would have For an innite number of mutually exclusive sets, A , i

AXIOM 2.3b:greatly confused the subject and we do not consider these pos- = i sibilities. FromA A axioms (plus = j, more to be presented shortly), the entire 1, 2, 3, . . . , these j = for all i one i 1

theory A can be developed. Before moving of probability Pr(A) 0 on There are also innite sets that are uncountable and that to that task, a corollary are not continuous, but to Axiom 2.3abeyond the scope of this book. So our purposes, we will consider is given. these sets areSample Space Pr(S) = 1 for 2

S Ai = Pr(Ai ). (2.2) Pr only the preceding two types of sample spaces. It is also possible to have a sample i=1 i=1 3A

COROLLARY 2.1: Consider M Pr(A)A1 Pr(B) . . . , AM that are mutually exclusive, A B = Pr(AB) = sets + , A2 , Ai Aj = for all i =Chapter 2 Introduction to Probability Theory j, It should Axiom 2.3a and 2.1 could be viewed as special

axiombe noted that Corollary cases of Axiom 2.3b. So, a more concise development could be obtained by starting 2.2 Axioms of Probability M with Axioms 2.1, 2.2, and 2.3b. This mayM more pleasing to some, but we believe be Ai = (2.1) probability of Athe approach given herePr a proof easier rigorous. Itfor possible to learning the B. We freely admit that this little is not Pr(Ai ). is to follow is the student i=1 prove Theoremmaterial for the rst to call on our sandi=1 2.1 without having time. analogy or even the use of AXIOM 2.3b: For proof will number of what we exclusive sets, Ai , i = Venn diagrams. The logic of thean inniteclosely follow mutuallyhave done here. probability of The preceding axioms do not tell us directly how to deal with the 3, . . , A Exercise 3B

1, 2,the.unioni Aj proof that i = j, that =setsin are not mutually The reader is ledthroughof for all 2.2. exclusive. This can be determined from two PROOF: This statement can be proved using mathematical induction. For those these axioms as follows. students who are results can also be obtained the idea basic axioms of Many other fundamental unfamiliar with this concept,from thebehind induction is to show Ai = Pr(Ai ). (2.2) Pr if the statement = m, then it will also hold for M probability.thatTHEOREM 2.1: is true for M i=1and B (notmust be developed laterm + 1. Once A few simple ones are presented here. More necessarily mutually = i=1 For any sets A exclusive), this is established, it is noted that by Axiom 2.3a, 2.1, it might help the in this chapter and in subsequent chapters. As with Theorem the statement applies for M = 2, and hencebe noted be true for 2.3a and diagram. for = = Pr(A) + Pr(B) It it must that Pr(A student to visualize these proofs by AxiomMB) =3. Since it is truePr(AM B). 3, it must also be (2.3) should A B drawing a Venn Corollary 2.1 could be viewed as special true for Axiom 2.3b. So, a more concise development could beCorollary by starting M = 4, and so on. In this way we can prove that obtained 2.1 is true cases of for any nite 1 Pr(A). visual proof of this important exercise for the Venn diagram are left pleasing to some, but we believe THEOREMwith Axioms M. The details of this proofbe more as an result using the reader (see 2.2: Pr(A) = 2.1, 2.2, and 2.3b. This may PROOF: We give a Exercise 2.1). Figure 2.1. Tois a the student in the type of reasoning needed to complete the shown in given here aid little easier to follow for the student learning the approach (by Axiom 2.2) proofs1 the rst = is A) A of this type, it P(B) P(A) Pr(A PROOF: material forB= Pr(S) time. helpful to think of a pile of sand lying in the sample space Unfortunately, axiomsThenot tell us directly how toA to showthenprobability of to the2.1. do probabilityis not sufcient would that Corollary 2.1 proof just outlined of the event deal with the be analogous The preceding shown in= Pr(A) + Pr(A) Figure (by Axiom 2.3a) is true for of two sets thatinnite number of sets. That has canbe accepted likewise for the the mass of thatof an are not mutually exclusive. This to be determined from union the case subset of the sand pile that is above the region A and on faith Assigningaxioms= 1follows. andthelisted here of the event B. For the union of the two events, if we simply added is Probabilities second part of Axiom 2.3. these Pr(A) as asPr(A). probability the the mass of the sand above A to the mass of the sand above B, we would double THEOREMTHEOREMB, then Pr(A) sets A and B (not necessarily mutually exclusive), 2.3: If A 2.1: For any Pr(B). count that region that is common to both sets. Hence, it is necessary to subtract the

PROOF: See Exercise 2.4. axiom Pr(A B) = Pr(A) + Pr(B) Pr(A B). (2.3) 2.3 Assigning Probabilities

PROOF: We give a visual proof of this important result using the Venn diagram S (fundamentalin Figure 2.1. atomic in the type reasoning needed to complete shown outcome) To aid the student outcome of requirements of ( 2 2this, ) three axioms that dene probability. see the thethis type, it is helpful to think of a pileTo sand we label sample space proofs of of lying in the M atomic outcomes ofA E as1 , 2 , atomic an experiment , M . These events are shown in Figure 2.1. The probability of the event A would then be analogous to taken to be mutually exclusive andThat is, i j= all = j, and M of exhaustive. is for i A and likewise for the mass of that subset the sand pile thatA above the region 1/M B 1 2 M = S. Then by Corollary 2.1 and Axiom 2.2, the section, probability was dened as a of the two the likelihood In the previousprobability of the event B. For the unionmeasure ofevents, if we simply added B axiom : of an event theevents of the sand above A to the mass of the sandprobabilities are or mass that satisfy the three Axioms 2.12.3. How above B, we would double Pr(particular events = Pr(common ) + Mathematically,it is = 1 (2.4) 1 2 M ) was not specied. sets. Hence, any assignment assigned to count that region that is 1 ) + Pr(2to both+ Pr(M ) = Pr(S)necessary to subtract the that satises the given axioms is acceptable. Practically speaking, we would like to If each atomic outcome is to be a way that the probability assignment actually assign probabilities to events in suchequally probable, then we must assign each a probability likelihood 1/M for there that equality in the preceding equation. Stochastic Midterm Summary by Mapun 2 represents theof Pr(i ) = of occurrence of to beevent. Two techniques are typically Figure 2.1 are diagram for proof of Theorem 2.1. Once this purpose and are described in the following paragraphs. used forthe probabilities of these outcomesVennassigned, the probabilities of some S more complicated eventsis possible to specify according outcomes of the exper-in In many experiments, it can be determined all of the to the rules set forth Section 2.2. This some fundamental outcomes, which we refer as as classical iment in terms of approach to assigning probabilities is referred to to the atomic

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2.3 Assigning Probabilities

Summer 2009

Stochastic Summary by Mapun

2.3