Ma Pun Summary

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<p>Summer 2009</p> <p>Stochastic Summary by Mapun</p> <p>Experiment, Sample Spaces, and Events Experiment: </p> <p> () E E5 5 Outcome: </p> <p> xi 1 E5</p> <p> HHTHT Event: </p> <p> C = Sample space: </p> <p> sample space discrete (nite) (innite) continuous continuum sample space S</p> <p>: 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) </p> <p> A = { 4} B = {} C = {}</p> <p> Sample Space Sample Space S1 S2 4. Sample Space n,n = 1,2,3, ... </p> <p> 1 = (T) 2 = (H,T) 3 = (H,H,T) ...</p> <p> Countably Innite Finite 5. (semi-closed) [0,1) Sample Space x 0 </p> <p>Stochastic Midterm Summary by Mapun</p> <p>1</p> <p>1 = (T), 2 = (H, T), 3 = (H, H, T), . . . .</p> <p>12Summer 2009</p> <p>Stochastic Summary by Mapun</p> <p>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:</p> <p>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 &lt; 1 x This is an example of an experiment with a continuous sampleSample 0 &lt; 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 &lt; 1/2}, AXIOM 2.3a:If A &lt; 1/4}, then Pr(A B) = Pr(A) + Pr(B). B = {|x 1/2| B = , C = {x = 1/2}. </p> <p> 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 </p> <p> 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 </p> <p> 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 </p> <p> 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 </p> <p> 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 </p> <p> 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</p> <p> 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 </p> <p> PROOF: See Exercise 2.4. axiom Pr(A B) = Pr(A) + Pr(B) Pr(A B). (2.3) 2.3 Assigning Probabilities </p> <p> 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</p> <p>13</p> <p>13</p> <p>15</p> <p>2.3 Assigning Probabilities</p> <p>Summer 2009</p> <p>Stochastic Summary by Mapun</p> <p>2.3 Assigning Probabilities events, 1 = H and 2 = T. Provided the coin is fair (again, not biased towards one side or the other), we have every reason to believe that these two events should be equally probable. These outcomes are mutually requirements of the we rule out that dene exclusive and collectively exhaustive (provided three axioms the pos- probability. To see this, we label th M its edge). According to our theory as sibility of the coin landing on atomic outcomes of an experiment E of 1 , 2 , , M . These atomic events a taken to be mutually exclusive and exhaustive. That is, i j = for all i = j, an probability, these events should be assigned probabilities of Pr(H) = 1 2 M = S. Then by Corollary 2.1 and Axiom 2.2, Pr(T) = 1/2. Pr(1 2 M ) = Pr(1 ) + Pr(2 ) + + Pr(M ) = Pr(S) = 1</p> <p>(2.</p> <p>16</p> <p>If each atomic outcome is to be equally EXAMPLE 2.7: Next consider the dice rolling experiment of Example probable, then we must assign each the six possible faces 1/M cubicle to </p> <p> 2.2. If the of Pr(i ) = of thefor there die be equality in the preceding equatio die is not loaded,probability Once the probabilities of these outcomes are assigned, the probabilities of som are reasonably taken to be equally likely to appear, in which case, forth 2.3 Assigning assignment is Pr(1) = Pr(2) = = Pr(6) = 1/6. From the probability Probabilitiesmore complicated events can be determined according to the rules set 17 Section 2.2. This approach assigning probabilities is referred to as the classic this assignment we can determine the probability of to axiom more complicated events, such as approach. : Chapter 2 Introduction to Probability Theory may occur. In Example 2.8, we may be tempted to dene the set of atomic outcomes Pr(evendifferent sums that= Pr(246) the two die faces. If we assign equally likely as the number is rolled) can occur on probability to each of these outcomes, then we (by Corollary 2.3) = Pr(2)+Pr(4)+Pr(6) arrive at the assignment of this procedure is the coin ipEXAMPLE 2.6: The simplest example ping experiment of Example 2.1. In this EXAMPLE 2.8: In 2) = 1/6+1/6+1/6ofdicePr(sum = assignment) case, there are only two atomic Pr(sum = Example 2.3, = pair (by= were rolled. In this exper- (2.5) = Pr(sum a 3) = probability 12) = 1/11. events, the H and 2 = T. Provided of iment, the most basic outcomes are 1 =36 different combinationsthe coin is fair (again, not biased = 1/2. towards one dice knows that the these the six atomic outcomes of the previous example.the other), we of likelihood of to believe that these Anyone with experience in games involvingside or Again, each have every reason atomic much lower the two eventsof of 1/36. equally 1/36 These outcomes are mutually outcomes assigned a probability rolling a 7.suppose here is that rolling a 2 is is than The problem we likelihood should be Next, probable. exclusive and most two dice = 5}. Then, the atomic events probability of the are not thecollectively exhaustive (provided want to nd thewe have assigned event A = {sum of basic outcomes and can be we rule out the possibility of the coin in Example 2.8. decomposed into simpler outcomes, as demonstratedlanding on its edge). According to our theory of Pr(A) = Pr((1, 4) problem(3, 2) (4, 1)) these events should be assigned probabilities of Pr(H) = (2, 3) encountered in the classical approach. Suppose probability, This is not the only Pr(T) 2)of measuring the height of an arbitrarily we consider= Pr(1, 4) + Pr(2, 3) + Pr(3, = + Pr(4, 1) (by Corollary 2.1) an experiment that consists 1/2. chosen student in your class and rounding that measurement to the nearest inch. = 1/36 + 1/36 + 1/36 + 1/36 (by probability assignment) The atomic outcomes of this experiment would consist of all the heights of the students in = 1/9. your class. However, it would not be reasonable to assign an equal probability to each height. Those EXAMPLE 2.7: Next consider the dice rolling experiment of Example heights corresponding to very tall or very short would be expected students to be less probable than those heights corresponding 2.2. If the die is not loaded, the six possible faces of the cubicle die So, probabilities to these events? to a medium height. how then do we assign The are reasonably taken to be equally likely to appear, in which case, associatedwith the classical problems approach to assigning probabilities can be the probability assignment is Pr(1) = Pr(2) EXAMPLE 2.9: In this example we will use the MATLAB com- = = Pr(6) = 1/6. From overcome using the relative frequency byapproach. relative this assignment of can determine the probability of more complicated mand rand to simulate the ipping we coins and the rolling of frequency relative frequency approach requires that the experiment we are concerned The events, such as dice. The command rand(m,n) creates a matrix of m rows and n with be repeatable, in which case, the probability of an event, A, can be assigned by </p> <p> matrix is a columns, where each element of the randomly selected repeating the experiment a large number of times and observing how many times number equally likely to fa...</p>