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23. Metric data structures Exhibit 23.4: The search for a record with key values (1903, v) starts with the scales and proceeds via the directory to the correct data bucket on disk. The dynamics of splitting and merging The dynamic behavior of the grid file is best explained by tracing an example: we show the effect of ... | algorithms and data structures_Page_261_Chunk4801 |
This book is licensed under a Creative Commons Attribution 3.0 License Exhibit 23.6: An overflowing bucket triggers a refinement of the space partition. Exhibit 23.7: Bucket A has been split into A and C, but the contents of B remain unchanged. Assuming that records keep arriving in the lower-left corner of the space, ... | algorithms and data structures_Page_262_Chunk4802 |
23. Metric data structures Simple geometric objects and their parameter spaces Consider a class of simple spatial objects, such as aligned rectangles in the plane (i.e. with sides parallel to the axes). Within its class, each object is defined by a small number of parameters. For example, an aligned rectangle is determ... | algorithms and data structures_Page_263_Chunk4803 |
This book is licensed under a Creative Commons Attribution 3.0 License Exhibit 23.9: A set of intervals represented in two different parameter spaces. Region queries of arbitrary shape Intersection is a basic component of other proximity queries, and thus deserves special attention. CAD design rules, for example, often... | algorithms and data structures_Page_264_Chunk4804 |
23. Metric data structures Exhibit 23.10: A set of aligned rectangles represented as a set of points in a four-dimensional parameter space. A point query is transformed into a cone-shaped region query. 3. Consider the class of circles in the plane. We represent a circle as a point in three-dimensional space by the coor... | algorithms and data structures_Page_265_Chunk4805 |
This book is licensed under a Creative Commons Attribution 3.0 License 1. Consider the class of intervals on a straight line. An interval i = (cx, dx) intersects a query interval Q = (cq, dq) if and only if its representing point lies in the shaded region shown in Exhibit 23.12; this region is given by the inequalities... | algorithms and data structures_Page_266_Chunk4806 |
23. Metric data structures Exhibit 23.13: Search region as a union of cones. Evaluating region queries with a grid file We have seen that proximity queries on spatial objects lead to search regions significantly more complex than orthogonal range queries. The grid file allows the evaluation of irregularly shaped search... | algorithms and data structures_Page_267_Chunk4807 |
This book is licensed under a Creative Commons Attribution 3.0 License Query preprocessing may involve a significant amount of computation based on small amounts of auxiliary data— the scales and the query—that are kept in central memory. The final access of data from disk is highly selective— data retrieved has a high... | algorithms and data structures_Page_268_Chunk4808 |
23. Metric data structures Exhibit 23.15: The location of congruent objects greatly affects the complexity of a quadtree representation. 2. Consider a grid file that stores points lying in a two-dimensional domain: the Cartesian product X1 × X2, where X1 = 0 .. 15 and X2 = 0 .. 15 are subranges of the integers. Buckets... | algorithms and data structures_Page_269_Chunk4809 |
This book is licensed under a Creative Commons Attribution 3.0 License Exhibit 23.16: This example shows bucket regions that cannot be merged pairwise. 3. Consider the class of circles in the plane represented as points in three-dimensional parameter space as proposed in chapter 23 in the section “Region queries of arb... | algorithms and data structures_Page_270_Chunk4810 |
This book is licensed under a Creative Commons Attribution 3.0 License Part VI: Interaction between algorithms and data structures: case studies in geometric computation Organizing and processing Euclidean space In Part III we presented a varied sample of algorithms that use simple, mostly static, data structures. Part... | algorithms and data structures_Page_271_Chunk4811 |
This book is licensed under a Creative Commons Attribution 3.0 License 24. Sample problems and algorithms Learning objectives: • The nature of geometric computation: three problems and algorithms chosen to illustrate the variety of issues encountered: • Convex hull yields to simple and efficient algorithms, straightfor... | algorithms and data structures_Page_272_Chunk4812 |
24. Sample problems and algorithms intersecting line segments, are special cases that often require special code. It is not always easy to envision all the kinds of degeneracies that may occur in a given problem. A configuration may be degenerate for a specific algorithm, whereas it may be nondegenerate for a different... | algorithms and data structures_Page_273_Chunk4813 |
This book is licensed under a Creative Commons Attribution 3.0 License Exhibit 24.1: The "gift-wrapping" approach to building the convex hull. 2. Divide-and-conquer comes to mind: Sort the points of S according to their x-coordinate, use the median x- coordinate to partition S into a left half SL and a right half SR, a... | algorithms and data structures_Page_274_Chunk4814 |
24. Sample problems and algorithms the parallel to PQ that passes through T. Having processed the side PQ, we extend the convex polygon to include T, and we now must process 2 additional sides,PT and TQ. The reader will observe a formal analogy between quicksort (“Sorting and its complexity”) and quickhull, which has g... | algorithms and data structures_Page_275_Chunk4815 |
This book is licensed under a Creative Commons Attribution 3.0 License Exhibit 24.6: Extending the partial upper hull U(P1, … , Pi–1) to the next point Pi The program 'ConvexHull' presented in “Algorithm animation” as an example for algorithm animation is written as an on-line algorithm: Rather than reading all the dat... | algorithms and data structures_Page_276_Chunk4816 |
24. Sample problems and algorithms dx[n] := x[n] – x[i]; dy[n] := y[n] – y[i]; MoveTo(px, py); Line(–dx[n], –dy[n]); b[n] := i end; MoveTo(px, py); PenSize(2, 2); Line(–dx[n], –dy[n]); PenNormal end; { ComputeTangent } The algorithm implemented by 'ConvexHull' is based on Graham's scan [Gra 72], where the points are or... | algorithms and data structures_Page_277_Chunk4817 |
This book is licensed under a Creative Commons Attribution 3.0 License Exhibit 24.7: Point-in polygon test by adding up all turning angles. The second algorithm computes the crossing number of G with respect to P. Draw a semi-infinite ray R from P in any direction (Exhibit 24.8). During the walk around the polygon G fr... | algorithms and data structures_Page_278_Chunk4818 |
24. Sample problems and algorithms the intersection of Q with one of the uncertain regions. As Q' has only about n / 3 vertices, the depth of recursion is O(log n). Actually, after the first comparison against the innermost triangular core of Q, we have no longer a general point-in-convex-polygon problem, but one with ... | algorithms and data structures_Page_279_Chunk4819 |
This book is licensed under a Creative Commons Attribution 3.0 License The complexity of this problem was unexpected until discovered in 1986 [WS 88]. Fortunately, this complexity is revealed not by requiring complicated algorithms, but in the analysis of the inherent complexity of the geometric problem. The example sh... | algorithms and data structures_Page_280_Chunk4820 |
24. Sample problems and algorithms Exhibit 24.12: The four line segments will be partitioned into subsets {1, 2} and {3, 4}. Exhibit 24.13: The min operation merges the solutions of this divide-and-conquer algorithm. The time complexity of this divide-and-conquer algorithm is obtained as follows. Given that at each lev... | algorithms and data structures_Page_281_Chunk4821 |
This book is licensed under a Creative Commons Attribution 3.0 License a given depth process at most V subsegments, where V is the total number of visible subsegments. Thus the total time is bounded by O(V · log n). How large can V be? Surprising theoretical results Let V(n) be the number of visible subsegments in a gi... | algorithms and data structures_Page_282_Chunk4822 |
24. Sample problems and algorithms left to right, and writing down the number of the line segment that is currently visible, we obtain a sequence of numbers (Exhibit 24.16). Exhibit 24.16: The Davenport-Schinzel sequence associated with a configuration of segments. A geometric configuration gives rise to a sequence u1,... | algorithms and data structures_Page_283_Chunk4823 |
This book is licensed under a Creative Commons Attribution 3.0 License Sequences with the properties 1 to 3, called Davenport-Schinzel sequences, have been studied in the context of linear differential equations. The maximal length of a Davenport-Schinzel sequence is k · n · α(n), where k is a constant and α(n) is the ... | algorithms and data structures_Page_284_Chunk4824 |
24. Sample problems and algorithms 7. Intersection test for line L and [convex] polygon Q If an (infinitely extended) line L intersects a polygon Q, it must intersect one of Q's edges. Thus a test for intersection of a given line L with a polygon can be reduced to repeated test of L for intersection with [some of] Q's ... | algorithms and data structures_Page_285_Chunk4825 |
This book is licensed under a Creative Commons Attribution 3.0 License 25. Plane-sweep: a general- purpose algorithm for two- dimensional problems illustrated using line segment intersection Learning objectives: • line segment intersection test • turning space dimensions into time dimensions • updating a y table and de... | algorithms and data structures_Page_286_Chunk4826 |
25. Plane-sweep: a general-purpose algorithm for two-dimensional problems illustrated using line segment intersection Exhibit 25.1: Two ways of applying divide-and-conquer to a set of objects embedded in the plane. In the first case, we hope for a separation into subsets S1 and S2 that permits an efficient test whether... | algorithms and data structures_Page_287_Chunk4827 |
This book is licensed under a Creative Commons Attribution 3.0 License suffices to test all segment pairs that become neighbors at some time during a left-to-right sweep of L - a number that is usually significantly smaller than n · (n – 1) / 2. As the sweep line L moves from left to right across the configuration, the... | algorithms and data structures_Page_288_Chunk4828 |
25. Plane-sweep: a general-purpose algorithm for two-dimensional problems illustrated using line segment intersection x-queue. For simplicity of presentation we assume that no two endpoints of line segments have equal x- or y- coordinates. The only operation to be performed on the x-queue is 'nextX': it returns the nex... | algorithms and data structures_Page_289_Chunk4829 |
This book is licensed under a Creative Commons Attribution 3.0 License if intersect(predY(s), s) or intersect (s, succY(s)) then terminate('intersection found') end else { e is right endpoint of s } begin if intersect(predY(s), succY(s)) then terminate('intersection found'); deleteY(s) end end; With at most 2 · n event... | algorithms and data structures_Page_290_Chunk4830 |
25. Plane-sweep: a general-purpose algorithm for two-dimensional problems illustrated using line segment intersection it can even be guaranteed that the size of the x-queue never exceeds O(n). The cost for exchanging two intersecting line segments in the y-table is O(log n), the costs for the other operations on the y-... | algorithms and data structures_Page_291_Chunk4831 |
This book is licensed under a Creative Commons Attribution 3.0 License 4. Delete u from the y-table 5. Delete s from the y-table 6. Delete t from the y-table Notice the calamity that struck at the critical step 3. The evaluation of a linear expression s(x) and the intersection routine for two segments both arrived at a... | algorithms and data structures_Page_292_Chunk4832 |
26. The closest pair 26. The closest pair Learning objectives: • Applying, implementing and analyzing plane sweep • Using plane sweep on three or more dimensions Sweep algorithms solve many kinds of proximity problems efficiently. We present a simple sweep that solves the two-dimensional closest pair problem elegantly ... | algorithms and data structures_Page_293_Chunk4833 |
This book is licensed under a Creative Commons Attribution 3.0 License involves an intricate argument for combining the solutions of two equally sized subsets, the iterative plane-sweep algorithm [HNS 88] uses a simple incremental update: Starting with the empty set of points, keep adding a single point until the final... | algorithms and data structures_Page_294_Chunk4834 |
26. The closest pair The program of the following section initializes the x-queue and y-table with the two leftmost points being active, with δ equal to their distance, and starts the sweep with the third point. The distinction between discarded and active points is motivated by the following argument. When a new point... | algorithms and data structures_Page_295_Chunk4835 |
This book is licensed under a Creative Commons Attribution 3.0 License The x-queue is initialized by procedure initX; 'initX' stores all the points into the x-queue, ordered by their x- coordinates. The empty y-table is created by procedure initY; A new point is inserted into the y-table by procedure insertY(p: point);... | algorithms and data structures_Page_296_Chunk4836 |
26. The closest pair if newDelta < delta then begin delta := newDelta; closestLeft := check; closestRight := current; end; until current.y – check.y > delta; end; { transition } Analysis We show that the algorithm described can be implemented so as to run in worst-case time O(n · log n) and space O(n). If the y-table i... | algorithms and data structures_Page_297_Chunk4837 |
This book is licensed under a Creative Commons Attribution 3.0 License Exhibit 26.4: Only few points at pairwise distance ≥ δ can populate a box of size 2 · δ by δ. We have shown that the bounding box can hold no more than eight points at pairwise distance ≥ δ when using the Manhattan metric d1. It is well known that f... | algorithms and data structures_Page_298_Chunk4838 |
26. The closest pair bounded by a constant. In the three-dimensional case, the corresponding two-dimensional orthogonal range query cannot in general be answered in time O(log n) (per retrieved point) using any of the known data structures. Straightforward search requires time O(n), resulting in an overall time O(n2) f... | algorithms and data structures_Page_299_Chunk4839 |
Grinstead and Snell’s Introduction to Probability The CHANCE Project1 Version dated 4 July 2006 1Copyright (C) 2006 Peter G. Doyle. This work is a version of Grinstead and Snell’s ‘Introduction to Probability, 2nd edition’, published by the American Mathematical So- ciety, Copyright (C) 2003 Charles M. Grinstead and J.... | prob_Page_1_Chunk4840 |
To our wives and in memory of Reese T. Prosser | prob_Page_2_Chunk4841 |
Contents Preface vii 1 Discrete Probability Distributions 1 1.1 Simulation of Discrete Probabilities . . . . . . . . . . . . . . . . . . . 1 1.2 Discrete Probability Distributions . . . . . . . . . . . . . . . . . . . . 18 2 Continuous Probability Densities 41 2.1 Simulation of Continuous Probabilities . . . . . . . . ... | prob_Page_3_Chunk4842 |
vi CONTENTS 9 Central Limit Theorem 325 9.1 Bernoulli Trials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325 9.2 Discrete Independent Trials . . . . . . . . . . . . . . . . . . . . . . . 340 9.3 Continuous Independent Trials . . . . . . . . . . . . . . . . . . . . . 356 10 Generating Functions 365 10.1 ... | prob_Page_4_Chunk4843 |
Preface Probability theory began in seventeenth century France when the two great French mathematicians, Blaise Pascal and Pierre de Fermat, corresponded over two prob- lems from games of chance. Problems like those Pascal and Fermat solved continued to influence such early researchers as Huygens, Bernoulli, and DeMoivr... | prob_Page_5_Chunk4844 |
viii PREFACE FEATURES Level of rigor and emphasis: Probability is a wonderfully intuitive and applicable field of mathematics. We have tried not to spoil its beauty by presenting too much formal mathematics. Rather, we have tried to develop the key ideas in a somewhat leisurely style, to provide a variety of interesting... | prob_Page_6_Chunk4845 |
PREFACE ix find an example, an application, or an exercise that you really like, it probably had its origin in Feller’s classic text, An Introduction to Probability Theory and Its Applications. We are indebted to many people for their help in this undertaking. The approach to Markov Chains presented in the book was deve... | prob_Page_7_Chunk4846 |
x PREFACE | prob_Page_8_Chunk4847 |
Chapter 1 Discrete Probability Distributions 1.1 Simulation of Discrete Probabilities Probability In this chapter, we shall first consider chance experiments with a finite number of possible outcomes ω1, ω2, . . . , ωn. For example, we roll a die and the possible outcomes are 1, 2, 3, 4, 5, 6 corresponding to the side th... | prob_Page_9_Chunk4848 |
2 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS to mean that the probability is 2/3 that a roll of a die will have a value which does not exceed 4. Let Y be the random variable which represents the toss of a coin. In this case, there are two possible outcomes, which we can label as H and T. Unless we have reason to sus... | prob_Page_10_Chunk4849 |
1.1. SIMULATION OF DISCRETE PROBABILITIES 3 .203309 .762057 .151121 .623868 .932052 .415178 .716719 .967412 .069664 .670982 .352320 .049723 .750216 .784810 .089734 .966730 .946708 .380365 .027381 .900794 Table 1.1: Sample output of the program RandomNumbers. Let X be a random variable with distribution function m(ω), w... | prob_Page_11_Chunk4850 |
4 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS We notice that when we tossed the coin 10,000 times, the proportion of heads was close to the “true value” .5 for obtaining a head when a coin is tossed. A math- ematical model for this experiment is called Bernoulli Trials (see Chapter 3). The Law of Large Numbers, which... | prob_Page_12_Chunk4851 |
1.1. SIMULATION OF DISCRETE PROBABILITIES 5 5 10 15 20 25 30 35 40 -10 -8 -6 -4 -2 2 4 6 8 10 Figure 1.1: Peter’s winnings in 40 plays of heads or tails. One can understand this calculation as follows: The probability that no 6 turns up on the first toss is (5/6). The probability that no 6 turns up on either of the first... | prob_Page_13_Chunk4852 |
6 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS A second interesting question about this game is the following: How many times in the 40 tosses will Peter be in the lead? Looking at the graph of his winnings (Figure 1.1), we see that Peter is in the lead when his winnings are positive, but we have to make some conventi... | prob_Page_14_Chunk4853 |
1.1. SIMULATION OF DISCRETE PROBABILITIES 7 Figure 1.2: Distribution of winnings. Figure 1.3: Distribution of number of times in the lead. | prob_Page_15_Chunk4854 |
8 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS 200 400 600 800 1000 1000 plays -50 -40 -30 -20 -10 0 10 20 Figure 1.4: Peter’s winnings in 1000 plays of heads or tails. 2000 4000 6000 8000 10000 10000 plays 0 50 100 150 200 Figure 1.5: Peter’s winnings in 10,000 plays of heads or tails. | prob_Page_16_Chunk4855 |
1.1. SIMULATION OF DISCRETE PROBABILITIES 9 of the time. A larger number of races would be necessary to have better agreement with the past experience. Therefore we ran the program to simulate 1000 races with our four horses. Although very tired after all these races, they performed in a manner quite consistent with ou... | prob_Page_17_Chunk4856 |
10 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS is of great importance in testing whether observed data fit a given probability dis- tribution. By the early 1900s it was clear that a better way to generate random numbers was needed. In 1927, L. H. C. Tippett published a list of 41,600 digits obtained by selecting numbe... | prob_Page_18_Chunk4857 |
1.1. SIMULATION OF DISCRETE PROBABILITIES 11 collision and so on.5 John von Neumann and Stanislas Ulam suggested that the problem be solved by modeling the experiment by chance devices on a computer. Their work being secret, it was necessary to give it a code name. Von Neumann chose the name “Monte Carlo.” Since that t... | prob_Page_19_Chunk4858 |
12 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS of its use in the early 1600s and says that its probable origin is the reference in Rabelais’s Book One, Chapter 20: Everything was done as planned, the only thing being that Gargantua doubted if they would be able to find, right away, breeches suitable to the old fellow’... | prob_Page_20_Chunk4859 |
1.1. SIMULATION OF DISCRETE PROBABILITIES 13 3 In the early 1600s, Galileo was asked to explain the fact that, although the number of triples of integers from 1 to 6 with sum 9 is the same as the number of such triples with sum 10, when three dice are rolled, a 9 seemed to come up less often than a 10—supposedly in the... | prob_Page_21_Chunk4860 |
14 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS red. If you win, delete the first and last numbers from your list. If you lose, add the amount that you last bet to the end of your list. Then use the new list and bet the sum of the first and last numbers (if there is only one number, bet that amount). Continue until your... | prob_Page_22_Chunk4861 |
1.1. SIMULATION OF DISCRETE PROBABILITIES 15 At the end of a year, which hospital will have the greater number of days on which more than 60 percent of the babies born were boys? (a) the large hospital (b) the small hospital (c) neither—the number of days will be about the same. Assume that the probability that a baby ... | prob_Page_23_Chunk4862 |
16 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS 0 1 2 3 -1 -2 -3 c. Random walk in three dimensions. b. Random walk in two dimensions. a. Random walk in one dimension. Figure 1.6: Random walk. | prob_Page_24_Chunk4863 |
1.1. SIMULATION OF DISCRETE PROBABILITIES 17 At the hotel there lived also some students with whom I usually took my meals and had friendly relations. On a certain day one of them expected the visit of his fianc´ee, what (sic) I knew, but I did not foresee that he and his fianc´ee would also set out for a stroll in the w... | prob_Page_25_Chunk4864 |
18 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS this question for dimensions one, two, and three. He established the remarkable result that the answer is yes in one and two dimensions and no in three dimensions. (c) Write a program to simulate a random walk in three dimensions and see whether, from this simulation and... | prob_Page_26_Chunk4865 |
1.2. DISCRETE PROBABILITY DISTRIBUTIONS 19 where each outcome i, for i = 1, . . . , 6, corresponds to the number of dots on the face which turns up. The event E = {2, 4, 6} corresponds to the statement that the result of the roll is an even number. The event E can also be described by saying that X is even. Unless ther... | prob_Page_27_Chunk4866 |
20 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS Let E ={HH,HT,TH} be the event that at least one head comes up. Then, the probability of E can be calculated as follows: P(E) = m(HH) + m(HT) + m(TH) = 1 4 + 1 4 + 1 4 = 3 4 . Similarly, if F ={HH,HT} is the event that heads comes up on the first toss, then we have P(F) =... | prob_Page_28_Chunk4867 |
1.2. DISCRETE PROBABILITY DISTRIBUTIONS 21 we see that 2m(C) + 2m(C) + m(C) = 1 , which implies that 5m(C) = 1. Hence, m(A) = 2 5 , m(B) = 2 5 , m(C) = 1 5 . Let E be the event that either A or C wins. Then E ={A,C}, and P(E) = m(A) + m(C) = 2 5 + 1 5 = 3 5 . 2 In many cases, events can be described in terms of other e... | prob_Page_29_Chunk4868 |
22 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS A B A B A B A A A B ∼ ⊃ ⊃ A B A B Figure 1.7: Basic set operations. Properties Theorem 1.1 The probabilities assigned to events by a distribution function on a sample space Ωsatisfy the following properties: 1. P(E) ≥0 for every E ⊂Ω. 2. P(Ω) = 1 . 3. If E ⊂F ⊂Ω, then P(... | prob_Page_30_Chunk4869 |
1.2. DISCRETE PROBABILITY DISTRIBUTIONS 23 Suppose next that A and B are disjoint subsets of Ω. Then every element ω of A ∪B lies either in A and not in B or in B and not in A. It follows that P(A ∪B) = P ω∈A∪B m(ω) = P ω∈A m(ω) + P ω∈B m(ω) = P(A) + P(B) , and Property 4 is proved. Finally, to prove Property 5, consid... | prob_Page_31_Chunk4870 |
24 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS Corollary 1.1 For any two events A and B, P(A) = P(A ∩B) + P(A ∩˜B) . 2 Property 4 can be generalized in another way. Suppose that A and B are subsets of Ωwhich are not necessarily disjoint. Then: Theorem 1.4 If A and B are subsets of Ω, then P(A ∪B) = P(A) + P(B) −P(A ∩... | prob_Page_32_Chunk4871 |
1.2. DISCRETE PROBABILITY DISTRIBUTIONS 25 First toss Second toss Third toss Outcome H H H H H H T T T T T T (Start) ω ω ω ω ω ω ω ω 1 2 3 4 5 6 7 8 H T Figure 1.8: Tree diagram for three tosses of a coin. Let A be the event “the first outcome is a head,” and B the event “the second outcome is a tail.” By looking at the... | prob_Page_33_Chunk4872 |
26 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS It is important to realize that when an experiment is analyzed to describe its possible outcomes, there is no single correct choice of sample space. For the ex- periment of tossing a coin twice in Example 1.2, we selected the 4-element set Ω={HH,HT,TH,TT} as a sample spa... | prob_Page_34_Chunk4873 |
1.2. DISCRETE PROBABILITY DISTRIBUTIONS 27 What is the probability of getting neither snakeeyes (double ones) nor boxcars (double sixes)? The event of getting either one of these two outcomes is the set E = {(1, 1), (6, 6)} . Hence, the probability of obtaining neither is given by P( ˜E) = 1 −P(E) = 1 −2 36 = 17 18 . 2... | prob_Page_35_Chunk4874 |
28 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS by seeing what kind of a bet you would be willing to make. For example, suppose that you are willing to make a 1 dollar bet giving 2 to 1 odds that Dartmouth will win. Then you are willing to pay 2 dollars if Dartmouth loses in return for receiving 1 dollar if Dartmouth ... | prob_Page_36_Chunk4875 |
1.2. DISCRETE PROBABILITY DISTRIBUTIONS 29 otherwise. Infinite sample spaces require new concepts in general (see Chapter 2), but countably infinite spaces do not. If Ω= {ω1, ω2, ω3, . . .} is a countably infinite sample space, then a distribution function is defined exactly as in Definition 1.2, except that the sum must no... | prob_Page_37_Chunk4876 |
30 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS Historical Remarks An interesting question in the history of science is: Why was probability not devel- oped until the sixteenth century? We know that in the sixteenth century problems in gambling and games of chance made people start to think about probability. But gamb... | prob_Page_38_Chunk4877 |
1.2. DISCRETE PROBABILITY DISTRIBUTIONS 31 be 0, 1, or 2, and hence we should assign equal probabilities for these three possible outcomes.16 Cardano chose the correct sample space for his dice problems and calculated the correct probabilities for a variety of events. Cardano’s mathematical work is interspersed with a ... | prob_Page_39_Chunk4878 |
32 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS a favorable bet. Pascal showed, by exact calculation, that 25 rolls are required for a favorable bet for a pair of sixes. The second problem was a much harder one: it was an old problem and con- cerned the determination of a fair division of the stakes in a tournament wh... | prob_Page_40_Chunk4879 |
1.2. DISCRETE PROBABILITY DISTRIBUTIONS 33 0 1 2 3 0 1 2 3 0 0 0 8 16 32 64 20 32 48 64 64 32 44 56 Number of games A has won Number of games B has won Figure 1.9: Pascal’s table. quicker and neater, which I would like to tell you here in a few words: for henceforth I would like to open my heart to you, if I may, as I ... | prob_Page_41_Chunk4880 |
34 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS then he needs two games to win and B needs three games to win; and so, if the tounament is called off, A should receive 44 pistoles. The letter in which Fermat presented his solution has been lost; but fortunately, Pascal describes Fermat’s method in a letter dated Monday... | prob_Page_42_Chunk4881 |
1.2. DISCRETE PROBABILITY DISTRIBUTIONS 35 We see that these two mathematicians arrived at two very different ways to solve the problem of points. Pascal’s method was to develop an algorithm and use it to calculate the fair division. This method is easy to implement on a computer and easy to generalize. Fermat’s method,... | prob_Page_43_Chunk4882 |
36 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS 8 A student must choose one of the subjects, art, geology, or psychology, as an elective. She is equally likely to choose art or psychology and twice as likely to choose geology. What are the respective probabilities that she chooses art, geology, and psychology? 9 A stu... | prob_Page_44_Chunk4883 |
1.2. DISCRETE PROBABILITY DISTRIBUTIONS 37 less than 85 percent lost one leg. What is the minimal possible percentage of those who simultaneously lost one ear, one eye, one hand, and one leg?22 *17 Assume that the probability of a “success” on a single experiment with n outcomes is 1/n. Let m be the number of experimen... | prob_Page_45_Chunk4884 |
38 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS 23 Let Ωbe the sample space Ω= {0, 1, 2, . . .} , and define a distribution function by m(j) = (1 −r)jr , for some fixed r, 0 < r < 1, and for j = 0, 1, 2, . . .. Show that this is a distribution function for Ω. 24 Our calendar has a 400-year cycle. B. H. Brown noticed tha... | prob_Page_46_Chunk4885 |
1.2. DISCRETE PROBABILITY DISTRIBUTIONS 39 26 Two cards are drawn successively from a deck of 52 cards. Find the probability that the second card is higher in rank than the first card. Hint: Show that 1 = P(higher) + P(lower) + P(same) and use the fact that P(higher) = P(lower). 27 A life table is a table that lists for... | prob_Page_47_Chunk4886 |
40 CHAPTER 1. DISCRETE PROBABILITY DISTRIBUTIONS (a) If r = 1/2, what is your chance of emerging from the interchange going west? (b) Find the value of r that maximizes your chance of a westward departure from the interchange. 30 (from Benkoski25) Consider a “pure” cloverleaf interchange in which there are no ramps for... | prob_Page_48_Chunk4887 |
Chapter 2 Continuous Probability Densities 2.1 Simulation of Continuous Probabilities In this section we shall show how we can use computer simulations for experiments that have a whole continuum of possible outcomes. Probabilities Example 2.1 We begin by constructing a spinner, which consists of a circle of unit circu... | prob_Page_49_Chunk4888 |
42 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES 0 x Figure 2.1: A spinner. To simulate this experiment on a computer is an easy matter. Many computer software packages have a function which returns a random real number in the in- terval [0, 1]. Actually, the returned value is always a rational number, and the values are... | prob_Page_50_Chunk4889 |
2.1. SIMULATION OF CONTINUOUS PROBABILITIES 43 1 x 1 y y = x2 E Figure 2.2: Area under y = x2. for this simple region we can find the exact area by calculus. In fact, Area of E = Z 1 0 x2 dx = 1 3 . We have remarked in Chapter 1 that, when we simulate an experiment of this type n times to estimate a probability, we can ... | prob_Page_51_Chunk4890 |
44 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES 1 1 1000 trials Estimate of area is .325 y = x2 E Figure 2.3: Computing the area by simulation. Buffon’s Needle Example 2.3 Suppose that we take a card table and draw across the top surface a set of parallel lines a unit distance apart. We then drop a common needle of unit ... | prob_Page_52_Chunk4891 |
2.1. SIMULATION OF CONTINUOUS PROBABILITIES 45 d 1/2 θ Figure 2.4: Buffon’s experiment. θ 0 1/2 0 d π/2 E Figure 2.5: Set E of pairs (θ, d) with d < 1 2 sin θ. Now the area of the rectangle is π/4, while the area of E is Area = Z π/2 0 1 2 sin θ dθ = 1 2 . Hence, we get P(E) = 1/2 π/4 = 2 π . The program BuffonsNeedle si... | prob_Page_53_Chunk4892 |
46 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES 0.00 5.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.00 10000 3.139 Figure 2.6: Simulation of Buffon’s needle experiment. Needle,”2 show that we can expect to have an error of not more than 5/√n about 95 percent of the time. Here n is the number of needles dropped. Thus... | prob_Page_54_Chunk4893 |
2.1. SIMULATION OF CONTINUOUS PROBABILITIES 47 0 0.5 1 1.5 2 0 0.2 0.4 0.6 0.8 1 Figure 2.7: Sum of two random numbers. [a, b] approximates the probability that a ≤X ≤b. But the sum of the areas of these bars also approximates the integral Z b a f(x) dx . This suggests that for an experiment with a continuum of possibl... | prob_Page_55_Chunk4894 |
48 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES 40 45 50 55 60 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 Figure 2.8: Sum of 100 random numbers. x y A B M θ β α Figure 2.9: Random chord. rectangular coordinate system with origin at the center of the circle (see Figure 2.9). We note that a chord of a circle is perpendicular to ... | prob_Page_56_Chunk4895 |
2.1. SIMULATION OF CONTINUOUS PROBABILITIES 49 1. To simulate this case, we choose values for x and y from [−1, 1] at random. Then we check whether x2 + y2 ≤1. If not, the point M = (x, y) lies outside the circle and cannot be the midpoint of any chord, and we ignore it. Oth- erwise, M lies inside the circle and is the... | prob_Page_57_Chunk4896 |
50 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES .0 1.0 .2 .4 .6 .8 1.0 .488 .227 .0 1.0 .2 .4 .6 .8 1.0 .0 1.0 .2 .4 .6 .8 1.0 .332 10000 10000 10000 Figure 2.10: Bertrand’s paradox. length L > √ 3 if its midpoint has distance d < 1/2 from the origin (see Figure 2.9). The following calculations determine the probability... | prob_Page_58_Chunk4897 |
2.1. SIMULATION OF CONTINUOUS PROBABILITIES 51 Length of Number of Number of Estimate Experimenter needle casts crossings for π Wolf, 1850 .8 5000 2532 3.1596 Smith, 1855 .6 3204 1218.5 3.1553 De Morgan, c.1860 1.0 600 382.5 3.137 Fox, 1864 .75 1030 489 3.1595 Lazzerini, 1901 .83 3408 1808 3.1415929 Reina, 1925 .5419 2... | prob_Page_59_Chunk4898 |
52 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES only the needle crossed a line 1808 times in 3408 casts) deduced that the length of the needle must have been 5/6. He calculated this from Buffon’s formula, assuming π = 355/113: L = πP(E) 2 = 1 2 355 113 1808 3408 = 5 6 = .8333 . Even with careful planning one would ... | prob_Page_60_Chunk4899 |
2.1. SIMULATION OF CONTINUOUS PROBABILITIES 53 7 For Buffon’s needle problem, Laplace9 considered a grid with horizontal and vertical lines one unit apart. He showed that the probability that a needle of length L ≤1 crosses at least one line is p = 4L −L2 π . To simulate this experiment we choose at random an angle θ be... | prob_Page_61_Chunk4900 |
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