Probability and stochastic processes yates 3rd edition pdf download

Probability and stochastic processes yates 3rd edition pdf download

PROBABILITY AND STOCHASTIC PROCESSES A Friendly Introduction for Electrical and Computer Engineers,Incrementa tu iq financiero resumen por capitulos de meditacion de la tecnica

Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers, 3rd Edition International Student Version Roy D. Yates, David J. Goodman ISBN: Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers, 3rd Edition | Wiley. This text introduces engineering students to probability theory 9/12/ · Probability and Stochastic Processes 3rd Edition Student Solutions Manual - [PDF Document] probability and stochastic processes 3rd edition student solutions 10/03/ · Probability and stochastic processes yates pdf 3rd edition solutions; Fundamentals of probability with stochastic processes 3rd edition; The paper in this book was manufactured 24/05/ · Probability Random Variables and Stochastic Processes, 3rd Probability and Stochastic Processes A Friendly Introduction for Electrical and Computer Engineers Second ... read more

With a few exceptions, the mathematical manipulations are not complex. You can go a long way solving problems with a four-function calculator. For many people, this apparent simplicity is dangerously misleading. The problem is that it is very tricky to apply the math to specific problems. A few of you will see things clearly enough to do everything right the first time. However, most people who do well in probability need to practice with a lot of examples to get comfortable with the work and to really understand what the subject is about. Most of the work in this course is that way, and the only way to do well is to practice a lot. We have the short quizzes to show you how well you are doing. Taking the midterm and final are similar to running in a five-mile race.

Most people can do it in a respectable time, provided they train for it. So, our advice to students is if this looks really weird to you, keep working at it. You will probably catch on. It may be harder than you think. xi I DJG will add one more personal remark. For many years, I have been paid to solve probability problems. This has been a major part of my research at Bell Labs and at Rutgers. Applying probability to engineering questions has been extremely helpful to me in my career, and it has led me to a few practical inventions. I hope you will find the material intrinsically interesting.

I hope you will learn to work with it. I think you will have many occasions to use it in future courses and throughout your career. We have worked hard to produce a text that will be useful to a large population of students and instructors. We welcome comments, criticism, and suggestions. Feel free to send us email at ryates winlab. edu or dgoodman winlab. FURTHER READING University libraries have hundreds of books on probability. Of course, each book is written for a particular audience and has its own emphasis on certain topics. We encourage you to spend an afternoon in the library examining a wide selection. For students using this text, a short reference list on page gives a sampling of the many books that may be helpful.

Texts on a similar mathematical level to this text include [LG92, Dra67, Ros96]. For an emphasis on data analysis and statistics, [MR94] is very readable. For those wishing to follow up on the random signal processing material introduced Chapter 10, we can recommend [Pap91, SW94, Vin98]. The material in Chapter 11 can be found in greatly expanded form in [Gal96, Ros96] and in a very accessible introduction in [Dra67]. The two volumes by Feller, [Fel68] and [Fel66] are classic texts in probability theory. For research engineers, [Pap91] is a valuable reference for stochastic processes. Over the three years in which we have used drafts of the book as the principal text for our courses, we have changed the book considerably in response to student comments and complaints. Two undergraduates merit special thanks. Ken Harris created MATLAB demon- strations of many of the concepts presented in the text. He also produced the diagram on the cover. Nisha Batra devised solutions for many of the homework prob- lems in early chapters.

xii Unique among our teaching assistants, Dave Famolari took the course as an under- graduate. Later as a teaching assistant, he did an excellent job writing homework so- lutions with a tutorial flavor. Other graduate students who provided valuable feedback and suggestions include Ricki Abboudi, Zheng Cai, Pi-Chun Chen, Sorabh Gupta, Amar Mahboob, Ivana Maric, David Pandian, Mohammad Saquib, Sennur Ulukus, and Aylin Yener. We also acknowledge with gratitude the support we received from the students and staff of WINLAB, the Wireless Information Network Laboratory at Rutgers Uni- versity. Ivan Seskar deserves special thanks for exercising his magic to make the WINLAB computers particularly hospitable to the electronic versions of the book and to the supporting material on the Word Wide Web.

The book also benefited from the reviews and suggestions conveyed to the pub- lisher by D. Clark of California State Polytechnic University at Pomona, Mark Clements of Georgia Institute of Technology, Gustavo de Veciana of the University of Texas at Austin, Fred Fontaine of Cooper Union University, Rob Frohne of Walla Walla College, Chris Genovese of Carnegie Mellon University, Simon Haykin of Mc- Master University, Ratnesh Kumar of the University of Kentucky, and our colleague Christopher Rose of Rutgers University. Finally, we acknowledge with respect and gratitude the inspiration and guidance of our teachers and mentors who conveyed to us when we were students the impor- tance and elegance of probability theory. We cite in particular Alvin Drake and Robert Gallager of MIT and the late Colin Cherry of Imperial College of Science and Tech- nology.

Roy D. Yates David J. Now you can be- gin. The title of this book is Probability and Stochastic Processes. We say and hear and read the word probability and its relatives possible, probable, probably in many contexts. Within the realm of applied mathematics, the meaning of probability is a question that has occupied mathematicians, philosophers, scientists, and social scien- tists for hundreds of years. Everyone accepts that the probability of an event is a number between 0 and 1. Some people interpret probability as a physical property like mass or volume or tem- perature that can be measured. This is tempting when we talk about the probability that a coin flip will come up heads.

This probability is closely related to the nature of the coin. Fiddling around with the coin can alter the probability of heads. Another interpretation of probability relates to the knowledge that we have about something. We might assign a low probability to the truth of the statement It is raining now in Phoenix, Arizona, because of our knowledge that Phoenix is in the desert. How- ever, our knowledge changes if we learn that it was raining an hour ago in Phoenix. This knowledge would cause us to assign a higher probability to the truth of the state- ment It is raining now in Phoenix. Both views are useful when we apply probability theory to practical problems. Whichever view we take, we will rely on the abstract mathematics of probability, which consists of definitions, axioms, and inferences theorems that follow from the axioms.

While the structure of the subject conforms to principles of pure logic, the terminology is not entirely abstract. Instead, it reflects the practical origins of prob- ability theory, which was developed to describe phenomena that cannot be predicted with certainty. The point of view is different from the one we took when we started studying physics. There we said that if you do the same thing in the same way over and over again — send a space shuttle into orbit, for example — the result will always be the same. To predict the result, you have to take account of all relevant facts. In this case, repetitions of the same procedure yield different results. The situation is not totally chaotic, however. While each outcome may be unpredictable, there are consistent patterns to be observed when you repeat the proce- dure a large number of times. Understanding these patterns helps engineers establish test procedures to ensure that a factory meets quality objectives.

In this repeatable procedure making and testing a chip with unpredictable outcomes the quality of in- dividual chips , the probability is a number between 0 and 1 that states the proportion of times we expect a certain thing to happen, such as the proportion of chips that pass a test. As an introduction to probability and stochastic processes, this book serves three purposes: It introduces students to the logic of probability theory. It helps students develop intuition into how the theory applies to practical situ- ations. It teaches students how to apply probability theory to solving engineering prob- lems. To exhibit the logic of the subject, we show clearly in the text three categories of theoretical material: definitions, axioms, and theorems.

Definitions establish the logic of probability theory, while axioms are facts that we have to accept without proof. Theorems are consequences that follow logically from definitions and axioms. Each theorem has a proof that refers to definitions, axioms, and other theorems. Although there are dozens of definitions and theorems, there are only three axioms of probability theory. These three axioms are the foundation on which the entire subject rests. To meet our goal of presenting the logic of the subject, we could set out the material as dozens of definitions followed by three axioms followed by dozens of theorems. Each theorem would be accompanied by a complete proof.

While rigorous, this approach would completely fail to meet our second aim of conveying the intuition necessary to work on practical problems. To address this goal, we augment the purely mathematical material with a large number of examples of practical phenomena that can be analyzed by means of probability theory. We also interleave definitions and theorems, presenting some theorems with complete proofs, others with partial proofs, and omitting some proofs altogether. We find that most engineering students study probability with the aim of using it to solve practical prob- lems, and we cater mostly to this goal. We also encourage students to take an interest in the logic of the subject — it is very elegant — and we feel that the material presented will be sufficient to enable these students to fill in the gaps we have left in the proofs. Therefore, as you read this book you will find a progression of definitions, axioms, theorems, more definitions, and more theorems, all interleaved with examples and comments designed to contribute to your understanding of the theory.

We also include brief quizzes that you should try to solve as you read the book. Each one will help you decide whether you have grasped the material presented just before the quiz. The problems at the end of each chapter give you more practice applying the material introduced in the chapter. They vary considerably in their level of difficulty. Some of them take you more deeply into the subject than the examples and quizzes do. Most people who study probability have already encountered set theory and are familiar with such terms as set, element, union, intersection, and complement.

For them, the following paragraphs will review material already learned and introduce the notation and terminology we use here. For people who have no prior acquaintance with sets, this material introduces basic definitions and the properties of sets that are important in the study of probability. A set is a collection of things. We use capital letters to denote sets. The things that together make up the set are elements. When we use mathematical notation to refer to set elements, we usually use small letters. Thus we can have a set A with elements x, y, and z. The symbol 2 denotes set inclusion. The def- inition allows someone to consider anything conceivable and determine whether that thing is an element of the set. There are many ways to define a set. The dots tell us to continue the sequence to the left of the dots.

Since there is no number to the right of the dots, we continue the sequence indefinitely, forming an infinite set. The definition of B implies that 9 2 B and 10 62 B. By definition, A is a subset of B if every member of A is also a member of B. This definition implies that a set is unaffected by the order of the elements in a definition. To work with sets mathematically it is necessary to define a universal set. This is the set of all things that we could possibly consider in a given context. In any study, all set operations relate to the universal set for that study.

The members of the universal set include all of the elements of all of the sets in the study. We will use the letter S to denote the universal set. By definition, every set is a subset of the universal set. The null set, which is also important, may seem like it is not a set at all. By definition it has no elements. The notation for the null set is φ. By definition φ is a subset of every set. It is customary to refer to Venn diagrams to display relationships among sets. By convention, the region enclosed by the large rectangle is the universal set S. Closed surfaces within this rectangle denote sets. There are three op- erations for doing this: union, intersection, and complement. Union and intersection combine two existing sets to produce a third set. The complement operation forms a new set from one existing set.

The notation and definitions are The union of sets A and B is the set of all elements A B that are either in A or in B, or in both. Another notation for intersection is AB. A The complement of a set A, denoted by A c , is the set of all elements in S that are not in A. The com- plement of S is the null set φ. Formally, Ac x 2 Ac if and only if x 62 A A fourth set operation is called the difference. It is a combination of intersection and complement. A-B The difference between A and B is a set A ; B that contains all elements of A that are not elements of B. In working with probability we will frequently refer to two important properties of collections of sets. Here are the definitions. As we see in the following theorem, this can be complicated to show. Theorem 1. Hence, x 62 A and x 62 B, which together imply x 2 Ac and x 2 Bc. In this case, x 2 Ac and x 2 Bc. Quiz 1. For the sets spec- ified in parts a — g below, shade the corresponding region of the Venn diagram.

i Are R, T , and M collectively exhaustive? j Are T and O mutually exclusive? State this condition in words. Probability is a number that describes a set. The higher the number, the more probability there is. In this sense probability is like a quantity that measures a physical phenomenon, for example, a weight or a temperature. However, it is not necessary to think about probability in physical terms. We can do all the math abstractly, just as we defined sets and set operations in the previous paragraphs without any reference to physical phenomena. Fortunately for engineers, the language of probability including the word prob- ability itself makes us think of things that we experience.

The basic model is a re- peatable experiment. An experiment consists of a procedure and observations. There is some uncertainty in what will be observed; otherwise, performing the experiment would be unnecessary. Some examples of experiments include 1. Flip a coin. Did it land on heads or tails? Walk to a bus stop. How long do you wait for the arrival of a bus? Give a lecture. How many students are seated in the fourth row? Transmit one of a collection of waveforms over a channel. What waveform arrives at the receiver?

Which waveform does the receiver identify as the transmitted waveform? For the most part, we will analyze models of actual physical experiments. We create models because real experiments generally are too complicated to analyze. For exam- ple, to describe all of the factors affecting your waiting time at a bus stop, you may consider The time of day. Is it rush hour? The speed of each car that passed by while you waited. The weight, horsepower, and gear ratios of each kind of bus used by the bus company. The psychological profile and work schedule of each bus driver. Some drivers drive faster than others.

The status of all road construction within miles of the bus stop. It should be apparent that it would be difficult to analyze the effect of each of these fac- tors on the likelihood that you will wait less than five minutes for a bus. Since we will focus on the model of the experiment almost exclusively, we often will use the word experiment to refer to the model of an experiment. Example 1. The result of each flip is unrelated to the results of previous flips. As we have said, an experiment consists of both a procedure and observations. It is im- portant to understand that two experiments with the same procedure but with different observations are different experiments. For example, consider these two experiments: Example 1. Flip a coin three times. Observe the sequence of heads and tails. Observe the number of heads. These two experiments have the same procedure: flip a coin three times. They are different experiments because they require different observations. We will describe models of experiments in terms of a set of possible experimental outcomes.

In the context of probability, we give precise meaning to the word outcome. Definition 1. Outcome: An outcome of an experiment is any possible observation of that experiment. Implicit in the definition of an outcome is the notion that each outcome is distinguish- able from any other outcome. As a result, we define the universal set of all possible outcomes. In probability terms, we call this universal set the sample space. Sample Space: The sample space of an experiment is the finest-grain, mutually exclusive, collectively exhaustive set of all possible outcomes. The finest-grain property simply means that all possible distinguishable outcomes are identified separately. The requirement that outcomes be mutually exclusive says that if one outcome occurs, then no other outcome also occurs.

For the set of outcomes to be collectively exhaustive, every outcome of the experiment must be in the sample space. The sample space in Example 1. Manufacture an integrated circuit and test it to determine whether it meets quality objectives. In common speech, an event is just something that occurs. In an experiment, we may say that an event occurs when a certain phenomenon is observed. To define an event mathematically, we must identify all outcomes for which the phenomenon is observed. That is, for each outcome, either the particular event occurs or it does not. In probability terms, we define an event in terms of the outcomes of the sample space. Event: An event is a set of outcomes of an experiment. The following table relates the terminology of probability to set theory: Set Algebra Probability set event universal set sample space element outcome All of this is so simple that it is boring.

While this is true of the definitions them- selves, applying them is a different matter. Defining the sample space and its outcomes are key elements of the solution of any probability problem. A probability problem arises from some practical situation that can be modeled as an experiment. To work on the problem, it is necessary to define the experiment carefully and then derive the sample space. Getting this right is a big step toward solving the problem. Suppose we roll a six sided die and observe the number of dots on the side facing upwards. Each subset of S is an event. Wait for someone to make a phone call and observe the duration of the call in minutes. An outcome x is a nonnegative real number. Consider three traffic lights encountered driving down a road. We say a light was red if the driver was required to come to a complete stop at that light; otherwise we call the light green. For the sake of simplicity, these definitions were carefully chosen to exclude the case of the yellow light.

An outcome of the experiment is a description of whether each light was red or green. We can denote the outcome by a sequence of r and g such as rgr, the outcome that the first and third lights were red but the second light was green. The event R2 would be the set of outcomes fgrg grr rrg rrrg. We can also denote an outcome as an intersection of events Ri and G j. For example, the event R1 G2 R3 is the set containing the single outcome frgrg. In Example 1. In this case, the set of events fG2 R2 g describes the events of interest. Moreover, for each possible outcome of the three light experiment, the second light was either red or green, so the set of events fG2 R2 g is both mutually exclusive and collectively exhaustive.

How- ever, fG2 R2 g is not a sample space for the experiment because the elements of the set do not completely describe the set of possible outcomes of the experiment. The set fG2 R2 g does not have the finest-grain property. Yet sets of this type are sufficiently useful to merit a name of their own. Event Space: An event space is a collectively exhaustive, mutually exclusive set of events. An event space and a sample space have a lot in common. The members of both are mutually exclusive and collectively exhaustive. They differ in the finest-grain property that applies to a sample space but not to an event space. Because it possesses the finest-grain property, a sample space contains all the details of an experiment. The members of a sample space are outcomes.

By contrast, the members of an event space are events. The event space is a set of events sets , while the sample space is a set of outcomes elements. Usually, a member of an event space contains many outcomes. Consider a simple example: Example 1. Flip four coins, a penny, a nickel, a dime, and a quarter. Examine the coins in order penny, then nickel, then dime, then quarter and observe whether each coin shows a head h or a tail t. What is the sample space? How many elements are in the sample space? The sample space consists of 16 four-letter words, with each letter either h or t. For example, the outcome tthh refers to the penny and the nickel showing tails and the dime and quarter showing heads. There are 16 members of the sample space.

For the four coins experiment of Example 1. Each Bi is an event containing one or more outcomes. Its members are mutually exclusive and collectively exhaustive. It is not a sample space because it lacks the finest-grain property. The experiment in Example 1. Mathematically, however, it is equivalent to many real engineering problems. For example, observe a modem transmit four bits from one telephone to another. Or, test four integrated circuits. For each one, observe whether the circuit is acceptable a , or a reject r. In all of these examples, the sample space contains 16 four-letter words formed with an alphabet containing two letters. If we are only interested in the number of times one of the letters occurs, it is sufficient to refer only to the event space B, which does not contain all of the information about the experiment but does contain all of the information we need.

The event space is simpler to deal with than the sample space because it has fewer members there are five events in the event space and 16 outcomes in the sample space. The simplification is much more significant when the complexity of the experiment is higher; for example, testing 10 circuits. The concept of an event space is useful because it allows us to express any event as a union of mutually exclusive events. We will observe in the next section that the entire theory of probability is based on unions of mutually exclusive events. In the coin tossing experiment of Example 1.

Many practical problems use the mathematical technique contained in the theorem. For example, find the probability that there are three or more bad circuits in a batch that come from a fabrication machine. Monitor three consecutive phone calls going through a telephone switching office. Clas- sify each one as a voice call v , if someone is speaking or a data call d if the call is carrying a modem or fax signal. Your observation is a sequence of three letters each letter is either v or d. For example, two voice calls followed by one data call corresponds to vvd. This leads to a set-theory representation with a sample space universal set S , outcomes s that are elements of S and events A that are sets of elements.

With respect to our physical idea of the experiment, the probability of an event is the proportion of the time that event is observed in a large number of runs of the experiment. This is the relative frequency notion of probability. Mathematically, this is expressed in the following axioms. Axiom 2. Axiom 3. Axioms 1 and 2 simply establish a probability as a number between 0 and 1. Axiom 3 states that the probability of the union of mutually exclusive events is the sum of the individual probabilities. We will use this axiom over and over in developing the theory of proba- bility and in solving problems. In fact, it is really all we have to work with. Everything else follows from Axiom 3. To use Axiom 3 to solve a practical problem, we analyze a complicated event in order to express it as the union of mutually exclusive events whose probabilities we can calculate. Then, we add the probabilities of the mutually exclusive events to find the probability of the complicated event we are interested in.

A useful extension of Axiom 3 applies to the union of two disjoint events. In fact, a simple proof of Theorem 1. It is a simple matter to extend Theorem 1. The correspondence refers to a sequential experiment consisting of n repetitions of the basic experiment. In these n trials, N A n is the number of times that event A occurs. In Chapter 7, we prove that lim n! Another consequence of the axioms can be expressed as the following theorem. The expression in the square brackets is an event. Note that fsi g is the formal notation for a set with the single element s i. Let Ti denote the duration in minutes of the ith phone call you place today. In such experiments, there are usually symmetry arguments that lead us to believe that no one outcome is any more likely than any other. As in Example 1. What is the probability of each outcome? A score of 90 to is an A, 80 to 89 is a B, 70 to 79 is a C, 60 to 69 is a D, and below 60 is a failing grade of F.

While we do not supply the proofs, we suggest that students prove at least some of these theorems in order to gain experience working with the axioms. The following theorem is a more complex consequence of the axioms. It is very useful. Full pdf package download full pdf package. Probability and stochastic processes yates 2nd edition keywords: Acces pdf probability and stochastic processes yates 2nd edition an introduction to stochastic modelingprobability and stochastic processes:. Probability And Stochastic Processes Yates Pdf Download August 19, Download now probability and stochastic processes: Probability and stochastic processes yates 3rd edition pdf from heavenlybells. However, We Have Yet To 1. Probability And Stochastic Processes A Friendly Introduction For Electrical And Computer Engineers Third Edition Quiz Solutions Roy D.

That is, simply reversing the labels of A and B proves the claim. Alter-natively, one can construct exactly the same proof as in part a withthe labels A and B reversed. c To prove that Ac and Bc are independent, we apply the result of part a to the sets A and Bc. Since we know from part a that A and Bc areindependent, part b says that Ac and Bc are independent. In the Venn diagram at right, assume the samplespace has area 1 corresponding to probability 1. The three way intersection ABChas zero probability, implying A, B, and C are notmutually independent since. Applying the ceiling function convertsthese random numbers to rndom integers in the set {1, 2,. Finally,we add 50 to produce random numbers between 51 and Problem 2.

The tree for the free throw experiment is. The P[ H ] is the probability that a person who has HIV tests negative forthe disease. This is referred to as a false-negative result. Thereason this probability is so low is that the a priori probability that a personhas HIV is very small. Thisresult should not be surprising since if the first flip is heads, it is likely thatcoin B was picked first. In this case, the second flip is less likely to be headssince it becomes more likely that the second coin flipped was coin A. a The primary difficulty in this problem is translating the words into thecorrect tree diagram. The tree for this problem is shown below. The reason for the dependence is that given H2 occurred,then we know there will be a third flip which may result in H3. That is,knowledge of H2 tells us that the experiment didnt end after the firstflip.

The starting point is to draw a tree of the experiment. We define the eventsW that the plant is watered, L that the plant lives, and D that the plantdies. The tree diagram is. In informal conversation, it can be confusing to distinguish between P[D W c]and P[W c D]; however, they are simple once you draw the tree. Technically, a gumball machine has a finite number of gumballs, but theproblem description models the drawing of gumballs as sampling from themachine without replacement. This is a reasonable model when the machinehas a very large gumball capacity and we have no knowledge beforehand ofhow many gumballs of each color are in the machine. Under this model, therequested probability is given by the multinomial probability. a Let Bi, Li, Oi and Ci denote the events that the ith piece is Berry,Lemon, Orange, and Cherry respectively.

Let F1 denote the event thatall three pieces draw are the same flavor. b Let Di denote the event that the ith piece is a different flavor from allthe prior pieces. Let Si denote the event that piece i is the same flavoras a previous piece. A tree for this experiment is. Alternatively, out of 11 pieces left,there are 3 colors each with 3 pieces that is, 9 pieces out of 11 that are different from the first piece. Given the first two pieces are different, there are 2 colors, each with3 pieces 6 pieces out of 10 remaining pieces that are a differentflavor from the first two pieces. c There are 6 red face cards J,Q,K of diamonds and hearts in a 52 carddeck. The number of codeswith exactly 3 zeros equals the number of ways of choosing the bits in whichthose zeros occur. Therefore there are. We can break down the experiment of choosing a starting lineup into a se-quence of subexperiments:.

Of the remaining 14 field players, choose 8 for the remaining field posi-tions. For the 9 batters consisting of the 8 field players and the designatedhitter , choose a batting lineup. ways to do this. So the total number of different starting lineups when the DH is selectedamong the field players is. Note that this overestimates the number of combinations the manager mustreally consider because most field players can play only one or two positions. Although these constraints on the manager reduce the number of possiblelineups, it typically makes the managers job more difficult. As for the count-ing, we note that our count did not need to specify the positions played by thefield players. Although this is an important consideration for the manager, itis not part of our counting of different lineups.

In fact, the 8 nonpitching fieldplayers are allowed to switch positions at any time in the field. For example,the shortstop and second baseman could trade positions in the middle of aninning. Although the DH can go play the field, there are some coomplicatedrules about this. Here is an excerpt from Major League Baseball Rule 6. The Designated Hitter may be used defensively, continuing tobat in the same position in the batting order, but the pitcher mustthen bat in the place of the substituted defensive player, unlessmore than one substitution is made, and the manager then mustdesignate their spots in the batting order. If we call landing on greeen a success, then G19 is theprobability of 19 successes in 40 trials.

A ticket is a winner if eachtime a box is scratched off, the box has the special mark. Continuing this argument, the probabilitythat a ticket is a winner is. a Since the probability of a zero is 0. Since there are always 5 lights, G, Y , and R obey themultinomial probability law:. a There are 3 group 1 kickers and 6 group 2 kickers. Using Gi to denotethat a group i kicker was chosen, we have. b To solve this part, we need to identify the groups from which the firstand second kicker were chosen. Let ci indicate whether a kicker waschosen from group i and let Cij indicate that the first kicker was chosenfrom group i and the second kicker from group j. The experiment tochoose the kickers is described by the sample tree:. If this is not clear,we derive this result by calculating P[K2 Cij] and using the law of totalprobability to calculate P[K2].

Thereason K1 and K2 are dependent is that if the first kicker is successful,then it is more likely that kicker is from group 1. This makes it morelikely that the second kicker is from group 2 and is thus more likely tomiss. c Once a kicker is chosen, each of the 10 field goals is an independenttrial. Out of 10kicks, there are 5 misses iff there are 5 successful kicks. Given the typeof kicker chosen, the probability of 5 misses is. To find the probability that the device works, we replace series devices 1, 2,and 3, and parallel devices 5 and 6 each with a single device labeled with theprobability that it works. In particular,. The probability P[W ] that the two devices in parallel work is 1 minus theprobability that neither works:. Thus, the probability the device works is. Note that each digit 0 through 9 is mapped to the 4 bit binary representa-tion of the digit.

That is, 0 corresponds to , 1 to , up to 9 whichcorresponds to Of course, the 4 bit binary numbers corresponding tonumbers 10 through 15 go unused, however this is unimportant to our prob-lem. the 10 digit number results in the transmission of 40 bits. For each bit,an independent trial determines whether the bit was correct, a deletion, oran error. In Problem 2. Since each of the 40 bit transmissions is an independent trial, the joint prob-ability of c correct bits, d deletions, and e erasures has the multinomial prob-ability. Rather than just solve the problem for 50 trials, we can write a function thatgenerates vectors C and H for an arbitrary number of trials n. The code forthis task is. The first line produces the n 1 vector C such that C i indicates whethercoin 1 or coin 2 is chosen for trial i. To test n 6-component devices, such that each component works with prob-ability q we use the following function:.

By applying these logical operators to the n 1 columnsof W, we simulate the test of n circuits. Lastly, we count the number N of working devices. As we see, the number of working devices is typically around 85 out of Solving Problem 2. For arbitrary number of trials n and failure probability q, the following func-tions evaluates replacing each of the six components by an ultrareliable device. This code is based on the code for the solution of Problem 2. By applying these logical opeators to the n 1 columns of W, wesimulate the test of n circuits. Note that in the code, we first generate the matrix W such that eachcomponent has failure probability q. Lastly, for eachcolumn replacement, we count the number N of working devices. From the above, we see, for example, that replacing the third component withan ultrareliable component resulted in 91 working devices.

The results arefairly inconclusive in that replacing devices 1, 2, or 3 should yield the sameprobability of device failure. In both cases, it is clear that replacing component 4 maximizes the devicereliability. The somewhat complicated solution of Problem 2. c We can view this as a sequential experiment: first we divide the file intoN packets and then we check that all N packets are received correctly. In the second stage, we could specify how many packets are receivedcorrectly; however, it is sufficient to just specify whether the N packetsare all received correctly or not. Using Cn to denote the event that npackets are transmitted and received correctly, we have. The tree for this experiment is. In Problem 3. The problem statement requiresthat 0. This implies n 4. Problem 3.

a In the setup of a mobile call, the phone will send the SETUP messageup to six times. Each time the setup message is sent, we have a Bernoullitrial with success probability p. Of course, the phone stops trying assoon as there is a success. Using r to denote a successful response, andn a non-response, the sample tree is. c Let B denote the event that a busy signal is given after six failed setupattempts. a Each paging attempt is an independent Bernoulli trial with successprobability p. The number of times K that the pager receives a messageis the number of successes in n Bernoulli trials and has the binomialPMF. b Let R denote the event that the paging message was received at leastonce.

The event R has probability. The the number of fish hooked, K, has the binomial PMF. a The paging message is sent again and again until a success occurs. That is, N has the geometric PMF. Hence, we must choose p to satisfy 1 1 p 3 0. a If each message is transmitted 8 times and the probability of a suc-cessful transmission is p, then the PMF of N , the number of successfultransmissions has the binomial PMF. The tree is. The packets are delay sensitive and can only be retransmitted d times. Further, the packet is transmittedd times if there are failures on the first d 1 transmissions, no matter what. the outcome of attempt d. So the random variable T , the number of timesthat a packet is transmitted, can be represented by the following PMF. a Since each day is independent of any other day, P[W33] is just the prob-ability that a winning lottery ticket was bought. Similarly for P[L87]and P[N99] become just the probability that a losing ticket was boughtand that no ticket was bought on a single day, respectively.

b Suppose we say a success occurs on the kth trial if on day k we buy aticket. Otherwise, a failure occurs. The random variable K is just the number of trials until the firstsuccess and has the geometric PMF. If we view buying a losingticket as a Bernoulli success, R, the number of losing lottery ticketsbought in m days, has the binomial PMF. Therefore D has thePascal PMF.

Embed Size px x x x x Matlab functions written as solutions to homework problems in this Stu-dents Solution Manual SSM can be found in the archive matsoln3student. Other Matlab functions used in the text or in these homework solutionscan be found in the archive matcode3e. The archives matcode3e. zipand matsoln3student. zip are available for download from the John Wileycompanion website. Two other documents of interest are also available fordownload:. This manual uses a page size matched to the screen of an iPad tablet.

If youdo print on paper and you have good eyesight, you may wish to print twopages per sheet in landscape mode. On the other hand, a Fit to Paperprinting option will create Large Print output. Based on the Venn diagram on the right, the complete Ger-landas pizza menu is Regular without toppings Regular with mushrooms Regular with onions Regular with mushrooms and onions Tuscan without toppings Tuscan with mushrooms. At Ricardos, the pizza crust is either Roman R or Neapoli-tan N. To draw the Venn diagram on the right, we makethe following observations:. Only Roman pizzas can be white. Hence W R. Only a Neapolitan pizza can have onions. Hence O N. Both Neapolitan and Roman pizzas can have mushrooms so that eventM straddles the {R,N} partition. The Neapolitan pizza can have both mushrooms and onions so M O. The problem statement does not preclude putting mushrooms on awhite Roman pizza.

Hence the intersection WM should not be empty. a An outcome specifies whether the connection speed is high h , medium m , or low l speed, and whether the signal is a mouse click c or atweet t. The sample space is. Problem 1. Here are four partitions. We can divide students into engineers or non-engineers. Let A1 equalthe set of engineering students and A2 the non-engineers. The pair{A1, A2} is a partition. We can also separate students by GPA. At Rutgers, {B1, B2,. Note that B5 is the set of all students with perfect 4. Of course, other schools use different scales for GPA. We can also divide the students by age. Let Ci denote the subset ofstudents of age i in years. At most universities, {C10, C11,. Since a university may have prodigies eitherunder 10 or over , we note that {C0, C1,.

Lastly, we can categorize students by attendance. Let D0 denote thenumber of students who have missed zero lectures and let D1 denote allother students. Although it is likely that D0 is an empty set, {D0, D1}is a well defined partition. An outcome is a pair i, j where i is the value of the first die and j is thevalue of the second die. The sample space is the set. The questionscan be answered using Theorem 1. Let Hidenote the event that the first card drawn is the ith heart where the first heartis the ace, the second heart is the deuce and so on. The event H that the first card is a heart can be written asthe mutually exclusive union. This is the answer you would expect since 13 out of 52 cards are hearts. Thepoint to keep in mind is that this is not just the common sense answer butis the result of a probability model for a shued deck and the axioms ofprobability.

The sample space is thencomposed of all the possible grades that she can receive. The probabilitythat the student gets an A is the probability that she gets a score of 9 orhigher. That is. In this case, by Theorem 1. Now we make our inductionhypothesis that the union-bound holds for any collection of n 1 subsets. Inthis case, given subsets A1,. Axiom 3 then implies. This problemis more challenging if you just use Axiom 3. We start by observing. Theprobability that a call with one handoff will be long is. The probabilitythat a long call will have one or more handoffs is. The first generation consists of two plants each with genotype yg or gy. A pea plant has yellow seeds if it. The sample outcomes can be written ijk where the first card drawn is i, thesecond is j and the third is k. Using this short-hand, the six unknowns p0, p1, p2, q0, q1, q2 fill the table as.

Thus, we have four equations and six unknowns, choosing p0 and p1 willspecify the other unknowns. Unfortunately, arbitrary choices for eitherp0 or p1 will lead to negative values for the other probabilities. In termsof p0 and p1, the other unknowns are. These extra facts uniquely specify the probabilities. In this case,. The works each phone sold is twice as likely to be anApricot than a Banana tells us that. Theprobability that two phones sold are the same is. This implies. b Proving that Ac and B are independent is not really necessary. SinceA and B are arbitrary labels, it is really the same claim as in part a. That is, simply reversing the labels of A and B proves the claim.

Alter-natively, one can construct exactly the same proof as in part a withthe labels A and B reversed. c To prove that Ac and Bc are independent, we apply the result of part a to the sets A and Bc. Since we know from part a that A and Bc areindependent, part b says that Ac and Bc are independent. In the Venn diagram at right, assume the samplespace has area 1 corresponding to probability 1. The three way intersection ABChas zero probability, implying A, B, and C are notmutually independent since. Applying the ceiling function convertsthese random numbers to rndom integers in the set {1, 2,. Finally,we add 50 to produce random numbers between 51 and Problem 2. The tree for the free throw experiment is. The P[ H ] is the probability that a person who has HIV tests negative forthe disease.

This is referred to as a false-negative result. Thereason this probability is so low is that the a priori probability that a personhas HIV is very small. Thisresult should not be surprising since if the first flip is heads, it is likely thatcoin B was picked first. In this case, the second flip is less likely to be headssince it becomes more likely that the second coin flipped was coin A. a The primary difficulty in this problem is translating the words into thecorrect tree diagram. The tree for this problem is shown below. The reason for the dependence is that given H2 occurred,then we know there will be a third flip which may result in H3. That is,knowledge of H2 tells us that the experiment didnt end after the firstflip. The starting point is to draw a tree of the experiment.

We define the eventsW that the plant is watered, L that the plant lives, and D that the plantdies. The tree diagram is. In informal conversation, it can be confusing to distinguish between P[D W c]and P[W c D]; however, they are simple once you draw the tree. Technically, a gumball machine has a finite number of gumballs, but theproblem description models the drawing of gumballs as sampling from themachine without replacement. This is a reasonable model when the machinehas a very large gumball capacity and we have no knowledge beforehand ofhow many gumballs of each color are in the machine.

Under this model, therequested probability is given by the multinomial probability. a Let Bi, Li, Oi and Ci denote the events that the ith piece is Berry,Lemon, Orange, and Cherry respectively. Let F1 denote the event thatall three pieces draw are the same flavor.

Probability And Stochastic Processes Yates Pdf Download,Menu Halaman Statis

Probability and Stochastic Processes A Friendly Introduction for Electrical and Computer Engineers SECOND EDITION Problem Solutions September 28, Draft Roy D. Yates, 10/03/ · Probability and stochastic processes yates pdf 3rd edition solutions; Fundamentals of probability with stochastic processes 3rd edition; The paper in this book was manufactured 24/05/ · Probability Random Variables and Stochastic Processes, 3rd Probability and Stochastic Processes A Friendly Introduction for Electrical and Computer Engineers Second 9/12/ · Probability and Stochastic Processes 3rd Edition Student Solutions Manual - [PDF Document] probability and stochastic processes 3rd edition student solutions 19/08/ · Probability and stochastic processes yates 3rd edition pdf from blogger.com This online pronouncement probability stochastic processes yates Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers, 3rd Edition International Student Version Roy D. Yates, David J. Goodman ISBN: ... read more

Menu Halaman Statis Beranda. Let the sample space of an experiment consist the probability of an A, which requires the student of all the undergraduates at a university. sample space of the experiment. Another interpretation of probability relates to the knowledge that we have about something. The correspondence refers to a sequential experiment consisting of n repetitions of the basic experiment. An example of a code word is A, B, C, D. The probabilities can be represented in the table in which the rows and columns are labeled by events and a table entry represents the probability of the intersection of the corresponding row and column events.

a The source continues to transmit packets until one is received correctly. detector of a pair is acceptable? In this case, by Theorem 1. Use a Venn diagram in which the event areas sion that genes determining different characteris- are proportional to their probabilities to illustrate tics are transmitted independently. They are different experiments because they require different observations.