概率論與數(shù)理統(tǒng)計(英文) 第三章

1、 3. Random Variables3.1 Definition of Random VariablesIn engineering or scientific problems, we are not only interested in the probability of events, but also interested in some variables depending on sample points. （定義在樣本點(diǎn)上的變量）For example, we maybe interested in the life of bulbs produced by a cert

2、ain company, or the weight of cows in a certain farm, etc. These ideas lead to the definition of random variables.1. random variable definitionDefinition 3.1.1 A random variable is a real valued function defined on a sample space; i.e. it assigns a real number to each sample point in the sample spac

3、e.Here are some examples.Example 3.1.1 A fair die is tossed. The number shown is a random variable, it takes values in the set . Example 3.1.2 The life of a bulb selected at random from bulbs produced by company A is a random variable, it takes values in the interval . Since the outcomes of a random

4、 experiment can not be predicted in advance, the exact value of a random variable can not be predicted before the experiment, we can only discuss the probability that it takes some value or the values in some subset of R.2. Distribution functionDefinition 3.1.2 Let be a random variable on the sample

5、 space . Then the function . is called the distribution function of Note The distribution function is defined on real numbers, not on sample space.Example 3.1.3 Let be the number we get from tossing a fair die. Then the distribution function of is (Figure 3.1.1) Figure 3.1.1 The distribution functio

6、n in Example 3.1.33. PropertiesThe distribution function of a random variable has the following properties:(1) is non-decreasing.In fact, if , then the event is a subset of the event ,thus (2), .(3)For any , .This is to say, the distribution function of a random variable is right continuous.Example

7、3.1.4 Let be the life of automotive parts produced by company A , assume the distribution function of is (in hours)Find ,.Solution By definition, . Question： What are the probabilities and ?Example 3.1.5 A player tosses two fair dice, if the total number shown is 6 or more, the player wins $1, other

8、wise loses $1. Let be the amount won, find the distribution function of .Solution Let be the total number shown, then the events contains sample points, . Thus , And so Thus Figure 3.1.2 The distribution function in Example 3.1.5The distribution function of random variables is a connection betweenpr

9、obability and calculus. By means of distribution function, the main tools in calculus, such as series, integrals are used to solve probability and statistics problems.3.2 Discrete Random Variables 離散型隨機(jī)變量In this book, we study two kinds of random variables.Definition 3.2.1 A random variable is calle

10、d a discrete random variable, if it takes values from a finite set or, a set whose elements can be written as a sequence Assume a discrete random variable takes values from the set . Let , (3.2.1)Then we have , . the probability distribution of the discrete random variable (概率分布) X a1 a2 anprobabili

11、ty p1 p2 pn注意隨機(jī)變量X的分布所滿足的條件(1) Pi 0(2) P1+P2+Pn=1離散型分布函數(shù)And the distribution function of is given by (3.2.2)In general, it is more convenient to use (3.2.1) instead of (3.2.2). Equation (3.2.1) is called the probability distribution of the discrete random variable .Example 1For an experiment in whic

12、h a coin is tossed three times (or 3 coins are tossed once), construct the distribution of X. (Let X denote the number of head occurrence)Solution n=3, p=1/2X pr 0 1/81 3/82 3/83 1/8Example 2在一次試驗中，事件A發(fā)生的概率為p, 不發(fā)生的概率為1p, 用X=0表示事件A沒有發(fā)生，X=1表示事件A發(fā)生，求X的分布。two-point distribution(兩點(diǎn)分布) X01P1-pp某學(xué)生參加考試得5分的

13、概率是p, X表示他首次得5分的考試次數(shù)，求X的分布。geometric distribution （幾何分布） X 1234kPpq1pq2pq3pqk1pExample 3 (射擊5發(fā)子彈) 某射手有5發(fā)子彈，射一次命中率為0.9,如果命中目標(biāo)就停止射擊，如果不命中則一直射到子彈用盡，求耗用子彈數(shù)x的概率分布。*Example 3.2.1 A die is tossed, by we denote the number shown, Assume that the probability is proportional to , . Find the probability distrib

14、ution of .Solution Assume that , constant, .Since the events , are mutually exclusive and their union is the certain event, i.e., the sample space , we have ,thus . The probability distribution of is (Figure 3.2.1) , . Figure 3.2.1 Probability distribution in Example 3.2.1Question. What is the diffe

15、rence between distribution functions and probability distributions例2 有一種驗血新方法：把k個人的血混在一起進(jìn)行化驗，如果結(jié)果是陰性，那么對這k 個人只作一次檢驗就夠了，如果結(jié)果是陽性，那么必須對這k個人再逐個分別化驗，這時k個人共需作k+1次檢驗。假設(shè)對所有人來說，化驗是陽性反應(yīng)的概率為p，而且這些人反映是獨(dú)立的。設(shè)表示每個人需要化驗的次數(shù)，求的分布（construct the distribution of ）Binomial distribution（二項分布）Example 3.2.2 A fair die is to

16、ssed 4 times. Let be the number of six got. Find the probability distribution of .Solution. The possible values of are .First we find the probability . Since means that no six occur in 4 tosses. The probability that six fails to occur in a single toss is , and all trials are independent, so .Now con

17、sider the probability , . Since means that six occurs exactly times, they may occur in any tosses of 4 tosses.The event that they occur in a special order (for example, the first tosses), has probability , and we have such combinations. Thus i.e. X01234PBinomial DistributionsAn experiment often cons

18、ists of repeated trials, each with two possible outcomes “success” and “failure”. The most useful application deals with the testing of items as they come off an assembly line, where each test or trial may indicate a defective or a non-defective item. We may choose to define either outcome as a succ

19、ess. The process is referred to a Bernoulli process. Each trial is called a Bernoulli trial.Consider an experiment consists of independent repeated trials, each trials result in two outcomes “success” and “failure”, and the probability of success, denote by , remains constant. Then this process is c

20、alled a Bernoulli process.Definition 3.4.1 The number of successes in Bernoulli trials is called a binomial random variable. The probability distribution of this discrete random variable is called the binomial distribution with parameters and , denoted by .The random variable in Example 3.2.2 is an

21、example of binomial random variable.Theorem 3.4.1 The probability distribution of the binomial distribution with parameters and is given by , (3.4.1)Proof First, consider the probability of obtaining consecutive successes, followed by consecutive failures. These events are independent, therefore the

22、 desired probability is .Since the successes and failures may occur in any order, and for any specific order, the probability is again . We must now determine the total number of sample points in the experiment that have successes and failures. This number is equal to the number of partition of outc

23、omes into two groups with in one group and in the other, i.e. . Because the partitions are mutually exclusive, thus we have , Let , the binomial expansion of the expression gives .Each term correspond to various values of binomial distribution, this is the reason that we called it “binomial distribu

24、tion”.Example 2For an experiment in which 9 coins are tossed, Let X denotes the number of head occurrence, construct the binomial distribution of Xwhat is the probability of obtaining between 3 and 6 successes.poisson distribution（泊松分布）Definition 3.5.1 A discrete random variable is called a Poisson

25、random variable, if it takes values from the set , and if , (3.5.1)Distribution (3.5.1) is called the Poisson distribution with parameter, denoted by .Note that .Here are some examples of Poisson random variables:(a) the number of radioactive particles passing through a counter in certain time perio

26、d;(b) the number of telephone calls received by an office in certain time period; (c) the number of bacteria in a given culture;（細(xì)菌，培養(yǎng)基）(d) the number of typing errors per page in a certain book.Example 3.5.1 From a laboratory experiments, it is known that the number of radioactive particles passing

27、（放射性粒子） through a counter in a given millisecond is a Poisson random variable with parameter . What is the probability that 6 particles enter the counter in a given millisecond?Solution The probability is .Example 3.5.2 The number of oil tankers arriving each day at a certain port is a Poisson rando

28、m variable with parameter 10. What is the probability that on a given day no more than 3 tankers having arrived?Solution. The probability is . HomeworkChapter 3 (P47) 1， 2， 3， 5，7， 21二項分布與泊松分布的關(guān)系Theorem 3.5.2 Let be a sequence of binomial random variables with probability distribution . If for some

29、constant , we have when , and , then , when , and .proof *Example 3.5.4 Suppose that, on average, 1 person in 1000 makes a numerical error in preparing his or her income tax return. If 5000 forms are selected at random and examined, find the probability that 6, 7 or 8 of the forms contain an error.S

30、olution Let be the number of forms contain an error, then has the binomial distribution of parameter and .Using Poisson distribution as approximations, we have ;. 二項分布的應(yīng)用例子Example 3.4.1 It is known that 15% of certain articles manufactured are defective, what is the probability that in a random samp

31、le of 5 articles(a ) exactly 2 are defective.(b) at least 2 are defective.Solution In this case, .(a) The probability is .(b)The desired probability is the sum of getting 2, 3, 4, 5 defective articles, or, we may first find the probability of the complement event, i.e., getting or 1 defective articl

32、e. So, if we denote the number of defective articles by , then we have. Example 3.4.2 A man is able to hit a target 7 times of 10 on the average.(a) Find the probability that he hits the target exactly 3 times in 6 shots;(b) In how many shots the probability that he hits the target at least one time

33、 is greater than 0.95?Solution (a) The probability is .(b) In shots, the probability that he hits at least one time is .Since when , we have ,so in 3 shots, the probability that he hits the target at least one time is .3.3. Expectation and Variance1Expectation (mean) 數(shù)學(xué)期望Suppose in the final exam, y

34、ou got 85 in calculus, 90 in algebra and 83 in statistics, then your average score is .Assume a player tossed a fair die 20 times. He won $11 when he get six and lost $1 otherwise. If he gets six 4 times. Then the average amount he gets per toss is Consider the future games. Since we cannot predict

35、the outcome of the game, we cannot predict the exact amount he will win in the game. But we can predict the average amount he will win. Assume he tosses the die 600 times, in average, six will occur 100 times, thus, the average amount he will win per toss would be We say that in average he will win

36、$1 per toss.Definition 3.3.1 Let be a discrete random variable. The expectation or mean of is defined as (3.3.1)NoticeIn the case that takes values from an infinite number set, (3.3.1) becomes an infinite series. If the series converges absolutely（級數(shù)絕對收斂）, we say the expectation exists, otherwise we

37、 say that the expectation of does not exist.Example 3.3.1 A fair die is tossed . Find the expectation of spots shown.Solution Since takes values from the set and the distribution is , .Thus,. If a discrete random variable assume each of its values with an equal probability, we say this probability d

38、istribution is a discrete uniform distribution（離散均勻分布）. The distribution in Example 3.3.1 is a discrete uniform distribution. Example 3.3.2 A player tosses a fair coin until a head occurs. If the first head occurs at -th time, the player wins dollars. Find the average amount the player wins.Solution

39、 Let be the amount the player wins. Then takes values from the set . The player wins dollars if and only if he gets tails first, and follows by a head. Thus So 二項分布的數(shù)學(xué)期望Now we give the expectation and variance of binomial distribution. Theorem 3.4.3 The expectation and variance of a binomial random

40、variable with parameters n and p are given by . (3.4.2)Proof Consider the identity.Regard as constants, as a variable. Differentiate both sides of this identity with respect to , we have . (3.4.3)Put , we get.But , thusTake differentiation on both sides of (3.4.3),.Put , we get . (3.4.4) Add (3.4.3)

41、 and (3.4.4) to get 泊松分布的期望Theorem 3.5.1 The expectation and variance of a Poisson random variable with parameter are, respectively, and . (3.5.2)Proof By the definition, .Homework chapter 3 8, 9, 10, 22, 27, 302008-3-19驗血問題驗血次數(shù)X 的數(shù)學(xué)期望為 N個人平均需化驗的次數(shù)為 . 由此可知，只要選擇使 , 則N個人平均需化驗的次數(shù).當(dāng)固定時,我們選取使得小于1且取到最小值,這

42、時就能得到最好的分組方法.例如,則,當(dāng)時, 取到最小值. 此時得到最好的分組方法.若,此時以分組,則按第二方案平均只需化驗. 這樣平均來說,可以減少40%的工作量. 補(bǔ)充例問題提出某工廠需要在五周內(nèi)采購1000噸原料，估計原料價格為500元的概率為0.3，600元的概率為0.3，700元的概率為0.4，試求最佳采購策略，使采購價格的期望值最小。思考題如果你能預(yù)先知道5周的原料價格，當(dāng)然是按最低價購買全部原料, 則此時價格的期望值是多少？The expectation of discrete random variables has the following properties數(shù)學(xué)期望性質(zhì)T

43、heorem 3.3.1 Let be a discrete random variable, then (assume all expectations exist):(a) If for some constant , then .(b) Let be a function of , then (3.3.2)(c) If are discrete random variables, then (3.3.3)(d) If , then .(e) If and , then .(f) For any constant , (3.3.4)(g) Schwarzs inequality. （許瓦慈

44、不等式）Let X,Y be random variables, then (3.3.5)The equality holds iff or for some constant a.Proof (a) By the definition.In the summation, for the term and for the other terms thus a.(b), (c) The proofs are given in advanced probability theory, so is omitted here.(d) If , then in the summation each te

45、rm is non-negative, thus .(e).If and , then in the summation the left side is and each term in right side is non-negative, so each term is . Thus, for the terms , we must have . This means .(f).Set in (b), we have (h) Assume , then . Consider the variable , s is a constant, then put we haveIf then ,

46、 the equality holds in (3.3.5). If , then so the equality holds. On the other hand, if the equality holds in (3.3.5), but , then we must have , 2Variance 方差Except the expectation of a random variable, we are interested in some other quantities related to a random variable. Lets consider an example.E

47、xampleA cigarette manufacturer tests tobaccos grown from two districts for nicotine contents, obtains the following resultsDistrict 1 24, 27, 25, 22, 22 (in milligrams)District 2 28 ,27, 25, 20, 20The average nicotine content for both district are the same： 24 milligrams. But the manufacturer prefer the tobaccos from district 1, because it has smaller dispersion than district 2, i.e., it is more stable.To measure the dispersion of a data set , whose ave