概率论与数理统计(英文) 第四章.doc

4. Continuous Random Variable 连续型随机变量Continuous random variables appear when we deal will quantities that are measured on a continuous scale. For instance, when we measure the speed of a car, the amount of alcohol in a persons blood, the tensile strength of new alloy.We shall learn how to determine and work with probabilities relating to continuous random variables in this chapter. We shall introduce to the concept of the probability density function. 4.1 Continuous Random Variable1. Definition Definition 4.1.1 A function f(x) defined on is called a probability density function （概率密度函数）if:(i) ;(ii) f(x) is intergrable (可积的) on and .Definition 4.1.2 Let f(x) be a probability density function. If X is a random variable having distribution function , (4.1.1)then X is called a continuous random variable having density function f(x). In this case,. (4.1.2)2. 几何意义 3. Note In most applications, f(x) is either continuous or piecewise continuous having at most finitely many discontinuities.Note 1 For a random variable X, we have a distribution function. If X is discrete, it has a probability distribution. If X is continuous, it has a probability density function. Note 2 Let X be a continuous random variable, then for any real number x,. 4. ExampleExample 4.1.2 Find k so that the following can serve as the probability density of a continuous random variable:Solution To satisfy the conditions (4.1.1), k must be nonnegative, and to satisfy the condition (4.1.2) we must haveso that . (Cauchy distribution 柯西分布) Example 4.1.3 Calculating probabilities from the probability density function If a random variable has the probability densityFind the probability from that it will take on value(a) between 0 and 2;(b) greater than 1.Solution Evaluating the necessary integrals, we get (a) (b) Example 4.1.4 Determining the distribution function of X, it is known Solution Performing the necessary integrations, we get 5. meanDefinition 4.1.2 Let X be a continuous random variable having probability density function f(x). Then the mean (or expectation) of X is defined by, (4.1.3)provided the integral converges absolutely.If the integral (4.1.3) does not converges absolutely（绝对收敛）, we say the mean of X does not exist. The mean of continuous random variable has the similar properties as discrete random variable.If g(X) is an integrable function of a continuous random variable X, having density function f(x), mean of g(X) isprovided the integral converges absolutely.Example 4.6.4 Let X be a random variable having Cauchy distribution, the probability density function is given by(a) Find E(X); (b) Let Find .Solution (a) Since the integral diverges（发散）, E(X) does not exist.(b) 6. varianceSimilarly, the variance and standard deviation of a continuous random variable X is defined by, (4.1.4)Where is the mean of X, is referred to as the standard deviation.We easily get. (4.1.5)Example 4.1.5 Determining the mean and variance using the probability density functionWith reference to the example 4.1.3: find the mean and variance of the given probability density.Solution Performing the necessary integrations, we getand.4.2 Uniform Distribution 均匀分布The uniform distribution, with the parameters a and b, has probability density functionwhose graph is shown in Figure abxf(x)Figure 4.2.1 The uniform probability density in the interval (a, b)To proof .To find the distribution function.The distribution function of the uniform distribution isNote that all values of x from a and b are “equally likely”, in the sense that the probability that x lies in an interval of width entirely contained in the interval from a to b is equal to , regardless of the exact location of the interval. To find the mean and variance of the uniform distributionAnd EX2 = Thus, 4.5 Exponential Distribution 指数分布Definition 4.5.1 A continuous variable X has an exponential distribution with parameter , if its density function is given by (4.5.1)Many random variables, such as the life of automotive parts, life of animals, time period between two calls arrives to an office, having a distribution called exponential distribution.Note Equation (4.5.1) really gives a density function, sinceTheorem 4.5.1 The mean and variance of a continuous random variable X having exponential distribution with parameter is given by.Proof Since the probability density function of X is (4.5.1), we have Example 4.5.1. Assume that the life Y of bulbs produced by company X has exponential distribution with mean .(a) Find the probability that a bulb selected at random from the product of company X has life longer than 450 hrs. (b) Select 5 bulbs randomly from the product of company X, what is the probability that at least 3 of them has life longer than 450 hrs.Solution (a) (b) The number Z of bulb of bulbs with life longer than 450 hrs has the binomial distribution B(n, p), where n=5, p=0.2231. Thus * homeworkP66 4.1, 4.3，4.5, 4.23, 4.24，4.264.3 Normal Distribution 正态分布The normal probability density usually referred to simply as the normal distribution, is by far the most important. It was studied first in the eighteenth century when scientists observed an astonishing degree of regularity in errors of measurement. They found that the patterns (distributions) they observed were closely approximated by a continuous distribution which they referred to as the “normal curve of errors” and attributed to the laws of chance. The normal distribution is one of the frequently used distributions in both applied and theoretical probability.1. DefinitionThe equation of the normal probability density, whose graph is shown in Figure 4.3.1, isFigure 4.3.1 The normal probability densityThe distribution is characterized by two parameters, traditionally designated and . (.)To proof . 。proof。It is often convenient to designate the face that X is normal with parameters and by the notation 2. the property of the normal probability density (1)The probability mass is distributed symmetrically about the point . The parameter , which can be any real number, thus determines the center of the distribution.(2)The parameter gives an indication of how the probability is spread around the center of the distribution. 3. The parameters and are indeed its mean and its standard deviation.EX=, DX= .4. standard normal distributionthe normal distribution with and , the normal probability density is the normal distribution function is is an even function: , and . To find the probability that a random variable having the standard normal distribution will take on a value between a and b, we use the equation as shown by the shaded are in Figure 4.3.3.Figure 4.3.3 5. find the probability Example 4.3.1 Calculating some standard normal distributionLet , take on a value(a) between 0.34 and 0.62; (b) greater than 0.65Solution (a) From the Appendix B (b), we refer to the corresponding standardized random variable,then . ( since () () =hence Also, , the probability .Example , find Solution = 0.7734(10.9878)=0.7612Example 4.3.2The actual amount of instant coffee that a filling machine puts into “4-ounce” jars may be looked upon as a random variable having a normal distribution with ounces. If only 2% of the jars to contain less than 4 ounces, what should be the mean fill of these jars?Solution To find such that,we look for the entry in Appendix closest to 0.02 and get 0.0202 corresponding to z = -2.05. As indicated in Figure 4.3.4.solving for , we find that ounces.Figure 4.3.4 There are also problems in which we are given probabilities relating to standard normal distributions and asked to find the corresponding values of z.Let be such that the probability is a that it will be exceeded by a random variable having the standard normal distribution. That is, as illustrated in Figure 4.3.5.Figure 4.3.5 The notation for a standard normal distributionThe results of the example that follows will be used extensively in subsequent chapters.Example 4.3.3 Two important values for . Find (a) z0.01; (b) z0.05.Solution (a) Since (z0.01)=1-0.01=0.99, we look for the entry in appendix which is closest to 0.99 and get 0.9901 corresponding to z=2.33. Thus z0.01=2.33.(b) Since (z0.05)=1-0.05=0.95, we look for the entry in Appendix which is closest to 0.95 and get 0.9495 and 0.9505 corresponding to z=1.64 and z=1.65. Thus, by interpolation, z0.05=1.645. 6 Theory , find , , P=0.6826, P=0.9544, P=0.99744.4 Normal Approximation to the Binomial Distribution, n is large (n30), p is close to 0.50,Theorem 4.4.1 If X is a random variable having the binomial distribution with the parameters n and p, i.e. the limiting form of the distribution function of the standardized random variable as , is given by the standard normal distributiongraph-Example For an experiment in which 9 coins are tossed, Let X denotes the number of head occurrence, construct the probability distribution of X.Solution X0123456789PNP193684126126843691N=512 直方图与正态曲线非常接近，所以正态分布是二项概率分布的一个极好的逼近，当n足够大时.Example 4.4.1 A normal approximation to binomial probabilitiesIf 20% of the memory chips made in a certain plant are defective, what are the probabilities that in a lot of l00 randomly chosen for inspection(a) at most 15 will be defective; (b) exactly 15 will be defective?Solution Since and for the binomial distribution with n=100 and p=0.20, i.e. , we find that the normal approximation to the binomial distribution yieldsFor part (a),.For part (b),If we do the exact binomial calculation on a computer instead of using normal approximation, we would have obtained 0.1285 instead of 0.1292 for part (a) and 0.0481 instead of 0.0454 for part (b). That both approximations are very close.Use the normal approximation to the binomial distribution only when np and n(1-p) are both greater than 15.Homework 4.13，4.17, 4.19, 4.22, 4.6 Function of Random VariablesIf the value of the random variable X can be observed in a trial, we are interested in a corresponding value Y=g(X) obtained by applying the rule for the function g to the observed value.We address the basic problem: Given the distribution for X, how can we assign probability to events determined by the new random variable Y?Example 4.6.1 The function of a discrete random variableSuppose the probability distribution of a discrete random variable X as in the following table, and . Determine the probability distribution of the random variable Y and Z.x -2 -1 0 1 2P(X=x)0.15 0.20 0.20 0.25 0.20Solution P(X=x)0.15 0.20 0.20 0.25 0.20 X -2 -1 0 1 27 4 3 4 7Z=2X1 -5 -3 -1 1 3When the values of X are equal to -2, -1, 0, 1, 2, the random variable Y are equal to 77,4, 3,4,7 respectively. Then,Y3 4 7P(Y=y)0.20 0.45 0.35 Thus, we get And Z -5 -3 -1 1 3P(Z=z)0.15 0.20 0.20 0.25 0.20Example 4.6.2 Suppose the random variable Y=aX+b, with . Determine FY(y).Solution We need consider two cases.(1) then, implies (2) thus,implies *If X is continuous, so is Y. In this case, Differentiating FY to obtain the density yields two cases that can be combined by the use of the absolute value to obtain . we know that if , then the corresponding standardized random variable, let us give the proof again. here , , , using the result of example 4.6.2, we get .Example 4.6.3 Square Root FunctionSuppose the random variable , and . Determine the distribution for Y.Solution The function is increasing on. Now iff . Thus for , , and if X is continuous, thus,. 4.7 Chebyshevs TheoremEarlier in this chapter we used examples to show how the standard measures the variation of a probability distribution, that is, how it reflects the concentration of probability in the neighborhood of the mean. If is large there is a correspondingly higher probability of getting values farther away mean. Formally, the idea is expressed by the following theorem.Theorem 4.7.1 If a probability distribution has mean and standard deviation , the probability of getting a value which deviates from by at least k is at most . Symbolically , .where is the probability associated with the set of outcomes for which x ,the value of a random variable having the given probability distribution, is such that .Then, the probability that a random variable will take on a value which deviates from the mean by at least 2 standard deviations is at most 1/4, the probability that it will take on a value which deviates from the mean by at least 6 standard deviations is at most 1/36.deviates from the mean the probability that a random variable k=2 P1/4k=6 P1/36Proof As an example, we consider the case that X is a continuous variable with probability density f(x). =So we have.-k+kFigure 4.7.1 Proof of Chebyshevs theoremxThis completes the proof of Theorem 4.7.1.To get an alternative form of Chebyshevs theorem, note that the is the complement of the event ; thus, . 对比正态分布的3定理。 Example 4.7.1 A probability bound using Chebyshevs theoremThe number of customers who visit a car dealers showroom on a Saturday morning is a random variable with =18 and =2.5. With what probability can we assert that there will be more than 8 but fewer than 28 customers?Solution Let X be the numbe