概率论与数理统计(英文) 第四章

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 person's 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 Variable 1. 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) f(x)0 for any xR; (ii) f(x) is intergrable (可积的) on (, )and f(x)dx1. Definition 4.1.2 Let f(x) be a probability density function. If X is a random variable having distribution function xF(x)P(Xx)f(t)dt, (4.1.1) then X is called a continuous random variable having density function f(x). In this case, x2P(x1Xx2)x1f(t)dt. (4.1.2) 2. 几何意义 xF(x)P(Xx)P((X,Y)|Xx, 0Yf(X))f(t)dt 46 x2 3. Note P(x1Xx2)x1f(t)dt 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, P(Xx)0. 0P(Xx)0P(Xx)xxxxf(x)dx f(x)dx0 lim0P(aXb)P(aXb)P(aXb)P(aXb) 4. Example Example 4.1.2 Find k so that the following can serve as the probability density of a continuous random variable: f(x)k1x2 (x) Solution To satisfy the conditions (4.1.1), k must be nonnegative, and to satisfy the condition (4.1.2) we must have f(x)dx1+xk2dxk1 so that k1. (Cauchy distribution 柯西分布) □ Example 4.1.3 Calculating probabilities from the probability density function If a random variable has the probability density 47 3e3x for x0 f(x)0 for x0Find 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 2(a) P(0x2)3e3xdx1e60.9975 0(b) P(x1)3e13xdxe30.0498 Example 4.1.4 Determining the distribution function of X, it is known 3e3x for x0 f(x) 0 for x0Solution Performing the necessary integrations, we get 0 for x0F(x)x3t 3x3edt1e for x00 P(x1)F(1)1e3 5 0 2 □ 0.95. mean If the integral (4.1.3) does not converges absolutely（绝对收敛）, we say the mean of X does not exist. Definition 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 E(X)xf(x)dx, (4.1.3) 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) is E(g(X))g(x)f(x)dx provided the integral converges absolutely. Example 4.6.4 48 Let X be a random variable having Cauchy distribution, the probability density function is given by f(x)1, x (1x)2(a) Find E(X); (b) Let X, 0X1 g(X)0, elsewhereFind E(g(X)). Solution (a) Since the integral (1x)|x|2dx diverges（发散）, E(X) does not exist. 1(b) E(g(X))6. variance g(x)f(x)dx(1x0x2)dxln22. Similarly, the variance and standard deviation of a continuous random variable X is defined by 2D(X)E((X)), (4.1.4) 2Where E(X)is the mean of X,  is referred to as the standard deviation. We easily get 2D(X)xf(x)dx. (4.1.5) 22Example 4.1.5 Determining the mean and variance using the probability density function 3e3x for x0With reference to the example 4.1.3: f(x) 0 for x0find the mean and variance of the given probability density. Solution Performing the necessary integrations, we get and xf(x)dxx3e03xdx13 2(x)f(x)dx2(x013)3e3xdx19. 4.2 Uniform Distribution 均匀分布 49 The uniform distribution, with the parameters a and b, has probability density function 1 for axb, f(x)ba0 elsewhere, whose graph is shown in Figure 4.2.1. f(x) To proof 0 1ba a b x Figure 4.2.1 The uniform probability density in the interval (a, b) f(x)dx1. To find the distribution function. The distribution function of the uniform distribution is 0 for xa,xaF(x) for axb, ba1 for xb. Note 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 xentirely contained in the interval from a to b is equal to x/(ba), regardless of the exact location of the interval. To find the mean and variance of the uniform distribution bax1badxab2 And b EX= 2 xa21badxaabb322 Thus, 22222aabb3(ba)ab 212 50 4.5 Exponential Distribution 指数分布 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. Definition 4.5.1 A continuous variable X has an exponential distribution with parameter (0), if its density function is given by x1e for x0 (4.5.1) f(x)0 for x0 Note Equation (4.5.1) really gives a density function, since xe01dx1. Theorem 4.5.1 The mean and variance of a continuous random variable X having exponential distribution with parameter is given by E(X), D(X). 2Proof Since the probability density function of X is (4.5.1), we have E(X)xf(x)dxxe01xdx. E(X)2xf(x)dx2x02212exdx2. 2D(X)E(X)[E(X)]. 2 Example 4.5.1. Assume that the life Y of bulbs produced by company X has exponential distribution with mean 300(hrs). (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) 51 P(Y450)1P(Y450)450101300ex300dxe1.50.2231 (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 P(Z3)C50.22310.7769C50.22310.77690.22310.0772. 332445****** homework P66 4.1, 4.3，4.5, 4.23, 4.24，4.26 4.3 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. Definition The equation of the normal probability density, whose graph is shown in Figure 4.3.1, is f(x)12e(x)/222 x (0.) To proof f(x)0. Figure 4.3.1 The normal probability density The distribution is characterized by two parameters, traditionally designated  and . f(x)dx1. 。。。proof。。。 It is often convenient to designate the face that X is normal with parameters  and  by the notation X~N(,) 2. the property of the normal probability density (1)The probability mass is distributed symmetrically about the point t. The parameter 52 2, 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= 2. 4. standard normal distribution the normal distribution with 0 and 1, the normal probability density is (x)12ex22 the normal distribution function is z(z)P(Zz)12ex/22dx (x) is an even function: (x)(x), and (z)1(z). 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 P(aZb)(b)(a) as shown by the shaded are in Figure 4.3.3. Figure 4.3.3 P(aZb) 53 5. find the probability Example 4.3.1 Calculating some standard normal distribution Let Z~N(0,1), take on a value (a) between －0.34 and 0.62; (b) greater than －0.65 Solution (a) From the Appendix B P(0.34Z0.62)(0.62)(0.34)(0.62)[1(0.34)]0.732410.63310.3655. (b) P(Z0.65)1(0,65)(0.65)0.7422. X~N(, ), 2we refer to the corresponding standardized random variable, XZ X~N(0,1). then Z ( since Xz)P(z)P(Xz ) Fz(z)P(Z z1212et12(t)222edt (t1tt) z2d1t (t1,dt11dt) =(z) hence ZX~N(0,1) x)P(Zx)(x)  P(Xx)P( XAlso, X~N(, 2), the probability P(aXb)P(aZbba).  54 Example X~N(1.5,, find P(X3) 431.52) Solution P(X3)P(3X3)P =P(31.52ZX31.5 231.52=(0.75)(2.25)(0.75)(1(2.25)) = 0.7734－(1－0.9878)=0.7612 Example 4.3.2 The 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 0.04ounces. 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 P(X4)P(Z40.04)(40.04)0.02, 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. 40.042.05 solving for , we find that 4.082 ounces. Figure 4.3.4 P(Z2.05) 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 z be such that the probability is a that it will be exceeded by a random variable having the standard normal distribution. That is, P(Zz) as illustrated in Figure 4.3.5. 55 The results of the example that follows will be used extensively in subsequent chapters. Example 4.3.3 Two important values for z. Find (a) z0.01; (b) z0.05. Solution (a) Since Ф(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 Ф(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. and z=1.65. Thus, by interpolation, z0.05=1.5. 6． 3 Theory X~N(, 2 Figure 4.3.5 The z notation for a standard normal distribution , find )P(X), P(2X2, ) P(3X3 ) P=0.6826, P=0.9544, P=0.9974 4.4 Normal Approximation to the Binomial Distribution X~B(n, p), n is large (n>30), p is close to 0.50, X~B(n, p)N(np,npq) Theorem 4.4.1 If X is a random variable having the binomial distribution with the parameters n 56 and p, i.e. X~B(n, p)the limiting form of the distribution function of the standardized random variable ZXnpnp(1p) as n, is given by the standard normal distribution z(z)12et/22dt z graph---- 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 X P NP 0 15121 95122 365123 845124 1265125 1265126 845127 365128 95129 1512 1 9 36 84 126 126 84 36 9 1 N=512 57 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 probabilities If 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 1000.2020and 1000.200.804for the binomial distribution with n=100 and p=0.20, i.e. X~B(100, 0.20), we find that the normal approximation to the binomial distribution yields For part (a), P(X15)P(X15.5)P(Z15.5204)(1.13)0.1292. For part (b), P(X15)P(14.5X15.5)P(14.5204Z15.5204) (1.13)(1.38)0.12920.08380.0454.If we do the exact binomial calculation on a computer instead of using normal approximation, 58 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 Variables If 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 variable Suppose the probability distribution of a discrete random variable X as in the following table, and YX23. Z2X1Determine the probability distribution of the random variable Y and Z. 2 1 0 1 2 P(X=x) 0.15 0.20 0.20 0.25 0.20 Solution P(X=x) X 0.15 0.20 0.20 0.25 0.20 2 1 0 1 2 x 2YX3 7 4 3 4 7 Z=2X－1 5 3 1 1 3 When the values of X are equal to 2, 1, 0, 1, 2, the random variable Y are equal to 7,4, 3,4,7 respectively. Then, P(Y3)P(X0)0.20, P(Y4)P(X1)P(X1)0.200.250.45, P(Y7)P(X2)P(X2)0.150.200.35, Thus, we get Y 3 4 7 P(Y=y) 0.20 0.45 0.35 And Z 5   1 3 59 P(Z=z) 0.15 0.20 0.20 0.25 0.20 Example 4.6.2 Suppose the random variable Y=aX+b, with a0. Determine FY(y). Solution We need consider two cases. (1) a0:Yy iff Xthen, P(Yy)P(aXby)P(Xyba) yba, implies FY(y)FX(yba). (2) a0:Yy iff Xthus, yba, P(Yy)P(aXby)P(Xyba)1P(Xyba), implies FY(y)1FX(yba)P(Xyba). ****** If X is continuous, so is Y. In this case, P(Xyba)0. Differentiating FY to obtain the density yields two cases that can be combined by the use of the absolute value to obtain fY(y)yb. fX() □ |a|a21we know that if X~N(, ), then the corresponding standardized random variable, ZX~N(0,1), let us give the proof again. 1,b here a, Y(x)2221X , fX(x)12fX(e, using the result of example 4.6.2, we get y2 fY(y) 1|a|yba)12e(y)/22212e2. Example 4.6.3 Square Root Function Suppose the random variable YX, and X0. Determine the distribution for Y. 60 Solution The function g(x)x is increasing on[0,). Now Xy iff Xy. 2Thus for y0, FY(y)P(Yy)P(Xy2)FX(y2), and if X is continuous, thus, dFY(y)dydFX(y)dy2fY(y)2yfX(y). 2 4.7 Chebyshev’s Theorem Earlier 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 σ 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 μ and standard deviation σ, the probability of getting a value which deviates from μ by at least kσ is at most P(|X|k)1k21k2. Symbolically , . where P(|X|k) 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 |X|k. 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 P≤1/4 k=6 P≤1/36 Proof As an example, we consider the case that X is a continuous variable with probability density f(x). P{|X|k}(x)k222|x|k1f(x)dx|x|kf(x)dx =1k22|x|k(x)f(x)dx21k22(x)f(x)dx21k2221k2 So we have P(|X|k)1k2. 61 μ-kσ μ μ+kσ x Figure 4.7.1 Proof of Chebyshev’s theorem This completes the proof of Theorem 4.7.1. To get an alternative form of Chebyshev’s theorem, note that the |x|k is the complement of the event |x|k; thus, P(|x|k)1 对比正态分布的3σ定理。 Example 4.7.1 A probability bound using Chebyshev’s theorem The number of customers who visit a car dealer’s showroom on a Saturday morning is a random variable with μ =18 and σ =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 number of customers. Since P(8X28)P(818X2818P)X(|1k2. / k10,k10P(|X|k)11k210/2., 5 and P(8X28)11421516. The most important feature of Chebyshe’s theorem is that it applies to any probability distribution for which μ and σ exist. 切比雪夫定理要求随机变量的期望和方差存在 However, it provides only an upper limit （上限）to the probability of getting a value that deviates from the mean by at least k standard deviations. For instance, we can assert in general that the probability P(|X|2)0.25, When a random variable X~B(16,0.50), 160.508,160.500.502, the value of probability which differs from the mean by at least 2 standard deviations is 62 P(|X|2)P(|X8|4)1P(|X8|4)111Ck5k160.51610.92320.0768. “At most 0.25” does not tell us that the actual probability may be as small as 0.0768. 问题思考 在P(|X|k) Homework jsj P69 27, 28, 29，30 4.29 Over the range of cylindrical parts manufactured on a computer controlled lathe, the standard deviation of the diameters is 0.002 mm. (a) What does Chebyshev’s theorem tell us about the probability that a new part will be within 0.006 units of the mean μ for that run? (b) If the 400 parts are made during the run, about what proportion do you expect will lie in the interval in Part (a)? 1k2，问P(X)=？ 例3.按规定，某车站每天8：00-9：00，9：00-10：00都恰有一辆客车到站，但到站的时刻是随机的，且两者到站的时间相互. 其规律为 8：10 8：30 8：50 到站时间 9：10 9：30 9：50 概率 16 36 26 一旅客8：20到车站，求他候车时间的数学期望. 解 设旅客的候车时间为X（以分计）. X 的分布律为 X 10 30 50 70 90 pk 3 2 11 13 12 66666666在上表中， P{X70}P(AB)P(A)P(B)1636, 其中A为事件“第一班车在8：10到站”，B 为“第二班车在9：30到站”. 候车时间的数学期望为 63 E(X)1036+3026+ 50136+ 70336+ 90236=27.22（分）. 某商店对某种家用电器的销售采用先使用后付款的方式. 记使用寿命为X（以年计），规定： 例4 X ≤1， 一台付款1500元； 1X≤2 ,一台付款2000元； 2X≤3,一台付款2500元；X3，一台付款3000元. 设寿命X服从指数分布，概率密度为 x110e, x0 f(x)10 0 , x≤0试求该商店卖出一台家电的平均收费（Y元）. 解 先求出寿命X落在各个时间区间的概率，即有 1P{X≤1}1001ex/10dx1e0.10.0952, P{1X≤2}211011ex10dxe0.2e0.30.0861, P{2X≤3}32103ex/10dxe0.2e0.30.0779, P{X3}110ex10dxe0.30.07408. 一台收费Y的分布律为 X pk 1500 0.0952 2000 0.0861 2500 0.0779 3000 0.7408 得E(X)2732.15,即平均一台收费2732.15元. □