概率论与数理统计(英文) 第五章.doc

5. Random vectors and Joint Probability Distributions 随机向量与联合概率分布5.1 Concept of Joint Probability Distributions(1) Discrete Variables Case 离散型Often, trials are conducted where two random variables are observed simultaneously in order to determine not only their individual behavior but also the degree of relationship between them.( X, Y)For two discrete random variables X and Y, we write the probability that X will take the value x and Y will take the value y as P(X=x, Y=y). Consequently, P(X=x, Y=y) is the probability of the intersection of the events X=x and Y=y. (X=x, Y=y) - (X=x)(Y=y)The distribution of probability is specified by listing the probabilities associated with all possible pairs of values x and y, either by formula or in a table. We refer to the function p(x, y)= P(X=x, Y=y) and the corresponding possible values (X, Y) as the joint probability distribution (联合分布)of X and Y. XYXy1 y2yjx1p11p12p1jx2p21p22p2jXipi1pi2pij They satisfy, where the sum is over all possible values of the variable.Example 5.1.1 Calculating probabilities from a discrete joint probability distributionLet X and Y have the joint probability distribution.X Y0 1 0120.1 0.2 0.4 0.2 0.1 0(a) Find ;(b) Find the probability distribution of the individual random variable X.Solution (a) The event is composed of the pairs of values (l,1), (2,0), and (2,l). Adding their corresponding probabilities(b) Since the event X=0 is composed of the two pairs of values (0,0) and (0,1), we add their corresponding probabilities to obtain.Continuing, we obtain and .In summary, , and is the probability distribution of X. Note that the probability distribution of appears in the lower margin of this enlarged table. The probability distribution of Y appears in the right-hand margin of the table. Consequently, the individual distributions are called marginal probability distributions.（边缘分布）X Y0 1 pX(x)0120.1 0.2 0.4 0.2 0.1 0pY(y)0.6 0.4 1.0From the example, we see that for each fixed value of x, the marginal probability distribution is obtained as,where the sum is over all possible values of the second variable. Continuing, we obtain.Example 3.5.3 Suppose the number of patent applications （专利申请）submitted by a company during a 1-year period is a random variable having the Poisson distribution with mean , ()and the various applications independently have probability of eventually being approved. Determine the distribution of the number of patent applications during the 1-year period that are eventually approved.先求联合分布密度，再求边缘分布Solution Let be the number of patent application being eventually approved during 1-year period. Then the event is the union of mutually exclusive events .If , then the random variable has the binomial distribution with parameter and : . Thus when kn, P(X=n, Y=k)=0, YX0k 0 0 0 1 n Hence the distribution of is Thus, has the Poisson distribution of mean .exercise从1，2，3，4，5五个数中不放回随机的接连地取3个，然后按大小排成，试求的联合分布，x1,x3 独立吗？ X31X13345 1 1/102/103/102 01/102/103 0 01/10HomeworkChap 5 1, (2) Continuous Variables Case 连续型随机向量There are many situations in which we describe an outcome by giving the values of several continuous random variables. For instance, we may measure the weight and the hardness of a rock, the pressure and the temperature of a gas. Suppose that X and Y are two continuous random variables. A function is called the joint probability density of these random variables, if the probability that is given by the multiple integralThus, a function can serve as a joint probability density if all of the following hold:for all values of x and y, f is integrable on R2 and .To extend the concept of a cumulative distribution function to the two variables case, we can define F(x, y), and we refer to the corresponding function F as the joint cumulative distribution function of the two random variables.（1）0（2）（3）若的分布函数F(x，y)在点（x, y）的某领域内存在且在点（x, y）处连续，则=（4）PD=, D为xoy平面内任一区域。Example 5.1.2 If the joint probability density of two random variables is given byFind the joint distribution function, and use it to find the probability .Solution By definition,Thus,.Hence,. ExampleIf the joint probability density of two random variables is given by (a)find the k; (b)find the probability solution since hence k=6. joint marginal densities 边缘密度Given the joint probability density of two random variables, the probability density of the X or Y can be obtained by integrating out another variable,.The functions fX and fY respectively are called the marginal density （边缘密度）of X and Y., Example The joint probability density of two random variables is given by find the marginal density from the joint densitywhen，hence exercises 求服从B上均匀分布的随机向量（X,Y）的分布密度及分布函数。其中B为x轴、y轴及直线y=2x+1所围成的三角形区域。If the pair (X, Y) has the joint probability density we call the pair (X,Y) has joint uniform distribution on the region D./*5.2 Conditional distribution 条件分布 Consistent with the definition of conditional probability of events when A is the event X=x and B is the event Y=y, the conditional probability distribution of X given Y=y is defined as for all x provided .Example 5.2.1 A conditional probability distributionWith reference to the example 5.1.1, find the conditional probability distribution of X given Y=l. X Y0 1 pX(x)0120.1 0.2 0.4 0.2 0.1 0pY(y)0.6 0.41Solution , and . Given two continuous random variables X and Y, we shall define the conditional probability density of the ant given that the second takes on the value y as,as before, where f(x,y) and fY(y) are the joint density of the two random variables and the marginal density of the second. Note that this definition parallels that of conditional probability.Example 5.2.2 Determining a conditional probability densityIf two random variables have the joint probability densityfind the conditional density of the first given that the second takes on the value y.Solution First we and the marginal density of the second random variab1e by integrating out x1, and we getand fY(y)=0 elsewhere. Hence, by definition, the conditional density of the first random variable given that the second takes on the value y is given byand for or and 0 y 0, y0, for elsewhere. So it can be seen that for all x and y.Then the random variables X and Y are independent. Homework 1, 7，8(1)(4), 9，11, 29 P84 5, 7，8(1)(4), 9, 10, 11随机向量的函数(P75)Given any two random variables X and Y, the function Z=g(X, Y) is also a random variable. 补充例例1 设的分布密度为 YX-112-15/202/206/2023/203/201/20求X+Y、XY的分布密度解 由的分布密度可得下表 P5/202/206/203/203/201/30(-1,-1)(-1, 1)(-1, 2)( 2,-1)( 2, 1)( 2, 2)X+Y-201134XY1-1-2-224从而得到(1) X+Y的分布密度为X+Y-20134 P5/202/209/203/201/20(2) XY的分布密度为XY-2 -1124 P9/20 2/105/203/201/20例2及的分布 设是两个相互独立的随机变量,它们的分布函数分别为和.现在来求及的分布函数.由于不大于等价与和不大于，故有.又由于和相互独立，得到的分布函数为即有 . 类似地，可得到的分布函数为.即 . 例3有2个相互独立工作的电子装置,它们的寿命 服从同一指数分布,其概率密度为 若将这2个电子装置串联联接组成整机,求整机寿命(以小时计)N的数学期望.解 的分布函数为由第三章5(5.8)式的分布函数为因而N的概率密度为于是N的数学期望为.Z=g(X, Y)The random variable Z has mean or expected value, in the continuous case, given by E(Z)=g(x, y) f(x, y)dxdy,In the discrete case, E(Z)= g(x, y) p(x, y)5.4 Covariance and Correlation 协方差和相关系数We now define two related quantities whose role in characterizing the interdependence of X and Y we want to examine.Definition 5.4.1 Suppose X and Y are random variables. The covariance of the pair X,Y is .The correlation coefficient of the pair X, Y is.Where Apparently, we have , and .When the X and Y are independent, , thus .Definition 5.4.2 The random variables X and Y are said to be uncorrelated iff . Then, we get the theorem 5.4.1.Theorem 5.4.1 If random variables X and Y are independent, they are uncorrelated.Theorem 5.4.2 Let is the correlation coefficient of the pair X,Y, then(1) ; (2) iff there are two constant a, b such that .Proof (1) We use Schwarzs inequality for expectation to assert.(2) , then.So .If , then .By Schwarzs inequality, there is a constant b such that Since , this means i.e. Example 5.4.1 Suppose the.Determine Cov(X, Y) and .Solution ,then, Example 5.4.2 Suppose two random variables X and Y are independent, . Let , where a, b are constants. Find the Cov(U,V) and (U,V).Solution Apparently, , so ,. Because X and Y are independent, , then ,then,. 5.5 Law of Large Numbers and Central Limit Theorem 中心极限定理We can find the steadily of the frequency of the events in large number of random phenomenon. And the average of large number of random variables are also steadiness. These results are the law of large numbers.Theorem 5.5.1 If a sequence of random variables is independent, with then. (5.5.1)Proof Since by Chebyshevs theorem() ）, we get ,thus,. Theorem 5.5.2 Let equals the number of the event A in n Bernoulli trials, and p is the probability of the event A on any one Bernoulli trial, then. (5.5.2)(频率具有稳定性)Proof Let then Since by theorem 5.5.1, . This theorem proved the steadily of the relative frequency of the events in large number of random trials. Therefore, in application, while the trials times n is a large number, we can use the frequency to replace the probability of a event. （此理论证明了频率的稳定性）Theorem 5.5.1 and 5.5.2 are known as the weak law of large numbers（弱大数定理）, or the law of averages. There is a companion to the weak law that is referred to the strong law of the large numbers. Under the same assumption of Theorem 5.5.1, we have. (5.5.3)In words, (5.5.3) states that with probability 1, the sample mean converge to the true mean as the sample size tends to infinity. Similarly, (5.5.2) can be enhanced to the following formula: . (5.5.4)This means that the relative frequency nA/n of successes converges to the probability p of successes with probability 1, i.e. nA/n almost always converges to p. This will help us to understand the precise meaning of Definition 2.3.1. （频率收敛于概率）Example 5.5.1 Show that for 40,000 flips of a balanced coin, the probability is at least 0.99 that the proportion of heads will fall between 0.475 and 0.525。（需证：，n=40,000, p=1/2, XB(n, p)）Solution Since and= from thus 100k=(0.025)(40,000), k=10from the alternative form of Chebyshevs theorem ,i.e. .Correspondingly, for 1000000 flips of a balanced coin the probability is at least 0.99 that the proportion of heads will fall between 0.495 and 0.505, thenThese results suggest that when n is large, the chances are that the proportion of heads will be very close to . When formulated for any binomial distribution with the parameters n and p, this result is the law of large numbers. The central limit theorem (CLT) 中心极限定理asserts that if a random variable X is sum of a large number of independent random variables, each with reasonable distributions, then X is approximately normally distributed. This celebrated theorem has been the object of prodigious research effort directed toward the discovery of the most general conditions under which it is valid. On the other hand this theorem serves as the basis of an extraordinary amount of applied work. In the statistics of large samples, the sample average is a constant times the sum of the random variables in the sampling process.Thus, for large samples, the sample average is approximately normal, whether or not the population distribution is normal. In much of the theory of errors of measurement, the observed error is the sum of a large number of independent random quantities that contribute to the result. Similarly, in the theory of noise, the noise signal is the sum of a large number of random components, independently produced. In such situation