工程数学(三)概率统计 离散数学（Engineering Mathematics (three) discrete mathematics of probability and statistics）

工程数学(三)概率统计 离散数学（Engineering Mathematics (three) discrete mathematics of probability and statistics）The third chapter random variable and its distributionIf we care about the event A=no defective, B=has at least 2 defective, C=no more than k defective, then A, B, C can be expressed as random variable Y respectivelyA=e|Y (E) =0, B=e|Y (E) = 2, C=e|Y = k. (E)For the sake of convenience, generally in the event that can save e, so it can be expressed as B=Y = 2 A=Y=0 exergy exergy, exergy exergy value, C=Y = k. random variables varies with the test results, it can not predict what value before the experiment, and it has a certain value for the random probability variable and ordinary function is essentially different.The introduction of random variables enables us to use random variables to describe various stochastic phenomena, so it is possible to study the results of random trials by calculusThe distribution function of 2. random variablesDefinition 3.2 let X be a random variable, X is the function of any real number and the function value on 0, 1F (x) =P (X = x), - - <x<+ -It is called the distribution function of random variable XFor arbitrary x 1, x 2 (x 1<x 2, a)P (x 1<X x = 2) =P (X = x 2) -P (X = x =F (1) x (x 2) -F 1 (3.1).Therefore, if the distribution function of the known X, X know any fall in the interval (x 1, 2 probability on the X in this sense, the distribution function of a complete description of the statistical regularity of random variables.The distribution function is a general function, and it can be used to study random variables by calculusIf the X is regarded as the axis of random point coordinates, then the distribution function F (x) x in the function value at that X falls in the range of X (probability on the infty,.3.3 cases of a random variable X of all possible values for X 1=-1, X 2=2, X 3=3. event X = x i (i=1,2,3) probability for exergy exergy P (X = -1 = 1 = 14) p,P (X = 2) = P 2 = 12,P (X = 3) = P 3 = 14, exergy exergy distribution function for X, and for PX = 12, P32<X = 52, P2 = X = 3.X only in X=-1, 2, 3 and three probability is not equal to 0, and F (x) is the value of the cumulative probability of X is less than or equal to the value of X, the probability of finite additivity of knowledge, it is less than or equal to X and the probability of P K x k. Therefore thereF (x) =0, x<-1,P (X=-1), -1 = x<2,P (X=-1) +P (X=2), 2 x<3,1, x = 3,That isGraph 3.1F (x) =0, x<-1,14, -1 = x<2,34, 2 = x<3,1, x = 3.The graph of F (x) is shown in Figure 3.1. It is a ladder shaped curve. There are jumping points at x=-1,2,3, and the jumping values are 14,12,14.alsoPX = 12=F12=14,P32<X = 52=F52-F32=34-14=12.P2 = X = 3=F (3) -F (2) +P (X=2) =1-34+12=34.From the distribution function and its graph of case 3.3, we can see that the distribution function has the property of right continuity and monotonicity, and we point out that the distribution function F (x) has the following basic properties:(1) 0 F (x) = 1 (- infty <x<+);(2) F (x) is monotone decreasing function, i.e. for any two X 1, x 2, when x 1<x 2 F (x 1) = F (x 2);(3) f Lim F X - infty F (x) =0, Lim x, K + - K F (x) =1;(4) Lim x, X r r + share 0F (x) =F (x (0) - <x - 0<+ -), any distribution function is a right continuous function.To prove (1) by F (x) =P (X = x), the nature of knowledge by probability 0 = F (x) = 1.(2) for any two X 1, x 2, when x 1<x 2, X 1 X = x = x 2. so the probability properties of 3.2P (X = x 1) = P (X = x 2), F (x 1) = F (x 2).For (3) we only illustrate from geometry. In Figure 3.2, X moves along the axis of the infinite interval endpoints (i.e. x, left - infinity), then random point X falls in the X left this event to impossible event, so the probability tends to 0, namely F (- 2) =0; and if the point x no limit right (x, + infinity),Then the random point X falls on the left side of X this event tends to a certain event, so its probability tends to 1, that is, F (+ +) =1.(4) the proof is omitted.3.2 discrete random variablesThere are some random variables, and all of them may take different values which are finite or countable infinite. Such random variable is called discrete random variable, and its distribution is called discrete distributionFor example, the city received 120 emergency telephone call is a circadian number of discrete random variable. If T remember certain elements of the life, it may take the value of a full range, is unable to enumerate them, so it is not a discrete random variable.To grasp the statistical law of a discrete random variable X, it is necessary to know only all the possible values of X and the probability of taking each possible valueThe distribution law of 1. discrete random variablesThe discrete distribution is expressed in terms of the distribution law defined belowA discrete random variable X the value of x k (k=1,2,. X=x i), probability is p I (i=1,2,. ), i.e.P (X=x I =p) I, i=1,2,. (3.2).We call the type (3.2) for the distribution of discrete random variables X. Distribution law can also be used to represent the form (see Table 3.1) table 2: 3.1Xx 1x. X n. P IP 1p 2. P n. The table above expressed.X probability of random variable X values from each value each accounted for some probability, the probability is 1. and we think of it as a probability of 1 with a certain distribution in all possible values, this is called the reason. By the probability distribution law of the definition of P I meet the following two conditions:(1) P (i=1,2 I = 0,. ).(2) - i=1p i=1. exergy exergy Sigma(2) because the X=x i u X=x 2 u. Is an inevitable event, and the X=x i X=x j=, I and j, the 1=P, i=1X=x i= Exergy for Sigma - i=1P (X=x I), namely i=1p i=1. Sigma for exergy3.4 cases of a car on the way to its destination through four groups of lights, each signal with probability 12. To allow or prohibit the car through the car stopped when X said for the first time, it has passed the signal lamp group number (with each signal lamp working independently of each other), distribution law for X.The solution expressed in P of each signal lamp prohibition car through the probability distribution law, namely X as shown in the following table: X01234p IP (1-p) P (1-p) 2p (1-p) 3P share share share (1-p) can be expressed in 4:P (X=i) = (1-p) IP share, i=0,1,2,3,P (X=4) = (1-p) 4. shareThe p=12 X01234p i0.50.250.1250.062 into 50.062 5 cases 3.5 bags in 5 balls, numbered 1, 2, 3, 4, 5, and 3 out of the ball from the X, to represent the minimum number out of the ball, and the distribution law of distribution function for X.解 由于X表示取出的3个球中的最小号码，因此X的所有可能取值为1, 2, 3, X=1表示3个球中的最小号码为1，那么另外两个球可在2, 3, 4, 5中任取2个，这样的可能取法有C24种；X=2表示3个球中的最小号码为2，那么另外两个球可在3, 4, 5中任取2个，这样的可能取法有C23种；X=3表示3个球中的最小号码为3，那么另外两个球只能是4与5，即此时只有一种取法，而在5个球中任取3个的所有可能取法共有C35种.由古典概率定义得 P (X=1) =C24C35=35, P (X=2) =C23C35=310, P (X=3) =1C35=110.Therefore, the distribution law of demand as shown in the following table: X123p i0.60.30.1 distribution function for X F (x).When x<1, X or x into the impossible event, so F (x) =0;When 1 = x<2, X = x=X=1, F (x) =P (X=1) =0.6.When 2 = x<3, X = x=X=1 or X=2, soF (x) =P (X=1) +P (X=2) =0.6+0.3=0.9.When x = 3,X = x for certain events, so the F (x) =1.ComprehensiveF (x) =0, x<1,0.6, 1 = x<2,0.9, 2 = x<3,1, x = 3.Example 3.5 shows that if we know the distribution law of discrete random variables, we can get the distribution function, and on the contrary, is it feasible? If we know the distribution function of discrete random variable, can we get the distribution law of random variable? Let's take example 3.5 as an exampleP (X=1) =P (x 1) =F (1) =0.6;P (X=2) =P (1<X 2) =F (2) -F (1) =0.9-0.6=0.3;P (X=3) =P (2<X 3) =F (3) -F (2) =1-0.9=0.1.The distribution law of X is: X123p i0.60.30.1 we see from the above analysis and the distribution law of distribution function is equivalent to the value of the description of discrete variables. Of course, for discrete random variables, using the distribution law to describe the value of law than the distribution function is convenient and intuitive.2. commonly used discrete distributionThree important discrete random variables are introduced below(1) (0-1) distributionThe random variable X only takes 0 and 1 two values, and its distribution law isP (X=k) =p share K share 1-k (1-p), k=0,1, 0<p<1,The distribution of X is called (0-1) distribution or two point distribution(0-1) distribution law distribution see table 3.2. table 3.2X01 1-PP general probability in a random trial while the results can be many, but if we are only concerned with some properties of the results, it can be divided into A and the sample space, and A, the definition of X=1; when the definition of X= 0, then X distribution is the distribution (0-1).For example, the registration of newborn infant sex, check whether the product is qualified, the power consumption of a factory is overloaded and the front repeatedly discussed "toss" test (0-1) can be used to describe the distribution of random variables. (0-1) distribution is a distribution often encountered.2) two item distributionThere are only two possible results for the trial E: A and E are called Bernoulli trialsLet P (A) =p (0<p<1), then P () =1-p. repeat E independently n times, then call this repeated independent test n heavy Bernoulli testHere the "repeat" refers to each test in P (A) =p remained unchanged; "independent" refers to the time the test results do not affect each other, if the C I I test results, C I for A or i=1,2,. "N." means "independence":P (C 1C 2. C n =P (C) P (1) C 2). P (C n) (3.3).In the N heavy Bernoulli test, if the random variable X is used to represent the number of events A occurring in the n test, the X may take the values of 0, 1, 2,. N, and by the two probability probability X K value for exergy exergy of P (X=k) = f C f k NP K share share share (1-p) n-k, k=0,1,2,. N., (3.4) the distribution of exergy exergy, X table 3.3. table 3.3X01. K. NP I (1-p) n C 1 f f share share share NP (1-p) n-1. F C f k NP K share share share (1-p) n-k. P Share N called the discrete distribution parameters for N, two of the distribution of P, denoted by XB (n, P), where 0<p<1, p=P (A).In probability theory, the two distribution is an important distribution of many random phenomena can be used to describe the distribution of two. For example in the defective rate is P in a batch of products back to any n product, with X representing the number of defective products out of the N in X. Obey the parameters N, P two B (n, P) distribution; if the batch batch is used, no back from n products, can also be considered to obey the X parameters N, P of the two distribution of B (n, P).A type of 3.6 cases according to the regulations, the service life of electronic components more than 1 500 h as a product, a large number of products known to the first grade product rate of 0.2, from a random sample of 20 only, ask the 20 element is only K (k=0,1,2,. (20) what is the probability of a grade?This is not put back sampling, but because the total number of these components is large, and the number of components checked is relatively small compared with the total number of components,It can be used as a replacement to deal with, so there will be errors but not error. We put a check element to see whether it is a product as a test, check the 20 element equivalent to 20. Only a few repeated trials with X a grade 20 element, X is a random variable, and the XB (20, 0.2).Therefore, the probability obtained by the formula (3.4) is immediatelyP (X=k) = f C f 20 (0.2) share K share K share 20-k, k=0,1,2 (0.8),. 20.In order to have an intuitive understanding of the results of this question, we make a diagram of the table above (Fig. 3.3)To see from the figure, when k increases, the probability of P (X=k) is the first increase until they reach a maximum value (in this case, when k=4 get the maximum value), then decreases monotonically. We pointed out that the general for N and P fixed, two distribution B (n, P) are have this property.In 3.7 cases with 80 sets of the same type of machine, each working independently of each other, the probability of failure is 0.01, and the failure of a machine can be handled by one person. The two methods with the maintenance workers considered. One is maintained by 4 people, each responsible for 20; the second is jointly by the 3 maintenance 80. Compare these two methods can not timely maintenance probability in machine fault size.According to the solution of the first method. X "units" first maintenance 20 Taichung the same time of failure, with A I (i=1,2,3,4) said the event "I maintenance 20 Taichung fault can not be repaired in time, so the 80 Taichung failure probability can not be timely maintenance forP (A 1 u A 2 u A 3 u A 4) = P (A 1) =P (X = 2).XB (20, 0.01), the exergy exergy of P (X = 2) =1- 1k=0P (X=k) =1- 1k=0 C R K sigma f share 20 share (0.01) K (0.99) 20-k=0.016 9. share with P A (exergy exergy 1, A 2, A 3, A 4) more than 0.016 9.According to the second methods to Y units 80 Taichung the same time failure. At this time, YB (80,0.01), so the 80 Taichung failure probability can not be timely maintenance for the exergy exergy of P (Y = 4) =1- 3k=0 C R K sigma f share 80 (0.01) K (0.99) share share 80-k=0.008 7. we see after an exergy exergy situation despite the heavy task (average maintenance about 27 units), but the work efficiency is not reduced but increased.3) Poisson distribution (Poisson 's distribution)Set the random variable X, and all possible values are 0,1,2,. And the probability of taking each value isP (X=k) = lambda KK f e f share share! Lambda, k=0,1,2,. ,Where the lambda >0 is a constant, X is said to obey the Poisson distribution of the parameter lambda, and the Poisson distribution is P (lambda)We know that P (X=k) = lambda f f - K share e share is greater than or equal to 0, k=0,1,2 K,. , and exergy Exergy for k=0P (X=k) = sigma sigma lambda - k=0 K E - share share r r r r = lambda K! E lambda sigma lambda - k=0 - share share KK! = f e f - F - f e share lambda lambda =1. share in the history of the Poisson distribution exergy exergy is introduced as the two distribution approximately. After many years of research found that many random phenomena are Poisson distribution, for example, telephone exchange in Taichung every moment received a phone call number, the number of car accidents happen every day on the highway, after the fall of the number of particles of radioactive material split in a region, can be described by Poisson the Poisson distribution is an important distribution. The distribution of the stochastic process, some people think that the Poisson distribution is the structure of random phenomena "elementary particles", it is one of the three important probability distribution, has good properties (e.g., countable additivity). For different lambda Poisson distribution, has a special table available Consult the relevant probabilityIn 3.8 cases (Poisson distribution data and the dead horse riding) f Borthiewicz f (1898) is a classic example of Poisson distribution,Observation of 10 cavalry was the number of dead horse riding has been a total of 200 records in 20 years, the following is the frequency distribution table (X represents a cavalry year by the number of horse riding dead): the death toll X frequency relative frequency frequency frequency 01090.545108.80.5441650.32566.20.3312220.11020.20.101330.0154.20.021 = 410.0050.60.003 fitting theory from the data obtained the average number of a cavalry year crushed into 0.61, such as the X quasi Poisson distribution synthesis of a =0.61 can beP k=P (X=k) = (0.61) share e share -0.61 KK! R r, k=0,1,2,.Calculate the theoretical frequency of the last row, then 200 by P K fitting frequency distribution is Poisson, table second last line data; can be seen from the fitting data and the actual data is consistent. In fact, a cavalry year not crushed is not crushed, generally assume that each cavalry was probability p Mata die are the same, and each is dead horse riding cavalry are independent of each other, so a year by the number of cavalry horse riding two dead obey distribution, but p is very small, and the number of big two cavalry, as the limit distribution, the Poisson distribution is well described this set of data.3.3 continuous random variablesThe discrete random variables discussed above may take only finitely many countable or multiple values, but there are some practical problems, random variables may take values can be filled with an interval (or several intervals), we define the random variable as a continuous random variable. For example, the aircraft landing at the airport in time and the service life of the product is the value of the random variable, not a continuous random variable may list, therefore cannot use the distribution of discrete random variables to describe the statistical law of them.1. probability density function and its propertiesWe give an example of continuous random variables and their distributions3.9 cases of a target is a radius of 2 m disc, set office hit the target point on the probability of a disc and the disc is proportional to the area, and can shoot the target, expressed as X impact point and center distance distribution function of random variable X.If the solution of x<0, X or x is not possible, soF (x) =P (X x =0.)If x = 0 2, by A. P (0 X x) =kx share 2, K is a constant. In order to determine the value of K, x=2, P (X = 0 2) =2 share 2K, but known P (X = 0 2) =1, it was k=14, P (0 X x) =x share 24.ThereforeF (x) =P (X = x) =P (X<0) +P (0 X x) =x share 24.If x>2, X is less than or equal to x is inevitable by the event, soF (x) =P (X x =1.)In Figure 3.4, the distribution function of X is assumed to be.F (x) =0, x<0,X share 24,0 = x = 2,1, x>2.In addition, you can see the distribution function in this case F (x), any x can be written in the form of exergy exergy of F (x) = x - F - share formula (T) f d f t, the exergy exergy exergy exergy of F (T) =t2,0 = t = 2,0, the other, that is F (x) exergy exergy can be expressed as a non negative function f (T) on the X variable upper limit integral.In general, if the distribution function of random variable X F (x), the existence of nonnegative function f (x), expressed as F of arbitrary real x exergy exergy can be (x) = x - F - share formula (T) f d f t, then X is called exergy exergy continuous random variable f (x), which is called the probability density function of X, referred to as the probability density or density function, and that the distribution of X for continuous distribution. The density function f (x) has the following properties:(1) f (x) = 0;(2) + - share exergy formula for f (x) d r r x=1; exergy(3) for any real x 1, x 2 (x 1 = x 2), P (x 1<X<x exergy exergy (2) =F x 2) -F (x 1) = x 2 - share formula for f (x) f d f x- share formula x 1 - - f (x) d x= x formula F R 2 share x 1F (x) F D F X; exergy exergy (4) if f (x) at point x is continuous, F '(x) =f (x);(5) if f (x) is defined on the real axis, except the finite points are continuous everywhere, and satisfy (1), (2),Is x share infty exergy formula F (T) d t k k is an exergy distribution function, f (x) is a density function.Intuitively, the interval on the X ax