概率论与数理统计英文

1、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 instanee, when we measure the speed of a car, the amount of alcohol in a pers ons blood, the ten sile stre ngth of new alloy.We shall lear n how to determ

2、 ine and work with probabilities relat ing to continu ous ran dom variables in this chapter. We shall introduce to the concept of the probability density function.4.1 Continuous Random Variable1. DefinitionDefinition 4.1.1 A function f(x) defined on (-：)is called a probability densityfunction (概率密度函

3、数)if:(i) f (x) _0 for any x R;oO(ii) f(x) is in tergrable(可积的)on (-:)a nd f (x)dx = 1.-nODefinition 4.1.2Let f(x) be a probability density function. If X is a random variable having distribution functionxF(x)二 P(X 乞 x)二 f(t)dt,(4.1.1)_oQthen X is called a continuous random variable having density fu

4、nction f(x). In this case,X2P(m : X : x2) = f (t)dt.(4.1.2)Xi2. 几何意义（）xF(x)二 P(X x)二 P(X,Y)|X zx, 0 乞丫 乞 f (X) = f (t)dt-oO69 / 17o 乞 p（x 二 X）空 limx * ;f (x)dx = 0 xx2P(x, :： X :：: x2) = f(t)dtx13. NoteIn most applications, f(x) is either continuous or piecewise continuous having at most fin itely m

5、any disc on ti nuities.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) * f (x

6、)dxP(a 乞 X 乞 b)二 P(a 乞 X : b)二 P(a ： X b) = P(a ： X : b)4. ExampleExample 4.1.2Find k so that the following can serve as the probability density of a continuous random variable:kf(xr FW)Solution To satisfy the conditions (4.1.1), k must be nonnegative, and to satisfy the condition(4.1.2) we must hav

7、eoaf (x)dx =joO,1 so that k .(Cauchy distribution 柯西分布)Example 4.1.3 Calculating probabilities from the probability density functionIf a ran dom variable has the probability den sityx for x 0f(x)= 3e0for x 乞 0Find the probability from that it will take on value(a) betwee n 0 and 2;(b) greater tha n

8、1.Solution Evaluating the necessary integrals, we get2(a) P(0 辽 x 乞 2) = j3exdx =1 - e =0.99750oO(b) P(x1) = 3eXdx = e =0.04981Example 4.1.4Determ ining the distributi on function of X, it is known工 3ex for x 0f(x)=0 for x 0Solution Performing the necessary integrations, we get(0for x _ 0lxx) 3edt =

9、 1ex for x 00P(x _1) = F(1) =1 -e =0.95025. meanIf the in tegral (4.1.3) does not con verges 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 functionf(x). Then themean (or expectation) of X is defined byE(X)二.xf

10、(x)dx ,(4.1.3)-oOThe 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) isooE(g(x)g(x)f(x)dx-joOprovided the in tegral con verges absolutely.Example

11、4.6.4Let X be a random variable having Cauchy distribution, the probability density function is give n by二(1x2)-：:x(a) Find E(X);(b) Letgwox：10, elsewhereFind E(g(X).00 |x|Solution (a) Since the integral 2 dx diverges(发散), E(X) does not exist.丿(1 + x )1(b) E(g(X)g(x)f (x)dx =_：:0xln22 dx-二(1 x )2 二6

12、. varianceSimilarly, the varianee and standard deviation of a continuous random variable X is defined by二2 =D(X) =E(X 7)2),(4.1.4)Where J = E(X) is the mean of X,二 is referred to as the standard deviation.We easily geto二2 二D(X) = x2f(x)dx-2. (4.1.5)Example 4.1.5Determining the mean and varianee usin

13、g the probability density funetion3ex for x 0With referenee to the example 4.1.3: f (x)- 0 for x 兰 0find the mean and varianee of the given probability density.Solution Performing the necessary integrations, we getoOoo.1 亠=xf(x)dx 二 x 3exdx =_ 0 3andc2：11=J (x_ A)2 f(x)dx = J(x_)3e：Xdx =_039均匀分布4.2

14、Uniform DistributionThe uniform distributi on, with the parameters a and b, has probability den sity fun eti on一 for a x : b, f(x)pb-a0elsewhere,whose graph is show n in Figure 4.2.1.f(x)1b aFigure 4.2.1 The uniform probability density in the interval (a, b)-beTo proof f (x)dx =1.*jtJdTo find the di

15、stribution function.The distribution function of the uniform distribution is9for x a,F(x) = x _a for a 兰xeb,b -a1 for x _ b.Note that all values of x from a and b are “ equally likely ”，in the sense that the probabixty thatlies in an interval of width lx entirely contained in the interval from a to

16、b is equal to =x/(b a), regardless of the exact location of the interval.To find the mean and variance of the uniform distributi onba1 dx =EXbx2aAnd1, a2 ab b2dx 二ba3Thus,2 a2+ab+b22 2()2-3I 2丿124.5 Exponential Distribution指数分布Many random variables, such as the life of automotive parts, life of anim

17、als, 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入(& A 0), if its density function is given byh仝、e 九 for x0f(x) = 5(4.5.1)0for x 兰 0Note Equati on (4.5

18、.1) really gives a den sity fun cti on, since(丄 e，dx = 1. 0 ZTheorem 4.5.1 The mean and variance of a continuous random variable X having exponential distribution with parameter丸 is given byProof Si nee the probability den sity fun ctio n of X is (4.5.1), we have 4 xE(X) = Jxf (x)dx = Jx edx = h.: 0

19、 1彳 xE(X2) = J x2 f (x)dx = Jx2 e 站x =2丸2.：:o2 2 2D(X)二E(X ) -E(X) = .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 ran dom from the product of compa ny X haslife Ion ger tha n 45

20、0 hrs.(b) Select 5 bulbs ran domly from the product of compa ny X, what is the probability that at least 3 of them has life Ion ger tha n 450 hrs.Solution (a)P(Y 450) = 1 - P(Y 乞 450)450x=1 _1 edx d5 =0.22310 300(b) The nu mber Z of bulb of bulbs with life Ion ger tha n 450 hrs has the bino mial dis

21、tributi onB( n, p), where n=5, p=0.2231. ThusP(Z _3) =C；30.223130.77692 C；0.223140.77690.22310.0772.* 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

22、first in the eighteenth century when scientists observed an ast onishing degree of regularity in errors of measureme nt. They found that the patter ns (distributi ons) they observed were closely approximated by a con ti nu ous distributi on which they referred to as the “ normacurve of errors and at

23、tributed to the laws of chanee. The normaldistribution is one of the frequently used distributions in both applied and theoretical probability.1. DefinitionThe equati on of the no rmal probability den sity, whose graph is show n in Figure 4.3.1, isf(x)二一1e“ 八:：x ：V2cFigure 4.3.1 The normal probabili

24、ty densityThe distribution is characterized by two parameters, traditionally designated 士 and 匚. (二 0.)To prooff (x) _0.f (x)dx =1.?proof?_oOIt is often convenient to designate the face that X is normal with parameters J and - by2the notation X )x* 2 3 4(x)二2二 e22. the property of the normal probabi

25、lity densitythe normal distribution function is(x) is an even function:( _x) = (x),and :( -z) =1：7 (z).To find the probability that a random variable having the standard normal distribution will take on a value betwee n a and b, we use the equatio nP(a7z2b)二(b)-(a)Figure 433 P(a 一 N(0,1).(sinceFz(z)

26、二 P(Z 布zFXlz )(t-J2F dtuCTdtje = t 一 ，dt11 dt )ffcr= G(z)X Ahence Z N(0,1)P(X 兰 x)= P(Zcrcr乞 x_)儿(x_)icr2Also, X N(.L,二),the probabilityP(a : X b) =P(a1:Z Example X N (1.5, 4) , find P(X 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 distrib

27、ution with the parametersnand p, i.e. X B(n, p) the limiting form of the distribution function of the standardized random variableas n = , is given by the standard normal distributiongraph-_ExampleFor an experiment in which 9 coins are tossed, Let X denotes the number of head occurrenee, con struct

28、the probabilitydistributi on of X.SolutionX0123456789P193684126126843691512512512512512512512512512512NP193684126126843691N=512r (or f)Figure 7-2Histogram and normal curve for expected frequencies of heads in tossing 9 coins 51 2 times.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

29、 approximation to binomial probabilitiesIf 20% of the memory chips made in a certa in pla nt are defective, what are the probabilities that in a lot of 100 ran domly chose n for in specti on(a) at most 15 will be defective;(b) exactly 15 will be defective?Solution Since J =100 0.20 = 20 and 二=100 0.

30、20 0.80 = 4 for the bi no mial 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 yieldsFor part (a),15 5_20P(X 乞 15) =P(X ：15.5) =P(Z)：(-1.13) = 0.1292.4For part (b),14.5 2015.5 20P(X =15) = P(14.5 ： X ：15.5) = P(Z)44=门(一1.13

31、) -门(一1.38) =0.1292 0.0838 =0.0454.If we do the exact bi no mial calculati on on a computer in stead of using no rmal approximati on, we would have obta ined 0.1285 in stead of 0.1292 for part (a) and 0.0481 in stead of 0.0454 for part (b). That both approximati ons are very close.Use the normal app

32、roximation to the binomial distribution only whennp and n(1-p) are bothgreater tha n 15.Homework4.13,4.17,4.19,4.22,4.6 Function of Random VariablesIf the value of the random variable X can be observed in a trial, we are interested in a corresp onding value Y=g(X) obta ined by appl ying the rule for

33、 the function g to the observed value. We address the basic problem: Given the distribution for X, how can we assign probability to events determ ined by the new ran dom variable Y?Example 4.6.1 The function of a discrete random variableSuppose the probability distribution of a discrete random varia

34、ble X as in the following table,2and Y = X 3 . Z - 2X -1 Determine the probability distribution of the random variable Yand Z.x-2-1012P(X=x)0.150.200.200.250.20SolutionP(X=x)0.150.200.200.250.20X-2-1012Y = X2 +374347Z=2X-1-5-3-113When the values ofX are equal to -2, -1,0, 1,2, the random variable Y

35、are equal t 匚斗3,4,7 respectively. Then,P(Y =3) =P(X =0) =0.20,P(Y =4)二 P(X = -1) P(X =1) =0.20 0.25 =0.45 ,P(Y =7)二 P(X 二-2) P(X =2) =0.15 0.20 =0.35,Thus, we getAndY347P(Y=y)0.200.450.35Z-3-113P(Z=z)0.150.200.200.250.20Example 462Suppose the random variable Y=aX+b, with a 0 . Determine FY(y).Soluti

36、on We n eed con sider two cases.y b(1) a 0: Y 二 y iff X,athen,y bP(Y y)二 P(aX b 乞 y)二 P(X _ ) aimpliesy bFY(y) = Fxr ).ay -b a ：0:Y _ y iff X ,athus,ybybP(Y y) = P(aX b 乞 y) = P(X _) =1 _P(X ：.),aaimpliesFY(y)=1 Fx(m)+ P(x =)aa*If X is continuous, so is Y. In this case,P(X=lb)=0. Differentiating aFy

37、 to obta in theden sity yields two cases that can be comb ined by the use of the absolute value to obta infY(y)fX(宁.XN(0,1)crwe know that if X N ( - 2), then the corresponding standardized random variable , let us give the proof aga in.1k1khere a ： 一，b = -, Y X -一aacra1J2LfX (x)e 2二 ,using the resul

38、t of example 4.6.2, we get1fY(y)a|fX( a)y .丄 ed1J2wExample 4.6.3 Square Root Fun ctionSuppose the random variable Y = X ,and X亠 0. Determine the distribution for Y.Solution The function g(x) = x is increasingon0, ：) . Now X -y 肝 X _ y2.22Thus for y _ 0, Fy( y)二 P(Y _ y) = P(X _ y ) = Fx (y ), and if

39、 X is continuous, thus,fY (y)二dFY(y)dydFx(y2)dy= 2yfx(y2).4.7 Chebyshev s TheoremEarlier in this chapter we used examples to show how the sta ndard measures the variati on of a probability distribution, that is, how it reflects the concentration of probability in the neighborhood of the mean. If cis

40、 large there is a correspondingly higher probability of getting values farther away mean. Formally, the idea is expressed by the follow ing theorem.Theorem 4.7.1 If a probability distribution has mean and standard deviation qthe probability of1getting a value which deviates from (iby at least k Hs a

41、t most 2 . Symbolically , k21P(|X -_2)乞kwhere 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 -|丄 k二.Then, the probability that a random variable will take on a value which devi

42、ates from the mean by at least 2 sta ndard deviati ons 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 meanthe probability that a ran dom variablek=2Pw 1/4k=6Pw 1/36Proof As an example, we c

43、onsider the case that X is a continuous variable with probability den sity f(x).P| X - 珥 k；(x)2k2；2f(x)dx2 1（x.S） f（x）dx_k2_2 . |xk 二k C 一:空11（x丄）f（x）dx=k2_2；- k2So we haveThis completes the proof of Theorem 4.7.1.To get an alter native form of Chebyshev theorem, n ote that the |x -|： k二 is the1 com

44、plement of the event | x - 打 _ k； thus, P（| x - 讣：k；） _ 1 2 .k对比正态分布的3 c定理？Example 4.7.1 A probability bound using Chebyshev s theoremThe number of customers who visit a car dealer showroom on a Saturday morning is a random variable with 卩=18 and c=2.5. With what probability can we assert that there

45、 will be more tha n 8 but fewer tha n 28 customers?Solution Let X be the nu mber of customers. Si nceP（8 ： X : 28） = P（8 一18 ： X -28 18） = P（| X -亠卜：10）B =1 0,k = 1 0/二1 0 /=2. 51115P（|X*）_12 and P（8 ： X ： 28）-12.k24216The most important feature of Chebyshe s theorem is that it applies to any probab

46、ilitydistribution for which and cexist.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 probabilityP（| X二|一 2匚）乞 0.25,When a random va

47、riable X B（16,0.50）, -16 0.50 =8,；丁八 16 0.50 0.50 =2,the value of probability which differs from the mean by at least 2 standard deviations isP(|X _i|_2；)= P(|X _8|_4)=仁 P(|X _8|： 4)11= 1 G；0.516 =1 0.9232 = 0.0768.k出“ At most 0.25” does not tell us that the actual probability may be as small as 0.0768.问题思考1在P(|X 卩彥畑)兰疋，问P(|X 片沁)=?Homework jsjP6927, 28,