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Chapter 3 Discrete Distributions 隨機變數(random variable,r.v.)之定義 Definition 3.1-1: Given a random experiment with an outcome space S0 , a function X that assigns to each element s in S0 one and only one real number X( s ) = x is called a random variable . The space of X is the set of real numbers S {x : X ( s) x, s S0} For convenience, we will think of the space S of X as being the outcome space. (1) If S is a countable set, we say that X is a r.v of discrete type. (2)If S is an interval (possibly unbounded) or a union of intervals, we say that X is a r.v of continuous type. Ex3.1.1: A rat is selected at random from a cage and its sex is determined . The set of possible outcomes is female and male . Thus the outcome space of this experiment is S0 ={ female , male } = { F , M }. Let X be a function defined on S0 such that X ( F ) = 0 and X( M ) =1. The space of the random variable X is S={0,1}. Ex3.1-2 Let the random experiment be the cast of a die, observing the number of spots on the side facing up. The outcome space S0 ={1,2,3,4,5.6}. Let X(s)=s. (X is the identity function.) Then the space of the random variable X is S={1,2,3,4,5,6}. Ex 4.1-1 Let the r.v. X be the length of time in minutes of waiting in line to buy tickets. The space of the random variable X is S {x : 0 x, x R} Notations Let X be a random variable and B be an event. The probability P(B) of event B will be also denoted by (1) P(a X b) if B {s : a X (s) b} (2) P( a X ) if B {s : a X (s)} (3) P( X a ) if B {s : X (s) a} p.m.f. of discrete type r.v. Def 3.1-2 The probability mass function (p,m,f) of discrete type variable X with space S is a function f(x) satisfying the following conditions: (a) f(x) >0 , x S . (b) f ( x) 1. xS (c) P( X A) f ( x), where A S. xA Discrete distributions • Uniform distribution(均勻分配) • Hypergeometric distribution (超幾何分配) • Bernoulli distribution(白努力分配) • Binomial distribution(二項分配) • Negative geometric distribution(負幾何分配) • Geometric distribution(幾何分配) • Poisson distribution(卜瓦松分配) p.d.f. of continuous type r.v. Def 4.1-2 The probability density function (p,d,f) of continuous type variable X with space S is an integrable function f(x) satisfying the following conditions: (a) f(x) >0 , x S . (b) S f ( x)dx 1 (c) The probability of the event X A is P( X A) f ( x)dx, where A S A Continuous distributions • Uniform distributions • Exponential distributions • Gamma distributions • Chi-square distributions • Normal distributions • Beta distributions Uniform distribution(均勻分配) Def: If X is a r.v. with a constant p.m.f., we will say that X has a uniform distribution. Ex: In Ex 3.1-2, the p.m.f. of X is 1 f ( x) , x 1,2,3,4,5,6. 6 Hypergeometric distribution (超幾何分配) Def: We say that a r.v. of discrete type X has a hypergeometric distribution if it’s p.m.f. is N1 N 2 f ( x) x n x , N1 N 2 n where x {x : x is an integer , 0 x n, x N1, and x N 2}, n is a positive integer . Ex3.1-6 A lots, consisting of 100 fuses, 20 of them are defective. Five fuses are chosen at random and tested. Let X be a r.v. of defective fuses in the sample of five, then the p.m.f. of X is 20 80 x 5 x f ( x) P ( X x) , x 0,1,2,3,4,5. 100 5 3.2 Mathematical Expectation (期望值) Ex3.2-1 An enterprising young man who needs a little extra money devises a game of chance in which some of his friends might wish to participate. The game that he proposes is to let the participant cast an unbiased die and then receive a payment according to the following schedule : If the event A={1,2,3} occurs, he receives 1¢ ; if B ={4,5} occurs, he receives 5¢; and if C={6} occurs, he receives 35¢. Q: How much should be charged for the opportunity of playing the game? Let X be the r.v. defined by the outcome of the cast of the die. The p.m.f. of X is f ( x) 16 , x 1,2,3,4,5,6. 1 x 1,2,3 The payment is u ( x) 5 x 4,5 35 x6 The mathematical expectation of payment is 6 u(x)f(x) x 1 1 1 1 1 1 1 1 1 1 5 5 35 8 6 6 6 6 6 6 Definition 3.2-1 If f(x) is the p.m.f of the r. v. X of the discrete type with space S and if u x f x exists, xs then the sum is called the mathematical expectation or the expected value of the function u(X) , and it is denoted by E[u(X)] . E u X u(x) f(x) That is , xs Remark If u(X)=X, E[ X ] .X If u( X ) ( X X )2 , where X is the population mean of X, E[( X X )2 ] X 2 Theorem3.2-1 When it exists, mathematical expectation E satisfies the following properties : (a) If c is a constant, E[c] = c . (b) If c is a constant and u is a function, E[cu(X)] = cE[u(X)] . (c) If c1 and c2 are constants and u1 and u2 are functions, then Ec1u1 X c2u2 X c1Eu1 X c2 Eu2 X Ex 3.2-4 Let u x x b 2 , where b is not a function of X, and suppose E[( X b) 2 ] exists. To find that value of b for Which E[( X b) 2 ] is a minimum. Mean of a hypergeometric distribution Ex 3.2-5 Let X have a hypergeometric distribution in which n objects are selected from N N1 N 2 objects. Then the mean of X is E X x n( N ) N1 N2 n x x 1 N xs n N The variance of a hypergeometric distribution Ex 3.2-9 Let X have a hypergeometric distribution in which n objects are selected from N N1 N 2 objects. The variance of X σ 2 E X μ 2 E X 2 μ 2 E[ X ( X 1)] E[ X ] μ 2 n(n 1)( N1 )( N1 1) nN1 nN1 2 ( ) N ( N 1) N N N1 N 2 N n n( )( )( ) N N N 1 Mean and Variance of a Uniform Distribution Ex 3.2-8 The mean of X, which has a uniform distribution on the first m positive integers, is given by m 1 μ E X 2 1 m 12m 1 2 m EX 2 x x 1 m 6 2 σ 2 Var X E X μ E X 2 μ2 m2 1 12 Remark Let X be a r. v. with mean X and variance X . 2 If Y=aX+b, where a and b are constant, is a r. v., too. The mean of Y is μY E Y E aX b aE X b aμ X b The variance of Y is 2 σY E Y μY 2 E aX b aμ X b 2 E a X μ a E[( X 2 X 2 2 X ) 2 ] a 2σ 2 X The standard deviation of Y is σ Y a σ X 3.3 Bernoulli Trials & The Binomial Distribution A Bernoulli experiment is a random experiment, the outcome of which can be classified in but one of two mutually exclusive and exhaustive ways, say, success or failure . A sequence of Bernoulli trials occurs when a Bernoulli experiment is performed several times so that the probability of success remain the same from trial to trial. Bernoulli 實驗 一實驗若滿足下述三特性, 則稱此實驗為Bernoulli 實驗. (1)實驗的outcome可非分為互斥的兩部分,一稱成功 (success),一稱失敗(failure). (2)成功機率(令為 p),在此實驗下恆固定不變. (3)試行間獨立 . Bernoulli distribution Def: Let X be a r. v. associated with a Bernoulli trial by defining it as follows: X ( success ) 1 and X ( failure) 0. (or X 一次Bernoulli實驗中,成功的次數) The p.m.f. of X can be written as f ( x) p x (1 p)1 x , x 0,1 we say that X has a Bernoulli distribution ( X ~ Ber( p) , X服從白努力分配 ). Mean and variance of a Bernoulli distribution Let X be a r. v. with a Bernoulli distribution, The mean of X is 1 E ( X ) xp x (1 p)1 x p x 0 The variance of X is 1 Var ( X ) 2 ( x p)2 p x (1 p)1 x pq x 0 The standard deviation of X is p(1 p) pq Binomial experiment A binomial experiment is a sequence of Bernoulli experiment which satisfies the following properties: 1. A Bernoulli ( success-failure ) experiment is performed n times . 2. The trials are independent . 3. The probability of success on each trial is a constant p ; the probability of failure is (1-p). 4.The random variable X equals the number of success in the n trials. Binomial distribution Def: Let X be the number of observed successes in n Bernoulli trials, the possible values of X are 0 , 1 , 2…….. , n. Then the p.m.f. of X is n x f ( x) p (1 p) n x , x 0,1,2,........, x n and the r. v. X is said to have a binomial distribution ( X ~ b(n, p) , X服從二項分配 ). The constants n and p are called the parameters of the binomial distribution. Remark Let n be a positive integer n n x x p (1 p ) n x [(1 p ) p ]n 1 x 0 n x f ( x) p (1 p ) n x , x 0,1,, n. x is a p.m.f. Example 3.3-5 In the instant lottery with 20% winning tickets, if X is equal to the number of winning tickets among n = 8 that are purchased, the probability of purchasing two winning tickets is 8 (0.2) 2 (0.8)6 0.2936 f (2) P( X 2) 2 Cumulative Distribution Function (c.d.f.) ( 累積分配函數) Def: The cumulative distribution function of a r.v. X is defined by F ( x) P( X x) Remark 1. f (t ) if X is discrete type t{ X x} F ( x) P( X x) x f(t)dt if X is continuous type 2. 0 F ( x) 1, x R 3. F(x) is nondecreasing. That is, x1 x2 F ( x1 ) F ( x2 ) The c.d.f. of binomial distribution Values of the c.d.f. of a r.v. X ~b(n,p) are given in Table II (p647-p651). x n! F ( x) P( X x) p (1 p) k nk k 0 k!(n k )! Ex 3.3-8 Let p=0.5 be the probability of a female chick hatching. Assuming independence, let X equal the number of female chicks out of newly hatched chicks selected at random. Then X ~ b(10,0.5) The probability of five or less female is 查表 P( X 5) F (5) 0.6230 The probability of exactly six female is P( X 6) P( X 6) P( X 5) 0.8281 0.6230 The probability of at least six female is P( X 6) 1 P( X 5) 1 0.6230 Ex3.3-9 Suppose that 65% of American public approve of the way the President of the U.S.A. is handling his job. Take a random sample of n=8 American and let Y equal the number who give approve. The Y~b(8,0.65). Find P(Y 6), P(Y 5) and P(Y 5). 查表 P(Y 6) P(8 Y 8 6) P( X 2) 0.4278 where X ~ b(8,0.35) P(Y 5) P(8 Y 8 5) P( X 3) 1 P( X 2) P(Y 5) P(8 Y 8 5) P( X 3) P( X 3) P( X 2) Remark Suppose that an urn contains N1 success balls and N 2 N1 p failure balls and we let N1 N 2 and X equal the number of success balls in a random sample of size n that is taken from this urn. If the sampling is done with replacement, the distribution of X is b(n,p); and if the sampling is done without replacement, X has a hypergeometric distribution. Ex: 若一箱中有8個紅球,4個白球. 現從此箱中隨機不 歸還取出3個球, 則此3個球中, 紅球的平均值為 何?變異數為何? Sol: 令X 3個球中,紅球的數目 X has a hypergeome tric distributi on with N 12, n 3. 8 12 3 2 1 6 E( X ) 3 2, Var ( X ) 3 . 12 12 1 3 3 11 3.4 The Moment-Generating Function (動差母函數) Def: Let X be a r. v. of the discrete type with p.m.f. f(x) and space S. If there is a positive number h such that E (e ) e f(x) tX tx xS exist and is finite for –h<t<h, h>0，then the function of t defined by M X (t ) E (etX ) etx f ( x) xS is called the moment-generating function of X . This is often abbreviated as m.g.f. Remark (1)If r.v. X and Y have the p.m.f. f(x) and g(y) respectively, the same space S={b1,b2,b3,…} and the same m.g.f. tb1 f ( ) tb2 f ( ) tb1 g ( ) tb2 g ( ) ... e b1 e b2 ... e b1 e b2 for all t (h, h) , Then mathematical theory requires that f (bi ) g (bi ), i 1,2,3, Remark (2)If X is a r.v. with space S={x1,x2,x3,…}, then p.m.f. of X is f ( xi ) P ( X xi ) pi , i 1,2, m.g.f .of X is pi etxi i 1 Example 3.4-2： et Suppose the m.g.f. of X is 2 , t ln 2. M (t ) 1 et 2 What is the p.m.f. of X? Find P(X > 3). Since M (t ) 1 e t e t et et e 2t e 3t 1 1 2 3 ... 2 2 2 2 2 2 2 3 1 1 1 (e t ) (e 2 t ) (e 3t ) 2 2 2 x 1 et 1. f ( x) P( X x) , x {1,2,}, when 2 2 2 3 1 1 1 P( X 3) 1 P( X 3) 1 2 2 2 Ex: Find the probability distribution of r.v.X in (1)and (2). (1)If the m.g.f. of X is given by 1 t 2 2t 2 4t M X (t ) e e e , t . 5 5 5 (1)If the m.g.f. of X is given by t M X (t ) e , t . Ex: If r.v.X has the m.g.f. M X (t ) 1 (2t 3t 5t ), t . 3 find the c.d.f. of X. Properties of m.g.f. Thm: Let X be a r.v. of discrete type, a and c are constant, a 0 , then (1) M a (t ) e at (2) M aX (t ) M X (at ) (3) M X c (t ) ect M X (t ) (4) M aX c (t ) ect M X (at ) (5) For any positive integer r , M Xr ) (0) E ( X r ). ( tr (6) When M X (t ) exists , M X (t ) M X (0) E ( X r ) . r! r 1 Example 3.4-7 Let the moments of X be defined by E ( X r ) 0.8, r 1,2,3,... Find the m.g.f.of X. The moment-generating function of X is tr M (t ) M (0) 0.8 0.2e 0t 0.8e1t r! r 1 Thm: Let X be a r.v. with m.g.f. M(t), then M (0) , 2 M (0) M (0).2 Pf: E ( X ) M (0) 2 E ( X ) [ E ( X )] M 2 2 (0) M (0)2 . Mean & Variance of Binomial Distribution Thm: Let X ~ b(n, p) , then . (1) M X (t ) ( pet 1 p ) n . (2) X np. (3) X np (1 p ). 2 Pf: The p.m.f. of X ~ b(n,p) is n x n x f ( x) p 1 p , x 0,1,2,, n. x the m.g.f. is e x p 1 p tx n n n x M X (t ) E e tX x x 0 n t 1 p n x n pe x x 0 x (1 p) pe t n , t t n 1 M (t ) n 1 p pe pe t M '' (t ) n(n 1)[(1 p ) pet ]n 2 ( pet ) 2 n[(1 p) pet ]n 1 ( pet ) X M (0) np X 2 M (0) M (0) np(1 p) 2 Ex: Find the probability distribution of X if the m.g.f. of X is given by 20 1 t 3 2t M X (t ) e e , t . 4 4 20 20 1 t 3 2t 20 t 1 3 t pf : M X (t ) e e e e 4 4 4 4 20 1 3 t 3 and e is the m.g.f .of b(20 ,p ) 4 4 4 by the property of m.g.f ., 3 we know that X Y 20, where Y ~ b(20, p ) 4 the p.m.f. of X is f ( x) P( X x) P(Y x 20) g ( x 20) 20 3 x 20 1 40 x , x 20,,40. x 20 4 4 Mean & variance of Bernoulli Distribution In the special case when n=1, X has a Bernoulli distribution and t M X (t ) (1 p) pe , t . X p X 2 p(1 p) Negative Binomial Distribution Observe a sequence of Bernoulli trials until exactly r success occur, where r is a fixed positive integer. If X denote the trial number on which the rth success is observed. Then the p.m.f.of X is x 1 r f ( x) p (1 p) x r , x r , r 1, r 1 We say that X has a negative binomial distribution Remark Let h(q) (1 q) r , r 0 h ( k ) (0) r k 1 k (1 q ) r r 1 q , for 1 q 1 k 0 k! k 0 ( let x k r ) x 1 x r r 1q xr x 1 r x r f ( x) p q p r (1 q) r 1, where q 1 p r 1 xr xr x 1 r f ( x) p (1 p ) x r , x r , r 1, is a p.m.f. r 1 Geometric Distribution When r =1. The p.m.f. of X is x 1 f ( x) p (1 p) , x 1,2, and We say that X has a geometric distribution. Remark p p(1 p) x 1 1 (1 p ) 1 x 1 f ( x) p (1 p ) x 1, x 1,2,is a p.m.f. Example 3.4-4 Some biology students were checking the eye color for a large number of fruit flies. For the individual fly,suppose that the probability of white eyes is 1/4 and the probability of red eyes is 3/4 , and that we may treat these observations as having independent Bernoulli trials.The probability that at least four flies have to be checked for eye color to observe a white-eyed fly is given by 3 3 27 P( X 4) P( X 3) q ( ) 3 0.4219 4 64 The probability that at most four flies have to be checked for eye color to observe a white-eyed fly is given by 3 4 175 P( X 4) 1 q 1 ( ) 4 0.6836 4 256 The probability that the first fly with white eyes is the forth fly considered is 41 3 3 1 27 P( X 4) q p( ) ( ) 0.1055 4 4 256 m.g.f. of the negative binomial distribution Let X be a r.v. with negative binomial distribution, then the m.g.f. of X is x 1 r M (t ) e tx r 1 p (1 p) x r xr x 1 ( pe ) t r r 1[(1 p )e t ]x r xr ( pet ) r , where (1 p )et 1 [1 (1 p )e t ]r Mean & variance of the negative binomial distribution Thm: Let X be a r.v. with negative binomial distribution, then r X p r (1 p ) 2 X p2 Pf: M ' (t ) r ( pet ) r [1 (1 p )et ] r 1 and M '' (t ) r ( pet ) r (r 1)[1 (1 p )et ] r 2 [(1 p )et ] r 2 ( pet ) r 1 ( pet )[1 (1 p )et ] r 1 M ' (0) rp 1 , M '' (0) rp 2 (r 1 p ) r r (1 p ) X M (0) , X M (0) [ M (0)] ' 2 '' ' 2 p p2 Example 3.4-5 Suppose that during practice, a basketball player can make a free throw 80% of the time. Furthermore, assume that a sequence of free-throw shooting can be thought of as independent Bernoulli trials. Let X equal the minimum number of free throws that this player must attempt to make a total of 10 shots. The p.m.f. of X is x 1 x 10 10 1(0.80) (0.20) g ( x) 10 , x 10,11,12... 1 10(0.20) 10 12.5, 2 3.125, 0.80 2 0.80 and 1.768 Mean & variance of the geometric distribution Thm: Let X be a r.v. with geometric distribution, then pet M X (t ) , t ln(1 p ) 1 (1 p )e t 1 X p (1 p ) X 2 p2 3-5 THE POISSON DISTRIBUTION Some experiments result in counting the number of times particular events occur in given times or on given physical objects. For example, (1) count the number of phone calls arriving at a switchboard between 9 and 10 A.M.. (2) count the number of flaws in 100 feet of wire. (3) count the number of customers that arrive at a ticket window between 12 noon and 2P.M.. (4) count the number of defects in a 100-foot roll of aluminum screen that is 2 feet wide. Poisson experiment Def: Let the number N(t) of changes that occur in a given continuous interval of length t be counted. We have a Poisson experiment with parameter 0 if the following are satisfied: (1)The probability of exactly one change in a sufficiently short interval of length h is approximately t . P( N (t ) 1) h o(t ), ast 0, o(t ) where o(t ) is a function of t such that lim 0. t 0 t (2) The probability of two or more changes in a sufficiently short interval is essentially zero. P( N (t ) 2) 0, as t 0. (3)The number of changes occurring in non- overlapping intervals are independent. p.m.f. of Poisson distributions. For a Poisson experiment, let N(t) denote the number of changes in an interval of length t. • Divide the interval of length t into n subintervals of equal length t n . • If n is sufficiently large, one change occurs in each of exactly y of these n subintervals. • Consider the occurrence or nonoccurrence of a change in each subinterval as a Bernoulli trial. By (1)-(3), we t t have a sequence of Bernoulli trial with p o . n n p.m.f. of Poisson distributions Then the p.m.f. of N (t ) is y n y n t t t t f ( y ) P ( N (t ) y ) lim o 1 o n y n n n n y n y n(n 1)(n y 1) t t t t t t lim o 1 o 1 o n y! n n n n n n n 1 2 y t n (t ) y y 1 t t t lim 1 1 1 1 o 1 o y! n n n n n n n n n e t (t ) y , y 0,1,2, y! p.m.f. of Poisson distributions. For a Poisson experiment, let X denote the number of changes in an interval of length 1. Since X=N(1), the p.m.f. of X is x e f ( x) x 0,1,2, ... where 0 x! Poisson distribution Def: We say that a r. v. X has a Poisson distribution with parameter 0, if its p.m.f is of the form x e f ( x) x 0,1,2, ... where 0 x! Remark (1) f x 0 for x 0,1, x e x (2) f(x) e x! e e 1 x 0 x 0 x! x 0 f is a p.m.f. Mean & variance of Poisson distribution Thm: Let X be a r. v. of Poisson distribution with parameter , then ( e t 1) M X (t ) e M ' 0 X X X 2 " 0 [ M ' 0]2 2 2 MX X Pf: x e M X t E e tX e tx x 0 x! e e e ee e e 1 t x t t x 0 x! ' t e t e et 1, M " t (e t ) 2 e et 1 et e et 1 MX X X M 0 X M 0 [ M 0]2 2 2 2 EX 3.5-1 Let X have a Poisson distribution with a mean of 5 . Then 6 x 5 ( 查表) 5 e P X 6 0.762 x 0 x! ( 查表) P X 5 1 P X 5 1 0.616 0.384 and P X 6 P(X 6 ) - P X 5 0.762 0.616 0.146 Remark • If events in a Poisson process occur at a mean rate of per unit , then the expected number of occurrences in an interval of length t is t . • The number of occurrences, say X ,in the interval of length t has the Poisson p.m.f. (t ) x e - t f ( x) , X 0,1,2, x! Ex: It is known that bacteria of a certain kind occur in water at the rate of three bacteria per cubic centimeter of water. Assume that this phenomeon obeys Poisson probability Law, what is the probability that a sample of two cubic centimeter of water contain (1) at most two bacteria. (2) at least three bacteria. Sol : N (1) ~ Po ( 3) N (2) ~ Po ( 6) 2 6 x e6 (1) P ( N (2) 2) 25e 6 x 0 x! (2) P ( N (2) 3) 1 P ( N (2) 2) 1 25e 6 . Example 3.5-5 Telephone calls enter a college switchboard on the average of two every 3 minutes. If one assumes a Poisson distribution, what is the probability of five or more calls arriving in a 9-minute period? Sol : N (3) ~ Po ( 2) N (9) ~ Po ( 6) Let X N (9) 4 6 x e 6 查表 P ( X 5) 1 P ( X 4) 1 1 0.285 x 0 x! Remark • If X has Poisson distribution with parameter 0. If n is large, then n x n x P( X x) 1 x n n • That is, if X ~b (n, p) with large n and small p, then n x (np ) x e np p (1 p) n x x x! • The approximation is quite accurate if n 20 and p 0.05 or if n 100 and p 0.10 Example 3.5-6 A manufacturer of Christmas tree light bulbs knows that 2% of its bulbs are defective. To approximate the probability that a box of 100 of there bulbs contains at most three defective bulbs , we use (1) Poisson distribution with 100(0.02) 2 which gives 3 x 2 2 e x! x 0 0.857 (2) binomial distribution 3 100 x 0.02x 0.98100 x 0.859 x 0 The Poisson approximation is extremely close to the true value, but much easier to find.

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