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6 間斷隨機變數及其常用的機率分配 學習目的1-1本章結構Introduction to Probability DistributionsnRandom VariablenRepresents a possible numerical value from an uncertain eventRandom VariablesDiscrete Random VariableContinuousRandom VariableCh.6Ch.7Probability DistributionsContinuous Probability DistributionsBinomialHypergeometricPoissonProbability DistributionsDiscrete Probability DistributionsNormalUniformExponentialCh.6Ch.7隨機變數的意義與種類Discrete Random VariablesnCan only assume a countable number of valuesExamples:nRoll a die twiceLet X be the number of times 4 comes up (then X could be 0,1,or 2 times)nToss a coin 5 times.Let X be the number of heads (then X =0,1,2,3,4,or 5)隨機變數間斷隨機變數間斷隨機變數Experiment:Toss 2 Coins.Let X=#heads.TTDiscrete Probability Distribution4 possible outcomesTTHHHHProbability Distribution 0 1 2 X X Value Probability 0 1/4=0.25 1 2/4=0.50 2 1/4=0.250.500.25 Probability 連續隨機變數連續隨機變數單一間斷隨機變數的機率分配傳真機故障的機率傳真機故障的機率銅板出現正面的次數與相對次數銅板出現正面的次數與相對次數銅板的機率分配銅板的機率分配信用卡持有數信用卡持有數信用卡持有數量的機率分配信用卡持有數量的機率分配信用卡持卡數的機率分配信用卡持卡數的機率分配全國大學生擁有電子郵件帳號個數，其機率分配如下：大學生每人擁有電子郵件帳號個數x(隨機變量)f(x)(隨機函數)F(x)(累加函數)10.04680.046820.17070.217530.23640.453940.28310.73750.17880.915860.04290.95877個以上0.04131期望值=3.6703 變異數=1.9538 離散機率分配例子單一間斷隨機變數的機率分配Discrete Random Variable Summary Measuresn Expected Value(or mean)of a discrete distribution(Weighted Average)nExample:Toss 2 coins,X=#of heads,compute expected value of X:E(X)=(0 x 0.25)+(1 x 0.50)+(2 x 0.25)=1.0 X P(X)0 0.25 1 0.50 2 0.25nVariance of a discrete random variablenStandard Deviation of a discrete random variablewhere:E(X)=Expected value of the discrete random variable X Xi =the ith outcome of XP(Xi)=Probability of the ith occurrence of XDiscrete Random Variable Summary MeasuresnExample:Toss 2 coins,X=#heads,compute standard deviation(recall E(X)=1)Discrete Random Variable Summary MeasuresPossible number of heads=0,1,or 2信用卡持有數的以下累加機率分配表信用卡持有數的以下累加機率分配表信用卡的機率分配單一間斷隨機變數的機率分配標準化隨機變數表標準化對話方塊單一間斷隨機變數函數的期望值與變異數單一間斷隨機變數函數的期望值與變異數單一間斷隨機變數函數的期望值與變異數離散分配n二項分配n卜瓦松分配n超幾何分配二項分配(Binomial Distribution)n重複執行n次獨立的實驗n實驗的可能結果只有兩種成功和失敗n已知每次實驗成功的機率固定為pn隨機變數x表示n次實驗中成功的次數n即稱x為二項隨機變數Binomial Probability DistributionnA fixed number of observations,nne.g.,15 tosses of a coin;ten light bulbs taken from a warehousenTwo mutually exclusive and collectively exhaustive categoriesne.g.,head or tail in each toss of a coin;defective or not defective light bulbnGenerally called“success”and“failure”nProbability of success is p,probability of failure is 1 pBinomial Probability DistributionnConstant probability for each observationne.g.,Probability of getting a tail is the same each time we toss the coinnObservations are independentnThe outcome of one observation does not affect the outcome of the othernTwo sampling methodsnInfinite population without replacementnFinite population with replacementPossible Binomial Distribution SettingsnA manufacturing plant labels items as either defective or acceptablenA firm bidding for contracts will either get a contract or notnA marketing research firm receives survey responses of“yes,I will buy”or“no,I will not”nNew job applicants either accept the offer or reject it二項機率分配Rule of CombinationsnThe number of combinations of selecting X objects out of n objects is where:n!=(n)(n-1)(n-2).(2)(1)X!=(X)(X-1)(X-2).(2)(1)0!=1 (by definition)P(X)=probability of X successes in n trials,with probability of success p on each trial X =number of successes in sample,(X=0,1,2,.,n)n =sample size(number of trials or observations)p =probability of“success”P(X)nX!nXp(1-p)XnX!()!=-Example:Flip a coin four times,let x=#heads:n=4p=0.51-p=(1-0.5)=0.5X=0,1,2,3,4Binomial Distribution Formula二項隨機實驗的樹枝圖二項隨機實驗的樹枝圖n賞鯨活動並不能向遊客保證每一次出海都能看的到鯨魚，根據經驗豐富的船長表示，能夠看到鯨魚(成功)的機率是0.1,只看得到海豚或其他魚類(失敗)的機率為0.9，如果此船某年要出船300趟。由於每次出海賞鯨成功與否並不會影響下一次出海結果，因此這些試驗相互獨立。試驗次數n=300，成功機率p=0.1，所以可以算出在一年船期中，預期這艘賞鯨船可以成功看到鯨魚的次數為 E(X)=300 0.1=30次，變異數為 Var(X)=300 0.1 0.9=27二項分配n 明新有40的畢業生會踏入社會工作，假設畢業生可被重複的抽選。今從中隨機選出15個畢業生，則恰有10個是社會新鮮人的機率為何？至多有10個執政黨員的機率？已知 n=15，p=0.4，1-p=0.6，xb(15，0.4)(1)求個別機率：(2)求累積機率：二項分配拜訪3個顧客2個會購買的機率二項分配二項機率分配圖Example:Calculating a Binomial ProbabilityWhat is the probability of one success in five observations if the probability of success is 0.1?X=1,n=5,and p=0.1n=5 p=0.1n=5 p=0.5Mean 0.2.4.6012345XP(X).2.4.6012345XP(X)0Binomial DistributionnThe shape of the binomial distribution depends on the values of p and nnHere,n=5 and p=0.1nHere,n=5 and p=0.5Binomial Distribution CharacteristicsnMeannVariance and Standard DeviationWheren=sample sizep=probability of success(1 p)=probability of failureUsing Binomial Tablesn=10 xp=.20p=.25p=.30p=.35p=.40p=.45p=.500123456789100.10740.26840.30200.20130.08810.02640.00550.00080.00010.00000.00000.05630.18770.28160.25030.14600.05840.01620.00310.00040.00000.00000.02820.12110.23350.26680.20010.10290.03680.00900.00140.00010.00000.01350.07250.17570.25220.23770.15360.06890.02120.00430.00050.00000.00600.04030.12090.21500.25080.20070.11150.04250.01060.00160.00010.00250.02070.07630.16650.23840.23400.15960.07460.02290.00420.00030.00100.00980.04390.11720.20510.24610.20510.11720.04390.00980.0010109876543210p=.80p=.75p=.70p=.65p=.60p=.55p=.50 xExamples:n=10,p=0.35,x=3:P(x=3|n=10,p=0.35)=0.2522n=10,p=0.75,x=2:P(x=2|n=10,p=0.75)=0.0004應用二項機率分配來辨別某些人言論的真假 n有位小姐宣布，她可以判別奶茶中的牛奶、紅茶哪一樣東西先倒進茶杯中。n這位小姐的宣稱，並不表示她能夠百分之百的加以判斷而只是錯的時候少，對的時候多，換句話說，她的判斷並非完全出於猜測。應用二項機率分配來辨別某些人言論的真假n為了試驗這位小姐是否吹牛，就準備了組茶杯，每組茶杯有一杯是茶先倒進去，而另一杯是牛奶先倒進去，且設組茶杯的安排無任何關連。n則依二項機率分配知，這位小姐正確地分辨出其中x 組的機率為 其中p表這位小姐能正確分辨每組的哪一杯是哪一樣先倒入的機率。應用二項機率分配來辨別某些人言論的真假n假若這位小姐能在10組奶茶中至少判別8組，是否可認為她所言不虛？事實還很難判定！n若這位小姐很老練，譬如假設p=0.85，則P(X 8)=0.82，也就是說，這位小姐證實自己所言的機會很大；n若這位小姐僅靠運氣、猜測，即p=0.5，則P(X 8)=0.055，換句話說，這位小姐能證實自己所言的機會很小。哪一位運氣不好n在日常生活中常常會遇到某些大家都不願做的事情，但是又不得不去做，因此我們會想出某些奇特的方法，來決定哪一位運氣不好的必須去做n例如大家同時丟銅板，若其中有一人的結果與其他人的結果都不相同，則此人即為算輸的人，必須去做大家不願做的事，譬如出錢請大家吃飯！哪一位運氣不好n設N個人同時丟同樣的銅板，且設彼此間不互相影響。若每個銅板出現正面的機率是p，試求輸的人出局的機率為何？n解：N的銅板中若有(N-1)個出現正面或(N-1)個出現背面，輸的人出局的遊戲就結束。哪一位運氣不好n4個人玩“輸的人出局”的丟銅板遊戲，如前p=1/2，試求在第n次結束遊戲的機率為何？且設每次的遊戲不互相影響。解:每次遊戲中只有兩種可能的結果出現，即結束遊戲與繼續遊戲，結束遊戲的機率為1/8，而繼續遊戲的機率為7/8，所謂在第n次結束遊戲，也就是最先的(n-1)次都失敗，一直到第n次才成功，因此第n次結束“輸的人出局”的遊戲的機率為超幾何實驗The Hypergeometric Distributionn“n”trials in a sample taken from a finite population of size NnSample taken without replacementnOutcomes of trials are dependentnConcerned with finding the probability of“x”successes in the sample where there are“K”successes in the population超幾何分配n 從52張紙牌中抽出10張而包含4張黑桃的機率為何？超幾何分配Using the Hypergeometric Distribution3 different computers are checked from 10 in the department.4 of the 10 computers have illegal software loaded.What is the probability that 2 of the 3 selected computers have illegal software loaded?The probability that 2 of the 3 selected computers have illegal software loaded is 0.30,or 30%.Poisson分配n在任何兩個等長的區間內，事件發生的機率相同n在任何區間內事件的發生與否，與其他任何區間內事件始否發生無關，相互獨立。n實例：某段時間內，高速公路上的車禍；n實例：書本一面之內的錯字個數，都屬Poisson分配。Poisson,Simeon-Denisn生於西元1781年於法國的皮蒂維埃 n卒於西元1840年於法國-巴黎nPoisson是一位數學家，力學家和物理學家，在科學上的著作量很大，內容包含數學、物理與天文 Poisson分配的用途nPoisson分配是品質管制的利器，它可以幫助我們決定生產過程是否出了毛病。nPoisson分配還有種種的用途：放射性物質的蛻變、細胞間因受x光照射而引起的染色體交換次數、細菌和血球的計數、交通事故數及死亡率等等莫不遵行Poisson分配。n其實，無論在自然科學、在工業、在農業、在商業、在醫藥、在交通、在社會或在軍事上無不可找到Poisson分配的應用。Apply the Poisson Distribution when:nYou wish to count the number of times an event occurs in a given area of opportunitynThe probability that an event occurs in one area of opportunity is the same for all areas of opportunity nThe number of events that occur in one area of opportunity is independent of the number of events that occur in the other areas of opportunitynThe probability that two or more events occur in an area of opportunity approaches zero as the area of opportunity becomes smallernThe average number of events per unit is (lambda)Poisson(泊松)分配Poisson Distribution Formulawhere:X=number of events in an area of opportunity =expected number of eventse=base of the natural logarithm system (2.71828.)Poisson Distribution CharacteristicsnMeannVariance and Standard Deviationwhere =expected number of events 一到假日便人潮擁擠的捷運動物園站，為了提供乘客更好的服務，想了解捷運票卡刷卡故障的情形作為日後改進參考。在尖峰時間故障事件平均每小時會出現2件，試問某星期日早上9:0011:00的尖峰期，發生4件以上票卡刷卡故障的機率多少？沒有票卡故障的機率又是多少？沒有票卡故障：發生四件以上：Poisson(泊松)分配某航空公司每5年會發生一起飛安意外，那麼10年內都沒有掉過一架飛機的機率是多少？又10年內發生三起以上不幸的機率是多少？(1)沒掉飛機 (2)三起以上 1-F(2)=1-0.677=0.323Poisson(泊松)分配Using Poisson TablesX 0.100.200.300.400.500.600.700.800.90012345670.90480.09050.00450.00020.00000.00000.00000.00000.81870.16370.01640.00110.00010.00000.00000.00000.74080.22220.03330.00330.00030.00000.00000.00000.67030.26810.05360.00720.00070.00010.00000.00000.60650.30330.07580.01260.00160.00020.00000.00000.54880.32930.09880.01980.00300.00040.00000.00000.49660.34760.12170.02840.00500.00070.00010.00000.44930.35950.14380.03830.00770.00120.00020.00000.40660.36590.16470.04940.01110.00200.00030.0000Example:Find P(X=2)if =0.50Graph of Poisson ProbabilitiesX =0.50012345670.60650.30330.07580.01260.00160.00020.00000.0000P(X=2)=0.0758 Graphically:=0.50 Poisson Distribution ShapenThe shape of the Poisson Distribution depends on the parameter :=0.50=3.00二項分配與泊松分配二項分配 泊松分配 二項分配與泊松分配事件機率1-1二元間斷隨機變數一般化的聯合機率分配表Expected ValueThe CovariancenThe covariance measures the strength of the linear relationship between two variablesnThe covariance:where:X=discrete variable XXi =the ith outcome of XY=discrete variable YYi =the ith outcome of YP(XiYi)=probability of occurrence of the ith outcome of X and the ith outcome of YComputing the Mean for Investment ReturnsReturn per$1,000 for two types of investmentsP(XiYi)Economic condition Passive Fund X Aggressive Fund Y 0.2 Recession-$25 -$200 0.5 Stable Economy+50 +60 0.3 Expanding Economy+100 +350 InvestmentE(X)=X=(-25)(0.2)+(50)(0.5)+(100)(0.3)=50E(Y)=Y=(-200)(0.2)+(60)(0.5)+(350)(0.3)=95Computing the Standard Deviation for Investment ReturnsP(XiYi)Economic condition Passive Fund X Aggressive Fund Y 0.2 Recession-$25 -$200 0.5 Stable Economy+50 +60 0.3 Expanding Economy +100 +350 InvestmentComputing the Covariance for Investment ReturnsP(XiYi)Economic condition Passive Fund X Aggressive Fund Y 0.2 Recession-$25 -$200 0.5 Stable Economy+50 +60 0.3 Expanding Economy+100 +350 InvestmentInterpreting the Results for Investment ReturnsnThe aggressive fund has a higher expected return,but much more risk Y=95 X=50 butY=193.71 X=43.30nThe Covariance of 8250 indicates that the two investments are positively related and will vary in the same directionThe Sum of Two Random VariablesnExpected Value of the sum of two random variables:nVariance of the sum of two random variables:nStandard deviation of the sum of two random variables:Portfolio Expected Return and Portfolio RisknPortfolio expected return(weighted average return):nPortfolio risk(weighted variability)Where w=portion of portfolio value in asset X (1-w)=portion of portfolio value in asset YPortfolio Example Investment X:X=50 X=43.30 Investment Y:Y=95 Y=193.21 XY=8250Suppose 40%of the portfolio is in Investment X and 60%is in Investment Y:The portfolio return and portfolio variability are between the values for investments X and Y considered individually
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