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Ch 7 實習 助教介紹 2 助教: 王珊彗 身分:博士班,6年級 聯絡方式: Shanhueiwang@gmail.com 工作:實習課、出作業、寄信~~ 作業問題 回信問題 Shan Huei, Wang 實習課時間(暫定,依老師進度修改) 星期四 日期 Random Variables and Discrete Probability Ch 7 Distributions 10月23日 Continuous Probability Distributions Ch 8 電腦(三) 日期 電腦(四) 10月16日 10月30日 Sampling Distributions 10月22日 (Ch7-8) 10月23日 (Ch7-8) Ch 9 11月5日 project 11月6日 Review 11月6日 project 11月13日 Mid-term Exam 實習(五) 作業 繳交時間 10月17日 V HW4 (Ch7) 10月24日 V HW5 (Ch8,電腦7-8) 10月30日 10月31日 V HW6 (Ch9)+複習題目 11月6日 X project X X 10月23日 11月27日 11月20日 Statistical Inference: Estimation Ch 10 11月21日 V X 11月27日 Statistical Inference: Estimation Ch 10 11月28日 V HW7 (ch 10) 12月4日 12月4日 Statistical Inference: Hypothesis Testing Ch 11 12月5日 V HW8 (ch11) 12月11日 12月11日 Statistical Inference: Hypothesis Testing Ch 11 12月12日 V HW9(Ch10-11電腦) 12月18日 12月19日 V HW10(ch12,電腦) 12月25日 12月26日 V X X X 12月10日 (Ch10-11) 12月11日 (Ch10-11) 12月18日 Statistical Inference: Inference about a Population Ch 12 12月25日 Statistical Inference: Inference about a Population Ch 12 1月8日 3 日期 Final Exam 12月24日 (Ch12)+office hour 12月25日 (Ch12)+office hour Shan Huei, Wang Agenda Random variable and Probability Distribution Overall review of Chapter 7 and 8 Poisson Distribution Binomial distribution 4 The Mean and the Variance Bivariate Distributions Marginal Probabilities Portfolio concept Shan Huei, Wang 1. Random Variables and Probability Distributions 骰子出現: 1,3,4,5,2,4,5,6,3,2, 1,1,2,3 若有100次結果，你要如何告訴別人? A random variable is a function or rule that assigns a numerical value to each simple event in a sample space. 為了降低分析的複雜性，將所有可能結果加以數值化 例如投銅板十次，正面出現次數的事件就是random variable 短期不知道是什麼，長期下來會呈現某種分配 5 Shan Huei, Wang 1. Random Variables and Probability Distributions 6 There are two types of random variables: Discrete random variable (C7) Continuous random variable (C 8) Shan Huei, Wang 1.1 Discrete Probability Distribution A table, formula, or graph that lists all possible values a discrete random variable can assume, together with associated probabilities, is called a discrete probability distribution To calculate the probability that the random variable X assumes the value x, P(X = x), 7 add the probabilities of all the simple events for which X is equal to x, or Use probability calculation tools (tree diagram), Apply probability definitions Shan Huei, Wang Example 1 The number of cars a dealer is selling daily were recorded in the last 100 days. This data was summarized in the table below. Estimate the probability distribution, and determine the probability of selling more than 2 cars a day. Daily sales Frequency 0 5 1 15 2 35 3 25 4 20 100 8 Shan Huei, Wang Solution From the table of frequencies we can calculate the relative frequencies, which becomes our estimated probability distribution Daily sales Relative Frequency 0 5/100=.05 1 15/100=.15 2 35/100=.35 3 25/100=.25 4 20/100=.20 1.00 9 .35 .25 .15 .20 .05 0 1 2 3 4 X P(X>2) = P(X=3) + P(X=4) = .25 + .20 = .45 Shan Huei, Wang Example 2 10 NTU Company tracks the number of desktop computer systems it sells over a year (360 days), assume they take the two-day order policy, please use the following information to decide the best order number so that they can satisfy 70% customers’ need. Note: For example, 26 days out of 360, 2 desktops were sold Shan Huei, Wang 1. 1 Describing the Population/ Probability Distribution 11 The probability distribution represents a population We’re interested in describing the population by computing various parameters. Specifically, we calculate the population mean and population variance. Shan Huei, Wang 1.1 Population Mean (Expected Value) and Population Variance Given a discrete random variable X with values xi, that occur with probabilities p(xi), the population mean of X is. 加權平均概念(權數是機率) E(X ) m x i p( x i ) all x i Let X be a discrete random variable with possible values xi that occur with probabilities p(xi), and let E(xi) = m. The variance of X is defined by V ( X ) 2 E[( X - m) 2 ] ( x i - m) 2 p( x i ) all xi 12 The s tan dard deviation is 2 Shan Huei, Wang 1.1 Laws of Expected Value and Variance Laws of Expected Value Laws of Variance 13 E(c) = c E(X + c) = E(X) + c E(cX) = cE(X) V(c) = 0 V(X + c) = V(X) V(cX) = c2V(X) Shan Huei, Wang 1.1 The Mean and the Variance The variance can also be calculated as follows: V ( X ) = s 2 = E( X 2 ) - m 2 = å xi2 p(xi ) - m 2 all xi Proof E [( X - m ) ] 2 E[ X 2 - 2X m m ] 2 E ( X ) - 2m E ( X ) E (m ) 2 E(X ) - m 2 14 2 2 Shan Huei, Wang Example 2 We are given the following probability distribution: x 0 P(x) 0.4 15 1 2 3 0.3 0.2 0.1 a. Calculate the mean, variance, and standard deviation b. Suppose that Y=3X+2. For each value of X, determine the value of Y. What is the probability distribution of Y? c. Calculate the mean, variance, and standard deviation from the probability distribution of Y. d. Use the laws of expected value and variance to calculate the mean, variance, and standard deviation of Y from the mean, variance, and standard deviation of X. Compare your answers in Parts c and d. Are they the same (except for rounding) Shan Huei, Wang Solution a. E(X) = 0(0.4) +1(0.3) + 2(0.2) +3(0.1) = 1 V(X)=E(X2)-{E(X)}2 =(0)2(0.4)+(1)2(0.3)+(2)2(0.2)+(3)2(0.1)-(1)2 =1 (X ) 1 b. x y P(y) 16 0 2 0.4 1 5 0.3 2 8 0.2 3 11 0.1 Shan Huei, Wang Solution c. E(Y) = 2(0.4) + 5(0.3) + 8(0.2) + 11(0.1) = 5 V(Y)=E(Y2)-{E(Y)}2 =(2)2(0.4)+(5)2(0.3)+(8)2(0.2)+(11)2(0.1)-(5)2 =9 d. 17 (Y ) 3 E(Y) = E(3X+2) = 3E(X)+2 = 5 V(Y) = V(3X+2) = 9V(X) = 9 Shan Huei, Wang 1.2 Bivariate Distributions The bivariate (or joint) distribution is used when the relationship between two random variables is studied. 也就是第六章所看到的聯合機率分配 The probability that X assumes the value x, and Y assumes the value y is denoted p(x,y) = P(X=x and Y = y) The joint probabilit y function satisfies the following conditions : 1. 0 p(x, y) 1 2. p(x, y) 1 18 all x all y Shan Huei, Wang 1.2 Bivariate Distributions 19 Example 7.5 Xavier and Yvette are two real estate agents. Let X and Y denote the number of houses that Xavier and Yvette will sell next week, respectively. The bivariate probability distribution is presented next. Shan Huei, Wang 1.2 Bivariate Distributions 0.42 p(x,y) Example 7.5 – continued Y 0 1 2 0.21 X 1 .42 .06 .02 0 .12 .21 .07 2 .06 .03 .01 0.12 0.06 0.06 0.07 0.02 0.01 Y 20 X=0 y=0 X 0.03 y=1 y=2 X=1 X=2 Shan Huei, Wang 1.3 Marginal Probabilities Example 7.5 – continued Sum across rows and down columns p(0,0) p(0,1) p(0,2) Y 0 1 2 p(x) 0 .12 .21 .07 .40 X 1 .42 .06 .02 .50 2 .06 .03 .01 .10 p(y) .60 .30 .10 1.00 P(Y=1), the marginal probability. The marginal probability P(X=0) 21 Shan Huei, Wang 1.3 Describing the Bivariate Distribution The joint distribution can be described by the mean, variance, and standard deviation of each variable. This is done using the marginal distributions. Y 0 1 2 p(x) 22 0 .12 .21 .07 .40 X 1 .42 .06 .02 .50 2 .06 .03 .01 .10 p(y) .60 .30 .10 1.00 x 0 1 2 p(x) E(X) = V(X) = y 0 1 2 p(y) E(Y) = V(Y) = Shan Huei, Wang 1.3 Describing the Bivariate Distribution The joint distribution can be described by the mean, variance, and standard deviation of each variable. This is done using the marginal distributions. Y 0 1 2 p(x) 23 0 .12 .21 .07 .40 X 1 .42 .06 .02 .50 2 .06 .03 .01 .10 p(y) .60 .30 .10 1.00 x 0 1 2 p(x) .4 .5 .1 E(X) = .7 V(X) = .41 y 0 1 2 p(y) .6 .3 .1 E(Y) = .5 V(Y) = .45 Shan Huei, Wang 1.3 Describing the Bivariate Distribution To describe the relationship between the two variables we compute the covariance and the coefficient of correlation Covariance: Coefficient of Correlation 24 COV(X,Y) = S(X – mx)(Y- my)p(x,y)=E(XY)-E(X)E(Y) COV(X,Y) xy Shan Huei, Wang 1.3 Describing the Bivariate Distribution Example 7.6 Calculate the covariance and coefficient of correlation between the number of houses sold by the two agents in Example 7.5 Solution = S(x-mx)(y-my)p(x,y) = (0-.7)(0-.5)p(0,0)+…(2-.7)(2-.5)p(2,2) = -.15 r=COV(X,Y)/xy = - .15/(.64)(.67) = -.35 COV(X,Y) Y 0 1 2 p(x) 25 0 .12 .21 .07 .40 X 1 .42 .06 .02 .50 2 .06 .03 .01 .10 p(y) .60 .30 .10 1.00 Shan Huei, Wang 1.3 Sum of Two Variables The probability distribution of X + Y is determined by Example 7.5 - continued 26 Determining all the possible values that X+Y can assume For every possible value C of X+Y, adding the probabilities of all the combinations of X and Y for which X+Y = C Find the probability distribution of the total number of houses sold per week by Xavier and Yvette.( X:0-2, Y: 0-2) Solution X+Y is the total number of houses sold. X+Y can have the values 0, 1, 2, 3, 4. Shan Huei, Wang 1.3 The Probability Distribution of X+Y The distribution of X+Y x+y p(x+y) 27 0 .12 1 .63 2 .19 3 .05 4 .01 如何求得Ｐ（Ｘ＋Ｙ）？ Shan Huei, Wang 1.3 The Probability Distribution of X+Y P(X+Y=0) = P(X=0 and Y=0) = .12 P(X+Y=1) = P(X=0 and Y=1)+ P(X=1 and Y=0) =.21 + .42 = .63 P(X+Y=2) = P(X=0 and Y=2)+ P(X=1 and Y=1)+ P(X=2 and Y=0) = .07 + .06 + .06 = .19 Y 0 1 2 p(x) 0 .12 .21 .07 .40 X 1 .42 .06 .02 .50 2 .06 .03 .01 .10 p(y) .60 .30 .10 1.00 The probabilities P(X+Y)=3 and P(X+Y) =4 are calculated the same way. The distribution follows 28 Shan Huei, Wang 1.3 The Expected Value and Variance of X+Y The distribution of X+Y x+y p(x+y) 1 .63 2 .19 3 .05 4 .01 The expected value and variance of X+Y can be calculated from the distribution of X+Y. 29 0 .12 E(X+Y)=0(.12)+ 1(63)+2(.19)+3(.05)+4(.01)=1.2 V(X+Y)=(0-1.2)2(.12)+(1-1.2)2(.63)+… =.56 Shan Huei, Wang 1.3 The Expected Value and Variance of X+Y The following relationship can assist in calculating E(X+Y) and V(X+Y) E(X+Y) =E(X) + E(Y); V(X+Y) = V(X) +V(Y) +2COV(X,Y) When X and Y are independent COV(X,Y) = 0, and V(X+Y) = V(X)+V(Y). Proof Var ( X Y ) E [( X Y - m x - m y ) ] 2 E [( X - m x ) (Y - m y ) 2 ( X - m x )( Y - m y )] 2 2 Var ( X ) Var (Y ) 2 Cov ( X , Y ) 30 Shan Huei, Wang Example 3 The bivariate distribution of X and Y is described here. x 31 y 1 2 1 0.28 0.42 2 0.12 0.18 a. Find the marginal probability distribution of X. b. Find the marginal probability distribution of Y c. Compute the mean and variance of X d. Compute the mean and variance of Y e. Compute the covariance and variance of Y Shan Huei, Wang Solution 32 a x P(x) 1 .4 2 .6 b y P(y) 1 .7 2 .3 c E(X) = 1(.4) + 2(.6) = 1.6 V(X) = (1–1.6)^2*(.4) + (2–1.6)^2*(.6) = .24 or (1^2)*0.4+(2^2)*0.6-(1.6)^2=.24 Shan Huei, Wang Solution d E(Y) = 1(.7) + 2(.3) = 1.3 V(Y) = (1–1.3)^2(.7) + (2–1.3)^2(.3) = .21 e E(XY) = (1)(1)(.28) + (1)(2)(.12) + (2)(1)(.42) + (2)(2)(.18) = 2.08 COV(X, Y) = E(XY)–E(X)E(Y) = 2.08 – (1.6)(1.3) = 0 r 33 COV ( X , Y ) xy =0 Shan Huei, Wang 2. Overall review of Chapter 7 and 8 • • Discrete Probability Distributions Binomial distribution Poisson Distribution • • • • • • 34 Continuous Probability Distributions Uniform Distribution Normal Distribution Standard Normal Distribution Exponential Distribution t Distribution F Distribution Shan Huei, Wang 3. The Binomial Distribution The binomial experiment can result in only one of two possible outcomes. Binomial Experiment Binomial Random Variable 35 There are n trials (n is finite and fixed). Each trial can result in a success or a failure. The probability p of success is the same for all the trials. All the trials of the experiment are independent The binomial random variable counts the number of successes in n trials of the binomial experiment. By definition, this is a discrete random variable (因為數的完). Shan Huei, Wang 3. Calculating the Binomial Probability In general, The binomial probability is calculated by: P ( X x ) p ( x ) C x p (1 - p ) n w here n Cx x n- x n! x ! ( n - x )! Mean and Variance of Binomial Variable E(X) m np V(X) s np(1 - p) 2 36 Shan Huei, Wang 3. Binomial distribution 37 Binomial distribution for n = 20 p = 0.1 (blue), p = 0.5 (green) and p = 0.8 (red) Shan Huei, Wang Example 4 a. b. 38 In the game of blackjack as played in casinos in Las Vegas, Atlantic City, Niagara Falls, as well as many other cities, the dealer has the advantages. Most players do not play very well. As a result, the probability that the average player wins a about 45%.Find the probability that an average player wins twice in 5 hands ten or more times in 25 hands Shan Huei, Wang Solution 5! (.45) (1 - .45) 2 5-2 a P(X = 2) = b Excel with n = 25 and p = .45: P(X 10) = 1 – P(X 9) = 1 – .2424 = .7576 2 !(5 - 2) ! = .3369 =BINOM.DIST(9,25,0.45,1) 39 Shan Huei, Wang 4. The Poisson Variable and Distribution The Poisson Random Variable (單位時間的來客數) The Poisson variable indicates the number of successes that occur during a given time interval or in a specific region in a Poisson experiment Probability Distribution of the Poisson Random Variable. -m x e m P(X x ) p( x ) x 0 , 1, 2 ... x! E(X ) V (X ) m 40 Shan Huei, Wang Poisson distribution A famous quote of Poisson is: "Life is good for only two things: Discovering mathematics and teaching mathematics." 41 Shan Huei, Wang Example 5 a. b. 42 The number of students who seek assistance with their statistics assignments is Poisson distributed with a mean of three per day. What is the probability that no student seek assistance tomorrow? (u? X?) Find the probability that 10 students seek assistance in a week. (u? X?) Shan Huei, Wang Solution 43 e -m m x a. P(X = 0 with μ = 3) = x! = = .0498 -m m e (3) 0 0! x b. P(X = 10 with μ = 21) = x! = = .0035 e -3 e - 21 ( 21 ) 10 10 ! Shan Huei, Wang