Discrete Random Variables and Probability Distributions

Transcription

1 Chapter 4 Discrete Random Variables and Probability Distributions 4.1 Random Variables A quantity resulting from an experiment that, by chance, can assume different values. A random variable is a variable that takes on numerical values determined by the outcome of a random experiment. 1

2 2CHAPTER 4. DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS 4.1 Introduction to Probability Distributions Random Variable Represents a possible numerical value from a random experiment Random Variables Discrete Random Variable Continuous Random Variable Ch. 4 Ch. 5 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hal l Ch. 4-3 Figure 4.1: There are two types of random variables: 1. Discrete Random Variables A random variable is discrete if it can take on no more than a countable number of values. eg: = number of heads in two flips of a coin. 2. Continuous Random Variables A random variable is continuous if it can take on any valueinaninterval. eg: = time required to run 100 metres. Notation Capital letters will denote random variables. Lower case letters denote the outcome of a random variable. ( = ) represents the probability of the random variable having the outcome.

3 4.2. PROBABILITY DISTRIBUTIONS FOR DISCRETE RANDOM VARIABLES Probability Distributions for Discrete Random Variables We can characterize the behavior of a discrete random variable by attaching probabilities to each possible value,, that can take on. The probability distribution function, P(x), of a discrete random variable X expresses the probability that X takes the value x, as a function of x. That is P(x)=P(X=x), for all values of x. 4.3 A Probability Distribution () For a discrete random variable is a table, graph, or formula that shows all possible values that can assume along with associated probabilities. It is a complete (probability) description of the random variable. Notes: 0 ( = ) 1. P ( = ) =1.

4 4CHAPTER 4. DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS Discrete Probability Distribution Experiment: Toss 2 Coins. Let X = # heads. Show P(x), i.e., P(X = x), for all values of x: 4 possible outcomes T T T H H T H H Probability Distribution x Value Probability 0 1/4 = /4 = /4 =.25 Probability x Statistics for Business and Economics, 6e 2007 Pearson Education, Inc. Chap 5-5 Figure 4.2:

5 4.4. CUMULATIVE DISTRIBUTION FUNCTION Cumulative Distribution Function Let be a random variable, then the cumulative distribution ( 0 ), is the function: ( 0 )= ( 0 ) ie. () is the probability that the random variable takes on a value less than or equal to 0. Let be a discrete random variable which can take on the values 1 2, and 1 2.Then(for ): ( )= ( )= P =1 ( = ). 0 ( ) 1. If then ( ) ( ). ( 1 )= ( = 1 ). ( )=1.

6 6CHAPTER 4. DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS 4.5 Descriptive Measures For Discrete Random Variables Expected Value of a Discrete Random Variable The expected value or mean of a discrete random variable, denoted[] or, is [] = = X ( = ) it is a weighted average of all possible values of, the weights being the associated probabilities. ( = ) The expected value is a measure of central tendency in that the probability distribution of will be centered around (or balanced at).

7 4.5. DESCRIPTIVE MEASURES FOR DISCRETE RANDOM VARIABLES 7 Expected Value Expected Value (or mean) of a discrete distribution (Weighted Average) μ E(x) xp(x) x Example: Toss 2 coins, x = # of heads, compute expected value of x: E(x) = (0 x.25) + (1 x.50) + (2 x.25) = 1.0 x P(x) Statistics for Business and Economics, 6e 2007 Pearson Education, Inc. Chap 5-8 Figure 4.3: Note the expected value of is not necessarily a value that can assume. Consider tossing a fair coin once and let equal the number of heads observed. =0 1. ( =0)= ( =1)=5 [] = =5 Expected Value as the balancing point of the distribution [Transparency 5.2] Variance and Standard Deviation Variance measures the dispersion of around it s expected value. If [] =, then the variance of is:

8 8CHAPTER 4. DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS [] = 2 = [( ) 2 ]= X ( ) 2 ( = ) Notes: [] 0 We can show [] =[( [ ]) 2 ]=[ 2 ] 2 = { []} 12 0 is the standard deviation of X. We often write [] =and [] = Example of Expected Value and Variance Let = the number of heads in two straight tosses of a coin. = ( = ) ( = ) ( ) 2 ( = ) 2 ( = ) =1 2 = 5 [ 2 ]=15 Note [ 2 ] 2 =15 1=5 = 2.

9 4.5. DESCRIPTIVE MEASURES FOR DISCRETE RANDOM VARIABLES 9 Probability Distributions Probability Distributions Ch. 5 Discrete Continuous Ch. 6 Probability Probability Distributions Distributions Binomial Hypergeometric and Poisson (omit) Uniform Normal Exponential (omit) Statistics for Business and Economics, 6e 2007 Pearson Education, Inc. Chap 5-14 Figure 4.4: Chapter 4 discrete and Chapter 5 is continuous

10 10CHAPTER 4. DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS 4.6 Properties of Expectations Let and be two constants and be a discrete random variable Expectation of a Constant [] = X ( = ) = X ( = ) = [[]] = [] since [] is a constant. [] = X ( = ) = X ( = ) =() Expectation of a function of a random variable Let be a random variable with probability distribution ( = ). Let = () then [] =[()] = X () ( = ) Note setting () =( ) 2 gives the formula for the variance Expectation of a linear function of = + [] =[ + ] =[]+

11 4.7. PROPERTIES OF VARIANCE Expectation of a sum of random variables [ + ]=[]+[ ] [ ]=[] [ ] Expectation of the Product of Independent Variables If and are independent, [ ]=[][ ] 4.7 Properties of Variance Variance of a Constant [] =([ ()] 2 )=([ ] 2 )=0 [] =[( []) 2 ]=[() 2 2 []+([]) 2 ] = 2 ([ 2 ] 2[]+([]) 2 )= 2 [] Variance of a linear function of a random variable Let = + then [] = [ + ] = 2 []

12 12CHAPTER 4. DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS Variance of Independent Random variables If and are independent since: [ + ]= []+ [ ] [ + ]=([ + ( + )] 2 ) = ([( ()] 2 2( ())( ( )) + [ ( )] 2 ) = ([ ()] 2 )+([ ( )] 2 ) (by independence the cross product terms are zero). Variance of the sum is the sum of the variances for independent random variables [ ]= [] + [ ] Variance of the difference of independent random variables is the sum of the variances Questions: NCT Examples of Discrete Probability Distributions Binomial Distibution In order to apply the binomial distribution three conditions must hold. 1. There are a fixed number of trials,, of an experiment with only two possible outcomes for each trial: success or failure. These trials are called Bernoulli trials. 2. The probability of success on any trial,, is constant. Note that in earlier versions of notes = 3. The outcome of each trial is independent of every other trial.

13 4.8. EXAMPLES OF DISCRETE PROBABILITY DISTRIBUTIONS 13 Bernoulli Distribution Consider only two outcomes: success or failure Let P denote the probability of success Let 1 P be the probability of failure Define random variable X: x = 1 if success, x = 0 if failure Then the Bernoulli probability function is P(0) (1P) and P(1) P Statistics for Business and Economics, 6e 2007 Pearson Education, Inc. Chap 5-16 Figure 4.5:

14 14CHAPTER 4. DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS Notation = Probability of a success on a single trial 1 = Probability of a failure on a single trial = Number of trials = Number of successes ( is the binomial variable) Bernoulli Distribution Mean and Variance The mean is µ = P μ E(X) X xp(x) (0)(1P) (1)P P The variance is s 2 = P(1 P) σ 2 E[(X μ) (0 P) 2 2 ] X (x μ) P(x) 2 (1P) (1P) P P(1 P) 2 Statistics for Business and Economics, 6e 2007 Pearson Education, Inc. Chap 5-17 Figure 4.6:

15 4.8. EXAMPLES OF DISCRETE PROBABILITY DISTRIBUTIONS 15 Sequences of x Successes in n Trials The number of sequences with x successes in n independent trials is: C n x n! x!(n x)! Where n! = n (n 1) (n 2)... 1 and 0! = 1 These sequences are mutually exclusive, since no two can occur at the same time Statistics for Business and Economics, 6e 2007 Pearson Education, Inc. Chap 5-18 Figure 4.7:

16 16CHAPTER 4. DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS Possible Binomial Distribution Settings A manufacturing plant labels items as either defective or acceptable A firm bidding for contracts will either get a contract or not A marketing research firm receives survey responses of yes I will buy or no I will not New job applicants either accept the offer or reject it Statistics for Business and Economics, 6e 2007 Pearson Education, Inc. Chap 5-20 Figure 4.8:

17 4.8. EXAMPLES OF DISCRETE PROBABILITY DISTRIBUTIONS 17 Binomial Distribution Formula P(x) n! P (1- P) x! ( n x )! X n X 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 Example: Flip a coin four times, let x = # heads: n = 4 P = P = (1-0.5) = 0.5 x = 0, 1, 2, 3, 4 Statistics for Business and Economics, 6e 2007 Pearson Education, Inc. Chap 5-21 Figure 4.9:

18 18CHAPTER 4. DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS The formula for binomial probability is: ( = ) =!!( )! (1 ) for =0 1 2 Notes Usually we use the binomial distribution when sampling is done: a. with replacement so that trials are independent and is constant. b. withoutreplacementwhenthepopulationislargerelativeto (so that the change in from trial to trial is not significant). (1 ) is the probability of one sequence (of trials) containing successes and failures. = = is the number of sequences that contain successes and failures. () = (mean of a binomial variable) () =(1 ) (variance of a binomial) Each different combination of and results in a different binomial distribution Example of a Binomial Calculation Toss a fair coin three times and let be a success and a failure. We have =3=05 and = number of heads observed. Then: (we observe only one ) = ( =1)= () = =3 05 =15 µ 3 (05) 1 (05) 2 = () =(1 ) =3 (05) (05) = 3 4

19 4.8. EXAMPLES OF DISCRETE PROBABILITY DISTRIBUTIONS Cumulative Binomial Probabilities Example: Calculating a Binomial Probability What is the probability of one success in five observations if the probability of success is 0.1? x = 1, n = 5, and P = 0.1 n! X nx P(x 1) P (1 P) x!(n x)! 5! 1 (0.1) (10.1) 1!(5 1)! 4 (5)(0.1)(0.9) Statistics for Business and Economics, 6e 2007 Pearson Education, Inc. Chap 5-22 Figure 4.10:

20 20CHAPTER 4. DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS The shape of the binomial distribution depends on the values of P and n Mean P(x) n = 5 P = 0.1 Here, n = 5 and P = 0.1 Binomial Distribution x Here, n = 5 and P = 0.5 P(x) n = 5 P = x Statistics for Business and Economics, 6e 2007 Pearson Education, Inc. Chap 5-23 Figure 4.11:

21 4.8. EXAMPLES OF DISCRETE PROBABILITY DISTRIBUTIONS 21 Binomial Distribution Mean and Variance Mean μ E(x) np Variance and Standard Deviation σ 2 np(1-p) σ np(1-p) Where n = sample size P = probability of success (1 P) = probability of failure Statistics for Business and Economics, 6e 2007 Pearson Education, Inc. Chap 5-24 Figure 4.12: The cumulative probability function for the binomial distribution is: ( ) = X =0!!( )! (1 ) For instance ( 2 =3= 2) = 2X =0 3!!(3 )! 2 (8) Calculating and summing individual binomial probabilities can take a great deal of work! Values of the binomial and cummulative function are computed in NCT Appendix Table 2 and 3 for various values of and arefromatableatthebackof this chapter.

22 22CHAPTER 4. DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS Using Binomial Tables N x p=.20 p=.25 p=.30 p=.35 p=.40 p=.45 p= Examples: n = 10, x = 3, P = 0.35: P(x = 3 n =10, p = 0.35) =.2522 n = 10, x = 8, P = 0.45: P(x = 8 n =10, p = 0.45) =.0229 Statistics for Business and Economics, 6e 2007 Pearson Education, Inc. Chap 5-26 Figure 4.13:

23 4.9. SHAPE OF THE BINOMIAL DISTRIBUTION 23 Table 3 does the cummulative binomial probabilities Note that marginal probabilities can be calculated from the Table as follows: ( = ) = ( ) ( 1) Note as well: ( ) =1 ( 1) Table 2 only lists a few values of and, so we can approximate using the nearest values Example of Cumulative Binomial Distribution Managers for the State Department of Transportation know that 70 % of the cars arrive at a toll for a bridge have the correct change. If 20 cars pass through the toll in the next 5 minutes, what is the probabiltiy that between 10 and 15 cars, inclusive, have the correct change? Answer Let be the number of people with the correct change. Clearly this is a binomial problem (verify that it satisfies the 3 conditions set out earlier). Clearly we have independence of trials since the fact that one driver has the correct change in no way influences the next driver. From the problem we are given =20and = 70. However our tables only have up to.5 so we need to redefine the problem. Let be the number of people with wrong change so that we are looking for the interval between 5 and 10 inclusive with the wrong change and the appropriate = 3 (5 10 =20= 30) = ( 10) ( 4) = = Shape of the Binomial Distribution 1. As approaches.5 the distribution becomes more symmetric. a. If 5 the distribution is skewed to the right. b. If 5 the distribution is skewed to the left. 2. As increases the distribution becomes more bell-shaped.

24 24CHAPTER 4. DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS Binomial Characteristics Examples Mean μ np (5)(0.1) 0.5 σ np(1- P) (5)(0.1)(1 0.1) P(x) n = 5 P = x σ μ np (5)(0.5) 2.5 np(1- P) (5)(0.5)(1 0.5) P(x) n = 5 P = x Statistics for Business and Economics, 6e 2007 Pearson Education, Inc. Chap 5-25 Figure 4.14: 4.10 The Binomial Fraction of Successes We denote the fraction of succeses. Notes: = As X takes one of the values ; f takes the corresponding value and the ( = ) = ( = ), so the probability distribution of f is easily derived from the probability distribution of X. We can use the formulas developed earlier for finding the expectation and variance of a linear function of a random variable to find the expectation and variance for the fraction of successes.

25 4.11. JOINTLY DISTRIBUTED DISCRETE RANDOM VARIABLES 25 Recall if then and = + [] =[]+ () = 2 [] Applying this logic to the binomial: () =[ ]= 1 [] = 1 = () = [ ]= 1 2 () = 1 ) 2(1 ) =( Jointly Distributed Discrete Random Variables In this section we study the distribution of 2 discrete wandom variables Similar to Chapter 4 where we considered the marginal, conditional and joint distribution of two events and We also will show how two variables are linearly related Joint and Marginal Probabilities Let and be two discrete random variables, we can denote the joint probability distribution as: ( ) = ( = = ) To obtain the marginal distribution of, we need to sum over all possible values of = () = X ( ) = X ( = = ) To get the marginal distribution of, we need to sum over all possible values of = () = X ( ) = X ( = = )

26 26CHAPTER 4. DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS Since these are probabilities we have the following: 0 ( ) 1 X X ( ) = 1 () 0, () 0 X () = X () = Conditional Probabilities and Independence Joint Probability Functions A joint probability function is used to express the probability that X takes the specific value x and simultaneously Y takes the value y, as a function of x and y P(x, y) P(X x Y y) The marginal probabilities are P(x) P(x, y) y P(y) x P(x,y) Statistics for Business and Economics, 6e 2007 Pearson Education, Inc. Chap 5-45 Figure 4.15:

27 4.11. JOINTLY DISTRIBUTED DISCRETE RANDOM VARIABLES 27 Again, as in Chapter 4, we can use the conditional probability formual to define: ( ) = ( ) () and ( ) = ( ) () Independence implies that ( ) = () () ( ) = () ( ) = ()

28 28CHAPTER 4. DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS Expected Value : Function of Jointly Distributed Ran- Conditional Probability Functions The conditional probability function of the random variable Y expresses the probability that Y takes the value y when the value x is specified for X. P(y x) P(x, y) P(x) Similarly, the conditional probability function of X, given Y = y is: P(x, y) P(x y) P(y) Statistics for Business and Economics, 6e 2007 Pearson Education, Inc. Chap 5-46 Figure 4.16:

29 4.11. JOINTLY DISTRIBUTED DISCRETE RANDOM VARIABLES 29 dom Variables Let and be two discrete random variables with joint probaility density functions ( ). The expectation of any function ( ) is defined as: [( )] = X X ( ) ( ) Example from Text: Suppose Charlotte Kind has 2 stocks, and Assume that there are only 4 possible returns for each of these stocks with joint probability 0% 5% 10% 15% Y returns () X returns 0% 5% 10% 15% 0% % % % () Clearly these returns for and are independent (why?) Suppose that each stock costs a dollar and we have 1 and and 2 what is the expected net return of the portfollio ( )= +2 3 then (notice that we first need to express the returns as gross returns: ie. 0% return is agross return of 1.0 which is =(1+) =10 5% is a gross return of =105 and so on:) [( )] = X X ( +2 3) ( ) = X X ( +2) () () 3 because of independence = [( ) 25 25] + [( ) ( ) = ([]+2[]) = 3 = ( ) 3=22 on a $3 investment you are expected to earn 22 cents

30 30CHAPTER 4. DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS 4.12 Covariance Covariancetellsushowvariablesmovetogetherrelativetotheirmeans Does one variable tend to be high when another is low? ( a negative covariance) Or do variable move together so both are high (relative to their means) together (a positive covarinace) Or is there no association relative to their means ( a zero covariance) Definition of a covariance.. Let and be 2 discrete random variables with population means of and repsectively. The expected value of the product of ( ) ( ) is the covariance: ( ) = [( ) ( )] = X X ( ) ( ) ( ) = [ ] = X X ( ) Notice that the above expression is in units of x times y.

31 4.12. COVARIANCE 31 Covariance Let X and Y be discrete random variables with means µ X and µ Y The expected value of (X - µ X )(Y - µ Y ) is called the covariance between X and Y For discrete random variables Cov(X, Y) An equivalent expression is Cov(X, Y) E[(X μ )(Y μ )] (x μ )(y μ )P(x, y) X Y x y E(XY) μ μ x y x y xyp(x, y) μ μ x y x y Statistics for Business and Economics, 6e 2007 Pearson Education, Inc. Chap 5-48 Figure 4.17:

32 32CHAPTER 4. DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS 4.13 Correlation We can define a measure, called the correlation coeffcient which is unit free and boundintheintervalof( 1 1) = ( )= this is a measure of linear association 1 1 ( ) To see that we measure linear assocation, imagine that linear relation exists for the variables = + We know the following from earlier work = + = Substituting this in for ( ) = [( ) ( )] = [ ) ( + ( + )] = [[( )[( )] 2 Now we use this and substitute into the expression for 2 = = () 1 that is the correlation is ±1, depending on the sign of 4.14 Independence, Covariance and Correlation If the discrete random variables and are independent, ( ) = X X ( ) ( ) ( ) = X X ( ) ( ) () () = X ( ) () X ( () = 0 0 0

33 4.14. INDEPENDENCE, COVARIANCE AND CORRELATION 33 so that = ( )= ( ) = 0 =0 Note that zero covariance does not imply independence. Covariance measures linear association and independence is about any association (nonlinear as well)