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Chap 5-1 Chapter 5 Discrete Random Variables and Probability Distributions EF 507 QUANTITATIVE METHODS FOR ECONOMICS AND FINANCE FALL 2008.

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Presentation on theme: "Chap 5-1 Chapter 5 Discrete Random Variables and Probability Distributions EF 507 QUANTITATIVE METHODS FOR ECONOMICS AND FINANCE FALL 2008."— Presentation transcript:

1 Chap 5-1 Chapter 5 Discrete Random Variables and Probability Distributions EF 507 QUANTITATIVE METHODS FOR ECONOMICS AND FINANCE FALL 2008

2 Chap 5-2

3 Chap 5-3 Introduction to Probability Distributions  Random Variable  Represents a possible numerical value from a random experiment Random Variables Discrete Random Variable Continuous Random Variable Ch. 5Ch. 6

4 Chap 5-4 Discrete Random Variables  Can only take on a countable number of values Examples:  Roll a die twice Let X be the number of times 4 comes up (then X could be 0, 1, or 2 times)  Toss a coin 5 times. Let X be the number of heads (then X = 0, 1, 2, 3, 4, or 5)

5 Chap 5-5 Experiment: Toss 2 Coins. Let X = # heads. T T Discrete Probability Distribution 4 possible outcomes T T H H HH Probability Distribution x x Value Probability 0 1/4 = /4 = /4 = Probability Show P(x), i.e., P(X = x), for all values of x:

6 Chap 5-6  P(x)  0 for any value of x  The individual probabilities sum to 1; (The notation indicates summation over all possible x values) Probability Distribution Required Properties

7 Chap 5-7 Cumulative Probability Function  The cumulative probability function, denoted F(x 0 ), shows the probability that X is less than or equal to x 0  In other words,

8 Chap 5-8 Expected Value  Expected Value (or mean) of a discrete distribution (Weighted Average)  Example: 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)

9 Chap 5-9 Variance and Standard Deviation  Variance of a discrete random variable X  Standard Deviation of a discrete random variable X

10 Chap 5-10 Standard Deviation Example  Example: Toss 2 coins, X = # heads, compute standard deviation (recall E(x) = 1) Possible number of heads = 0, 1, or 2

11 Chap 5-11 Functions of Random Variables  If P (x) is the probability function of a discrete random variable X, and g(X) is some function of X, t hen the expected value of function g is

12 Chap 5-12 Linear Functions of Random Variables  Let a and b be any constants.  a) i.e., if a random variable always takes the value a, it will have mean a and variance 0  b) i.e., the expected value of b·X is b·E(x)

13 Chap 5-13 Linear Functions of Random Variables  Let random variable X have mean µ x and variance σ 2 x  Let a and b be any constants.  Let Y = a + bX  Then the mean and variance of Y are  so that the standard deviation of Y is (continued)

14 Chap 5-14 Probability Distributions Continuous Probability Distributions Binomial Hypergeometric Poisson Probability Distributions Discrete Probability Distributions Uniform Normal Exponential Ch. 5Ch. 6

15 Chap 5-15 The Binomial Distribution Binomial Hypergeometric Poisson Probability Distributions Discrete Probability Distributions

16 Chap 5-16 Bernoulli Distribution  Consider only two outcomes: “success” or “failure”  Let P denote the probability of success (1)  Let 1 – P be the probability of failure (0)  Define random variable X: x = 1 if success, x = 0 if failure  Then the Bernoulli probability function is

17 Chap 5-17 Bernoulli Distribution Mean and Variance  The mean is µ = P  The variance is σ 2 = P(1 – P)

18 Chap 5-18 Sequences of x Successes in n Trials  The number of sequences with x successes in n independent trials is : 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

19 Chap 5-19 Binomial Probability Distribution 1-A fixed number of observations, n  e.g., 15 tosses of a coin; ten light bulbs taken from a warehouse 2-Two mutually exclusive and collectively exhaustive categories  e.g., head or tail in each toss of a coin; defective or not defective light bulb  Generally called “success” and “failure”  Probability of success is P, probability of failure is 1 – P 3-Constant probability for each observation  e.g., Probability of getting a tail is the same each time we toss the coin 4-Observations are independent  The outcome of one observation does not affect the outcome of the other

20 Chap 5-20 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

21 Chap 5-21 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) n x ! nx P(1- P) X n X ! ( ) !    Example: Flip a coin four times, let x = # heads: n = 4 P = P = ( ) = 0.5 x = 0, 1, 2, 3, 4 Binomial Distribution Formula

22 Chap 5-22 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

23 Chap 5-23 n = 5 P = 0.1 n = 5 P = 0.5 Mean x P(x) x P(x) 0 Binomial Distribution  The shape of the binomial distribution depends on the values of P and n  Here, n = 5 and P = 0.1  Here, n = 5 and P = 0.5

24 Chap 5-24 Binomial Distribution Mean and Variance  Mean  Variance and Standard Deviation Wheren = sample size P = probability of success (1 – P) = probability of failure

25 Chap 5-25 n = 5 P = 0.1 n = 5 P = 0.5 Mean x P(x) x P(x) 0 Binomial Characteristics Examples

26 Chap 5-26 Using Binomial Tables (Table 2, p.844) Nx…p=.20p=.25p=.30p=.35p=.40p=.45p= ………………………………………………………… Examples: n = 10, x = 3, P = 0.35: P(x = 3|n =10, p = 0.35) = n = 10, x = 8, P = 0.45: P(x = 8|n =10, p = 0.45) =

27 Chap 5-27 Using PHStat  Select PHStat / Probability & Prob. Distributions / Binomial…

28 Chap 5-28 Using PHStat  Enter desired values in dialog box Here:n = 10 p =.35 Output for x = 0 to x = 10 will be generated by PHStat Optional check boxes for additional output (continued)

29 Chap 5-29 P(x = 3 | n = 10, P =.35) =.2522 PHStat Output P(x > 5 | n = 10, P =.35) =.0949

30 Chap 5-30 The Hypergeometric Distribution Binomial Poisson Probability Distributions Discrete Probability Distributions Hypergeometric

31 Chap 5-31 The Hypergeometric Distribution  “n” trials in a sample taken from a finite population of size N  Sample taken without replacement  Outcomes of trials are dependent  Concerned with finding the probability of “X” successes in the sample where there are “S” successes in the population

32 Chap 5-32 Hypergeometric Distribution Formula Where N = population size S = number of successes in the population N – S = number of failures in the population n = sample size x = number of successes in the sample n – x = number of failures in the sample

33 Chap 5-33 Using the Hypergeometric Distribution ■Example: 3 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? N = 10n = 3 S = 4 x = 2 The probability that 2 of the 3 selected computers have illegal software loaded is 0.30, or 30%.

34 Chap 5-34 Hypergeometric Distribution in PHStat  Select: PHStat / Probability & Prob. Distributions / Hypergeometric …

35 Chap 5-35 Hypergeometric Distribution in PHStat  Complete dialog box entries and get output … N = 10 n = 3 S = 4 x = 2 P(X = 2) = 0.3 (continued)

36 Chap 5-36 The Poisson Distribution Binomial Hypergeometric Poisson Probability Distributions Discrete Probability Distributions

37 Chap 5-37 The Poisson Distribution  Apply the Poisson Distribution when: 1-You wish to count the number of times an event occurs in a given continuous interval 2-The probability that an event occurs in one subinterval is very small and is the same for all subintervals 3-The number of events that occur in one subinterval is independent of the number of events that occur in the other subintervals 4-There can be no more than one occurrence in each subinterval 5-The average number of events per unit is (lambda)

38 Chap 5-38 Poisson Distribution Formula where: x = number of successes per unit = expected number of successes per unit e = base of the natural logarithm system ( )

39 Chap 5-39 Poisson Distribution Characteristics  Mean  Variance and Standard Deviation where = expected number of successes per unit

40 Chap 5-40 Using Poisson Tables (Table 5, p.853) X Example: Find P(X = 2) if = 0.50

41 Chap 5-41  Example 1: The average number of trucks arriving on any one day at a truck depot in a certain city is known to be 12. What is the probability that on a given day fewer than nine trucks will arrive at this depot?  X: number of trucks arriving on a given day.  Lambda=12  P(X<9) = SUM[p(x;12)] from 0 to 8 =  Example 2: The number of complaints that a call center receives per day is a random variable having a Poisson distribution with lambda=3.3. Find the probability that the call center will receive only two complaints on any given day.

42 Chap 5-42 Graph of Poisson Probabilities X = P(X = 2) = Graphically: = 0.50

43 Chap 5-43 Poisson Distribution Shape  The shape of the Poisson Distribution depends on the parameter : = 0.50 = 3.00

44 Chap 5-44 Poisson Distribution in PHStat  Select: PHStat / Probability & Prob. Distributions / Poisson…

45 Chap 5-45 Poisson Distribution in PHStat  Complete dialog box entries and get output … P(X = 2) = (continued)

46 Chap 5-46 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  The marginal probabilities are

47 Chap 5-47 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.  Similarly, the conditional probability function of X, given Y = y is:

48 Chap 5-48 Independence  The jointly distributed random variables X and Y are said to be independent if and only if their joint probability function is the product of their marginal probability functions: for all possible pairs of values x and y  A set of k random variables are independent if and only if

49 Chap 5-49 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  An equivalent expression is

50 Chap 5-50 Covariance and Independence  The covariance measures the strength of the linear relationship between two variables  If two random variables are statistically independent, the covariance between them is 0  The converse is not necessarily true Example: mean_x=0; mean_y= -1/3; E(xy)=0; sigma_xy=0 So the covariance is 0, but for x=-1 and y=-1 f(x,y) is not equal to g(x)h(y)

51 Chap 5-51 Correlation  The correlation between X and Y is:  ρ = 0  no linear relationship between X and Y  ρ > 0  positive linear relationship between X and Y  when X is high (low) then Y is likely to be high (low)  ρ = +1  perfect positive linear dependency  ρ < 0  negative linear relationship between X and Y  when X is high (low) then Y is likely to be low (high)  ρ = -1  perfect negative linear dependency

52 Chap 5-52 Portfolio Analysis  Let random variable X be the price for stock A  Let random variable Y be the price for stock B  The market value, W, for the portfolio is given by the linear function (a is the number of shares of stock A, b is the number of shares of stock B)

53 Chap 5-53 Portfolio Analysis  The mean value for W is  The variance for W is or using the correlation formula (continued)

54 Chap 5-54 Portfolio Example Investment X: μ x = 50 σ x = Investment Y: μ y = 95 σ y = σ xy = 8250 Suppose 40% of the portfolio (P) 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

55 Chap 5-55 Interpreting the Results for Investment Returns  The aggressive fund has a higher expected return, but much more risk μ y = 95 > μ x = 50 but σ y = > σ x =  The Covariance of 8,250 indicates that the two investments are positively related and will vary in the same direction

56 Chap 5-56 Example: Investment Returns Return per $1,000 for two types of investments P(x i y i ) Economic condition Passive Fund X Aggressive Fund Y 0.2 Recession- $ 25 - $ Stable Economy Expanding Economy Investment E(x) = μ x = (-25)(0.2) +(50)(0.5) + (100)(0.3) = 50 E(y) = μ y = (-200)(0.2) +(60)(0.5) + (350)(0.3) = 95

57 Chap 5-57 Covariance for Investment Returns P(x i y i ) Economic condition Passive Fund X Aggressive Fund Y 0.2 Recession- $ 25 - $ Stable Economy Expanding Economy Investment

58 Chap 5-58 Computing the Standard Deviation for Investment Returns P(x i y i ) Economic condition Passive Fund X Aggressive Fund Y 0.2 Recession- $ 25 - $ Stable Economy Expanding Economy Investment


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