Learning Objec0ves. Statistics for Business and Economics. Discrete Probability Distribu0ons

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1 Statistics for Business and Economics Discrete Probability Distribu0ons Learning Objec0ves In this lecture, you learn: The proper0es of a probability distribu0on To compute the expected value and variance of a probability distribu0on To calculate the covariance and understand its use in finance To compute probabili0es from binomial, hypergeometric, and Poisson distribu0ons How to use the binomial, hypergeometric, and Poisson distribu0ons to solve business problems

2 Defini0ons Random Variables A random variable represents a possible numerical value from an uncertain event. Discrete random variables produce outcomes that come from a coun0ng process (e.g. number of classes you are taking). Con8nuous random variables produce outcomes that come from a measurement (e.g. your annual salary, or your weight). Definitions Defini0ons Random Variables Random Variables Discrete Random Variable Continuous Random Variable

3 Discrete Random Variables Can only assume a countable number of values Examples: Roll a die twice Let X be the number of 8mes 4 occurs (then X could be 0, 1, or 8mes) Toss a coin 5 8mes. Let X be the number of heads (then X = 0, 1,, 3, 4, or 5) Probability Distribu0on For A Discrete Random Variable A probability distribu8on for a discrete random variable is a mutually exclusive lis0ng of all possible numerical outcomes for that variable and a probability of occurrence associated with each outcome. Number of Classes Taken Probability

4 Experiment: Toss Coins. 4 possible outcomes T T H H Example of a Discrete Random Variable Probability Distribu0on T H T H Let X = # heads. Probability Distribution X Value Probability Probability 0 1/4 = /4 = /4 = X Discrete Random Variables Expected Value (Measuring Center) Expected Value (or mean) of a discrete random variable (Weighted Average) µ = E(X) = N i= 1 x P( X = i x i ) Example: Toss coins, X = # of heads, compute expected value of X: E(X) = ((0)(0.5) + (1)(0.50) + ()(0.5)) = 1.0 X P(X=x i )

5 Discrete Random Variables Measuring Dispersion Variance of a discrete random variable σ = N i= 1 [x i E(X)] Standard Devia0on of a discrete random variable P(X = x ) i σ = σ = N i= 1 [x i E(X)] P(X = x ) i where: E(X) = Expected value of the discrete random variable X x i = the i th outcome of X P(X=x i ) = Probability of the i th occurrence of X Discrete Random Variables Measuring Dispersion (continued) Example: Toss coins, X = # heads, compute standard devia0on (recall E(X) = 1) σ = xi E(X)] P(X = x i ) σ = (0 1) (0.5) + (1 1) (0.50) + ( 1) (0.5) = 0.50 = Possible number of heads = 0, 1, or

6 Covariance The covariance measures the strength of the linear rela0onship between two discrete random variables X and Y. A posi0ve covariance indicates a posi0ve rela0onship. A nega0ve covariance indicates a nega0ve rela0onship. The Covariance Formula The covariance formula: σ XY = N i= 1 [ x i E( X )][( y i E( Y)] P( x y ) i i where: X = discrete random variable X X i = the i th outcome of X Y = discrete random variable Y Y i = the i th outcome of Y P(X i Y i ) = probability of occurrence of the i th outcome of X and the i th outcome of Y

7 Investment Returns The Mean Consider the return per $1000 for two types of investments. Prob. Economic Condition Passive Fund X Investment Aggressive Fund Y 0. Recession - $5 - $ Stable Economy + $50 + $ Expanding Economy + $100 + $350 Investment Returns The Mean E(X) = µ X = (-5)(.) +(50)(.5) + (100)(.3) = 50 E(Y) = µ Y = (-00)(.) +(60)(.5) + (350)(.3) = 95 Interpretation: Fund X is averaging a $50.00 return and fund Y is averaging a $95.00 return per $1000 invested.

8 Investment Returns Standard Devia0on σ σ X Y = (-5 50) (.) + (50 50) (.5) + (100 50) (.3) = = (-00 95) (.) + (60 95) (.5) + (350 95) (.3) = Interpretation: Even though fund Y has a higher average return, it is subject to much more variability and the probability of loss is higher. Investment Returns Covariance σ XY = (-5 50)(-00 95)(.) + (50 50)(60 95)(.5) + (100 50)(350 95)(.3) = 8,50 Interpretation: Since the covariance is large and positive, there is a positive relationship between the two investment funds, meaning that they will likely rise and fall together.

9 The Sum of Two Random Variables Expected Value of the sum of two random variables: E(X + Y) = E(X) + E(Y) Variance of the sum of two random variables: Var(X = σ + σ + + Y) = σx+ Y X Y σ XY Standard devia0on of the sum of two random variables: σ X+ Y = σx+ Y PorWolio Expected Return and Expected Risk Investment porwolios usually contain several different funds (random variables) The expected return and standard devia0on of two funds together can now be calculated. Investment Objec0ve: Maximize return (mean) while minimizing risk (standard devia0on).

10 PorWolio Expected Return and PorWolio Risk PorWolio expected return (weighted average return): E(P) = we(x) + (1 w)e(y) PorWolio risk (weighted variability) σ P = w σx + (1 w) σy + w(1- w)σ XY Where w = propor0on of porwolio value in asset X (1 - w) = propor0on of porwolio value in asset Y PorWolio Example Investment X: μ X = 50 σ X = Investment Y: μ Y = 95 σ Y = σ XY = 850 Suppose 40% of the porwolio is in Investment X and 60% is in Investment Y: E(P) = 0.4(50) + (0.6)(95) = 77 σ P = (0.4) = (43.30) + (0.6) (193.71) + (0.4)(0.6)(8,50) The porwolio return and porwolio variability are between the values for investments X and Y considered individually

11 Probability Distributions Probability Distributions Discrete Probability Distributions Binomial Poisson Hypergeometric Continuous Probability Distributions Normal Uniform Exponential Binomial Probability Distribu0on n n n A fixed number of observations, n n e.g., 15 tosses of a coin; ten light bulbs taken from a warehouse Each observation is categorized as to whether or not the event of interest occurred n e.g., head or tail in each toss of a coin; defective or not defective light bulb n Since these two categories are mutually exclusive and collectively exhaustive n When the probability of the event of interest is represented as π, then the probability of the event of interest not occurring is 1 - π Constant probability for the event of interest occurring (π) for each observation n Probability of getting a tail is the same each time we toss the coin

12 Binomial Probability Distribu0on n Observations are independent (continued) n The outcome of one observation does not affect the outcome of the other n Two sampling methods deliver independence n n Infinite population without replacement Finite population with replacement Possible Applica0ons for the Binomial Distribu0on A manufacturing plant labels items as either defec0ve or acceptable A firm bidding for contracts will either get a contract or not A marke0ng 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

13 The Binomial Distribu0on Coun0ng Techniques Suppose the event of interest is obtaining heads on the toss of a fair coin. You are to toss the coin three 0mes. In how many ways can you get two heads? Possible ways: HHT, HTH, THH, so there are three ways you can gekng two heads. This situa0on is fairly simple. We need to be able to count the number of ways for more complicated situa0ons. Coun0ng Techniques Rule of Combina0ons The number of combina0ons of selec0ng X objects out of n objects is C x n = n! x!(n x)! where: n! =(n)(n - 1)(n - )... ()(1) X! = (X)(X - 1)(X - )... ()(1) 0! = 1 (by definition)

14 Coun0ng Techniques Rule of Combina0ons How many possible 3 scoop combina0ons could you create at an ice cream parlor if you have 31 flavors to select from? The total choices is n = 31, and we select X = 3. 31! 31! ! 31 C 3 = = = = = 4,495 3!(31 3)! 3!8! 3 1 8! Binomial Distribu0on Formula P(X=x n,π) = n! x! ( n - x )! x n - x π (1-π) P(X=x n,π) = probability of x events of interest in n trials, with the probability of an event of interest being π for each trial x = number of events of interest in sample, (x = 0, 1,,..., n) n π = sample size (number of trials or observations) = probability of event of interest Example: Flip a coin four times, let x = # heads: n = 4 π = π = (1-0.5) = 0.5 X = 0, 1,, 3, 4

15 Example: Calcula0ng a Binomial Probability What is the probability of one success in five observations if the probability of an event of interest is 0.1? x = 1, n = 5, and π = 0.1 P(X = 1 5,0.1) = n! x π (1 π ) x!(n x)! 5! 1 = (0.1) (1 0.1) 1!(5 1)! = (5)(0.1)(0.9) = n x 5 1 The Binomial Distribu0on Example Suppose the probability of purchasing a defective computer is 0.0. What is the probability of purchasing defective computers in a group of 10? x =, n = 10, and π = 0.0 P(X = 10, 0.0) n! x = π (1 π ) x!(n x)! 10! = (.0) (1.0)!(10 )! = (45)(.0004)(.8508) = n x 10

16 The Binomial Distribu0on Shape The shape of the binomial distribu0on depends on the values of π and n n Here, n = 5 and π = P(X=x 5, 0.1) x n Here, n = 5 and π =.5 P(X=x 5, 0.5) x Binomial Distribu0on Characteris0cs Mean µ = E(X) = nπ n Variance and Standard Deviation σ σ = = nπ(1 -π) nπ(1-π) Where n = sample size π = probability of the event of interest for any trial (1 π) = probability of no event of interest for any trial

17 The Binomial Distribu0on Characteris0cs P(X=x 5, 0.1) x µ = nπ = (5)(.5) =.5 Examples µ = nπ = (5)(.1) = 0.5 σ = nπ(1-π) = (5)(.1)(1.1) = σ = nπ(1-π) = (5)(.5)(1.5) = P(X=x 5, 0.5) x The Poisson Distribu0on Defini0ons You use the Poisson distribu8on when you are interested in the number of 0mes an event occurs in a given area of opportunity. An area of opportunity is a con0nuous unit or interval of 0me, volume, or such area in which more than one occurrence of an event can occur. The number of scratches in a car s paint The number of mosquito bites on a person The number of computer crashes in a day

18 The Poisson Distribu0on Apply the Poisson Distribu0on when: You wish to count the number of 0mes an event occurs in a given area of opportunity The probability that an event occurs in one area of opportunity is the same for all areas of opportunity The number of events that occur in one area of opportunity is independent of the number of events that occur in the other areas of opportunity The probability that two or more events occur in an area of opportunity approaches zero as the area of opportunity becomes smaller The average number of events per unit is λ (lambda) Poisson Distribu0on Formula P( X = x λ) = e x λ x! λ where: x = number of events in an area of opportunity λ = expected number of events e = base of the natural logarithm system ( )

19 Poisson Distribu0on Characteris0cs Mean µ = λ n Variance and Standard Deviation σ = λ σ = λ where λ = expected number of events Further details on the Poisson distribu0on Study Section 4.8 from the text book

20 Graph of Poisson Probabili0es Graphically: λ = 0.50 X λ = P(X = λ=0.50) = Poisson Distribu0on Shape The shape of the Poisson Distribu0on depends on the parameter λ : λ = 0.50 λ = 3.00

21 The Hypergeometric Distribu0on The binomial distribu8on is applicable when selec0ng from a finite popula0on with replacement or from an infinite popula0on without replacement. The hypergeometric distribu8on is applicable when selec0ng from a finite popula0on without replacement. The Hypergeometric Distribu0on n trials in a sample taken from a finite popula0on of size N Sample taken without replacement Outcomes of trials are dependent Concerned with finding the probability of X=x i items of interest in the sample where there are A items of interest in the popula0on

22 Hypergeometric Distribu0on Formula P(X = x n,n,f) = [ FC x ][ N F C n x ] NC n = " F $ % # x & ' " N F % $ ' # n x & " N% $ ' # n & Where N = population size F = number of items of interest in the population N F = number of events not of interest in the population n = sample size x = number of items of interest in the sample n x = number of events not of interest in the sample Proper0es of the Hypergeometric Distribu0on The mean of the hypergeometric distribu0on is µ = E(X) = nf N The standard devia0on is σ = nf(n-f) N N-n N -1 Where N- n N -1 is called the Finite Popula0on Correc0on Factor used when sampling without replacement from a finite popula0on

23 Using the Hypergeometric Distribu0on Example: 3 different computers are selected from 10 in the department. 4 of the 10 computers have illegal soqware loaded. What is the probability that of the 3 selected computers have illegal soqware loaded? N = 10 n = 3 F = 4 x =! F # $ " x% &! N F $! 4 # & " n x % " # $ P(X = 3,10,4) = = % &! # 6$ " 1 % &! N$! 10$ # & # & " n % " 3 % = (6)(6) 10 = 0.3 The probability that of the 3 selected computers have illegal software loaded is 0.30, or 30%. First Summary Addressed the probability distribu0on of a discrete random variable Defined covariance and discussed its applica0on in finance Discussed the Binomial distribu0on Discussed the Poisson distribu0on Discussed the Hypergeometric distribu0on

24 (Addi8onal) Topic The Poisson Approxima0on To The Binomial Distribu0on Learning Objec0ves In this topic, you learn: When to use the Poisson distribu0on to approximate the binomial distribu0on How to use the Poisson distribu0on to approximate the binomial distribu0on

25 The Poisson Distribu0on Can Be Used To Approximate The Binomial Distribu0on Most useful when n is large and π is very small The approxima0on gets berer as n gets larger and π gets smaller The Formula For The Approxima0on e P( X ) nπ ( nπ ) X! X Where P(X) = probability of X events of interest given the parameters n and π n = sample size π = probability of an event of interest e = mathematical constant approximated by.7188 X = number of events of interest in the sample (X = 0, 1,,..., n)

26 The Mean & Standard Devia0on Of The Poisson Distribu0on µ = E( X ) = λ = nπ σ = λ = nπ Calcula0ng Probabili0es Suppose π = 0.08 and n = 0 then μ = 1.6 P( X = 1) = e 0*0.08 (0*0.08) 1! 1 = 0.330

27 Topic Summary In this topic, you learned: When to use the Poisson distribu0on to approximate the binomial distribu0on How to use the Poisson distribu0on to approximate the binomial distribu0on