1. MGMT 242 Spring, 1999 Random Variables and Probability Distributions Chapter 4 “Never draw to an inside straight.” from Maxims Learned at My Mother’s Knee

2. MGMT 242 Spring, 1999 Goals for Chapter 4 Define Random Variable, Probability Distribution Understand and Calculate Expected Value Understand and Calculate Variance, Standard Deviation Two Random Variables--Understand –Statistical Independence of Two Random Variables –Covariance & Correlation of Two Random Variables Applications of the Above

3. MGMT 242 Spring, 1999 Random Variables Refers to possible numerical values that will be the outcome of an experiment or trial and that will vary randomly;Refers to possible numerical values that will be the outcome of an experiment or trial and that will vary randomly; Notation: uppercase letters for last part of alphabet--X, Y, ZNotation: uppercase letters for last part of alphabet--X, Y, Z Derived from hypothetical population (infinite number), the members of which will have different values of the random variableDerived from hypothetical population (infinite number), the members of which will have different values of the random variable Example--let Y = height of females between 18 and 20 years of age; population is infinite number of females between 18 and 20Example--let Y = height of females between 18 and 20 years of age; population is infinite number of females between 18 and 20 Actual measurements are carried out on a sample, which is randomly chosen from population.Actual measurements are carried out on a sample, which is randomly chosen from population.

4. MGMT 242 Spring, 1999 Probability Distributions A probability distribution gives the probability for a specific value of the random variable:A probability distribution gives the probability for a specific value of the random variable: –P(Y= y) gives the probability distribution when values for P are specified for specific values of y –example: tossed a coin twice(fair coin); Y is the number of heads that are tossed; possible events: TT, TH, HT, HH; each event is equally probable if coin is a fair coin, so probability of Y=0 (event: TT) is 1/4; probability of Y =2 is 1/4; probability of Y = 1 is 1/4 +1/4 = 1/2; –Thus, for example: P(Y=0) = 1/4 P(Y=1) = 1/2 P(Y=2) = 1/4

5. MGMT 242 Spring, 1999 Probability Distributions--Discrete Variables Notation: P(Y=y) or, occasionally, P Y (y) (with y being some specific value;Notation: P(Y=y) or, occasionally, P Y (y) (with y being some specific value; P Y (y) is some value when Y=y and 0 otherwiseP Y (y) is some value when Y=y and 0 otherwise Example--(Ex. 4.1) 8 people in a business group, 5 men, 3 women. Two people sent out on a recruiting trip. If people randomly chosen, find P Y (y) for Y being the number of women sent out on the recruiting trip.Example--(Ex. 4.1) 8 people in a business group, 5 men, 3 women. Two people sent out on a recruiting trip. If people randomly chosen, find P Y (y) for Y being the number of women sent out on the recruiting trip. Solution (see board work): P Y (2) = 3/28; P Y (1) = 15/28; P Y (0) = 10 /28Solution (see board work): P Y (2) = 3/28; P Y (1) = 15/28; P Y (0) = 10 /28

6. MGMT 242 Spring, 1999 Cumulative Probability Distribution-- Discrete Variables Notation: P(Y  y) means the probability that Y is less than or equal to y; F Y (y) is the same.Notation: P(Y  y) means the probability that Y is less than or equal to y; F Y (y) is the same. Complement rule: P(Y > y) = 1 - F Y (y).Complement rule: P(Y > y) = 1 - F Y (y). Example (previous scenario, 5 men, 3 women, interview trips). What is the probability that at least one woman will be sent on an interview trip? (Note: “at least” means 1 or greater than 1)Example (previous scenario, 5 men, 3 women, interview trips). What is the probability that at least one woman will be sent on an interview trip? (Note: “at least” means 1 or greater than 1) –= P (Y> 0) = 1 - F Y (0) = 1 - P Y (0) = 1 - 10/28 = 18/28 –In this example, it’s just as easy to calculate P(Y  1) = P(Y=1) + P(Y=2) = 15 /28 + 3 / 28, but many times it’s not as easy.

7. MGMT 242 Spring, 1999 Expectation Values, Mean Values The expectation value of a discrete random variable Y is defined by the relation ( y i ) y i, that is to say, the sum of all possible values of Y, with each value weighted by the probability of the value.The expectation value of a discrete random variable Y is defined by the relation E(Y) =  i P Y ( y i ) y i, that is to say, the sum of all possible values of Y, with each value weighted by the probability of the value. Note that E(Y) is the mean value of Y; E(Y) is also denoted as.Note that E(Y) is the mean value of Y; E(Y) is also denoted as. Example (for previous case, 8 people, 5 men, 3 women,…) If Y is the number of women on an interview trip (Y= 0, 1, or 2), then E(Y) = (3/28) x 2 + (15/28) x 1 + (10/28)x0 =21/28 = 3/4Example (for previous case, 8 people, 5 men, 3 women,…) If Y is the number of women on an interview trip (Y= 0, 1, or 2), then E(Y) = (3/28) x 2 + (15/28) x 1 + (10/28)x0 =21/28 = 3/4

8. MGMT 242 Spring, 1999 Variance of a Discrete Random Variable The variance of a discrete random variable, V(Y), is the expectation of the square of the deviation from the mean (expected value); V(Y) is also denoted as Var(Y)The variance of a discrete random variable, V(Y), is the expectation of the square of the deviation from the mean (expected value); V(Y) is also denoted as Var(Y) V(Y) = E[(Y- E(Y)) 2 ] = ( y i ) [y i - E(Y)] 2 ;V(Y) = E[(Y- E(Y)) 2 ] =  i P Y ( y i ) [y i - E(Y)] 2 ; V(Y) = E(Y 2 ) - (E(Y)) 2V(Y) = E(Y 2 ) - (E(Y)) 2 The second formula for V(Y) is derived as follows:The second formula for V(Y) is derived as follows: –V(Y) =  i P Y ( y i ) [y i 2 - 2 y i E(y) + (E(Y)) 2 ] –or V(Y) = E(Y 2 ) - 2 E(y) E(y) + (E(Y)) 2 = E(Y 2 ) - (E(Y)) 2 Example (previous, 5 men, 3 women, etc..)Example (previous, 5 men, 3 women, etc..) –V(Y) = {(3/28)(2- 3/4) 2 + (15/28) (1- 3/4) 2 + (10/28) (0- 3/4) 2, or V(Y) = (3/28) 2 2 + (15/28) 1 2 + 0 - (3/4) 2 = 27/28

9. MGMT 242 Spring, 1999 Expectation Values--Another Example Exercise 4.25, Investment in 2 Apartment Houses

10. MGMT 242 Spring, 1999 Continuous Random Variables Reasons for using a continuous variable rather than discreteReasons for using a continuous variable rather than discrete –Many, many values (e.g. salaries)--too many to take as discrete; –Model for probability distribution makes it convenient or necessary to use a continuous variable-- Uniform Distribution (any value between set limits equally likely) Exponential Distribution (waiting times, delay times) Normal (Gaussian) Distribution, the “Bell Shaped Curve” (many measurement values follow a normal distribution either directly or after an appropriate transformation of variables; also mean values of samples follow a normal distribution, generally.)

11. MGMT 242 Spring, 1999 Probability Density and Cumulative Density Functions for Continuous Variables Probability density function, f X (x) defined:Probability density function, f X (x) defined: –P(x  X  x+dx) = f X (x) dx, that is, the probability that the random variable X is between x and x+dx is given by f X (x) dx Cumulative density function, F X (x), defined:Cumulative density function, F X (x), defined: –P(X  x ) = F X (x) –F X (x’) =  f X (x)dx, where the integral is taken from the lowest possible value of the random variable X to the value x’.

12. MGMT 242 Spring, 1999 Probability Density and Cumulative Density Functions for Continuous Variables, Example: Exercise 4.12 Model for time, t, between successive job applications to a large computer is given by F T (t) = 1 - exp(-0.5t).Model for time, t, between successive job applications to a large computer is given by F T (t) = 1 - exp(-0.5t). Note that F T (t) = 0 for t =0 and that F T (t) approaches 1 for t approaching infinity.Note that F T (t) = 0 for t =0 and that F T (t) approaches 1 for t approaching infinity. Also, f T (t) = the derivative of F T (t), or f T (t) = 0.5exp(-0.5t)Also, f T (t) = the derivative of F T (t), or f T (t) = 0.5exp(-0.5t)

13. MGMT 242 Spring, 1999 Probability Density and Cumulative Density Functions for Continuous Variables-- Example, Problem 4.12

14. MGMT 242 Spring, 1999 Expectation Values for Continuous Variables The expectation value for a continuous variable is taken by weighting the quantity by the probability density function, f Y (y), and then integrating over the range of the random variableThe expectation value for a continuous variable is taken by weighting the quantity by the probability density function, f Y (y), and then integrating over the range of the random variable E(Y) =  y f Y (y) dy;E(Y) =  y f Y (y) dy; –E(Y), the mean value of Y, is also denoted as  Y E(Y 2 ) =  y f Y (y) dy;E(Y 2 ) =  y f Y (y) dy; The variance is given by V(Y) = E(Y 2 ) - (  Y ) 2The variance is given by V(Y) = E(Y 2 ) - (  Y ) 2

15. MGMT 242 Spring, 1999 Continuous Variables--Example Ex. 4.18, text. An investment company is going to sell excess time on its computer; it has determined that a good model for the its own computer usage is given by the probability density function f Y (y) = 0.0009375[40-0.1(y-100) 2 ] for 80 < y < 120 f Y (y) = 0, otherwise. The important things to note about this distribution function can be determined by inspection –there is a maximum in f Y (y) at y = 100 –f Y (y) is 0 at y=80 and y = 120 –f Y (y) is symmetric about y=100 (therefore E(y) = 100 and F Y (y)=1/2 at y = 100). –f Y (y) is a curve that looks like a symmetric hump.

16. MGMT 242 Spring, 1999 Two Random Variables The situation with two random variables, X and Y, is important because the analysis will often show if there is a relation between the two, for example, between height and weight; years of education and income; blood alcohol level and reaction time.The situation with two random variables, X and Y, is important because the analysis will often show if there is a relation between the two, for example, between height and weight; years of education and income; blood alcohol level and reaction time. We will be concerned primarily with the quantities that show how strong (or weak) the relation is between X and Y:We will be concerned primarily with the quantities that show how strong (or weak) the relation is between X and Y: –The covariance of X and Y is defined by Cov(X,Y) = E[(X-  X )(Y-  Y )] = E(XY) -  X  Y –The correlation of X and Y is defined by Cor(X,Y) = Cov(X,Y) /  (V(X)V(Y) = Cov(X,Y)/(  X  Y )

17. MGMT 242 Spring, 1999 Properties of Covariance, Correlation Cov(X,Y) is positive (>0) if X and Y both increase or both decrease together; Examples:Cov(X,Y) is positive (>0) if X and Y both increase or both decrease together; Examples: – height and weight; years of education and salary; Cov(X,Y) is zero if X and Y are statistically independent: [ Cov(X,Y) = E(XY) -  X  Y = E(X)E(Y)-  X  Y =  X  Y - = 0] Examples: adult hat size and IQ;Cov(X,Y) is zero if X and Y are statistically independent: [ Cov(X,Y) = E(XY) -  X  Y = E(X)E(Y)-  X  Y =  X  Y -  X  Y = 0] Examples: adult hat size and IQ; Cov(X,Y) is negative if Y increases while X decreases; Example:Cov(X,Y) is negative if Y increases while X decreases; Example: –annual income, number of bowling games per year. (no disrespect meant to bowlers).

18. MGMT 242 Spring, 1999 Statistical Independence of Two Random Variables If two random variables, X and Y, are statistically independent, thenIf two random variables, X and Y, are statistically independent, then –P(X=x, Y=y) = P(X=x) P(Y=y), that is to say, the joint probability density function can be written as the product of probability density functions for X and Y. –Cov(X, Y) = 0 This follows from the above relation: Cov(X,Y) = E[(X-µ X ) (Y-µ Y )] =[E(X-µ X )][E(Y-µ Y )] = (µ X -µ X ) (µ Y -µ Y ) = 0