Random Variables and Probability Distributions Chapter 4 “Never draw to an inside straight.” from

Maxims Learned at My Mother’s Knee

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 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; • Notation: 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 variable • Example--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.

MGMT 242 Spring, 1999

Probability Distributions

• 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 MGMT 242 Spring, 1999

Probability Distributions--Discrete Variables

• 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 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.

• Solution (see board work): P Y (2) = 3/28; P Y (1) = 15/28; P Y (0) = 10 /28 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.

• 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) – = 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. MGMT 242 Spring, 1999

Expectation Values, Mean Values

• 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 < Y > .

• 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 = 3/4 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) • V(Y) = E[(Y- E(Y)) 2 ] =  i P Y ( y i ) [y i - E(Y)] 2 ; • V(Y) = E(Y 2 ) (E(Y)) 2 • 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..) – V(Y) = {(3/28)(2- 3/4) or V(Y) = (3/28) 2 2 2 + (15/28) (1- 3/4) 2 + (10/28) (0- 3/4) 2 , + (15/28) 1 2 + 0 - (3/4) 2 = 27/28 MGMT 242 Spring, 1999

Expectation Values--Another Example

Exercise 4.25, Investment in 2 Apartment Houses Return -50 0 50 100 150 200 250 exp.val.

variance st.dev.

0.02

0.03

0.2

0.5

0.2

0.03

0.02

0.15

0.1

0.1

0.1

0.3

0.2

0.05

100 2500 105 8475 50 92.05976

-1 0 10 50 30 6 5 450 300 500 0 500 300 450 sum exp1 sum var1 100 2500 MGMT 242 Spring, 1999

Continuous Random Variables

• Reasons 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.) MGMT 242 Spring, 1999

Probability Density and Cumulative Density Functions for Continuous Variables

• 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: – P(X  x ) = F X (x) – F X (x’) = value x’.  f X (x)dx, where the integral is taken from the lowest possible value of the random variable X to the 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).

• 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) MGMT 242 Spring, 1999

Probability Density and Cumulative Density Functions for Continuous Variables--

Example, Problem 4.12

FsubY 1.2

1 0.8

0.6

0.4

0.2

0 0 2 4 6 8 10 12 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 variable • 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; • The variance is given by V(Y) = E(Y 2 ) - (  Y ) 2 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 f Y (y) = 0, otherwise.

] for 80 < y < 120 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.

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.

• 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 ) 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: – 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  X  Y = 0] Examples: adult hat size and IQ; • 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).

MGMT 242 Spring, 1999

Statistical Independence of Two Random Variables

• If 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 MGMT 242 Spring, 1999