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Random Variables and Probability Distributions Chapter 4. “Never draw to an inside straight.” from Maxims Learned at My Mother’s Knee. Goals for Chapter 4. Define Random Variable, Probability Distribution Understand and Calculate Expected Value
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Random Variables and Probability DistributionsChapter 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, PY(y) (with y being some specific value; • PY(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 PY(y) for Y being the number of women sent out on the recruiting trip. • Solution (see board work): PY(2) = 3/28; PY(1) = 15/28; PY(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; FY(y) is the same. • Complement rule: P(Y > y) = 1 - FY(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 - FY(0) = 1 - PY(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) = iPY( yi) yi , 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] = iPY( yi) [yi - E(Y)]2 ; • V(Y) = E(Y2) - (E(Y))2 • The second formula for V(Y) is derived as follows: • V(Y) = iPY( yi) [yi2 - 2 yi E(y) + (E(Y))2] • or V(Y) = E(Y2) - 2 E(y) E(y) + (E(Y))2 = E(Y2) - (E(Y))2 • 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) 22 + (15/28) 12 + 0 - (3/4)2 = 27/28 MGMT 242 Spring, 1999
Expectation Values--Another ExampleExercise 4.25, Investment in 2 Apartment Houses 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, fX(x) defined: • P(x X x+dx) = fX(x) dx, that is, the probability that the random variable X is between x and x+dx is given by fX(x) dx • Cumulative density function, FX(x), defined: • P(X x ) = FX(x) • FX(x’) = fX(x)dx, where the integral is taken from the lowest possible value of the random variable X to the value x’. 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 FT(t) = 1 - exp(-0.5t). • Note that FT(t) = 0 for t =0 and that FT(t) approaches 1 for t approaching infinity. • Also, fT(t) = the derivative of FT(t), or fT(t) = 0.5exp(-0.5t) MGMT 242 Spring, 1999
Probability Density and Cumulative Density Functions for Continuous Variables--Example, Problem 4.12 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, fY(y), and then integrating over the range of the random variable • E(Y) = y fY(y) dy; • E(Y), the mean value of Y, is also denoted as Y • E(Y2) = y fY(y) dy; • The variance is given by V(Y) = E(Y2) - (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 fY(y) = 0.0009375[40-0.1(y-100)2 ] for 80 < y < 120 fY(y) = 0, otherwise. The important things to note about this distribution function can be determined by inspection • there is a maximum in fY(y) at y = 100 • fY(y) is 0 at y=80 and y = 120 • fY(y) is symmetric about y=100 (therefore E(y) = 100 and FY(y)=1/2 at y = 100). • fY(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) -XY • 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) -XY= E(X)E(Y)-XY = XY - XY= 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-µX) (Y-µY)] =[E(X-µX)][E(Y-µY)] = (µX-µX) (µY-µY) = 0 MGMT 242 Spring, 1999