4 1 Mathematical Expectation q Example Repair costs

4. 1 Mathematical Expectation q Example: Repair costs for a particular machine are represented by the following probability distribution: x $50 $200 $350 P(X = x) 0. 3 0. 2 0. 5 q What is the expected value of the repairs? Ø That is, over time what do we expect repairs to cost on average? JMB Chapter 4 Lecture 1 EGR 252 2012 Slide 1

Expected Value – Repair Costs q μ = E(X) Ø μ = mean of the probability distribution q For discrete variables, μ = E(X) = ∑ x f(x) So, for our example, q. E(X) = 50(0. 3) + 200(0. 2) + 350(0. 5) = $230 JMB Chapter 4 Lecture 1 EGR 252 2012 Slide 2

Another Example – Investment q By investing in a particular stock, a person can take a profit in a given year of $4000 with a probability of 0. 3 or take a loss of $1000 with a probability of 0. 7. What is the investor’s expected gain on the stock? X $4000 P(X) 0. 3 -$1000 0. 7 E(X) = $4000 (0. 3) -$1000(0. 7) = $500 JMB Chapter 4 Lecture 1 EGR 252 2012 Slide 3

Expected Value - Continuous Variables q For continuous variables, μ = E(X) = ∫ x f(x) dx q Vacuum cleaner example: problem 7 pg. 88 f(x) = { x, 2 -x, 0, 0<x<1 1≤x<2 elsewhere (in hundreds of hours. ) = 1 * 100 = 100. 0 hours of operation annually, on average JMB Chapter 4 Lecture 1 EGR 252 2012 Slide 4

Functions of Random Variables q Ex 4. 4. pg. 111: Probability of X, the number of cars passing through a car wash in one hour on a sunny Friday afternoon, is given by x P(X = x) 4 5 1/12 6 7 8 9 1/4 1/6 q Let g(X) = 2 X -1 represent the amount of money paid to the attendant by the manager. What can the attendant expect to earn during this hour on any given sunny Friday afternoon? E[g(X)] = Σ g(x) f(x) = Σ (2 X-1) f(x) = (2*4 -1)(1/12) +(2*5 -1)(1/12) …+(2*9 -1)(1/6) = $12. 67 JMB Chapter 4 Lecture 1 EGR 252 2012 Slide 5

4. 2 Variance of a Random Variable q Recall our example: Repair costs for a particular machine are represented by the following probability distribution: x $50 200 350 P(X = x) 0. 3 0. 2 0. 5 q What is the variance of the repair cost? – That is, how might we quantify the spread of costs? JMB Chapter 4 Lecture 1 EGR 252 2012 Slide 6

Variance – Discrete Variables q For discrete variables, Ø σ2 = E [(X - μ)2] = ∑ (x - μ)2 f(x) = E (X 2) - μ 2 q Recall, for our example, μ = E(X) = $230 Ø Preferred method of calculation: Ø σ2 = [E(X 2)] – μ 2 = 502 (0. 3) + 2002 (0. 2) + 3502 (0. 5) – 2302 = $17, 100 Ø Alternate method of calculation: Ø σ2 = E(X- μ)2 f(x) = (50 -230)2 (0. 3) + (200 -230)2 (0. 2) + (350 -230)2 (0. 5) = $17, 100 JMB Chapter 4 Lecture 1 EGR 252 2012 Slide 7

Variance - Investment Example q By investing in a particular stock, a person can take a profit in a given year of $4000 with a probability of 0. 3 or take a loss of $1000 with a probability of 0. 7. What are the variance and standard deviation of the investor’s gain on the stock? Ø E(X) = $4000 (0. 3) -$1000 (0. 7) = $500 Ø σ2 = [∑(x 2 f(x))] – μ 2 = (4000)2(0. 3) + (-1000)2(0. 7) – 5002 = $5, 250, 000 Ø σ = $2291. 29 JMB Chapter 4 Lecture 1 EGR 252 2012 Slide 8

Variance of Continuous Variables q For continuous variables, σ2 = E [(X - μ)2] =[∫ x 2 f(x) dx] – μ 2 q Recall our vacuum cleaner example pr. 7 pg. 88 { f(x) = x, 2 -x, 0, 0<x<1 1≤x<2 elsewhere (in hundreds of hours of operation. ) q What is the variance of X? The variable is continuous, therefore we will need to evaluate the integral. JMB Chapter 4 Lecture 1 EGR 252 2012 Slide 9

Variance Calculations for Continuous Variables q (Preferred calculation) q What is the standard deviation? Ø σ = 0. 4082 hours JMB Chapter 4 Lecture 1 EGR 252 2012 Slide 10

Covariance/ Correlation q A measure of the nature of the association between two variables q Describes a potential linear relationship q Positive relationship Ø Large values of X result in large values of Y q Negative relationship Ø Large values of X result in small values of Y q “Manual” calculations are based on the joint probability distributions q Statistical software is often used to calculate the sample correlation coefficient (r) JMB Chapter 4 Lecture 1 EGR 252 2012 Slide 11

What if the distribution is unknown? q Chebyshev’s theorem: The probability that any random variable X will assume a value within k standard deviations of the mean is at least 1 – 1/k 2. That is, P(μ – kσ < X < μ + kσ) ≥ 1 – 1/k 2 q “Distribution-free” theorem – results are weak q If we believe we “know” the distribution, we do not use Chebyshev’s theorem to characterize variability JMB Chapter 4 Lecture 1 EGR 252 2012 Slide 12