4.1 Mathematical Expectation Example: Repair costs for a particular machine are represented by the following probability distribution: What is the expected value of the repairs? That is, over time what do we expect repairs to cost on average? x $50 $200 $350 P(X = x) 0.3 0.2 0.5 The expected value or mean of a probability distribution is the long run theoretical average. JMB Chapter 4 Lecture 1 EGR 252 2011

Expected Value – Repair Costs μ = mean of the probability distribution For discrete variables, μ = E(X) = ∑ x f(x) So, for our example, E(X) = 50(0.3) + 200(0.2) + 350(0.5) = $230 E(x) <weighted average> E(X) = 50(0.3) + 200(0.2) + 350(0.5) = $230 JMB Chapter 4 Lecture 1 EGR 252 2011

Another Example – Investment 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 -$1000 P(X) 0.3 0.7 E(X) = $4000 (0.3) -$1000(0.7) = $500 X 4000 -1000 P(X) 0.3 0.7 E(X) = 4000 (0.3) -1000(0.7) = 500 JMB Chapter 4 Lecture 1 EGR 252 2011

Expected Value - Continuous Variables For continuous variables, μ = E(X) = E(X) = ∫ x f(x) dx Vacuum cleaner example: problem 7 pg. 88 x, 0 < x < 1 f(x) = 2-x, 1 ≤ x < 2 0, elsewhere (in hundreds of hours.) { = 1 * 100 = 100.0 hours of operation annually, on average JMB Chapter 4 Lecture 1 EGR 252 2011

Functions of Random Variables 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 Let g(X) = 2X -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 (2X-1) f(x) = (2*4-1)(1/12) +(2*5-1)(1/12) …+(2*9-1)(1/6) = $12.67 x 4 5 6 7 8 9 P(X = x) 1/12 1/4 1/6 Σ (2x-1) f(x) = 7(1/12) + 9(1/12) … + 17(1/6) = $12.67 JMB Chapter 4 Lecture 1 EGR 252 2011

4.2 Variance of a Random Variable Recall our example: Repair costs for a particular machine are represented by the following probability distribution: What is the variance of the repair cost? That is, how might we quantify the spread of costs? x $50 200 350 P(X = x) 0.3 0.2 0.5 JMB Chapter 4 Lecture 1 EGR 252 2011

Variance – Discrete Variables For discrete variables, σ2 = E [(X - μ)2] = ∑ (x - μ)2 f(x) = E (X2) - μ2 Recall, for our example, μ = E(X) = $230 Preferred method of calculation: σ2 = [E(X2)] – μ2 = 502 (0.3) + 2002 (0.2) + 3502 (0.5) – 2302 = $17,100 Alternate method of calculation: σ2 = E(X- μ)2 f(x) = (50-230)2 (0.3) + (200-230)2 (0.2) + (350-230)2 (0.5) = $17,100 E(X2) – μ2 = 502 (0.3) + 2002 (0.2) + 3502 (0.5) – 2302 = $17,100 JMB Chapter 4 Lecture 1 EGR 252 2011

Variance - Investment Example 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? E(X) = $4000 (0.3) -$1000 (0.7) = $500 σ2 = [∑(x2 f(x))] – μ2 = (4000)2(0.3) + (-1000)2(0.7) – 5002 = $5,250,000 σ = $2291.29 X 4000 -1000 P(X) 0.3 0.2 E(X) = 4000 (0.3) -1000(0.7) = 500 σ2 = ∑(x2 f(x)) –μ2 = (4000)2(0.3) + (-1000)2(0.7) – 5002 = $5,250,000 σ =$2291.29 JMB Chapter 4 Lecture 1 EGR 252 2011

Variance of Continuous Variables For continuous variables, σ2 = E [(X - μ)2] =[∫ x2 f(x) dx] – μ2 Recall our vacuum cleaner example pr. 7 pg. 88 x, 0 < x < 1 f(x) = 2-x, 1 ≤ x < 2 0, elsewhere (in hundreds of hours of operation.) What is the variance of X? The variable is continuous, therefore we will need to evaluate the integral. { E(X) = ∫x2 f(x)dx – μ2 JMB Chapter 4 Lecture 1 EGR 252 2011

Variance Calculations for Continuous Variables (Preferred calculation) What is the standard deviation? σ = 0.4082 hours [∫01 x3 dx + ∫12x2 (2-x)dx] – μ2 = x4/4 |10 + (2x3/3 – x4/4)|12 - 12 = 0.1667 σ = 0.4082 JMB Chapter 4 Lecture 1 EGR 252 2011

Covariance/ Correlation A measure of the nature of the association between two variables Describes a potential linear relationship Positive relationship Large values of X result in large values of Y Negative relationship Large values of X result in small values of Y “Manual” calculations are based on the joint probability distributions Statistical software is often used to calculate the sample correlation coefficient (r) JMB Chapter 4 Lecture 1 EGR 252 2011

What if the distribution is unknown? 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/k2. That is, P(μ – kσ < X < μ + kσ) ≥ 1 – 1/k2 “Distribution-free” theorem – results are weak If we believe we “know” the distribution, we do not use Chebyshev’s theorem to characterize variability JMB Chapter 4 Lecture 1 EGR 252 2011