Chapter 4. Discrete Probability Distributions Section 4.2. Expected Values of Random Variables

1. Chapter 4. Discrete Probability DistributionsSection 4.2. Expected Values of Random Variables Jiaping Wang Department of Mathematical Science 02/06/2013, Monday

2. Outline Expected Values Variance and Standard Deviation Some Properties Tchebysheff’s Theorem Homework #4

3. Part 1. Expected Values

4. Long-run Relative Frequency of Occurrence Suppose you and your friend are matching balanced coins. Each of you flips a coin. If the upper faces match, you win $1; if they do not match, you lose $1. So on average, how much will you win per game in a long run? From these flipping coins, you win +1 and -1 (lose money) in half of the time, so the average is (-1)(1/2)+(1)(1/2) = 0.

5. Definition 4.4 Definition 4.4 The expected value of a discrete random variable X with probability distribution p(x) is given as (The sum is over all values of x for which p(x)>0) We sometimes use the notation E(X)=μ for this equivalence. Note: Not all expected values exist, the sum above must converge absolutely, ∑|x|p(x)<∞.

6. Theorem 4.1 Now if we change the wining function, g(x)=3x, which means, if coins are matched, then you win $3, otherwise, you lose $3. So the expected value becomes (3)(1/2)+(-3)(1/2)=0 Theorem 4.1 If X is a discrete random variable with probability p(x) and if g(x) is any real-valued function of X, then E(g(x)=∑g(x)p(x).

7. Part 2. Variance and Standard Deviation

8. Definitions 4.5 and 4.6 The variance of a random variable X with expected value μ is given by V(X)=E[(X- μ)2] Sometimes we use the notation σ2 = E[(X- μ)2] For this equivalence. The standard deviation is a measure of variation that maintains the original units of measure. The standard deviation of a random variable is the square root of the variance and is given by

9. Cont. Case I: μ = (-1.0)(1/2)+(1)(1/2)=0; σ2=(-1)2(1/2)+(1)2(1/2)=1. Case II: μ = (-1.5)(1/2)+(1)(1/4)+ (2)(1/4)=0; σ2=(-1.5)2(1/2)+(1)2(1/4)+ 1)2(1/4)=2.375.

10. Cont. Consider the percentages as probabilities, then for 2000 μ = 3(0.069)+8(0.073)+…+(90)(0.033) =36.6, and σ2 = (3-36.6)2(0.069)+ (8-36.6)2(0.073) + … + (90-36.6)2(0.033) = 510.76 Similarly, we can find for 2100, μ = 42.5 and σ2 = 691.69

11. Example 4.3 A department supervisor is considering purchasing a photocopy machine. One consideration is how often the machine will need repairs. Let X denote the number of repairs during one year. Base on past performance, the distribution of X is shown as follows What is the expected number of repairs during a year? What are the variance and standard deviation of the number of repairs during a year?

12. Solution • E(X)=∑xp(x) = (0)(0.2)+(1)(0.3)+(2)(0.4)+(3)(0.1)=1.4 • V(X)= ∑(x-E(X))2p(x) • = (0-1.4) 2(0.2)+(1-1.4) 2(0.3)+(2-1.4) 2(0.4)+(3-1.4) 2(0.1) • = 0.84. • So the standard deviation is 0.92.

13. Part 3. Some Properties

14. Theorem 4.2 For any random variable X and constants a and b. E(aX + b) = aE(X) + b V(aX + b) = a2V(X) Standardized random variable: If X has mean μ and standard deviation σ, then Y=(X – μ)/ σ has E(Y)=0 and V(Y)=1, thus Y can be called the standardized random variable of X. Theorem 4.3 If X is a random variable with mean μ, then V(X)= E(X2) – μ2

15. Example 4.4 The department supervisor in Example 4.3 wants to consider the cost of maintenance before purchasing the photocopy machine. The cost of maintenance consists of the expense of a service agreement and the cost of repairs. The service agreement can be purchased for $200. With the agreement, the cost of each repair is $50. Find the mean, variance and standard deviation of the annual costs of repair for the photocopy machine. Solution: Now the annual cost of the maintenance is 50X + 200. Recall that E(X)=1.4 and V(X)=0.84 from Example 4.3, so we have E(50X+200)=50E(X)+200=(50)(1.4)+200=270; V(50X+200)=2500V(X)= (2500)(0.84)=2100, so the standard deviation is $45.83.

16. Example 4.5 Use the results of Theorem 4.3 to compute the variance of X as given in Example 4.3. Solution: E(X)=1.4, E(X2)=∑x2p(x)=02(0.2)+12(0.3)+22(0.4)+32(0.1)=2.8 Then V(X)= E(X2)-E2(X) =2.8-1.56=0.84.

17. Part 4. Tchebysheff’s Inequality

18. Theorem 4.4 Tchebysheff’s Theorem. Let X be a random variable with mean μ and standard deviation σ. Then for any positive k, P(|X – μ|/ σ < k) ≥ 1/k2 Consider Y=(X- μ)/ σ, then E(Y)=0 and V(Y)=1.

19. Example 4.6 The daily production of electric motors at a certain factory averaged 120 with a standard deviation of 10. What can be said about the fraction of days on which the production level falls between 100 and 140? Find the shortest interval certain to contain at least 90% of the daily production levels. Solution: The interval from 100 to 140 means k=2 so that |X-μ|<2σ for μ=120 and σ=10. So based on Chebysheff’s inequality, we have 1-1/4=0.75, which mean at least 75% of all days will have a total production values that falls in this interval. TO find k, we set (1-1/k2)=0.9 , then k=3.16, the interval will be from 120-(3.16)(10) to 120+(3.16)(10), or 88.4 to 151.6.

20. Example 4.7 The annual cost of maintenance for a certain photocopy machine has a mean of $270 and a variance of $2100 (see example 4.4). The manager wants to budget enough for maintenance that he is likely to go over the budget amount. He is considering budgeting $400 for maintenance. How often will the maintenance cost exceed this amount? Solution: Let X=400, so k=(400-270)/45.8=2.84, then the interval 270-(2.84)(45.8) to 270 + (2.84)(45.8) or 140 to 400 contains at least 1-1/(2.84*2.84) = 0.88, thus at most 0.12 of the probability mass can exceed $400.

21. Example 4.8 Suppose the random variable X has the probability mass function given in the table Evaluate Tchebysheff’s inequality for k=2. Solution: Find mean E(X)=0, find V(X)=1/8 + 1/8=0.25, so σ = 0.5, thus P(|X-E(X)|<2(0.5))=P(|X-E(X)|<1)=P(X=0)=3/4. By Chebysheff’s inequality, P(|X-E(X)|<2 σ) ≥1-1/4=0.75.

22. Homework #4 Page 100: 4.2 Page 101: 4.8 Page 102: 4.12 Page 116: 4.17 Page 118: 4.26 Page 119: 4.33 Page 120: 4.42