Mathematical Expectation and Chebyshev's Theorem: Engineering Data Analysis, Slides of Data Analysis & Statistical Methods

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4: Mathematical

Expectation

EM 7: Engineering Data Analysis

Second Semester, 2019-

Pamantasan ng Lungsod ng Valenzuela

Lecture Outline

• What is mathematical expectation?
• Mean of random variables
• Variance and covariance of random variables
• Means and variances of linear combination of random variables
• Chebyshev’s Theorem 2

Example Assuming that fair coins were tossed, we find the probability distribution as followed: The mean of the random variable X or the mean of the probability distribution of X is: 𝜇 = 0 1 4

• 1 1 2
• 2 1 4 = 1 4 Sample points Probability 𝑃(𝑋 = 0) 1/ 𝑃(𝑋 = 1) 1/ 𝑃(𝑋 = 2) 1/ This means that a person who tosses 2 coins over and over again will, on the average, get 1 head per toss.

Mean of Random Variables The mean or expected value of a random variable X describes where the probability distribution is centered. 5 DEFINITION: Let X be a random variable with probability distribution 𝑓(𝑥). The mean or expected value of X is: 𝜇 = 𝐸 𝑋 = ෍ 𝑥𝑓(𝑥) ௫ if X is discrete. If X is continuous, 𝜇 = 𝐸 𝑋 = න 𝑥𝑓 𝑥 𝑑𝑥 ஶ ିஶ

Example In a gambling game, a man is paid Php 5.00 if he gets all heads or all tails when 3 coins are tossed, and he will pay out Php 3.00 if either one of two heads show. What is his expected gain? 7 E1 (gambler will win) = {HHH, TTT} + E2 (gambler will lose) = {HHT, HTH, THH, HTT, THT, TTH} - Y - random variable that the gambler will win (expected gain) μ = E(Y) = (5)(1/4) + (-3)(3/4) = - In this game, the gambler will, on average, lose Php 1.00 per toss of the 3 coins. A game is “fair” if the gambler on the average, come out even.

Example Let X be the random variable that denotes the life in hours of a certain electronic device. The probability density function is: 𝑓 𝑥 = ቐ 20, 𝑥ଷ^ , 𝑥 > 100 0, elsewhere Find the expected life of this type of device. 8 μ = E(X) = int(-ifty,ifty) x(20,000/x3)dx = int(100,ifty) 20,000/x2 dx = 200.

Example Suppose that the number of cars X that pass through a car wash between 4:00pm and 5:00pm on any sunny Friday has the following probability distribution: Let 𝑔 𝑋 = 10𝑋 − 1 represent the amount of money in pesos, paid to the attendant by the manager. Find the attendant’s expected earnings for this particular time period. 10 x 4 5 6 7 8 9 𝑃(𝑋 = 𝑥) 1/12 1/12 1/4 1/4 1/6 1/ E[g(X)] = E(10X-1)

Example Let X be a random variable with density function: 𝑓 𝑥 = ൞ 𝑥ଶ 3 , −1 < 𝑥 < 2 0, elsewhere Find the expected value of 𝑔 𝑋 = 4𝑋 + 3. 11 E(4X + 3) = int(-1,2) (4x+3)(x^2/3) dx = 8

Example Two refills for a ballpoint pen are selected at random from a box that contains 3 blue refills, 2 red refills, and 3 green refills. If X is the number of blue refills and Y is the number of red refills selected, find the expected value of 𝑔 𝑋, 𝑌 = 𝑋𝑌. 13 f(x,y) 0 x 1 2 Row total y 0 3/28 9/28 3/28 15/ 1 3/14 3/14 3/ 2 1/28 1/ Column total 5/14 15/28 3/28 1 E(XY) = (0)(0)f(0,0) + (0)(1)f(0,1) + (0)(2)f(0,2) + (1)(0)f(1,0) + (1)(1)f(1,1) + (2)(0)f(2,0) = f(1,1) = 3/

Variance of Random Variables The mean does not give adequate description of the shape of the distribution. The most important measure of variability of a random variable X is obtained by letting 𝑔 𝑋 = 𝑋 − 𝜇 ଶ^ in 𝐸[𝑔 𝑋 ]. 14

Example Let the random variable X represent the number of automobiles that are used for official business purposes on any given workday. The probability distribution for company A is: and for company B is: Find the variance for the probability distributions of both companies. 16 x 1 2 3 𝑓(𝑥) 0.30 0.40 0. X 0 1 2 3 4 𝑓(𝑥) 0.20 0.10 0.30 0.30 0. A: mu = E(X) = 2. sigma2 = sum(1,3) (x-2)^2 f(x) = 0. B: mu = E(X) = 2. sigma2 = sum(0,4) (x-2)^2 f(x) = 1.

Variance of Random Variables Example: Let the random variable X represent the number of defective parts for a machine when 3 parts are sampled from a production line and tested. The following is the probability distribution of X. Using previous definition, calculate its variance. 17 DEFINITION: The variance of a random variable X is: 𝜎ଶ^ = 𝐸 𝑋ଶ^ − 𝜇ଶ x 1 2 3 4 𝑓(𝑥) 0.51 0.38 0.10 0. Solution: mu = 00.51 + 10.38 + 20.10 + 30.01 = 0. E(X2) = 00.51 + 10.38 + 40.10 + 90.01 = 0. Therefore Variance = 0.87 – 0.61^2 = 0.

Variance of Random Variables The variance or standard deviation has meaning only when two or more distributions that have the same units of measurement are compared. It would not be meaningful to compare the variance of a distribution of different quantities. We shall now extend the concept of variance of a random variable X to include random variables related to X. 19

Variance of Random Variables 20 DEFINITION: Let X be a random variable with probability distribution 𝑓 𝑥. The variance of random variable 𝑔(𝑋) is: 𝜎௚^ ଶ(^ ௑)= 𝐸 𝑔 𝑋 − 𝜇௚ ௑^ ଶ^ = ෍[𝑔 𝑥 − 𝜇௚(௑)]ଶ𝑓(𝑥) ௫ if X is discrete. If X is continuous, 𝜎௚^ ଶ(^ ௑)= 𝐸 𝑔 𝑋 − 𝜇௚ ௑^ ଶ^ = න 𝑔 𝑥 − 𝜇௚ ௑^ ଶ𝑓 𝑥 𝑑𝑥 ஶ ିஶ Example: Calculate the variance of 𝑔 𝑋 = 2𝑋 + 3 , where X is a random variable with probability distribution, x 0 1 2 3 𝑓(𝑥) ¼ 1/8 ½ 1/ Mu_2x+3 = E(2X+3) = sum_0^3 (2x+3)f(x) = 6 Using above definition, var_2x+3 = E{[(2X+3) – mu_2X+3]^2} = E[(2X+3-6)^2] = E(4X^2 – 12X + 9) = sum_0^ (4x^2 – 12x + 9)f(x) = 4