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Suppose an ordinary, fair six-sided die is rolled (i.e., for i = 1, 2, 3, 4, 5, 6, there is one side with i spots), and X = “the number of spots facing upward” is observed. What is the average value for X ?. 1 + 2 + 3 + 4 + 5 + 6 6.
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Suppose an ordinary, fair six-sided die is rolled (i.e., for i = 1, 2, 3, 4, 5, 6, there is one side with i spots), and X = “the number of spots facing upward” is observed. What is the average value for X? 1 + 2 + 3 + 4 + 5 + 6 6 1 1 1 1 1 1 = (1) + (2) + (3) + (4) + (5) + (6) = 3.5 6 6 6 6 6 6 Suppose a fair six-sided die has three sides with 1 spot, two sides with 2 spots, and one side with 3 spots. This die is rolled, and Y = “the number of spots facing upward” is observed. What is the average value for Y ? 3 2 1 5 (1) + (2) + (3) = 6 6 6 3 3 2 1 10 (12) + (22) + (32) = 6 6 6 3 What is the average value for Y 2? What is the average value for 10Y 2 + 5Y 15 ?
What is the average value for 10Y 2 + 5Y 15 ? 3 [10(12) + 5(1) 15] + 6 2 [10(22) + 5(2) 15] + 6 1 80 [10(32) + 5(3) 15] = 6 3
Section 2.2 Important definition and theorem in the text: The definition of mathematical expectation or expected value Definition 2.2-1 If c is any constant, E(c) = c If c is any constant, and u(x) is a function, then E[cu(X)] = c E[u(X)] . If c1 and c2 are any constants, and u1(x) and u2(x) are functions, then E[c1u1(X) + c2u2(X)] = c1E[u1(X)] + c2E[u2(X)] . Theorem 2.2-1 1. An urn contains one red chip labeled with the integer 1, two blue chips labeled distinctively with the integers 1 and 2, and three white chips labeled distinctively with the integers 1, 2, and 3. Two chips are randomly selected without replacement and the random variable X = "sum of the observed integers" is recorded. (a) Find the p.m.f. of X. The space of X is {2, 3, 4, 5} .
3/15 = 1/5 if x = 2 6/15 = 2/5 if x = 3 The p.m.f. of X is f(x) = 4/15 if x = 4 2/15 if x = 5 Find each of the following: E(X) = E(5) = E(9X) = E(X2) = E(4 + 3X– 10X2) = E[(X + 3)2] = (b) 3 (2) — + 15 6 (3) — + 15 4 (4) — + 15 2 (5) — = 15 50 10 — = — 15 3 5 10 (9) — = 3 30 3 (22) — + 15 6 (32) — + 15 4 (42) — + 15 2 (52) — = 15 180 —– = 12 15 4 + 10 – 120 = –106 E(4) + 3E(X) – 10E(X2) = 41 E[X2 + 6X + 9] = E(X2) + 6E(X) + E(9) =
1 ——— if x = 1, 2, 3, … . x(x + 1) 2. The random variable X has p.m.f. f(x) = Verify that f(x) is a p.m.f., and show that E(X) does not exist. n 1 ——— = x(x + 1) n —— . n + 1 By induction, we can show that x = 1 1 ——— = x(x + 1) It follows that 1 x = 1 1 1 1 — + — + — + … 2 3 4 1 1 1 E(X) = (1) —— + (2) —— + (3) —— + … = (1)(2) (2)(3) (3)(4) This is the well-known harmonic series, which is known not to converge. Therefore E(X) does not exist.
3. An urn contains 4 white chips and 6 black chips, and it costs one dollar to play a game involving the urn. The player selects 2 chips at random and without replacement. If at least one of the chips is white, the player's dollar is returned, and the player receives an additional d dollars; if neither of the two chips is white, the player loses the dollar paid. What is the value of d for which the expected winnings is zero? Let X be the dollar value of the winnings. The p.m.f. of X is f(x) = 2 — 3 if x = d 2d– 1 ——— 3 E(X) = 1 — 3 if x = – 1 E(X) = 0 if and only if d = $0.50
