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)  = 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 ? (1)  + (2)  + (3)  =  What is the average value for Y 2 ? (1 2 )  + (2 2 )  + (3 2 )  =  What is the average value for 10Y 2 + 5Y  15 ?

3 [10(1 2 ) + 5(1)  15]  + 6 What is the average value for 10Y 2 + 5Y  15 ? 2 [10(2 2 ) + 5(2)  15]  [10(3 2 ) + 5(3)  15]  =  6 3

Section 2.2 Important definition and theorem in the text: The definition of mathematical expectation or expected value Definition If c is any constant, E(c) = c If c is any constant, and u(x) is a function, then E[c u(X)] = c E[u(X)]. If c 1 and c 2 are any constants, and u 1 (x) and u 2 (x) are functions, then E[c 1 u 1 (X) + c 2 u 2 (X)] = c 1 E[u 1 (X)] + c 2 E[u 2 (X)]. Theorem 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. Find the p.m.f. of X.The space of X is{2, 3, 4, 5}. (a)

(b) The p.m.f. of X is f(x) = if x = 2 if x = 3 if x = 4 if x = 5 3/15 = 1/5 6/15 = 2/5 4/15 2/15 Find each of the following: E(X) = E(5) = E(9X) = E(X 2 ) = E(4 + 3X – 10X 2 ) = E[(X + 3) 2 ] = 3 (2) — (3) — (4) — (5) — = — = — (9) — = (2 2 ) — (3 2 ) — (4 2 ) — (5 2 ) — = —– = E(4) + 3E(X) – 10E(X 2 ) = – 120 = –106 E[X 2 + 6X + 9] =E(X 2 ) + 6E(X) + E(9) = 41

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. 1 ——— if x = 1, 2, 3, …. x(x + 1) By induction, we can show that  x = 1 n 1 ——— = x(x + 1) n ——. n + 1 It follows that  x = 1  1 ——— = x(x + 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) = ifx = d ifx = – 1 2 — 3 1 — 3 E(X) = 2d – 1 ——— 3 E(X) = 0 if and only if d =$0.50