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

What is the average value for Y 2? 3 2 1 10(12) + (22) + (32) = 6 6 6 3

What is the average value for 10Y 2 + 5Y 15 ?

3[10(12) + 5(1) 15] +

6

What is the average value for 10Y 2 + 5Y 15 ?

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 valueDefinition 2.2-1

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 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.

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 = 2if x = 3if x = 4if x = 5

3/15 = 1/56/15 = 2/54/152/15

Find each of the following:

E(X) =

E(5) =

E(9X) =

E(X2) =

E(4 + 3X – 10X2) =

E[(X + 3)2] =

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

E(4) + 3E(X) – 10E(X2) = 4 + 10 – 120 = –106

E[X2 + 6X + 9] = E(X2) + 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)

1

1 1 1E(X) = (1) —— + (2) —— + (3) —— + … =

(1)(2) (2)(3) (3)(4)

1 1 1— + — + — +… 2 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) =

if x = d

if x = – 1

2— 3

1— 3

E(X) = 2d – 1——— 3

E(X) = 0 if and only if d = $0.50