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