Expectations. Definition Let X be a discrete rv with set of possible values D and pmf p(x). The expected value or mean value of X, denoted by E(X ) or

Transcription

1

2 Definition Let X be a discrete rv with set of possible values D and pmf p(x). The expected value or mean value of X, denoted by E(X ) or µ X, is E(X ) = µ X = x D x p(x)

3 Definition Let X be a discrete rv with set of possible values D and pmf p(x). The expected value or mean value of X, denoted by E(X ) or µ X, is E(X ) = µ X = x D x p(x) e.g (Problem 30) A group of individuals who have automobile insurance from a certain company is randomly selected. Let Y be the number of moving violations for which the individual was cited during the last 3 years. The pmf of Y is y Then the expected value of p(y) moving violations for that group is µ Y = E(Y ) = = 0.60

4

5 y p(y) Assume the total number of individuals in that group is 100, then there are 60 individuals without moving violation, 25 with 1 moving violation, 10 with 2 moving violations and 5 with 3 moving violations.

6 y p(y) Assume the total number of individuals in that group is 100, then there are 60 individuals without moving violation, 25 with 1 moving violation, 10 with 2 moving violations and 5 with 3 moving violations. The population mean is calculated as µ = = 0.60

7 y p(y) Assume the total number of individuals in that group is 100, then there are 60 individuals without moving violation, 25 with 1 moving violation, 10 with 2 moving violations and 5 with 3 moving violations. The population mean is calculated as µ = = µ = = = 0.60

8 y p(y) Assume the total number of individuals in that group is 100, then there are 60 individuals without moving violation, 25 with 1 moving violation, 10 with 2 moving violations and 5 with 3 moving violations. The population mean is calculated as µ = = µ = = = 0.60 The population size is irrevelant if we know the pmf!

9

10 Examples: Let X be a Bernoulli rv with pmf 1 p x = 0 p(x) = p x = 1 0 x 0, or 1

11 Examples: Let X be a Bernoulli rv with pmf 1 p x = 0 p(x) = p x = 1 0 x 0, or 1 Then the expected value for X is E(X ) = 0 p(0) + 1 p(1) = p

12 Examples: Let X be a Bernoulli rv with pmf 1 p x = 0 p(x) = p x = 1 0 x 0, or 1 Then the expected value for X is E(X ) = 0 p(0) + 1 p(1) = p We see that the expected value of a Bernoulli rv X is just the probability that X takes on the value 1.

13

14 Examples: Consider the cards drawing example again and assume we have infinitely many cards this time. Let X = the number of drawings until we get a. If the probability for getting a is α, then the pmf for X is { α(1 α) x 1 x = 1, 2, 3,... p(x) = 0 otherwise

15 Examples: Consider the cards drawing example again and assume we have infinitely many cards this time. Let X = the number of drawings until we get a. If the probability for getting a is α, then the pmf for X is { α(1 α) x 1 x = 1, 2, 3,... p(x) = 0 otherwise The expected value for X is E(X ) = D x p(x) = xα(1 α) x 1 = α [ d dα (1 α)x ] x=1 x=1

16 Examples: Consider the cards drawing example again and assume we have infinitely many cards this time. Let X = the number of drawings until we get a. If the probability for getting a is α, then the pmf for X is { α(1 α) x 1 x = 1, 2, 3,... p(x) = 0 otherwise The expected value for X is E(X ) = D x p(x) = xα(1 α) x 1 = α [ d dα (1 α)x ] x=1 x=1 E(X ) = α{ d dα [ (1 α) x ]} = α{ d dα (1 α α )} = 1 α x=1

17

18 Examples 3.20 Let X be the number of interviews a student has prior to getting a job. The pmf for X is { k x = 1, 2, 3,... p(x) = x 2 0 otherwise where k is chosen so that x=1 (k/x 2 ) = 1. (It can be showed that x=1 (1/x 2 ) <, which implies that such a k exists.) The expected value of X is µ = E(X ) = x=1 The expected value is NOT finite! x k x 2 = k 1 x =! x=1

19 Examples 3.20 Let X be the number of interviews a student has prior to getting a job. The pmf for X is { k x = 1, 2, 3,... p(x) = x 2 0 otherwise where k is chosen so that x=1 (k/x 2 ) = 1. (It can be showed that x=1 (1/x 2 ) <, which implies that such a k exists.) The expected value of X is µ = E(X ) = x=1 The expected value is NOT finite! Heavy Tail: x k x 2 = k 1 x =! x=1

20 Examples 3.20 Let X be the number of interviews a student has prior to getting a job. The pmf for X is { k x = 1, 2, 3,... p(x) = x 2 0 otherwise where k is chosen so that x=1 (k/x 2 ) = 1. (It can be showed that x=1 (1/x 2 ) <, which implies that such a k exists.) The expected value of X is µ = E(X ) = x=1 x k x 2 = k 1 x =! x=1 The expected value is NOT finite! Heavy Tail: distribution with a large amount of probability far from µ

21

22 Example (Problem 38) Let X = the outcome when a fair die is rolled once. If before the 1 die is rolled you are offered either 3.5 dollars or 1 X dollars, would you accept the guaranteed amount or would you gamble?

23 Example (Problem 38) Let X = the outcome when a fair die is rolled once. If before the 1 die is rolled you are offered either 3.5 dollars or 1 X dollars, would you accept the guaranteed amount or would you gamble? x p(x)

24 Example (Problem 38) Let X = the outcome when a fair die is rolled once. If before the 1 die is rolled you are offered either 3.5 dollars or 1 X dollars, would you accept the guaranteed amount or would you gamble? x p(x) x

25 Example (Problem 38) Let X = the outcome when a fair die is rolled once. If before the 1 die is rolled you are offered either 3.5 dollars or 1 X dollars, would you accept the guaranteed amount or would you gamble? x p(x) x Then the expected dollars from gambling is E( 1 X ) = 6 x=1 1 x p( 1 x ) = = < 1 3.5

26

27 Proposition If the rv X has a set of possible values D and pmf p(x), then the expected value of any function h(x ), denoted by E[h(X )] or µ hx, is computed by E[h(X )] = h(x) p(x) D

28

29 Example 3.23 A computer store has purchased three computers of a certain type at $500 apiece. It will sell them for $1000 apiece. The manufacturer has agreed to repurchase any computers still unsold after a specified period at $200 apiece.

30 Example 3.23 A computer store has purchased three computers of a certain type at $500 apiece. It will sell them for $1000 apiece. The manufacturer has agreed to repurchase any computers still unsold after a specified period at $200 apiece. Let X denote the number of computers sold, and suppose that p(0) = 0.1, p(1) = 0.2, p(2) = 0.3, p(3) = 0.4. Let h(x ) denote the profit associated with selling X units,

31 Example 3.23 A computer store has purchased three computers of a certain type at $500 apiece. It will sell them for $1000 apiece. The manufacturer has agreed to repurchase any computers still unsold after a specified period at $200 apiece. Let X denote the number of computers sold, and suppose that p(0) = 0.1, p(1) = 0.2, p(2) = 0.3, p(3) = 0.4. Let h(x ) denote the profit associated with selling X units, then h(x ) = revenue cost = 1000X + 200(3 X ) 1500 = 800X 900.

32 Example 3.23 A computer store has purchased three computers of a certain type at $500 apiece. It will sell them for $1000 apiece. The manufacturer has agreed to repurchase any computers still unsold after a specified period at $200 apiece. Let X denote the number of computers sold, and suppose that p(0) = 0.1, p(1) = 0.2, p(2) = 0.3, p(3) = 0.4. Let h(x ) denote the profit associated with selling X units, then h(x ) = revenue cost = 1000X + 200(3 X ) 1500 = 800X 900. The expected profit is E[h(X )] = h(0) p(0) + h(1) p(1) + h(2) p(2) + h(3) p(3) = ( 900)(0.1) + ( 100)(0.2) + (700)(0.3) + (1500)(0.4) = 700

33

34 Proposition E(aX + b) = a E(X ) + b (Or, using alternative notation, µ ax +b = a µ X + b.)

35 Proposition E(aX + b) = a E(X ) + b (Or, using alternative notation, µ ax +b = a µ X + b.) e.g. for the previous example, E[h(X )] = E(800X 900) = 800 E(X ) 900 = 700

36 Proposition E(aX + b) = a E(X ) + b (Or, using alternative notation, µ ax +b = a µ X + b.) e.g. for the previous example, E[h(X )] = E(800X 900) = 800 E(X ) 900 = 700 Corollary 1. For any constant a, E(aX ) = a E(X ). 2. For any constant b, E(X + b) = E(X ) + b.

37

38 Definition Let X have pmf p(x) and expected value µ. Then the variance of X, denoted by V (X ) or σ 2 X, or just σ2 X, is V (X ) = D (x µ) 2 p(x) = E[(X µ) 2 ] The stand deviation (SD) of X is σ X = σx 2

39

40 Example: For the previous example, the pmf is given as x p(x)

41 Example: For the previous example, the pmf is given as x p(x) then the variance of X is V (X ) = σ 2 = 3 (x 2) 2 p(x) x=0 = (0 2) 2 (0.1) + (1 2) 2 (0.2) + (2 2) 2 (0.3) + (3 2) 2 (0.4) = 1

42

43 Recall that for sample variance s 2, we have s 2 = S xx n 1 = x 2 i ( x i ) 2 n 1 n

44 Recall that for sample variance s 2, we have s 2 = S xx n 1 = x 2 i ( x i ) 2 n 1 n Proposition V (X ) = σ 2 = [ D x 2 p(x)] µ 2 = E(X 2 ) [E(X )] 2

45 Recall that for sample variance s 2, we have s 2 = S xx n 1 = x 2 i ( x i ) 2 n 1 n Proposition V (X ) = σ 2 = [ D x 2 p(x)] µ 2 = E(X 2 ) [E(X )] 2 e.g. for the previous example, the pmf is given as x p(x) Then V (X ) = E(X 2 ) [E(X )] 2 = (2) 2 = 1

46

47 Proposition If h(x ) is a function of a rv X, then V [h(x )] = σ 2 h(x ) = D {h(x) E[h(X )]} 2 p(x) = E[h(X ) 2 ] {E[h(X )]} 2 If h(x ) is linear, i.e. h(x ) = ax + b for some nonrandom constant a and b, then V (ax + b) = σ 2 ax +b = a2 σ 2 X and σ ax +b = a σ X In particular, σ ax = a σ X, σ X +b = σ X

48

49 Example 3.23 continued A computer store has purchased three computers of a certain type at $500 apiece. It will sell them for $1000 apiece. The manufacturer has agreed to repurchase any computers still unsold after a specified period at $200 apiece. Let X denote the number of computers sold, and suppose that p(0) = 0.1, p(1) = 0.2, p(2) = 0.3, p(3) = 0.4. Let h(x ) denote the profit associated with selling X units, then h(x ) = revenue cost = 1000X + 200(3 X ) 1500 = 800X 900.

50 Example 3.23 continued A computer store has purchased three computers of a certain type at $500 apiece. It will sell them for $1000 apiece. The manufacturer has agreed to repurchase any computers still unsold after a specified period at $200 apiece. Let X denote the number of computers sold, and suppose that p(0) = 0.1, p(1) = 0.2, p(2) = 0.3, p(3) = 0.4. Let h(x ) denote the profit associated with selling X units, then h(x ) = revenue cost = 1000X + 200(3 X ) 1500 = 800X 900. The variance of h(x ) is V [h(x )] = V [800X 900] = V [X ] = 640, 000 And the SD is σ h(x ) = V [h(x )] = 800.