Selection on Observables

1.
Peter M. Lance,PhD MEASURE Evaluation University of North Carolina at Chapel Hill September 13 and 15, 2016 Selection on Observables

2.
Global, five-year, $180Mcooperative agreement Strategic objective: To strengthen health information systems – the capacity to gather, interpret, and use data – so countries can make better decisions and sustain good health outcomes over time. Project overview

3.
Improved country capacityto manage health information systems, resources, and staff Strengthened collection, analysis, and use of routine health data Methods, tools, and approaches improved and applied to address health information challenges and gaps Increased capacity for rigorous evaluation Phase IV Results Framework

4.
Global footprint (morethan 25 countries)

5.
• The programimpact evaluation challenge • Randomization • Selection on observables • Within estimators • Instrumental variables

6.
• The programimpact evaluation challenge • Randomization • Selection on observables • Within estimators • Instrumental variables

9.
X Y P

10.
X Y P 𝑌 = 𝑃∙ 𝑌1 + 1 − 𝑃 ∙ 𝑌0

11.
X Y P

12.
X Y P

13.
𝐸 𝑌1 − 𝑌0 =𝐸 𝑌1 − 𝐸 𝑌0 Average Y across sample of participants − Average Y across sample of non−participants

14.

15.

16.

17.

18.

19.

20.
𝐸 𝑌1 𝑃= 1 = 𝐸 𝑌1 𝑃 = 0 = 𝐸 𝑌1

21.
X Y P

22.
𝐸 𝑋 𝑃= 1 ≠ 𝐸 𝑋 𝑃 = 0

23.
X Y P

24.
X Y P

25.
𝐸 𝑋 𝑃= 1 ≠ 𝐸 𝑋 𝑃 = 0

26.
𝐸 𝑋 𝑃= 1 ≠ 𝐸 𝑋 𝑃 = 0 𝐸 𝑌1 𝑃 = 1 ≠ 𝐸 𝑌1 𝑃 = 0

27.
𝐸 𝑋 𝑃= 1 ≠ 𝐸 𝑋 𝑃 = 0 𝐸 𝑌0 𝑃 = 1 ≠ 𝐸 𝑌0 𝑃 = 0

28.

29.
𝐸 𝑌1 𝑃 =1 ≠ 𝐸 𝑌1 𝑃 = 0 𝑋 = 𝑥∗ 𝐸 𝑌1 𝑃 = 1, 𝑋 = 𝑥∗ = 𝐸 𝑌1 𝑃 = 0, 𝑋 = 𝑥∗ 𝐸 𝑌0 𝑃 = 1, 𝑋 = 𝑥∗ = 𝐸 𝑌0 𝑃 = 0, 𝑋 = 𝑥∗

30.

31.
X Y P

32.

33.

34.

35.

36.
𝐸 𝑌1 − 𝑌0 |𝑋= 𝑥∗ = 𝐸 𝑌1 |𝑋 = 𝑥∗ − 𝐸 𝑌0 |𝑋 = 𝑥∗ Average Y across sample of participants for whom 𝑋 = 𝑥∗ − Average Y across sample of participants of non−participants 𝑋 = 𝑥∗

37.
𝐸 𝑌1 − 𝑌0 =𝐸 𝑌1 − 𝐸 𝑌0 Population Participation Rate Poor .4 .7 Middle .5 .3 Rich .1 .1

38.

39.

40.
𝐸 𝑌1 − 𝑌0 =𝐸 𝑌1 − 𝐸 𝑌0 Participants (𝑷 = 𝟏) Non- participants (𝑷 = 𝟎) Poor .64 .21 Middle .34 .63 Rich .02 .16

41.

42.

43.
𝐸 𝑌1 − 𝑌0 =𝐸 𝑌1 − 𝐸 𝑌0 Participants (𝑷 = 𝟏) Weight Poor .64 .63 Middle .34 1.47 Rich .02 4.4

44.

45.

46.

47.
𝐸 𝑌1 − 𝑌0 =𝐸 𝑌1 − 𝐸 𝑌0 Average of Y across a sample of participants = 𝑖=1 𝑛 𝑃 𝑤𝑖 ∗ 𝑌𝑖 𝑛 𝑃 𝒘𝒊=.63 if individual i is poor 𝒘𝒊=1.47 if individual i is middle class 𝒘𝒊=4.4 if individual i is rich

48.
𝐸 𝑌1 − 𝑌0 =𝐸 𝑌1 − 𝐸 𝑌0 Average of Y across a sample of participants = 𝑖=1 𝑛 𝑃 𝑌𝑖 𝑛 𝑃 𝒘𝒊=.63 if individual i is poor 𝒘𝒊=1.47 if individual i is middle class 𝒘𝒊=4.4 if individual i is rich

49.
𝐸 𝑌1 − 𝑌0 =𝐸 𝑌1 − 𝐸 𝑌0 Average of Y across a sample of participants = 𝑖=1 𝑛 𝑃 𝑤𝑖 ∙ 𝑌𝑖 𝑖=1 𝑛 𝑃 𝑤𝑖 𝒘𝒊=.63 if individual i is poor 𝒘𝒊=1.47 if individual i is middle class 𝒘𝒊=4.4 if individual i is rich

50.
𝐸 𝑌1 − 𝑌0 =𝐸 𝑌1 − 𝐸 𝑌0 Average of Y across a sample of participants = 𝑖=1 𝑛 𝑃 𝒘𝒊 ∙ 𝑌𝑖 𝑖=1 𝑛 𝑃 𝑤𝑖 𝒘𝒊=.63 if individual i is poor 𝒘𝒊=1.47 if individual i is middle class 𝒘𝒊=4.4 if individual i is rich

51.

52.

53.
𝐸 𝑌1 − 𝑌0 =𝐸 𝑌1 − 𝐸 𝑌0 . 4 ∗ Average Y across sample of participants who are poor + .5 ∗ Average Y across sample of non−participants who are middle class +.1 ∗ Average Y across sample of participants who are rich

54.

56.
𝑌0 = 𝛽0+ 𝜖 𝑌1 = 𝛽0 + 𝛽1 + 𝜖

57.
𝑌1 − 𝑌0 =𝛽0 + 𝛽1 + 𝜖 − 𝛽0 + 𝜖 = 𝛽0 + 𝛽1 + 𝜖 − 𝛽0 − 𝜖 = 𝛽1

58.
𝑌1 − 𝑌0 =𝛽0 + 𝛽1 + 𝜖 − 𝛽0 + 𝜖 = 𝛽0 + 𝛽1 + 𝜖 − 𝛽0 − 𝜖 = 𝛽1

59.
𝑌1 − 𝑌0 =𝛽0 + 𝛽1 + 𝜖 − 𝛽0 + 𝜖 = 𝛽0 + 𝛽1 + 𝜖 − 𝛽0 − 𝜖 = 𝛽1

60.
𝑌1 − 𝑌0 =𝛽0 + 𝛽1 + 𝜖 − 𝛽0 + 𝜖 = 𝛽0 + 𝛽1 + 𝜖 − 𝛽0 − 𝜖 = 𝛽1

61.
𝑌 = 𝑃∙ 𝑌1 + 1 − 𝑃 ∙ 𝑌0 = 𝑃 ∗ 𝛽0 + 𝛽1 + 𝜖 + 1 − 𝑃 ∗ 𝛽0 + 𝜖 = 𝑃 ∗ 𝛽0 + 𝑃 ∗ 𝛽1 + 𝑃 ∗ 𝜖 +𝛽0 + 𝜖 − 𝑃 ∗ 𝛽0 − 𝑃 ∗ 𝜖 = 𝛽0 + 𝑃 ∗ 𝛽1 + 𝜖

62.
𝑌 = 𝑃∙ 𝑌1 + 1 − 𝑃 ∙ 𝑌0 = 𝑃 ∙ 𝛽0 + 𝛽1 + 𝜖 + 1 − 𝑃 ∙ 𝛽0 + 𝜖 = 𝑃 ∗ 𝛽0 + 𝑃 ∗ 𝛽1 + 𝑃 ∗ 𝜖 +𝛽0 + 𝜖 − 𝑃 ∗ 𝛽0 − 𝑃 ∗ 𝜖 = 𝛽0 + 𝑃 ∗ 𝛽1 + 𝜖

63.
𝑌 = 𝑃∙ 𝑌1 + 1 − 𝑃 ∙ 𝑌0 = 𝑃 ∙ 𝛽0 + 𝛽1 + 𝜖 + 1 − 𝑃 ∙ 𝛽0 + 𝜖 = 𝑃 ∙ 𝛽0 + 𝑃 ∙ 𝛽1 + 𝑃 ∙ 𝜖 +𝛽0 + 𝜖 − 𝑃 ∙ 𝛽0 − 𝑃 ∙ 𝜖 = 𝛽0 + 𝑃 ∗ 𝛽1 + 𝜖

64.

65.

66.
𝑌 = 𝑃∙ 𝑌1 + 1 − 𝑃 ∙ 𝑌0 = 𝑃 ∙ 𝛽0 + 𝛽1 + 𝜖 + 1 − 𝑃 ∙ 𝛽0 + 𝜖 = 𝑃 ∙ 𝛽0 + 𝑃 ∙ 𝛽1 + 𝑃 ∙ 𝜖 +𝛽0 + 𝜖 − 𝑃 ∙ 𝛽0 − 𝑃 ∙ 𝜖 = 𝛽0 + 𝑃 ∙ 𝛽1 + 𝜖

67.
𝑌 = 𝛽0+ 𝑃 ∙ 𝛽1 + 𝜖

68.
𝑌 = 𝛽0+ 𝛽1 ∙ 𝑃 + 𝜖

69.
𝑌0 = 𝛽0 +𝜖 𝑌1 = 𝛽0 + 𝛽1 + 𝜖 𝑌 = 𝑃 ∙ 𝑌1 + 1 − 𝑃 ∙ 𝑌0

70.
𝑌0 = 𝛽0 +𝜖 𝑌1 = 𝛽0 + 𝛽1 + 𝜖 𝑌 = 𝑃 ∙ 𝑌1 + 1 − 𝑃 ∙ 𝑌0

71.
𝑌0 = 𝛽0 +𝜖 𝑌1 = 𝛽0 + 𝛽1 + 𝜖 𝑌 = 𝑃 ∙ 𝑌1 + 1 − 𝑃 ∙ 𝑌0

72.
𝑌 = 𝛽0+ 𝛽1 ∙ 𝑃 + 𝜖

73.
𝑌0 = 𝛽0 +𝜖 𝑌1 = 𝛽0 + 𝛽1 + 𝜖 𝑌 = 𝑃 ∙ 𝑌1 + 1 − 𝑃 ∙ 𝑌0

74.
𝑌 = 𝛽0+ 𝛽1 ∙ 𝑃 + 𝜖

75.
𝑌𝑖 = 𝛽0+ 𝛽1 ∙ 𝑃𝑖 + 𝜖𝑖 𝑖 = 1, … , 𝑛

76.
𝜖𝑖 = 𝑌𝑖− 𝛽0 − 𝛽1 ∙ 𝑃𝑖 𝑚𝑖𝑛 𝛽0, 𝛽1 𝑖=1 𝑛 𝜀𝑖 2 = 𝑚𝑖𝑛 𝛽0, 𝛽1 𝑖=1 𝑛 𝑌𝑖 − 𝛽0 − 𝛽1 ∙ 𝑃𝑖 2 𝜖𝑖

77.
𝑌 = 𝛽0+ 𝛽1 ∙ 𝑃 + 𝜖

78.
𝑌 = 𝛽0+ 𝛽1 ∙ 𝑃 + 𝜖

79.
𝜖𝑖 = 𝑌𝑖− 𝛽0 − 𝛽1 ∙ 𝑃𝑖 𝑚𝑖𝑛 𝛽0, 𝛽1 𝑖=1 𝑛 𝜀𝑖 2 = 𝑚𝑖𝑛 𝛽0, 𝛽1 𝑖=1 𝑛 𝑌𝑖 − 𝛽0 − 𝛽1 ∙ 𝑃𝑖 2 𝜖𝑖

80.
𝛽1 = 𝑖=1 𝑛 𝑃𝑖 −𝑃 ∙ 𝑌𝑖 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 where 𝑃 = 𝑖=1 𝑛 𝑃𝑖 𝑛

81.
𝐸 𝛽1 =𝛽1

82.
An expected valuefor a random variable is the average value from a large number of repetitions of the experiment that random variable represents An expected value is the true average of a random variable across a population Expected value

83.
An expected valuefor a random variable is the average value from a large number of repetitions of the experiment that random variable represents An expected value is the true average of a random variable across a population Expected value

84.
An expected valueis the true average of a random variable across a population 𝐸 𝑋 = some true value Expected value

85.
𝑌 = 𝛽0+ 𝛽1 ∙ 𝑃 + 𝜖

86.
𝐸 𝑐 =𝑐 𝐸 𝑐 ∙ 𝑊 = 𝑐 ∙ 𝐸 𝑊 𝐸 𝑊 + 𝑍 = 𝐸 𝑊 + 𝐸 𝑍 𝐸 𝑊 − 𝑍 = 𝐸 𝑊 − 𝐸 𝑍 𝐸 𝑎 ∙ 𝑊 ± 𝑏 ∙ 𝑍 = 𝑎 ∙ 𝐸 𝑊 ± 𝑏 ∙ 𝐸 𝑍 Expectations: Properties

87.
𝑬 𝒄 =𝒄 𝐸 𝑐 ∙ 𝑊 = 𝑐 ∙ 𝐸 𝑊 𝐸 𝑊 + 𝑍 = 𝐸 𝑊 + 𝐸 𝑍 𝐸 𝑊 − 𝑍 = 𝐸 𝑊 − 𝐸 𝑍 𝐸 𝑎 ∙ 𝑊 ± 𝑏 ∙ 𝑍 = 𝑎 ∙ 𝐸 𝑊 ± 𝑏 ∙ 𝐸 𝑍 Expectations: Properties

88.
𝐸 𝑐 =𝑐 𝑬 𝒄 ∙ 𝑾 = 𝒄 ∙ 𝑬 𝑾 𝐸 𝑊 + 𝑍 = 𝐸 𝑊 + 𝐸 𝑍 𝐸 𝑊 − 𝑍 = 𝐸 𝑊 − 𝐸 𝑍 𝐸 𝑎 ∙ 𝑊 ± 𝑏 ∙ 𝑍 = 𝑎 ∙ 𝐸 𝑊 ± 𝑏 ∙ 𝐸 𝑍 Expectations: Properties

89.
𝐸 𝑐 =𝑐 𝐸 𝑐 ∙ 𝑊 = 𝑐 ∙ 𝐸 𝑊 𝑬 𝑾 + 𝒁 = 𝑬 𝑾 + 𝑬 𝒁 𝐸 𝑊 − 𝑍 = 𝐸 𝑊 − 𝐸 𝑍 𝐸 𝑎 ∙ 𝑊 ± 𝑏 ∙ 𝑍 = 𝑎 ∙ 𝐸 𝑊 ± 𝑏 ∙ 𝐸 𝑍 Expectations: Properties

90.
𝐸 𝑐 =𝑐 𝐸 𝑐 ∙ 𝑊 = 𝑐 ∙ 𝐸 𝑊 𝐸 𝑊 + 𝑍 = 𝐸 𝑊 + 𝐸 𝑍 𝑬 𝑾 − 𝒁 = 𝑬 𝑾 − 𝑬 𝒁 𝐸 𝑎 ∙ 𝑊 ± 𝑏 ∙ 𝑍 = 𝑎 ∙ 𝐸 𝑊 ± 𝑏 ∙ 𝐸 𝑍 Expectations: Properties

91.
𝐸 𝑐 =𝑐 𝐸 𝑐 ∙ 𝑊 = 𝑐 ∙ 𝐸 𝑊 𝐸 𝑊 + 𝑍 = 𝐸 𝑊 + 𝐸 𝑍 𝐸 𝑊 − 𝑍 = 𝐸 𝑊 − 𝐸 𝑍 𝑬 𝒂 ∙ 𝑾 ± 𝒃 ∙ 𝒁 = 𝒂 ∙ 𝑬 𝑾 ± 𝒃 ∙ 𝑬 𝒁 Expectations: Properties

92.
𝐸 𝑊 ∙𝑍 ≠ 𝐸 𝑊 ∙ 𝐸 𝑍 𝐸 𝑊 𝑍 ≠ 𝐸 𝑊 𝐸 𝑍 𝐸 𝑓 𝑊 ≠ 𝑓 𝐸 𝑊 Expectations: Properties

93.
𝑬 𝑾 ∙𝒁 ≠ 𝑬 𝑾 ∙ 𝑬 𝒁 𝐸 𝑊 𝑍 ≠ 𝐸 𝑊 𝐸 𝑍 𝐸 𝑓 𝑊 ≠ 𝑓 𝐸 𝑊 Expectations: Properties

94.
𝐸 𝑊 ∙𝑍 ≠ 𝐸 𝑊 ∙ 𝐸 𝑍 𝑬 𝑾 𝒁 ≠ 𝑬 𝑾 𝑬 𝒁 𝐸 𝑓 𝑊 ≠ 𝑓 𝐸 𝑊 Expectations: Properties

95.
𝐸 𝑊 ∙𝑍 ≠ 𝐸 𝑊 ∙ 𝐸 𝑍 𝐸 𝑊 𝑍 ≠ 𝐸 𝑊 𝐸 𝑍 𝑬 𝒇 𝑾 ≠ 𝒇 𝑬 𝑾 Expectations: Properties

96.
𝑌𝑖 = 𝛽0+ 𝛽1 ∙ 𝑃𝑖 + 𝜖𝑖 𝛽1 = 𝑖=1 𝑛 𝑃𝑖 − 𝑃 ∙ 𝑌𝑖 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

97.

98.

99.
𝐸 𝛽1 =𝛽1

101.
𝐸 𝛽1 =𝛽1

103.
𝛽1 = 𝑖=1 𝑛 𝑃𝑖 −𝑃 ∙ 𝑌𝑖 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

104.
𝐸( 𝛽1) =𝐸 𝑖=1 𝑛 𝑃𝑖 − 𝑃 ∙ 𝑌𝑖 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

105.
𝑌𝑖 = 𝛽0+ 𝛽1 ∙ 𝑃𝑖 + 𝜖𝑖

106.
𝐸( 𝛽1) =𝐸 𝑖=1 𝑛 𝑃𝑖 − 𝑃 ∙ 𝑌𝑖 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

107.
𝐸 𝛽1 = 𝐸 𝑖=1 𝑛 𝑃𝑖− 𝑃 ∙ 𝛽0 + 𝛽1 ∙ 𝑃𝑖 + 𝜖𝑖 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

108.
𝐸 𝛽1 = 𝐸 𝑖=1 𝑛 𝛽0∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 + 𝑖=1 𝑛 𝛽1 ∙ 𝑃𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

109.
𝐸 𝑊 +𝑍 = 𝐸 𝑊 + 𝐸 𝑍 𝐸 𝑊 + 𝑍 + 𝑄 = 𝐸 𝑊 + 𝐸 𝑍 + 𝐸 𝑄

110.
𝐸 𝑊 +𝑍 = 𝐸 𝑊 + 𝐸 𝑍 𝐸 𝑊 + 𝑍 + 𝑄 = 𝐸 𝑊 + 𝐸 𝑍 + 𝐸 𝑄

111.

112.

113.

114.

115.

116.
𝐸 𝛽1 = 𝐸 𝑖=1 𝑛 𝛽0∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 + 𝐸 𝑖=1 𝑛 𝛽1 ∙ 𝑃𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 + 𝐸 𝑖=1 𝑛 𝜖𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

117.

118.
𝐸 𝑐 ∙𝑊 = 𝑐 ∙ 𝐸 𝑊

119.

120.
𝐸 𝛽1 = 𝛽0∙ 𝐸 𝑖=1 𝑛 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 + 𝐸 𝑖=1 𝑛 𝛽1 ∙ 𝑃𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 + 𝐸 𝑖=1 𝑛 𝜖𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

121.
𝐸 𝛽1 = 𝛽0∙ 𝐸 𝑖=1 𝑛 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 + 𝐸 𝑖=1 𝑛 𝛽1 ∙ 𝑃𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 + 𝐸 𝑖=1 𝑛 𝜖𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

122.
𝐸 𝛽1 = 𝛽0∗ 𝐸 0 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 + 𝐸 𝑖=1 𝑛 𝛽1 ∙ 𝑃𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 + 𝐸 𝑖=1 𝑛 𝜖𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

123.
𝐸 𝛽1 = 𝛽0∗ 0 +𝐸 𝑖=1 𝑛 𝛽1 ∙ 𝑃𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 + 𝐸 𝑖=1 𝑛 𝜖𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

124.
𝐸 𝛽1 = 𝐸 𝑖=1 𝑛 𝛽1∙ 𝑃𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 + 𝐸 𝑖=1 𝑛 𝜖𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

125.
𝐸 𝛽1 = 𝐸 𝑖=1 𝑛 𝛽1∙ 𝑃𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 + 𝐸 𝑖=1 𝑛 𝜖𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

126.
𝐸 𝛽1 = 𝛽1∙ 𝐸 𝑖=1 𝑛 𝑃𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 + 𝐸 𝑖=1 𝑛 𝜖𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

127.
𝐸 𝛽1 = 𝛽1∙ 𝐸 𝑖=1 𝑛 𝑃𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 + 𝐸 𝑖=1 𝑛 𝜖𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

128.
𝑖=1 𝑛 𝑃𝑖 ∙ 𝑃𝑖− 𝑃 = 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

129.
𝐸 𝛽1 = 𝛽1∗ 𝐸 𝑖=1 𝑛 𝑃𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 + 𝐸 𝑖=1 𝑛 𝜖𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

130.
𝐸 𝛽1 = 𝛽1∙ 𝐸 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 + 𝐸 𝑖=1 𝑛 𝜖𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

131.
𝐸 𝛽1 = 𝛽1∙ 𝐸 1 +𝐸 𝑖=1 𝑛 𝜖𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

132.
𝐸 𝛽1 = 𝛽1∙ 1 +𝐸 𝑖=1 𝑛 𝜖𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

133.
𝐸 𝛽1 = 𝛽1 +𝐸 𝑖=1 𝑛 𝜖𝑖∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

134.

135.
𝐸 𝛽1 = 𝛽1 +0

136.
𝐸 𝛽1 = 𝛽1

137.

138.
𝐸 𝑊 ∙𝑍 ≠ 𝐸 𝑊 ∙ 𝐸 𝑍 Expectations: Properties

139.

140.
𝐸 𝛽1 = 𝛽1 + 𝑖=1 𝑛 𝐸 𝜖𝑖∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

141.
𝐸 𝛽1 = 𝛽1 + 𝑖=1 𝑛 𝐸 𝑖=1 𝑛 𝑃𝑖− 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 ∙ 𝐸 𝜖𝑖

142.
𝐸 𝛽1 = 𝛽1 + 𝑖=1 𝑛 𝐸 𝑖=1 𝑛 𝑃𝑖− 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 ∙ 0

143.
𝐸 𝛽1 = 𝛽1 +0

144.
𝐸 𝛽1 = 𝛽1

145.
𝐸 𝛽1 = 𝛽1 𝑌𝑖= 𝛽0 + 𝛽1 ∙ 𝑃𝑖 + 𝜖𝑖

146.
𝐸 𝛽1 = 𝛽1 𝑌𝑖= 𝛽0 + 𝛽1 ∙ 𝑃𝑖 + 𝜖𝑖

147.
𝐸 𝛽1 = 𝛽1 𝑌𝑖= 𝛽0 + 𝛽1 ∙ 𝑃𝑖 + 𝜖𝑖

148.
Peter M. Lance,PhD MEASURE Evaluation University of North Carolina at Chapel Hill September 13 and 15, 2016 Selection on Observables: Part Deux

149.
𝑌0 = 𝛽0 +𝜖 𝑌1 = 𝛽0 + 𝛽1 + 𝜖 𝑌 = 𝑃 ∙ 𝑌1 + 1 − 𝑃 ∙ 𝑌0 𝑌 = 𝛽0 + 𝛽1 ∙ 𝑃 + 𝜖

150.

151.

152.

153.

154.
𝐸 𝛽1 = 𝛽1 𝑌𝑖= 𝛽0 + 𝛽1 ∙ 𝑃𝑖 + 𝜖𝑖

155.
𝐸 𝛽1 = 𝛽1 𝑌𝑖= 𝛽0 + 𝛽1 ∙ 𝑃𝑖 + 𝜖𝑖

156.
𝐸 𝛽1 = 𝛽1 𝑌𝑖= 𝛽0 + 𝛽1 ∙ 𝑃𝑖 + 𝜖𝑖

157.
𝑌0 = 𝛽0+ 𝜀 𝑌1 = 𝛽0 + 𝛽1 + 𝜀

158.
𝑌0 = 𝛽0+ 𝛽2 ∙ 𝑥 + 𝜀 𝑌1 = 𝛽0 + 𝛽1 + 𝛽2 ∙ 𝑥 + 𝜀

159.
𝑌1 − 𝑌0 =𝛽0 + 𝛽1 + 𝜖 − 𝛽0 + 𝜖 = 𝛽0 + 𝛽1 + 𝜖 − 𝛽0 − 𝜖 = 𝛽1

160.
𝑌1 − 𝑌0 =𝛽0 + 𝛽1 + 𝛽2 ∗ 𝑥 + 𝜀 − 𝛽0 + 𝛽2 ∗ 𝑥 + 𝜀 = 𝛽0 + 𝛽1 + 𝜖 − 𝛽0 − 𝜖 = 𝛽1

161.
𝑌1 − 𝑌0 =𝛽1

162.

163.
𝑌 = 𝑃∙ 𝑌1 + 1 − 𝑃 ∙ 𝑌0 = 𝑃 ∙ 𝛽0 + 𝛽1 + 𝛽2 ∙ 𝑥 + 𝜀 + 1 − 𝑃 ∙ 𝛽0 + 𝛽2 ∙ 𝑥 + 𝜀 = 𝑃 ∗ 𝛽0 + 𝑃 ∗ 𝛽1 + 𝑃 ∗ 𝜖 +𝛽0 + 𝜖 − 𝑃 ∗ 𝛽0 − 𝑃 ∗ 𝜖

164.
𝑌 = 𝛽0+ 𝛽1 ∙ 𝑃 + 𝛽2 ∙ 𝑥 + 𝜀

165.
Cost of participation 𝐶= 𝜌0 + 𝜌1 ∙ 𝑥

166.
Cost of participation 𝐶= 𝜌0 + 𝜌1 ∙ 𝑥

167.
Benefit-Cost>0 𝑌1 − 𝑌0 − C> 0 𝛽1 − 𝐶 > 0 𝛽1 − 𝛾0 + 𝛾1 ∗ 𝑥 > 0

168.
Benefit-Cost>0 𝑌1 − 𝑌0− C > 0 𝛽1 − 𝐶 > 0 𝛽1 − 𝛾0 + 𝛾1 ∗ 𝑥 > 0

169.
Benefit-Cost>0 𝑌1 − 𝑌0− C > 0 𝛽1 − 𝐶 > 0 𝛽1 − 𝛾0 + 𝛾1 ∗ 𝑥 > 0

170.
Benefit-Cost>0 𝑌1 − 𝑌0− C > 0 𝛽1 − 𝐶 > 0 𝛽1 − 𝜌0 + 𝜌1 ∙ 𝑥 > 0

171.
𝑥 𝑃

172.
X Y P

173.
Benefit-Cost>0 𝛽1 − 𝜌0+ 𝜌1 ∙ 𝑥 > 0 𝜀

174.

175.

176.
P and 𝜺are independent

177.
True model: 𝑌 =𝛽0 + 𝛽1 ∙ 𝑃 + 𝛽2 ∙ 𝑥 + 𝜀 We actually attempt to estimate: 𝑌 = 𝜏0 + 𝜏1 ∙ 𝑃 + 𝜖 Error term now contains: 𝛽2 ∙ 𝑥

178.

179.

180.
𝐸( 𝜏1) =𝐸 𝑖=1 𝑛 𝑃𝑖 − 𝑃 ∙ 𝑌𝑖 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

181.
𝐸 𝜏1 = 𝐸 𝑖=1 𝑛 𝑃𝑖− 𝑃 ∙ 𝛽0 + 𝛽1 ∙ 𝑃𝑖 + 𝛽2 ∙ 𝑥𝑖 + 𝜀𝑖 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

182.
𝐸 𝜏1 = 𝐸 𝑖=1 𝑛 𝛽0∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 + 𝑖=1 𝑛 𝛽1 ∙ 𝑃𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 + 𝑖=1 𝑛 𝛽2 ∙ 𝑥𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 + 𝑖=1 𝑛 𝜀𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

183.
𝐸 𝜏1 = 𝐸 𝑖=1 𝑛 𝛽0∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 + 𝐸 𝑖=1 𝑛 𝛽1 ∙ 𝑃𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 +𝐸 𝑖=1 𝑛 𝛽2 ∙ 𝑥𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 +𝐸 𝑖=1 𝑛 𝜀𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

184.
𝐸 𝜏1 = 𝐸 𝑖=1 𝑛 𝛽0∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 + 𝐸 𝑖=1 𝑛 𝛽1 ∙ 𝑃𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 +𝐸 𝑖=1 𝑛 𝛽2 ∙ 𝑥𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 +𝐸 𝑖=1 𝑛 𝜀𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

185.
𝐸 𝜏1 = 0 +𝐸 𝑖=1 𝑛 𝛽1∙ 𝑃𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 +𝐸 𝑖=1 𝑛 𝛽2 ∙ 𝑥𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 +𝐸 𝑖=1 𝑛 𝜀𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

186.
𝐸 𝜏1 = 𝐸 𝑖=1 𝑛 𝛽1∙ 𝑃𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 +𝐸 𝑖=1 𝑛 𝛽2 ∙ 𝑥𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 +𝐸 𝑖=1 𝑛 𝜀𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

187.
𝐸 𝜏1 = 𝐸 𝑖=1 𝑛 𝛽1∙ 𝑃𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 +𝐸 𝑖=1 𝑛 𝛽2 ∙ 𝑥𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 +𝐸 𝑖=1 𝑛 𝜀𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

188.
𝐸 𝜏1 = 𝛽1 +𝐸 𝑖=1 𝑛 𝛽2∙ 𝑥𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 +𝐸 𝑖=1 𝑛 𝜀𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

189.
𝐸 𝜏1 = 𝛽1 +𝐸 𝑖=1 𝑛 𝛽2∙ 𝑥𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 +𝐸 𝑖=1 𝑛 𝜀𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

190.
𝐸 𝜏1 = 𝛽1 +𝐸 𝑖=1 𝑛 𝛽2∙ 𝑥𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 +0

191.
𝐸 𝜏1 = 𝛽1 +𝐸 𝑖=1 𝑛 𝛽2∙ 𝑥𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

192.
𝐸 𝜏1 = 𝛽1 +𝐸 𝑖=1 𝑛 𝛽2∙ 𝑥𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

193.
𝐸 𝑐 ∙𝑊 = 𝑐 ∙ 𝐸 𝑊

194.
𝐸 𝜏1 = 𝛽1 +𝛽2∙ 𝐸 𝑖=1 𝑛 𝑥𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

195.
𝐸 𝜏1 = 𝛽1 +𝛽2∙ 𝐸 𝑖=1 𝑛 𝑥𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

196.
𝐸 𝜏1 = 𝛽1 +𝛽2∙ 𝐸 𝑖=1 𝑛 𝑥𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2 𝑥𝑖= 𝛾0 + 𝛾1 ∙ 𝑃𝑖 + 𝜗𝑖

197.

198.

199.
𝐸 𝜏1 = 𝛽1 +𝛽2∙ 𝛾1 𝑖=1 𝑛 𝑥 𝑖=1 𝑛 𝑥 𝑥𝑖= 𝛾0 + 𝛾1 ∙ 𝑃𝑖 + 𝜗𝑖

200.

201.
𝐸 𝜏1 =𝛽1 + 𝛽2 ∙ 𝐸 𝑖=1 𝑛 𝑥𝑖 ∙ 𝑃𝑖 − 𝑃 𝑖=1 𝑛 𝑃𝑖 − 𝑃 2

202.
𝐸 𝜏1 =𝛽1 + 𝛽2 ∙ 𝛾1

203.
𝐸 𝜏1 =𝛽1 + 𝛽2 ∙ 𝛾1 The Actual Causal Effect of P on y

204.
𝐸 𝜏1 =𝛽1 + 𝛽2 ∙ 𝛾1 The actual causal effect of P on y The actual causal effect of the omitted variable X on Y

205.
𝐸 𝜏1 =𝛽1 + 𝛽2 ∙ 𝛾1 The actual causal effect of P on y Th”Effect” of P on x: 𝑥𝑖= 𝛾0 + 𝛾1 ∙ 𝑃𝑖 + 𝜗𝑖 The actual causal effect of the omitted variable X on Y

206.

207.

208.

209.

210.
True model: 𝑌 =𝛽0 + 𝛽1 ∙ 𝑃 + 𝛽2 ∙ 𝑥 + 𝜀 We actually attempt to estimate: 𝑌 = 𝜏0 + 𝜏1 ∙ 𝑃 + 𝜖 𝑬 𝝉 𝟏 ≠ 𝜷

211.

212.

213.
𝑌0 = 𝛽0 +𝛽2 ∙ 𝑥1 + 𝛽3 ∙ 𝑥2 + 𝛽4 ∙ 𝑥2 + 𝜀 𝑌1 = 𝛽0 + 𝛽1 + 𝛽2 ∙ 𝑥1 + 𝛽3 ∙ 𝑥2 + 𝛽4 ∙ 𝑥2 + 𝜀

214.
𝑌0 = 𝛽0 +𝛽2 ∙ 𝑥1 + 𝛽3 ∙ 𝑥2 + 𝛽4 ∙ 𝑥2 + 𝜀 𝑌1 = 𝛽0 + 𝛽1 + 𝛽2 ∙ 𝑥1 + 𝛽3 ∙ 𝑥2 + 𝛽4 ∙ 𝑥3 + 𝜀

215.
𝑌1 − 𝑌0 =𝛽0 + 𝛽1 + 𝜖 − 𝛽0 + 𝜖 = 𝛽0 + 𝛽1 + 𝜖 − 𝛽0 − 𝜖 = 𝛽1

216.
𝑌1 − 𝑌0 = 𝛽0+ 𝛽1 + 𝛽2 ∙ 𝑥1 + 𝛽3 ∙ 𝑥2 + 𝛽4 ∙ 𝑥2 + 𝜀 − 𝛽0 + 𝛽2 ∙ 𝑥1 + 𝛽3 ∙ 𝑥2 + 𝛽4 ∙ 𝑥2 + 𝜀 = 𝛽0 + 𝛽1 + 𝜖 − 𝛽0 − 𝜖 = 𝛽1

217.
𝑌1 − 𝑌0 =𝛽1

218.

219.
𝑌 = 𝑃∙ 𝑌1 + 1 − 𝑃 ∙ 𝑌0 = 𝑃 ∙ 𝛽0 + 𝛽1 + 𝛽2 ∙ 𝑥1 + 𝛽3 ∙ 𝑥2 + 𝛽4 ∙ 𝑥3 + 𝜀 + 1 − 𝑃 ∙ 𝛽0 + 𝛽2 ∙ 𝑥 + 𝛽2 ∙ 𝑥1 + 𝛽3 ∙ 𝑥2 + 𝛽4 ∙ 𝑥3 + 𝜀

220.
𝑌 = 𝛽0+ 𝛽1 ∙ 𝑃 + 𝛽2 ∙ 𝑥1 + 𝛽3 ∙ 𝑥2 + 𝛽4 ∙ 𝑥3 + 𝜀

221.
Cost of participation 𝐶= 𝜌0 + 𝜌1 ∙ 𝑥1 + 𝜌2 ∙ 𝑥2 + 𝜌3 ∙ 𝑥3

222.
True model: 𝑌 =𝛽0 + 𝛽1 ∙ 𝑃 + 𝛽2 ∙ 𝑥1 + 𝛽3 ∙ 𝑥2 + 𝛽4 ∙ 𝑥3 + 𝜀 We actually attempt to estimate: 𝑌 = 𝜏0 + 𝜏1 ∙ 𝑃 + 𝜖

223.
𝐸 𝜏1 =𝛽1 + 𝛽2 ∙ 𝛾11 + 𝛽3 ∙ 𝛾21 + 𝛽4 ∙ 𝛾31

224.
𝐸 𝜏1 =𝛽1 + 𝛽2 ∙ 𝛾11 + 𝛽3 ∙ 𝛾21 + 𝛽4 ∙ 𝛾31 The actual causal effect of P on y

225.
𝐸 𝜏1 =𝛽1 + 𝛽2 ∙ 𝛾11 + 𝛽3 ∙ 𝛾21 + 𝛽4 ∙ 𝛾31 The actual causal effect of P on y Th”Effect” of P on x1: 𝑥1𝑖= 𝛾10 + 𝛾11 ∙ 𝑃𝑖 + 𝜗1𝑖 The actual causal effect of the omitted variable X1 on Y

226.
𝐸 𝜏1 =𝛽1 + 𝛽2 ∙ 𝛾11 + 𝛽3 ∙ 𝛾21 + 𝛽4 ∙ 𝛾31 The actual causal effect of P on y ”Effect” of P on x1: 𝑥1𝑖= 𝛾10 + 𝛾11 ∙ 𝑃𝑖 + 𝜗1𝑖 The actual causal effect of the omitted variable X1 on Y The actual causal effect of the omitted variable X2 on Y Th”Effect” of P on x2: 𝑥2𝑖= 𝛾20 + 𝛾21 ∙ 𝑃𝑖 + 𝜗2𝑖

227.
𝐸 𝜏1 =𝛽1 + 𝛽2 ∙ 𝛾11 + 𝛽3 ∙ 𝛾21 + 𝛽4 ∙ 𝛾31 The actual causal effect of P on y ”Effect” of P on x1: 𝑥1𝑖= 𝛾10 + 𝛾11 ∙ 𝑃𝑖 + 𝜗1𝑖 The actual causal effect of the omitted variable X1 on Y The actual causal effect of the omitted variable X2 on Y ”Effect” of P on x2: 𝑥2𝑖= 𝛾20 + 𝛾21 ∙ 𝑃𝑖 + 𝜗2𝑖 Th”Effect” of P on x1: 𝑥3𝑖= 𝛾30 + 𝛾31 ∙ 𝑃𝑖 + 𝜗3𝑖 The actual causal effect of the omitted variable X3 on Y

228.
𝐸 𝜏1 =𝛽1 + 𝛽2 ∙ 𝛾11 + 𝛽3 ∙ 𝛾21 + 𝛽4 ∙ 𝛾31 The actual causal effect of P on y ”Effect” of P on x1: 𝑥1𝑖= 𝛾10 + 𝛾11 ∙ 𝑃𝑖 + 𝜗1𝑖 The actual causal effect of the omitted variable X1 on Y The actual causal effect of the omitted variable X2 on Y ”Effect” of P on x2: 𝑥2𝑖= 𝛾20 + 𝛾21 ∙ 𝑃𝑖 + 𝜗2𝑖 Th”Effect” of P on x3: 𝑥3𝑖= 𝛾30 + 𝛾31 ∙ 𝑃𝑖 + 𝜗3𝑖 The actual causal effect of the omitted variable X3 on Y

229.
𝐸 𝜏1 =𝛽1 + 𝛽2 ∙ 𝛾11 + 𝛽2 ∙ 𝛾21 + 𝛽3 ∙ 𝛾31

230.
Potential outcomes 𝑌0 = 𝛽0+ 𝛽2 ∙ 𝑥1 + 𝛽3 ∙ 𝑥2 + 𝜀 𝑌1 = 𝛽0 + 𝛽1 + 𝛽2 ∙ 𝑥1 + 𝛽3 ∙ 𝑥2 + 𝜀 Costs of participation 𝐶 = 𝜌0 + 𝜌2 ∙ 𝑥2 + 𝜌3 ∙ 𝑥3

231.
X2 Y P X1 X3

232.
𝐸 𝜏1 =𝛽1 + 𝛽2 ∙ 𝛾11 + 𝛽2 ∙ 𝛾21 + 𝛽3 ∙ 𝛾31

233.
𝐸 𝜏1 =𝛽1 + 𝛽2 ∙ 𝛾11 + 𝛽2 ∙ 𝛾21 + 𝛽3 ∙ 𝛾31

234.
𝐸 𝜏1 =𝛽1 + 𝛽2 ∙ 𝛾21 + 𝛽3 ∙ 𝛾31

235.
X2 Y P X1 X3

236.
𝐸 𝜏1 =𝛽1 + 𝛽2 ∙ 𝛾21 + 𝛽3 ∙ 𝛾31

237.
𝐸 𝜏1 =𝛽1 + 𝛽2 ∙ 𝛾21

238.
X2 Y P X1 X3

239.
𝐸 𝜏1 =𝛽1 + 𝛽2 ∙ 𝛾21

240.
X2 Y P X1 X3

241.
𝑌 = 𝛽0+ 𝛽1 ∙ 𝑃 + 𝛽2 ∙ 𝑥1 + 𝛽3 ∙ 𝑥2 + 𝛽4 ∙ 𝑥3 + 𝜀

243.
𝑌 = 𝛽0+ 𝛽1 ∙ 𝑃 + 𝛽2 ∙ 𝑥 + 𝜀 𝐸 𝛽1 = 𝛽1

244.
𝑌 = 𝛽0+ 𝛽1 ∙ 𝑃 + 𝛽2 ∙ 𝑥 + 𝜀 Regress 𝑌 on 𝑃 and 𝑥 𝐸 𝛽1 = 𝛽1

245.
𝑌 = 𝛽0+ 𝛽1 ∙ 𝑃 + 𝛽2 ∙ 𝑥 + 𝜀 Regress 𝑌 on 𝑃 and 𝑥 𝐸 𝛽1 = 𝛽1

246.
𝑌 = 𝛽0+ 𝛽1 ∙ 𝑃 + 𝛽2 ∙ 𝑥 + 𝜀 Regress 𝑌 on 𝑃 and 𝑥 and 𝑍 𝐸 𝛽1 ≠ 𝛽1

247.
𝑌 = 𝛽0+ 𝛽1 ∙ 𝑃 + 𝛽2 ∙ 𝑥 + 𝜀 Regress 𝑌 on 𝑃 and 𝑥 and 𝑍 𝐸 𝛽1 ≠ 𝛽1

248.
Bad controls

249.
X2 Y P X1 X3

250.
X2 Y P X1 X3 Z

251.
𝑌 = 𝛽0+ 𝛽1 ∙ 𝑃 + 𝛽𝑧 ∙ 𝑍 + 𝜔 1.Is correlated with the regressor of interest 2.Is correlated with the error term

252.

253.

254.

255.
Matching

256.
X Y P

258.
𝑋 = 0 ifmale 1 if female

259.
𝐸 𝑌1 |𝑋 𝐸𝑌0 |𝑋 𝐸 𝑌1 |𝑋 − 𝐸 𝑌 0 |𝑋 𝑋 = 0 6 4 2 𝑋 = 1 5 1 4

260.
𝐸 𝑌1 |𝑋 𝐸𝑌0 |𝑋 𝐸 𝑌1 |𝑋 − 𝐸 𝑌 0 |𝑋 𝑋 = 0 6 4 2 𝑋 = 1 5 1 4 𝐸 𝑌1 − 𝑌0 |𝑋 = 0 ∙ 𝑃𝑟 𝑋 = 0 +𝐸 𝑌1 − 𝑌0 |𝑋 = 1 ∙ 𝑃𝑟 𝑋 = 1 = 2 ∙ .5 + 4 ∙ .5 = 3

261.
𝐸 𝑌1 |𝑋 𝐸𝑌0 |𝑋 𝐸 𝑌1 |𝑋 − 𝐸 𝑌 0 |𝑋 𝑋 = 0 6 4 2 𝑋 = 1 5 1 4 𝐸 𝑌1 |𝑋 = 0 ∙ 𝑃𝑟 𝑋 = 0 +𝐸 𝑌1 |𝑋 = 1 ∙ 𝑃𝑟 𝑋 = 1 = 6 ∙ .5 + 5 ∙ .5 = 5.5

262.
𝐸 𝑌1 |𝑋 𝐸𝑌0 |𝑋 𝐸 𝑌1 |𝑋 − 𝐸 𝑌 0 |𝑋 𝑋 = 0 6 4 2 𝑋 = 1 5 1 4 𝐸 𝑌0 |𝑋 = 0 ∙ 𝑃𝑟 𝑋 = 0 +𝐸 𝑌0 |𝑋 = 1 ∙ 𝑃𝑟 𝑋 = 1 = 4 ∙ .5 + 1 ∙ .5 = 2.5

263.
𝐸 𝑃 =1|𝑋 𝐸 𝑃 = 0|𝑋 𝑋 = 0 .1 .9 𝑋 = 1 .5 .5 𝐸 𝑃 = 1|𝑋 = 0 ∙ 𝑃𝑟 𝑋 = 0 +𝐸 𝑃 = 1|𝑋 = 1 ∙ 𝑃𝑟 𝑋 = 1 = .1 ∙ .5 + .5 ∙ .5 = .3

264.
𝐸 𝑃 =1|𝑋 𝐸 𝑃 = 0|𝑋 𝑋 = 0 .1 .9 𝑋 = 1 .5 .5 𝐸 𝑃 = 1|𝑋 = 0 ∙ 𝑃𝑟 𝑋 = 0 +𝐸 𝑃 = 1|𝑋 = 1 ∙ 𝑃𝑟 𝑋 = 1 = .1 ∙ .5 + .5 ∙ .5 = .3

265.
𝑛 = 1000 𝑖=1 𝑛 𝑋𝑖 𝑛 = 𝑖=1 1000 𝑋𝑖 1000 ≈500 𝑖=1 𝑛 𝑃𝑖 𝑛 = 𝑖=1 1000 𝑃𝑖 1000 ≈ 300

266.
𝑛 = 1000 𝑖=1 𝑛 𝑋𝑖 𝑛 = 𝑖=1 1000 𝑋𝑖 1000 ≈.5 𝑖=1 𝑛 𝑃𝑖 𝑛 = 𝑖=1 1000 𝑃𝑖 1000 ≈ 300

267.
𝑛 = 1000 𝑖=1 𝑛 𝑋𝑖 𝑛 = 𝑖=1 1000 𝑋𝑖 1000 ≈.5 𝑖=1 𝑛 𝑃𝑖 𝑛 = 𝑖=1 1000 𝑃𝑖 1000 ≈ .3

268.
𝑗=1 300 𝑌𝑗 300 : 6 ∙.16667 + 5 ∙ .8333 = 5.16652

269.
𝑗=1 300 𝑌𝑗 300 : 6 ∙.16667 + 5 ∙ .8333 = 5.16652 𝑗=1 700 𝑌𝑗 700 : 4 ∙ .6428 + 1 ∙ .3571 = 2.9286

270.
?

271.
𝑖 = 237 𝑋237= 0 𝑃237 = 1 𝑌237 = 𝑌237 1 = 5.3

272.
𝑖 = 237 𝑋237= 0 𝑃237 = 1 𝑌237 = 𝑌237 1 = 5.3

273.
𝑖 = 237 𝑋237= 0 𝑃237 = 1 𝑌237 = 𝑌237 1 = 5.3

274.
𝑖 = 237 𝑋237= 0 𝑃237 = 1 𝑌237 = 𝑌237 1 = 5.3

275.
𝑖 = 237 𝑋237= 0 𝑃237 = 1 𝑌237 = 𝑌237 1 = 6.3

276.
𝐼237 = 𝑌237 1 −𝑌237 0 = 6.3−?

277.
𝑌237 0

278.
𝑌237 0 = 𝑌766 0

279.
𝑌237 0 = 𝑌48 0 + 𝑌109 0 + 𝑌418 0 +𝑌505 0 + 𝑌919 0 5

280.
𝐼𝑖 = 𝑌𝑖 1 −𝑌𝑖 0

281.
𝐴𝑇𝐸 = 𝑖=1 𝑛 𝐼𝑖 𝑛

282.
𝑘=0 𝐾 𝐸 𝐼|𝑋 =𝑘 ∙ 𝑃𝑟 𝑋 = 𝑘 = 𝑘=0 𝐾 𝐸 𝑌1 − 𝑌0 |𝑋 = 𝑘 ∙ 𝑃𝑟 𝑋 = 𝑘

283.
𝑘=0 𝐾 𝐸 𝐼|𝑋 =𝑘, 𝑃 = 1 ∙ 𝑃𝑟 𝑋 = 𝑘|𝑃 = 1 = 𝑘=0 𝐾 𝐸 𝑌1 − 𝑌0 |𝑋 = 𝑘, 𝑃 = 1 ∙ 𝑃𝑟 𝑋 = 𝑘|𝑃 = 1

284.
𝐼237 = 𝑌237 1 −𝑌237 0 = 6.3−?

285.
𝑘=0 𝐾 𝐸 𝐼|𝑋 =𝑘 ∙ 𝑃𝑟 𝑋 = 𝑘 = 𝑘=0 𝐾 𝐸 𝑌1 − 𝑌0 |𝑋 = 𝑘 ∙ 𝑃𝑟 𝑋 = 𝑘

286.
X Y P

288.
2X Gender-male/female 5X SES-5quintiles 4X Age-broken up into 4 categories 2X Sector-Urban/Rural 6X Religion-Christian, Muslim, Hindu, Buddhist, Jewish, Traditional. 6X Occupation-6 occupation types 2 Insurance-Insured/not insured ____ 5,760 potential “types”

289.
2X Gender-male/female 5X SES-5quintiles 4X Age-broken up into 4 categories 2X Sector-Urban/Rural 6X Religion-Christian, Muslim, Hindu, Buddhist, Jewish, Traditional. 6X Occupation-6 occupation types 2 Insurance-Insured/not insured ____ 2 potential “types”

290.

291.

292.

293.

294.

295.

296.

297.
Pr(𝑃 = 1|𝑋)

298.
𝐼237 = 𝑌237 1 −𝑌237 0 = 6.3−?

299.
𝐼237 = 𝑌237 1 −𝑌237 0 = 6.3−?

300.
Pr(𝑃 = 1|𝑋)

301.
𝑃𝑟(𝑃 = 1|𝑋)

302.
𝑃𝑟 𝑃 =1 𝑋 = 𝑒𝑥𝑝 𝛽0 + 𝛽1 ∙ 𝑋 1 + 𝑒𝑥𝑝 𝛽0 + 𝛽1 ∙ 𝑋

303.
The basics ofthe method 1. Pool your sample of participants and non-participants and define various characteristics 𝑋. 2. Estimate the probability of program participation conditional on Pr 𝑃 = 1|𝑋 3. Compute the propensity score using the fitted binary participation model. 𝑃𝑟 𝑃𝑖 = 1|𝑋𝑖 4. Find a counterfactual outcome for each individual by identifying some individual who did experience the counterfactual conditions and had a similar propensity score

304.
Matching approaches 1. Nearestneighbor 2. Caliper 3. (Budget) Caliper 4. Weighting

305.
Matching approaches 1. Nearestneighbor 2. Caliper 3. (Budget) Caliper 4. Weighting

306.
Balancing property Pr 𝑃= 1|𝑋

307.
Propensity score 0 1 P=0 P=1 Common support

308.
Propensity score 0 1 P=0 P=1 Failureof common support

309.
Other applications ofthe propensity score 1.Weighting 2.Regression

310.

311.
𝐸 𝑃 =1|𝑋 𝐸 𝑃 = 0|𝑋 𝑋 = 0 .1 .9 𝑋 = 1 .5 .5 𝐸 𝑃 = 1|𝑋 = 0 ∙ 𝑃𝑟 𝑋 = 0 +𝐸 𝑃 = 1|𝑋 = 1 ∙ 𝑃𝑟 𝑋 = 1 = .1 ∙ .5 + .5 ∙ .5 = .3 1,000 observations

312.
𝐸 𝑃 =1|𝑋 𝐸 𝑃 = 0|𝑋 𝑋 = 0 .1 .9 𝑋 = 1 .5 .5 𝐸 𝑃 = 1|𝑋 = 0 ∙ 𝑃𝑟 𝑋 = 0 +𝐸 𝑃 = 1|𝑋 = 1 ∙ 𝑃𝑟 𝑋 = 1 = .1 ∙ .5 + .5 ∙ .5 = .3 1,000 observations 300 participants

313.
𝐸 𝑃 =1|𝑋 𝐸 𝑃 = 0|𝑋 𝑋 = 0 .1 .9 𝑋 = 1 .5 .5 𝐸 𝑃 = 1|𝑋 = 0 ∙ 𝑃𝑟 𝑋 = 0 +𝐸 𝑃 = 1|𝑋 = 1 ∙ 𝑃𝑟 𝑋 = 1 = .1 ∙ .5 + .5 ∙ .5 = .3 1,000 observations 300 participants 700 non-participants

314.
𝐸 𝑃 =1|𝑋 𝐸 𝑃 = 0|𝑋 𝑋 = 0 .1 .9 𝑋 = 1 .5 .5 𝐸 𝑃 = 1|𝑋 = 0 ∙ 𝑃𝑟 𝑋 = 0 +𝐸 𝑃 = 1|𝑋 = 1 ∙ 𝑃𝑟 𝑋 = 1 = .1 ∙ .5 + .5 ∙ .5 = .3 1,000 observations 300 participants 700 non-participants 50 men 250 women

315.
𝐸 𝑃 =1|𝑋 𝐸 𝑃 = 0|𝑋 𝑋 = 0 .1 .9 𝑋 = 1 .5 .5 𝐸 𝑃 = 1|𝑋 = 0 ∙ 𝑃𝑟 𝑋 = 0 +𝐸 𝑃 = 1|𝑋 = 1 ∙ 𝑃𝑟 𝑋 = 1 = .1 ∙ .5 + .5 ∙ .5 = .3 1,000 observations 300 participants 700 non-participants 50 men 250 women 450 men 250 women

316.

317.

318.
𝑊𝑌𝑖 = 𝑃𝑖 ∙𝑌𝑖 𝑃𝑟 𝑃𝑖 = 1|𝑋𝑖 − 1 − 𝑃𝑖 ∙ 𝑌𝑖 1 − 𝑃𝑟 𝑃𝑖 = 1|𝑋𝑖 𝐴𝑇𝐸 = 𝑖=1 𝑛 𝑊𝑌𝑖 𝑖=1 𝑛 𝑃𝑖 𝑃𝑟 𝑃𝑖 = 1|𝑋𝑖 + 1 − 𝑃𝑖 1 − 𝑃𝑟 𝑃𝑖 = 1|𝑋𝑖

319.
𝑊𝑌𝑖 = 𝑃𝑖 ∙𝑌𝑖 𝑃𝑟 𝑃𝑖 = 1|𝑋𝑖 − 1 − 𝑃𝑖 ∙ 𝑌𝑖 1 − 𝑃𝑟 𝑃𝑖 = 1|𝑋𝑖 𝐴𝑇𝐸 = 𝑖=1 𝑛 𝑊𝑌𝑖 𝒊=𝟏 𝒏 𝑷𝒊 𝑷𝒓 𝑷𝒊 = 𝟏|𝑿𝒊 + 𝟏 − 𝑷𝒊 𝟏 − 𝑷𝒓 𝑷𝒊 = 𝟏|𝑿𝒊

320.
𝑊𝑌𝑖 = 𝑃𝑖 ∙𝑌𝑖 𝑃𝑟 𝑃𝑖 = 1|𝑋𝑖 − 1 − 𝑃𝑖 ∙ 𝑌𝑖 1 − 𝑃𝑟 𝑃𝑖 = 1|𝑋𝑖 𝐴𝑇𝐸 = 𝑖=1 𝑛 𝑊𝑌𝑖 𝑖=1 𝑛 𝑃𝑖 𝑃𝑟 𝑃𝑖 = 1|𝑋𝑖 + 1 − 𝑃𝑖 1 − 𝑃𝑟 𝑃𝑖 = 1|𝑋𝑖

321.

322.
Regress 𝑌𝑖 on𝑃𝑖 and 𝑃𝑟 𝑃𝑖 = 1|𝑋𝑖

323.
Regress 𝑌𝑖 on𝑃𝑖 and 𝑃𝑟 𝑃𝑖 = 1|𝑋𝑖

324.
Conclusion

325.
Links: The manual: http://www.measureevaluation.org/resources/publications/ms- 14-87-en The webinarintroducing the manual: http://www.measureevaluation.org/resources/webinars/metho ds-for-program-impact-evaluation My email: pmlance@email.unc.edu

326.
MEASURE Evaluation isfunded by the U.S. Agency for International Development (USAID) under terms of Cooperative Agreement AID-OAA-L-14-00004 and implemented by the Carolina Population Center, University of North Carolina at Chapel Hill in partnership with ICF International, John Snow, Inc., Management Sciences for Health, Palladium Group, and Tulane University. The views expressed in this presentation do not necessarily reflect the views of USAID or the United States government. www.measureevaluation.org

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