Bivariate Normal Distribution and Regression

Application to Galton’s Heights of Adult Children and Parents

Sources: Galton, Francis (1889). Natural Inheritance, MacMillan, London.Galton, F.; J.D. Hamilton Dickson (1886). “Family Likeness in Stature”, Proceedings of the Royal Society of London, Vol. 40, pp.42-73.

Data – Heights of Adult Children and Parents

• Adult Children Heights are reported by inch, in a manner so that the median of the grouped values is used for each (62.2”,…,73.2” are reported by Galton). – He adjusts female heights by a multiple of 1.08– We use 61.2” for his “Below” – We use 74.2” for his “Above”

• Mid-Parents Heights are the average of the two parents’ heights (after female adjusted). Grouped values at median (64.5”,…,72.5” by Galton)– We use 63.5” for “Below”– We use 73.5” for “Above”

Adult Child vs Mid-Parent Height

60

61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

63 64 65 66 67 68 69 70 71 72 73

Mid-Parent

Ad

ult

Ch

ild

Mid-Parent Height

0

50

100

150

200

250

63.5 64.5 65.5 66.5 67.5 68.5 69.5 70.5 71.5 72.5

Height

Fre

qu

en

cy

Adult Child Heights

0

20

40

60

80

100

120

140

160

180

61.2 62.2 63.2 64.2 65.2 66.2 67.2 68.2 69.2 70.2 71.2 72.2 73.2 74.2

Height

Fre

qu

en

cy

Joint Density Function

21

2211

22222

12111

2122

222

21

221121

211

2222

21

21

)()(

)()(

:where

,2

12

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12

1),(

YYE

YVYE

YVYE

yyyyyy

yyf

0

0.05

0.1

0.15

0.2

-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3

x1

Bivariate Normal Density

0.15-0.2

0.1-0.15

0.05-0.1

0-0.05

Marginal Distribution of Y1 (P. 1)

222

22

1121

222212211

222

2112

221

2

222

21

222

22

1122

22

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22

2112

221

2

222

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222

211

221

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22

2112

221

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Marginal Distribution of Y1 (P. 2)

21

211

21

2

2

22

2

1

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22

221

211

21

11

22

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1

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2

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Conditional Distribution of Y2 Given Y1=y1 (P. 1)

21

2211

21

221122

222

2222

22

222

21

221121

2211

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Conditional Distribution of Y2 Given Y1=y1 (P. 2)

222

1

2112112

2

1

211222

2222

2

2

1

211222

2222

2

21

22

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: offunction a then ,square"perfect " theforming then exponent, theofr denominato in the out Pulling

yNyYY

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This is referred to as the REGRESSION of Y2 on Y1

Summary of Results

221

2

1221221

222

1

2112112

1

2

2

122112

1222

1

21

2

2

1

211222

2222

2

12

2222

2111

222

222

22

22

121

211

21

11

2122

222

21

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211

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21

21

1,~|1,~|

12

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12

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12

1exp

12

1|

:onsDistributi lConditiona

,~,~

2exp

2

1

2exp

2

1

:onsDistributi onal) Unconditi(aka Marginal

,2

12

1exp

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1),(

:onDistributiJoint

yNyYY

yNyYY

yy

yyyf

yy

yyyf

NYNY

yy

yf

yy

yf

yyyyyy

yyf

Heights of Adult Children and Parents

• Empirical Data Based on 924 pairs (F. Galton)

• Y2 = Adult Child’s Height

– Y2 ~ N(68.1,6.39) 2=2.53

• Y1 = Mid-Parent’s Height

– Y1 ~ N(68.3,3.18) 1=1.78

• COV(Y1,Y2) = 2.02 2 = 0.20

• Y2|Y1=y1 is Normal with conditional mean and variance:

26.211.511.5)20.1(39.61|

638.05.246.43638.01.6818.3

39.6)45.0(3.681.68|

12 |22

2112

111

1

2112112

yYyYYV

yyyy

yYYE

y1Unconditional 63.5 66.5 69.5 72.5

E[Y2|y1] 68.1 65.0 66.9 68.8 70.8

Y2|y1 2.53 2.26 2.26 2.26 2.26

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

62.96

64.206

65.452

66.698

67.944

69.19

70.436

71.682

72.928

y1

Joint Density Function

0.035-0.04

0.03-0.035

0.025-0.03

0.02-0.025

0.015-0.02

0.01-0.015

0.005-0.01

0-0.005

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

62.96

64.206

65.452

66.698

67.944

69.19

70.436

71.682

72.928

y1

Joint Density Function

0.035-0.04

0.03-0.035

0.025-0.03

0.02-0.025

0.015-0.02

0.01-0.015

0.005-0.01

0-0.005

Distributions of Heights of Adult Children

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

59.5 60.5 61.5 62.5 63.5 64.5 65.5 66.5 67.5 68.5 69.5 70.5 71.5 72.5 73.5 74.5 75.5 76.5

y2

f(y

2)

uncond

y1=63.5

y1=66.5

y1=69.5

y1=72.5

E(Child)=

Parent+constant

Galton’s Finding

E(Child) independent of parent

Regression to the Mean

63.5

64.5

65.5

66.5

67.5

68.5

69.5

70.5

71.5

72.5

63.5 64.5 65.5 66.5 67.5 68.5 69.5 70.5 71.5 72.5

y1

E(Y

2) E(Y2|y1)=24.5+.638y1

E(Y2|y1)=0.21+y1

E(Y2|y1)=E(Y2)

Expectations and Variances

• E(Y1) = 68.3 V(Y1) = 3.18

• E(Y2) = 68.1 V(Y2) = 6.39

• E(Y2|Y1=y1) = 24.5+0.638y1

• EY1[E(Y2|Y1=y1)] = EY1[24.5+0.638Y1] = 24.5+0.638(68.3) = 68.1 = E(Y2)

• V(Y2|Y1=y1) = 5.11 EY1[V(Y2|Y1=y1)] = 5.11

• VY1[E(Y2|Y1=y1)] = VY1[24.5+0.638Y1] = (0.638)2

V(Y1) = (0.407)3.18 = 1.29

• EY1[V(Y2|Y1=y1)]+VY1[E(Y2|Y1=y1)] = 5.11+1.29=6.40 = V(Y2) (with round-off)