Bivariate Normal Distribution and Regression - PowerPoint PPT Presentation

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Bivariate Normal Distribution and Regression

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Bivariate Normal Distribution and Regression Application to Galton s Heights of Adult Children and Parents Sources: Galton, Francis (1889). Natural Inheritance ... – PowerPoint PPT presentation

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Title: Bivariate Normal Distribution and Regression


1
Bivariate Normal Distribution and Regression
  • Application to Galtons 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.

2
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

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Joint Density Function
m1m20 s1s21 r0.4
7
Marginal Distribution of Y1 (P. 1)
8
Marginal Distribution of Y1 (P. 2)
9
Conditional Distribution of Y2 Given Y1y1 (P. 1)
10
Conditional Distribution of Y2 Given Y1y1 (P. 2)
This is referred to as the REGRESSION of Y2 on Y1
11
Summary of Results
12
Heights of Adult Children and Parents
  • Empirical Data Based on 924 pairs (F. Galton)
  • Y2 Adult Childs Height
  • Y2 N(68.1,6.39) s22.53
  • Y1 Mid-Parents Height
  • Y1 N(68.3,3.18) s11.78
  • COV(Y1,Y2) 2.02 ? r 0.45, r2 0.20
  • Y2Y1y1 is Normal with conditional mean and
    variance

y1 Unconditional 63.5 66.5 69.5 72.5
EY2y1 68.1 65.0 66.9 68.8 70.8
sY2y1 2.53 2.26 2.26 2.26 2.26
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16
E(Child) Parentconstant Galtons
Finding E(Child) independent of parent
17
Expectations and Variances
  • E(Y1) 68.3 V(Y1) 3.18
  • E(Y2) 68.1 V(Y2) 6.39
  • E(Y2Y1y1) 24.50.638y1
  • EY1E(Y2Y1y1) EY124.50.638Y1
    24.50.638(68.3) 68.1 E(Y2)
  • V(Y2Y1y1) 5.11 ? EY1V(Y2Y1y1) 5.11
  • VY1E(Y2Y1y1) VY124.50.638Y1 (0.638)2
    V(Y1) (0.407)3.18 1.29
  • EY1V(Y2Y1y1)VY1E(Y2Y1y1)
    5.111.296.40 V(Y2) (with round-off)
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