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Bivariate Normal Distribution and Regression

Bivariate Normal Distribution and Regression. Application to Galton’s Heights of Adult Children and Parents Sources: Galton, Francis (1889). Natural Inheritance, MacMillan, London.

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Bivariate Normal Distribution and Regression

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  1. 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.

  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”

  3. Joint Density Function m1=m2=0 s1=s2=1 r=0.4

  4. Marginal Distribution of Y1 (P. 1)

  5. Marginal Distribution of Y1 (P. 2)

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

  7. Conditional Distribution of Y2 Given Y1=y1 (P. 2) This is referred to as the REGRESSION of Y2 on Y1

  8. Summary of Results

  9. 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) s2=2.53 • Y1 = Mid-Parent’s Height • Y1 ~ N(68.3,3.18) s1=1.78 • COV(Y1,Y2) = 2.02  r = 0.45, r2 = 0.20 • Y2|Y1=y1is Normal with conditional mean and variance:

  10. E(Child)= Parent+constant Galton’s Finding E(Child) independent of parent

  11. 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)

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