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

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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, 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”

Joint Density Function         

Marginal Distribution of Y 1 (P. 1)

Marginal Distribution of Y 1 (P. 2)

Conditional Distribution of Y 2 Given Y 1 =y 1 (P. 1)

Conditional Distribution of Y 2 Given Y 1 =y 1 (P. 2) This is referred to as the REGRESSION of Y 2 on Y 1

Summary of Results

Heights of Adult Children and Parents Empirical Data Based on 924 pairs (F. Galton) Y 2 = Adult Child’s Height –Y 2 ~ N(68.1,6.39)  2 =2.53 Y 1 = Mid-Parent’s Height –Y 1 ~ N(68.3,3.18)  1 =1.78 COV(Y 1,Y 2 ) = 2.02   2 = 0.20 Y 2 |Y 1 =y 1 is Normal with conditional mean and variance: y1y1 Unconditional 63.566.569.572.5 E[Y 2 |y 1 ]68.165.066.968.870.8  Y2|y1 2.532.26

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

Expectations and Variances E(Y 1 ) = 68.3 V(Y 1 ) = 3.18 E(Y 2 ) = 68.1 V(Y 2 ) = 6.39 E(Y 2 |Y 1 =y 1 ) = 24.5+0.638y 1 E Y1 [E(Y 2 |Y 1 =y 1 )] = E Y1 [24.5+0.638Y 1 ] = 24.5+0.638(68.3) = 68.1 = E(Y 2 ) V(Y 2 |Y 1 =y 1 ) = 5.11  E Y1 [V(Y 2 |Y 1 =y 1 )] = 5.11 V Y1 [E(Y 2 |Y 1 =y 1 )] = V Y1 [24.5+0.638Y 1 ] = (0.638) 2 V(Y 1 ) = (0.407)3.18 = 1.29 E Y1 [V(Y 2 |Y 1 =y 1 )]+V Y1 [E(Y 2 |Y 1 =y 1 )] = 5.11+1.29=6.40 = V(Y 2 ) (with round-off)

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