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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,

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Presentation on theme: "Bivariate Normal Distribution and Regression Application to Galton’s Heights of Adult Children and Parents Sources: Galton, Francis (1889). Natural Inheritance,"— Presentation transcript:

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

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

7 Marginal Distribution of Y 1 (P. 1)

8 Marginal Distribution of Y 1 (P. 2)

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

10 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

11 Summary of Results

12 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 E[Y 2 |y 1 ]  Y2|y

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

17 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 ) = y 1 E Y1 [E(Y 2 |Y 1 =y 1 )] = E Y1 [ Y 1 ] = (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 [ Y 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 )] = =6.40 = V(Y 2 ) (with round-off)


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