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hw1_sol - Assignment 1 Sample Solutions 1 Ho Exercise 2.1 p...

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Assignment 1 – Sample Solutions 1. Ho ff – Exercise 2.1 p. 225 The joint distribution of occupational categories for fathers and sons, p ( Y 1 , Y 2 ), is given by: son’s occupation father’s occupation farm operatives craftsmen sales professional farm 0.018 0.035 0.031 0.008 0.018 operatives 0.002 0.112 0.064 0.032 0.069 craftsmen 0.001 0.066 0.094 0.032 0.084 sales 0.001 0.018 0.019 0.010 0.051 professional 0.001 0.029 0.032 0.043 0.130 (a) Marginal probability distribution of father’s occupation: p ( Y 1 = y 1 ) = y 2 p ( Y 1 = y 1 , Y 2 = y 2 ) Y 1 farm operatives craftsmen sales professional p ( y 1 ) 0.110 0.279 0.277 0.099 0.235 (b) Marginal probability distribution of son’s occupation: p ( Y 2 = y 2 ) = y 1 p ( Y 1 = y 1 , Y 2 = y 2 ) Y 2 farm operatives craftsmen sales professional p ( y 2 ) 0.023 0.260 0.240 0.125 0.352 (c) Conditional distribution of a son’s occupation, given the father is a farmer: p ( Y 2 = y 2 | Y 1 = farmer) = p ( Y 2 = y 2 , Y 1 = farmer) p ( Y 1 = farmer) Y 2 farm operatives craftsmen sales professional p ( y 2 | Y 1 = farmer) 0.164 0.318 0.282 0.073 0.164 (d) Conditional distribution of a father’s occupation, given the son is a farmer: p ( Y 1 = y 1 | Y 2 = farmer) = p ( Y 1 = y 1 , Y 2 = farmer) p ( Y 2 = farmer) Y 1 farm operatives craftsmen sales professional p ( y 1 | Y 2 = farmer) 0.783 0.087 0.043 0.043 0.043
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2. Ho ff – Exercise 3.1 p. 227 (a) Y 1 , . . . , Y n | θ iid Bernoulli( θ ). The joint distribution of p ( Y 1 = y 1 , . . . , Y 100 = y 100 ) is given by: p ( Y 1 = y 1 , . . . , Y 100 = y 100 | θ ) = 100 i =1 θ y i (1 θ ) 1 y i = θ 100 i =1 y i (1 θ ) 100 100 i =1 y i . 100 i =1 Y i represents the number of successes in 100 i.i.d. trials, so 100 i =1 Y i Binomial(100 , θ ) p 100 i =1 Y i = y | θ = 100 y θ y (1 θ ) 100 y . (b) Suppose 100 i =1 Y i = 57 and θ { 0 . 0 , 0 . 1 , . . . , 0 . 9 , 1 . 0 } . We can use the following R code to compute and plot p ( 100 i =1 Y i = 57 | θ ) for the 11 values of θ : theta = seq(0, 1, 0.1); n=100; sumy = 57 dbinom(sumy, n, theta) plot(theta, dbinom(sumy, n, theta), type="h", xlab=expression(theta), ylab=expression(paste("p(",sum(y),"=57|",theta,")"))) θ 0.0 0.1 0.2 0.3 0.4 0.5 p ( 100 i =1 Y i = 57 | θ ) 0.000 4 . 1 × 10 31 3 . 74 × 10 16 1 . 31 × 10 8 2 . 29 × 10 4 0.030 θ 0.6 0.7 0.8 0.9 1.0 p ( 100 i =1 Y i = 57 | θ ) 0 . 067 × 10 2 1 . 85 × 10 3 1 . 00 × 10 7 9 . 40 × 10 18 0.000 (c) Suppose p ( θ = 0 . 0) = p ( θ = 0 . 1) = . . . = p ( θ = 0 . 9) = p ( θ = 1 . 0) = 1 11 .
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