Exercise 2.1 Solution Example - Hoff, A First Course in Bayesian Statistical Methods
標準ベイズ統計学 演習問題 2.1 解答例

\begin{align*} Pr (Y_1 = \text{farm}) & = Pr(Y_1 = \text{farm} | Y_2 = \text{farm})+ Pr(Y_1 = \text{farm} | Y_2 = \text{operatives})\\ &+ Pr(Y_1 = \text{farm} | Y_2 = \text{craftsman}) \\&+ Pr(Y_1 = \text{farm} | Y_2 = \text{sales}) \\&+ Pr(Y_1 = \text{farm} | Y_2 = \text{professinonal})\\ &= 0.018 + 0.035 + 0.031 + 0.008 + 0.018 \\ &= 0.110 \end{align*}

同様にして、

\begin{align*} & Pr (Y_1 = \text{operatives}) = 0.002 + 0.112 + 0.064 + 0.032 + 0.069 = 0.279\\ & Pr (Y_1 = \text{craftsman}) = 0.001 + 0.066 + 0.094 + 0.032 + 0.084 = 0.277\\ & Pr (Y_1 = \text{sales}) = 0.001 + 0.018 + 0.019 + 0.010 + 0.051 = 0.099\\ & Pr (Y_1 = \text{professional}) = 0.001 + 0.029 + 0.032 + 0.043 + 0.130 = 0.235 \end{align*}

\begin{align*} &\text{Pr}(Y_2 = \text{farm}) = 0.018 + 0.002 + 0.001 + 0.001 + 0.001 = 0.023\\ &\text{Pr}(Y_2 = \text{operatives}) = 0.035 + 0.112 + 0.066 + 0.018 + 0.029 = 0.26\\ &\text{Pr}(Y_2 = \text{craftsman}) = 0.031 + 0.064 + 0.094 + 0.019 + 0.032 = 0.24\\ &\text{Pr}(Y_2 = \text{sales}) = 0.008 + 0.032 + 0.032 + 0.010 + 0.043 = 0.125\\ &\text{Pr}(Y_2 = \text{professional}) = 0.018 + 0.069 + 0.084 + 0.051 + 0.130 = 0.352 \end{align*}

the conditional distribution of a son’s occupation, given that the father is a farmer;

\begin{align*} &\text{Pr}(Y_2 = \text{farm} | Y_1 = \text{farm}) = \frac{ \text{Pr}(Y_1 = \text{farm} , Y_2 = \text{farm}) }{ \text{Pr}(Y_1 = \text{farm}) } = \frac{0.018}{0.110} = 0.164\\ &\text{Pr}(Y_2 = \text{operatives} | Y_1 = \text{farm}) = \frac{ \text{Pr}(Y_1 = \text{farm} , Y_2 = \text{operatives}) }{ \text{Pr}(Y_1 = \text{farm}) } = \frac{0.035}{0.110} = 0.318\\ &\text{Pr}(Y_2 = \text{craftsman} | Y_1 = \text{farm}) = \frac{ \text{Pr}(Y_1 = \text{farm} , Y_2 = \text{craftsman}) }{ \text{Pr}(Y_1 = \text{farm}) } = \frac{0.031}{0.110} = 0.282\\ &\text{Pr}(Y_2 = \text{sales} | Y_1 = \text{farm}) = \frac{ \text{Pr}(Y_1 = \text{farm} , Y_2 = \text{sales}) }{ \text{Pr}(Y_1 = \text{farm}) } = \frac{0.008}{0.110} = 0.073\\ &\text{Pr}(Y_2 = \text{professional} | Y_1 = \text{farm}) = \frac{ \text{Pr}(Y_1 = \text{farm} , Y_2 = \text{professional}) }{ \text{Pr}(Y_1 = \text{farm}) } = \frac{0.018}{0.110} = 0.164 \end{align*}

the conditional distribution of a father’s occupation, given that the son is a farmer;

\begin{align*} &\text{Pr}(Y_1 = \text{farm} | Y_2 = \text{farm}) = \frac{ \text{Pr}(Y_1 = \text{farm} , Y_2 = \text{farm}) }{ \text{Pr}(Y_2 = \text{farm}) } = \frac{0.018}{0.023} = 0.783\\ &\text{Pr}(Y_1 = \text{operatives} | Y_2 = \text{farm}) = \frac{ \text{Pr}(Y_1 = \text{operatives} , Y_2 = \text{farm}) }{ \text{Pr}(Y_2 = \text{farm}) } = \frac{0.002}{0.023} = 0.087\\ &\text{Pr}(Y_1 = \text{craftsman} | Y_2 = \text{farm}) = \frac{ \text{Pr}(Y_1 = \text{craftsman} , Y_2 = \text{farm}) }{ \text{Pr}(Y_2 = \text{farm}) } = \frac{0.001}{0.023} = 0.043\\ &\text{Pr}(Y_1 = \text{sales} | Y_2 = \text{farm}) = \frac{ \text{Pr}(Y_1 = \text{sales} , Y_2 = \text{farm}) }{ \text{Pr}(Y_2 = \text{farm}) } = \frac{0.001}{0.023} = 0.043\\ &\text{Pr}(Y_1 = \text{professional} | Y_2 = \text{farm}) = \frac{ \text{Pr}(Y_1 = \text{professional} , Y_2 = \text{farm}) }{ \text{Pr}(Y_2 = \text{farm}) } = \frac{0.001}{0.023} = 0.043 \end{align*}