Exercise 2.5 Solution Example - Hoff, A First Course in Bayesian Statistical Methods
標準ベイズ統計学 演習問題 2.5 解答例

	X = 0	X = 1
Y = 0	0.2	0.3
Y = 1	0.3	0.2

By using the table above (上の表を用いて),

\begin{align*} \text{E}\left[Y\right] &= \text{Pr}(Y = 0) \cdot 0 + \text{Pr}(Y = 1) \cdot 1 \\ &= (0.2 + 0.3) \cdot 0 + (0.3 + 0.2) \cdot 1 \\ &= 0.5 \end{align*}

\begin{align*} \text{Var}\left[Y \mid X = 0\right] &= \text{Pr}(Y = 0 \mid X = 0) \cdot (0 - \text{E}\left[Y \mid X = 0\right])^2 + \text{Pr}(Y = 1 \mid X = 0) \cdot (1 - \text{E}\left[Y \mid X = 0\right])^2 \\ &= 0.4 \cdot (0- 0.6)^2 + 0.6 \cdot (1 - 0.6)^2 \\ &= 0.24 \\ \\ \text{Var}\left[Y \mid X = 1\right] &= \text{Pr}(Y = 0 \mid X = 1) \cdot (0 - \text{E}\left[Y \mid X = 1\right])^2 + \text{Pr}(Y = 1 \mid X = 1) \cdot (1 - \text{E}\left[Y \mid X = 1\right])^2 \\ &= 0.6 \cdot (0- 0.4)^2 + 0.4 \cdot (1 - 0.4)^2 \\ &= 0.24 \\ \\ \text{Var}\left[Y\right] &= \text{Pr}(Y = 0) \cdot (0 - \text{E}\left[Y\right])^2 + \text{Pr}(Y = 1) \cdot (1 - \text{E}\left[Y\right])^2 \\ &= 0.5 \cdot (0- 0.5)^2 + 0.5 \cdot (1 - 0.5)^2 \\ &= 0.25 \end{align*}

The reason why \(Var[Y]\) is greater than \(Var[Y \mid X = 0]\) and \(Var[Y \mid X = 1]\) is that the uncertainty increases because it is not determined from which urn the ball is chosen.

Var[Y]が Var[\(Y \mid X = 0\)] と Var[\(Y \mid X = 1\)]よりも大きいのは、どちらの壺からボールを選ぶか決まっていない分不確実性が大きくなるから。

\begin{align*} \text{Pr}(X = 0 \mid Y = 1) &= \frac{\text{Pr}(Y = 1 \mid X = 0) \text{Pr}(X = 0)}{\text{Pr}(Y = 1)} \\ &= \frac{0.6 \cdot 0.5}{0.5} \\ &= 0.6 \end{align*}