情報理論第3回　条件付き確率とエントロピー練習問題解答例

X

x_{m}

x_{f}

Y

y_{a}

y_{c}

X = [\begin{matrix} x_{m} & x_{f} \\ p_{m} & p_{f} \end{matrix}] = [\begin{matrix} x_{m} & x_{f} \\ \frac{3}{5} & \frac{2}{5} \end{matrix}]

Y = [\begin{matrix} y_{a} & y_{c} \\ p_{a} & p_{c} \end{matrix}] = [\begin{matrix} y_{a} & y_{c} \\ \frac{1}{2} & \frac{1}{2} \end{matrix}]

Y (x_{m}) = [\begin{matrix} y_{a}|x_{m} & y_{c}|x_{m} \\ \frac{1}{3} & \frac{2}{3} \end{matrix}]

Y (x_{f}) = [\begin{matrix} y_{a}|x_{f} & y_{c}|x_{f} \\ \frac{3}{4} & \frac{1}{4} \end{matrix}]

X

x_{m}

Y

H (Y | x_{m}) = - \frac{1}{3} \log \frac{1}{3} - \frac{2}{3} \log \frac{2}{3} = 0.91

X

x_{f}

Y

H (Y | x_{f}) = - \frac{3}{4} \log \frac{3}{4} - \frac{1}{4} \log \frac{1}{4} = 0.82

X

Y

p_{m} = \frac{3}{5}

p_{f} = \frac{2}{5}

H (Y | X) = p_{m} \times H (Y | x_{m}) + p_{f} \times H (Y | x_{f}) = \frac{3}{5} \times 0.91 + \frac{2}{5} \times 0.82 = 0.874 ≅ 0.87

情報理論 第3回 条件付き確率とエントロピー 練習問題解答例