《神经网络与深度学习》习题答案

文章目录

前言
第 11 章
- 11-1

前言

自己写的或找的习题答案，其中自己写的答案都会同步更新到邱锡鹏老师在 github 上关于这本书的 solutions 项目中（具体题解都在 issues 中）。

第 11 章

11-1

(1)

已知p(x1,x2,x3)=p(x3)⋅p(x2∣x3)⋅p(x1∣x2)已知\space p(x_1,x_2,x_3) = p(x_3)·p(x_2|x_3)·p(x_1|x_2) 已知 p(x1,x2,x3)=p(x3)⋅p(x2∣x3)⋅p(x1∣x2)

独立性
p(x1,x2,x3)=p(x2∣x1,x3)⋅p(x1,x3)p(x3)⋅p(x2∣x3)⋅p(x1∣x2)=p(x2∣x1,x3)⋅p(x1,x3)p(x1,x3)=p(x3)⋅p(x2∣x3)⋅p(x1∣x2)p(x2∣x1,x3)≠p(x1)⋅p(x3)∴X1与X3不独立\begin{aligned} p(x_1,x_2,x_3) &= p(x_2|x1,x3)·p(x_1,x_3) \\ p(x_3)·p(x_2|x_3)·p(x_1|x_2) &= p(x_2|x1,x3)·p(x_1,x_3) \\ p(x_1,x_3) &= \frac{p(x_3)·p(x_2|x_3)·p(x_1|x_2)}{p(x_2|x1,x3)} \\ &≠ p(x_1)·p(x_3) \\ ∴ X_1 与 X_3 不独立 \end{aligned} p(x1,x2,x3)p(x3)⋅p(x2∣x3)⋅p(x1∣x2)p(x1,x3)∴X1与X3不独立=p(x2∣x1,x3)⋅p(x1,x3)=p(x2∣x1,x3)⋅p(x1,x3)=p(x2∣x1,x3)p(x3)⋅p(x2∣x3)⋅p(x1∣x2)=p(x1)⋅p(x3)
条件独立性

p(x1,x3∣x2)=p(x1,x2,x3)p(x2)=p(x3)⋅p(x2∣x3)⋅p(x1∣x2)p(x2)=p(x2,x3)⋅p(x1∣x2)p(x2)=p(x2)⋅p(x3∣x2)⋅p(x1∣x2)p(x2)=p(x1∣x2)⋅p(x3∣x2)∴X1⊥⁣ ⁣ ⁣⊥X3∣X2\begin{aligned} p(x_1,x_3|x2) &= \frac{p(x_1,x_2,x_3)}{p(x_2)} \\ &= \frac{p(x_3)·p(x_2|x_3)·p(x_1|x_2)}{p(x_2)} \\ &= \frac{p(x_2,x_3)·p(x_1|x_2)}{p(x_2)} \\ &= \frac{p(x2)·p(x_3|x_2)·p(x_1|x_2)}{p(x_2)} \\ &= p(x_1|x_2)·p(x_3|x_2) \\ ∴X_1 \perp\!\!\!\perp X_3|X_2 \end{aligned} p(x1,x3∣x2)∴X1⊥⊥X3∣X2=p(x2)p(x1,x2,x3)=p(x2)p(x3)⋅p(x2∣x3)⋅p(x1∣x2)=p(x2)p(x2,x3)⋅p(x1∣x2)=p(x2)p(x2)⋅p(x3∣x2)⋅p(x1∣x2)=p(x1∣x2)⋅p(x3∣x2)

(2)

与 (1) 镜像，不再赘述。

(3)

已知p(x1,x2,x3)=p(x2)⋅p(x1∣x2)⋅p(x3∣x2)已知\space p(x_1,x_2,x_3) = p(x_2)·p(x_1|x_2)·p(x_3|x_2) 已知 p(x1,x2,x3)=p(x2)⋅p(x1∣x2)⋅p(x3∣x2)

独立性
p(x1,x2,x3)=p(x2∣x1,x3)⋅p(x1,x3)p(x2)⋅p(x1∣x2)⋅p(x3∣x2)=p(x2∣x1,x3)⋅p(x1,x3)p(x1,x3)=p(x2)⋅p(x1∣x2)⋅p(x3∣x2)p(x2∣x1,x3)≠p(x1)⋅p(x3)∴X1与X3不独立\begin{aligned} p(x_1,x_2,x_3) &= p(x_2|x1,x3)·p(x_1,x_3) \\ p(x_2)·p(x_1|x_2)·p(x_3|x_2) &= p(x_2|x1,x3)·p(x_1,x_3) \\ p(x_1,x_3) &= \frac{p(x_2)·p(x_1|x_2)·p(x_3|x_2)}{p(x_2|x1,x3)} \\ &≠ p(x_1)·p(x_3) \\ ∴ X_1 与 X_3 不独立 \end{aligned} p(x1,x2,x3)p(x2)⋅p(x1∣x2)⋅p(x3∣x2)p(x1,x3)∴X1与X3不独立=p(x2∣x1,x3)⋅p(x1,x3)=p(x2∣x1,x3)⋅p(x1,x3)=p(x2∣x1,x3)p(x2)⋅p(x1∣x2)⋅p(x3∣x2)=p(x1)⋅p(x3)
条件独立性

p(x1,x3∣x2)=p(x1,x2,x3)p(x2)=p(x2)⋅p(x1∣x2)⋅p(x3∣x2)p(x2)=p(x1∣x2)⋅p(x3∣x2)∴X1⊥⁣ ⁣ ⁣⊥X3∣X2\begin{aligned} p(x_1,x_3|x2) &= \frac{p(x_1,x_2,x_3)}{p(x_2)} \\ &= \frac{p(x_2)·p(x_1|x_2)·p(x_3|x_2)}{p(x_2)} \\ &= p(x_1|x_2)·p(x_3|x_2) \\ ∴X_1 \perp\!\!\!\perp X_3|X_2 \end{aligned} p(x1,x3∣x2)∴X1⊥⊥X3∣X2=p(x2)p(x1,x2,x3)=p(x2)p(x2)⋅p(x1∣x2)⋅p(x3∣x2)=p(x1∣x2)⋅p(x3∣x2)

(4)

已知p(x1,x2,x3)=p(x1)⋅p(x3)⋅p(x2∣x1,x3)已知\space p(x_1,x_2,x_3) = p(x_1)·p(x_3)·p(x_2|x_1,x_3) 已知 p(x1,x2,x3)=p(x1)⋅p(x3)⋅p(x2∣x1,x3)

独立性
p(x1,x2,x3)=p(x2∣x1,x3)⋅p(x1,x3)p(x1)⋅p(x3)⋅p(x2∣x1,x3)=p(x2∣x1,x3)⋅p(x1,x3)p(x1,x3)=p(x1)⋅p(x3)⋅p(x2∣x1,x3)p(x2∣x1,x3)=p(x1)⋅p(x3)∴X1与X3独立\begin{aligned} p(x_1,x_2,x_3) &= p(x_2|x1,x3)·p(x_1,x_3) \\ p(x_1)·p(x_3)·p(x_2|x_1,x_3) &= p(x_2|x1,x3)·p(x_1,x_3) \\ p(x_1,x_3) &= \frac{p(x_1)·p(x_3)·p(x_2|x_1,x_3)}{p(x_2|x1,x3)} \\ &= p(x_1)·p(x_3) \\ ∴ X_1 与 X_3 独立 \end{aligned} p(x1,x2,x3)p(x1)⋅p(x3)⋅p(x2∣x1,x3)p(x1,x3)∴X1与X3独立=p(x2∣x1,x3)⋅p(x1,x3)=p(x2∣x1,x3)⋅p(x1,x3)=p(x2∣x1,x3)p(x1)⋅p(x3)⋅p(x2∣x1,x3)=p(x1)⋅p(x3)
条件独立性

p(x1,x3∣x2)=p(x1,x2,x3)p(x2)=p(x1)⋅p(x3)⋅p(x2∣x1,x3)p(x2)≠p(x1∣x2)⋅p(x3∣x2)∴X1̸⁣⊥⁣ ⁣ ⁣⊥X3∣X2\begin{aligned} p(x_1,x_3|x2) &= \frac{p(x_1,x_2,x_3)}{p(x_2)} \\ &= \frac{p(x_1)·p(x_3)·p(x_2|x_1,x_3)}{p(x_2)} \\ &≠ p(x_1|x_2)·p(x_3|x_2) \\ ∴X_1 \not \! \perp \!\!\! \perp X_3|X_2 \end{aligned} p(x1,x3∣x2)∴X1⊥⊥X3∣X2=p(x2)p(x1,x2,x3)=p(x2)p(x1)⋅p(x3)⋅p(x2∣x1,x3)=p(x1∣x2)⋅p(x3∣x2)