标签：aligned frac end 神经网络 深度 x2 x3 习题 x1
文章目录
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 ( x 1 , x 2 , x 3 ) = p ( x 3 ) ⋅ p ( x 2 ∣ x 3 ) ⋅ p ( x 1 ∣ x 2 ) 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)
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独立性
p ( x 1 , x 2 , x 3 ) = p ( x 2 ∣ x 1 , x 3 ) ⋅ p ( x 1 , x 3 ) p ( x 3 ) ⋅ p ( x 2 ∣ x 3 ) ⋅ p ( x 1 ∣ x 2 ) = p ( x 2 ∣ x 1 , x 3 ) ⋅ p ( x 1 , x 3 ) p ( x 1 , x 3 ) = p ( x 3 ) ⋅ p ( x 2 ∣ x 3 ) ⋅ p ( x 1 ∣ x 2 ) p ( x 2 ∣ x 1 , x 3 ) ≠ p ( x 1 ) ⋅ p ( x 3 ) ∴ X 1 与 X 3 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 ( x 1 , x 3 ∣ x 2 ) = p ( x 1 , x 2 , x 3 ) p ( x 2 ) = p ( x 3 ) ⋅ p ( x 2 ∣ x 3 ) ⋅ p ( x 1 ∣ x 2 ) p ( x 2 ) = p ( x 2 , x 3 ) ⋅ p ( x 1 ∣ x 2 ) p ( x 2 ) = p ( x 2 ) ⋅ 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 ⊥ ⊥ X 3 ∣ X 2 \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 ( x 1 , x 2 , x 3 ) = p ( x 2 ) ⋅ p ( x 1 ∣ x 2 ) ⋅ p ( x 3 ∣ x 2 ) 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)
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独立性
p ( x 1 , x 2 , x 3 ) = p ( x 2 ∣ x 1 , x 3 ) ⋅ p ( x 1 , x 3 ) p ( x 2 ) ⋅ p ( x 1 ∣ x 2 ) ⋅ p ( x 3 ∣ x 2 ) = p ( x 2 ∣ x 1 , x 3 ) ⋅ p ( x 1 , x 3 ) p ( x 1 , x 3 ) = p ( x 2 ) ⋅ p ( x 1 ∣ x 2 ) ⋅ p ( x 3 ∣ x 2 ) p ( x 2 ∣ x 1 , x 3 ) ≠ p ( x 1 ) ⋅ p ( x 3 ) ∴ X 1 与 X 3 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 ( x 1 , x 3 ∣ x 2 ) = p ( x 1 , x 2 , x 3 ) p ( x 2 ) = 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 ⊥ ⊥ X 3 ∣ X 2 \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 ( x 1 , x 2 , x 3 ) = p ( x 1 ) ⋅ p ( x 3 ) ⋅ p ( x 2 ∣ x 1 , x 3 ) 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)
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独立性
p ( x 1 , x 2 , x 3 ) = p ( x 2 ∣ x 1 , x 3 ) ⋅ p ( x 1 , x 3 ) p ( x 1 ) ⋅ p ( x 3 ) ⋅ p ( x 2 ∣ x 1 , x 3 ) = p ( x 2 ∣ x 1 , x 3 ) ⋅ p ( x 1 , x 3 ) p ( x 1 , x 3 ) = p ( x 1 ) ⋅ p ( x 3 ) ⋅ p ( x 2 ∣ x 1 , x 3 ) p ( x 2 ∣ x 1 , x 3 ) = p ( x 1 ) ⋅ p ( x 3 ) ∴ X 1 与 X 3 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 ( x 1 , x 3 ∣ x 2 ) = p ( x 1 , x 2 , x 3 ) p ( x 2 ) = 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 ̸ ⊥ ⊥ X 3 ∣ X 2 \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)
标签：aligned,frac,end,神经网络,深度,x2,x3,习题,x1 来源： https://blog.csdn.net/wtfloser/article/details/120605840
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