P8 保凸运算2

P8 保凸运算

保凸运算

透视函数 Perspative Function

P:Rn+1Rndom  P=RnR++P: R^{n+1} \mapsto R^n \quad dom \; P=R^n*R_{++}P:Rn+1↦RndomP=Rn∗R++​

P(z,t)=ztzRntR++P(z,t) = \frac{z}{t} \quad z \in R^n \quad t \in R_{++}P(z,t)=tz​z∈Rnt∈R++​

降维过程,将前所有元素,除以最后一维,并去掉最后一维。

P8 保凸运算2
任何一个函数,通过透视函数之后,还是凸集。

例:考虑 Rn+1R^{n+1}Rn+1 线段。x=(x~,xn+1)x~Rnxn+1R++x=(\tilde x,x_{n+1}) \quad \tilde x \in R^n \quad x_{n+1} \in R_{++}x=(x~,xn+1​)x~∈Rnxn+1​∈R++​; y=(y~,yn+1)y~Rnyn+1R++y=(\tilde y,y_{n+1}) \quad \tilde y \in R^n \quad y_{n+1} \in R_{++}y=(y~​,yn+1​)y~​∈Rnyn+1​∈R++​

θ0\theta \geq 0θ≥0 线段维 θx+(1θ)y\theta x + (1-\theta) yθx+(1−θ)y

证明:线段经过透视函数P后,仍然时凸集。

xPP(x)x \rightarrow ^ P P(x)x→PP(x) \quadyPP(y)y \rightarrow ^ P P(y)y→PP(y)
θx+(1θ)yPP(θx+(1θ)y)\theta x + (1-\theta) y \rightarrow ^ P P(\theta x + (1-\theta) y)θx+(1−θ)y→PP(θx+(1−θ)y)

μ=θxn+1θxn+1+(1θ)yn+1\mu =\frac{\theta x_{n+1}}{ \theta x_{n+1} + (1-\theta) y_{n+1} }μ=θxn+1​+(1−θ)yn+1​θxn+1​​
P(θx+(1θ)y)=θx~+(1θ)y~θxn+1+(1θ)yn+1=θxn+1θxn+1+(1θ)yn+1x~xn+1+(1θ)yn+1θxn+1+(1θ)yn+1y~yn+1=μP(x)+(1μ)P(y) \begin{aligned} P(\theta x + (1-\theta) y) &= \frac{\theta \tilde {x} + (1-\theta) \tilde {y}}{ \theta x_{n+1} + (1-\theta) y_{n+1} } \\ &=\frac{\theta x_{n+1}}{ \theta x_{n+1} + (1-\theta) y_{n+1} } \frac{\tilde {x}}{x_{n+1}} + \frac{(1-\theta) y_{n+1}}{ \theta x_{n+1} + (1-\theta) y_{n+1} } \frac{\tilde {y}}{y_{n+1}} \\ &=\mu P(x) + (1-\mu) P(y) \end{aligned} P(θx+(1−θ)y)​=θxn+1​+(1−θ)yn+1​θx~+(1−θ)y~​​=θxn+1​+(1−θ)yn+1​θxn+1​​xn+1​x~​+θxn+1​+(1−θ)yn+1​(1−θ)yn+1​​yn+1​y~​​=μP(x)+(1−μ)P(y)​

P8 保凸运算2

任意凸集的反透视映射仍是凸集。

P1(C)={(x,t)Rn+1xtc,t>0}P^{-1}(C)=\lbrace (x,t) \in R^{n+1} | \frac{x}{t} \in c, t > 0 \rbraceP−1(C)={(x,t)∈Rn+1∣tx​∈c,t>0}

考虑(x,t)P1(C)(y,s)P1(C)0θ1(x,t) \in P^{-1}(C) \quad (y,s) \in P^{-1}(C) \quad 0 \leq \theta \leq 1(x,t)∈P−1(C)(y,s)∈P−1(C)0≤θ≤1

θx+(1θ)yθt+(1θ)s?C\frac{\theta x+(1-\theta)y}{\theta t+(1-\theta)s} \in^? Cθt+(1−θ)sθx+(1−θ)y​∈?C

θtθt+(1θ)sxt+(1θtθt+(1θ)s)ys \begin{aligned} \frac{\theta t}{\theta t+(1-\theta)s}\frac{x}{t}+(1-\frac{\theta t}{\theta t+(1-\theta)s})\frac{y}{s} \end{aligned} θt+(1−θ)sθt​tx​+(1−θt+(1−θ)sθt​)sy​​
xt\frac{x}{t}tx​ 与 ys\frac{y}{s}sy​ 都是C中的一点,C是凸集。所以结论成立。

线性分数函数

g:RnRm+1g:R^n \mapsto R^{m+1}g:Rn↦Rm+1 为伪仿射函数
g(x)=[ACT]x+[bd]ARm+nbRmCRndRg(x)=\begin{bmatrix} A \\ C^T \\ \end{bmatrix} x + \begin{bmatrix} b \\ d \\ \end{bmatrix} \quad A \in R^{m+n} \quad b\in R^{m} \quad C \in R^{n} \quad d \in Rg(x)=[ACT​]x+[bd​]A∈Rm+nb∈RmC∈Rnd∈R

P:Rm+1RmP:R^{m+1} \mapsto R^{m}P:Rm+1↦Rm
定义:
f:RnRm=Pgf:R^n \mapsto R^m = P \circ gf:Rn↦Rm=P∘g 先经过ggg,在经过PPP。
本质是仿射变换+透视变换。

f(x)=Ax+bCT+ddom  f{xCTx+d>0}f(x) = \frac{Ax+b}{C^T+d} \quad dom \;f \lbrace x | C^Tx + d > 0 \rbracef(x)=CT+dAx+b​domf{x∣CTx+d>0}

例:两个随机变量的联合概率 \rightarrow→ 条件概率

u{1..n},v{1...m}u \in \lbrace {1..n} \rbrace,v \in \lbrace {1...m} \rbraceu∈{1..n},v∈{1...m}

Pij=P(u=i,v=j)P_{ij}=P(u=i,v=j)Pij​=P(u=i,v=j)
fij=P(u=iv=j)f_{ij}=P(u=i|v=j)fij​=P(u=i∣v=j)

fij=Pijk=1nPkjf_{ij}= \frac{P_{ij}}{\sum_{k=1}^n P_{kj}}fij​=∑k=1n​Pkj​Pij​​

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