全部笔记的汇总贴(视频也有传送门):中科大-凸优化
一、椭球是球的仿射映射
ε
=
{
x
∣
(
x
−
x
c
)
T
P
−
1
(
x
−
x
c
)
}
P
∈
S
+
+
n
{
u
∣
∣
∣
u
∣
∣
2
≤
1
}
P
1
2
P
1
2
=
P
\varepsilon=\{x|(x-x_c)^TP^{-1}(x-x_c)\}\;\;P\in S_{++}^n\\\{u|\;||u||_2\le1\}\;\;\;P^{\frac12}P^{\frac12}=P
ε={x∣(x−xc)TP−1(x−xc)}P∈S++n{u∣∣∣u∣∣2≤1}P21P21=P
f
(
u
)
=
P
1
2
u
+
x
c
{
f
(
u
)
∣
∣
∣
u
∣
∣
2
≤
1
}
=
{
P
1
2
u
+
x
c
∣
∣
∣
u
∣
∣
2
≤
1
}
f(u)=P^{\frac12}u+x_c\\\{f(u)|\;||u||_2\le1\}=\{P^{\frac12}u+x_c|\;||u||_2\le1\}
f(u)=P21u+xc{f(u)∣∣∣u∣∣2≤1}={P21u+xc∣∣∣u∣∣2≤1}
我们令
x
=
Δ
P
1
2
u
+
x
c
u
=
P
−
1
2
(
x
−
x
c
)
x\overset{\Delta}{=}P^{\frac12}u+x_c\;\;\;u=P^{-\frac12}(x-x_c)
x=ΔP21u+xcu=P−21(x−xc)
所以,
{
f
(
u
)
∣
∣
∣
u
∣
∣
2
≤
1
}
=
{
x
∣
∣
∣
P
−
1
2
(
x
−
x
c
)
∣
∣
2
≤
1
}
=
{
x
∣
(
x
−
x
c
)
T
P
−
1
(
x
−
x
c
)
≤
1
}
\{f(u)|\;||u||_2\le1\}=\{x|\;||P^{-\frac12}(x-x_c)||_2\le1\}\\=\{x|(x-x_c)^TP^{-1}(x-x_c)\le1\}
{f(u)∣∣∣u∣∣2≤1}={x∣∣∣P−21(x−xc)∣∣2≤1}={x∣(x−xc)TP−1(x−xc)≤1}
二、透视函数(Perspective Function)
P
:
R
n
+
1
→
R
n
d
o
m
P
=
R
n
∗
R
+
+
P:\R^{n+1}\rightarrow\R^n\;\;\;\;\;dom \;P=R^n*R_{++}
P:Rn+1→RndomP=Rn∗R++
P
(
z
,
t
)
=
z
t
z
∈
R
n
,
t
∈
R
+
+
P(z,t)=\frac zt\;\;\;\;z\in \R^n,t\in\R_{++}
P(z,t)=tzz∈Rn,t∈R++
考虑
R
n
+
1
\R^{n+1}
Rn+1内线段,
x
=
(
x
~
∈
R
n
,
x
n
+
1
∈
R
+
+
)
y
=
(
y
~
∈
R
n
,
y
n
+
1
∈
R
+
+
)
x=(\underset{\in\R^n}{\widetilde{x}},\underset{\in\R_{++}}{x_{n+1}})\;\;\;y=(\underset{\in\R^n}{\widetilde{y}},\underset{\in\R_{++}}{y_{n+1}})
x=(∈Rnx
,∈R++xn+1)y=(∈Rny
,∈R++yn+1)
θ
≥
0
\theta\ge0
θ≥0,线段为
θ
x
+
(
1
−
θ
)
y
\theta x+(1-\theta)y
θx+(1−θ)y
证明 线段
→
P
\overset{P}{\rightarrow}
→P线段
x
→
P
P
(
x
)
\;\;\;\;\;\;\;\;x\overset{P}{\rightarrow}P(x)
x→PP(x)
y
→
P
P
(
y
)
\;\;\;\;\;\;\;\;y\overset{P}{\rightarrow}P(y)
y→PP(y)
θ
x
+
(
1
−
θ
)
y
→
P
P
(
θ
x
+
(
1
−
θ
)
y
)
\theta x+(1-\theta)y\overset{P}{\rightarrow}P(\theta x+(1-\theta)y)
θx+(1−θ)y→PP(θx+(1−θ)y)
P
(
θ
x
+
(
1
−
θ
)
y
)
=
θ
x
~
+
(
1
−
θ
)
y
~
θ
x
n
+
1
+
(
1
−
θ
)
y
n
+
1
=
θ
x
n
+
1
θ
x
n
+
1
+
(
1
−
θ
)
y
n
+
1
μ
∈
[
0
,
1
]
⋅
x
~
x
n
+
1
P
(
x
)
+
(
1
−
θ
)
y
n
+
1
θ
x
n
+
1
+
(
1
−
θ
)
y
n
+
1
1
−
μ
⋅
y
~
y
n
+
1
P
(
y
)
=
μ
P
(
x
)
+
(
1
−
μ
)
P
(
y
)
P(\theta x+(1-\theta)y)=\frac{\theta \widetilde{x}+(1-\theta)\widetilde{y}}{\theta x_{n+1}+(1-\theta)y_{n+1}}\\=\underset{\color{blue}\mu\in[0,1]}{\frac{\theta x_{n+1}}{\theta x_{n+1}+(1-\theta)y_{n+1}}}\cdot\underset{\color{blue}P(x)}{\frac{\widetilde{x}}{x_{n+1}}}+\underset{\color{blue}1-\mu}{\frac{(1-\theta)y_{n+1}}{\theta x_{n+1}+(1-\theta)y_{n+1}}}\cdot\underset{\color{blue}P(y)}{\frac{\widetilde{y}}{y_{n+1}}}\\=\mu P(x)+(1-\mu)P(y)
P(θx+(1−θ)y)=θxn+1+(1−θ)yn+1θx
+(1−θ)y
=μ∈[0,1]θxn+1+(1−θ)yn+1θxn+1⋅P(x)xn+1x
+1−μθxn+1+(1−θ)yn+1(1−θ)yn+1⋅P(y)yn+1y
=μP(x)+(1−μ)P(y)
θ → μ \theta\rightarrow\mu θ→μ是透视映射
任意凸集的反透视映射仍是凸集
P
−
1
(
C
)
=
{
(
x
,
t
)
∈
R
n
+
1
∣
x
t
∈
C
,
t
>
0
}
P^{-1}(C)=\{(x,t)\in\R^{n+1}|\frac xt\in C,t>0\}
P−1(C)={(x,t)∈Rn+1∣tx∈C,t>0}
考虑
(
x
,
t
)
∈
P
−
1
(
C
)
(
y
,
S
)
∈
P
−
1
(
C
)
,
0
≤
θ
≤
1
(x,t)\in P^{-1}(C)\;\;(y,S)\in P^{-1}(C),0\le\theta\le1
(x,t)∈P−1(C)(y,S)∈P−1(C),0≤θ≤1
θ
x
+
(
1
−
θ
)
y
θ
t
+
(
1
−
θ
)
S
=
θ
t
θ
t
+
(
1
−
θ
)
S
μ
⋅
x
t
∈
C
+
(
1
−
θ
)
S
θ
t
+
(
1
−
θ
)
S
1
−
μ
⋅
y
S
∈
C
∈
C
\frac{\theta x+(1-\theta)y}{\theta t+(1-\theta)S}=\underset{\color{blue}\mu}{\frac{\theta t}{\theta t+(1-\theta)S}}\cdot\underset{\color{blue}\in C}{\frac xt}+\underset{\color{blue}1-\mu}{\frac{(1-\theta)S}{\theta t+(1-\theta)S}}\cdot\underset{\color{blue}\in C}{\frac yS}{\color{red}\in C}
θt+(1−θ)Sθx+(1−θ)y=μθt+(1−θ)Sθt⋅∈Ctx+1−μθt+(1−θ)S(1−θ)S⋅∈CSy∈C
三、线性分数函数
线性分数函数 f f f(由点 x x x先做个仿射变换,再做个透视得到)
g
:
R
n
→
R
m
+
1
g:\R^n\rightarrow\R^{m+1}
g:Rn→Rm+1为仿射映射,
g
(
x
)
=
(
A
C
T
)
x
+
(
b
d
)
A
∈
R
m
∗
n
,
C
∈
R
n
,
b
∈
R
m
,
d
∈
R
g(x)=\begin{pmatrix} A\\C^T\end{pmatrix}x+\begin{pmatrix} b\\d\end{pmatrix}\;\;\;\;\;A\in\R^{m*n},C\in\R^n,b\in\R^m,d\in\R
g(x)=(ACT)x+(bd)A∈Rm∗n,C∈Rn,b∈Rm,d∈R
P
:
R
m
+
1
→
R
m
P:\R^{m+1}\rightarrow\R^m
P:Rm+1→Rm为透视变换
f
:
R
n
→
R
m
=
Δ
P
⋅
g
f:\R^n\rightarrow\R^m\overset{\Delta}{=}P\cdot g
f:Rn→Rm=ΔP⋅g(复合函数)线性分数函数
P
(
g
(
C
)
)
P(g(C))
P(g(C))
f ( x ) = A x + b C T x + d , d o m f = { x ∣ C T x + d > 0 } f(x)=\frac{Ax+b}{C^Tx+d},\;\;dom\;f=\{x|C^Tx+d>0\} f(x)=CTx+dAx+b,domf={x∣CTx+d>0}
- 线性分数变换是一个保凸变换,但是它其实是一个非线性变换,所以很特殊的。
两个随机变量的联合概率 ⟶ 映 射 \overset{映射}{\longrightarrow} ⟶映射条件概率(保凸映射)
u = { 1 , ⋯ , n } v = { 1 , ⋯ , m } u=\{1,\cdots,n\}\;\;v=\{1,\cdots,m\} u={1,⋯,n}v={1,⋯,m}
联合概率: P i j = P ( u = i , v = j ) P_{ij}=P(u=i,v=j) Pij=P(u=i,v=j)
条件概率: f i j = P ( u = i ∣ v = j ) f_{ij}=P(u=i|v=j) fij=P(u=i∣v=j)
由条件概率的定义可知: f i j = P i j ∑ k = 1 n P k j f_{ij}=\frac{P_{ij}}{\sum_{k=1}^nP_{kj}} fij=∑k=1nPkjPij
这是一个线性分式映射,因为可以写成 f i j = ( 0 , ⋯ , 1 , ⋯ , 0 ) ( P 1 j , ⋯ , P n j ) T 1 T ( P 1 j , ⋯ , P n j ) T = A x + 0 C T x + 0 f_{ij}=\frac{(0,\cdots,1,\cdots,0)(P_{1j},\cdots,P_{nj})^T}{1^T(P_{1j},\cdots,P_{nj})^T}=\frac{Ax+0}{C^Tx+0} fij=1T(P1j,⋯,Pnj)T(0,⋯,1,⋯,0)(P1j,⋯,Pnj)T=CTx+0Ax+0
成立的前提是证明 { P i j } \{P_{ij}\} {Pij}是一个凸集。
下一章传送门:中科大-凸优化 笔记(lec9)-广义不等式、分离与支撑超平面、对偶锥与广义不等式