Web Pages
→
Nodes of Graph
Link
→
Edges of Graph
\text{Web Pages} \rightarrow \text{Nodes of Graph} \\ \text{Link} \rightarrow \text{Edges of Graph}
Web Pages→Nodes of GraphLink→Edges of Graph
一开始,人们考虑用入度来表示结点的重要性。因为如果一个网页很重要,必然会有很多的网站引用他。
Importance
⇔
in degree
\text{Importance} \Leftrightarrow \text{in degree}
Importance⇔in degree
Simulation
→
Random Effect
→
Random Variables
→
Distribution
→
Simulation
\text{Simulation} \rightarrow \text{Random Effect} \rightarrow \text{Random Variables} \rightarrow \text{Distribution} \rightarrow \text{Simulation}
Simulation→Random Effect→Random Variables→Distribution→Simulation
所谓仿真,就是产生一组伪随机样本,这些伪随机样本可以产生某种分布
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Pseudo-Random Sample
Generate \quad Z_{1},...,Z_{n} \sim f(x) \\ \text{Pseudo-Random Sample}
GenerateZ1,...,Zn∼f(x)Pseudo-Random Sample
我们对这些分布并不了解,但是我们可以转化为对样本的研究。
比如我们需要计算期望,但是积分非常难算,我们就可以用样本均值替代。
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E(Z) = \int_{-\infty}^{+\infty} xf(x)dx \\ \rightarrow \frac{1}{n} \sum_{k=1}^n Z_k
E(Z)=∫−∞+∞xf(x)dx→n1k=1∑nZk
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Irreducible + Non-Perdic
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P = (P_{ij})_{ij} \\ b_i = P(Z_0 =i) \\ \text{Irreducible + Non-Perdic} \Rightarrow \pi = \pi P
P=(Pij)ijbi=P(Z0=i)Irreducible + Non-Perdic⇒π=πP
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\pi_j=(\pi P)_j = \sum_i \pi_i P_{ij} = \sum_i \pi_j P_{ji} \\ =\pi_j \sum_i P_{ji} = \pi_j *1 = \pi_j
πj=(πP)j=i∑πiPij=i∑πjPji=πji∑Pji=πj∗1=πj
因此细致平衡成立,并且是方程的充分条件,细致平衡成立,方程就成立。
2.3.2 第二步:构成一步转移矩阵P
其实MCMC的一步转移矩阵P开始是任取的,但是一般不满足细致平衡条件
∀
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\forall P(Proposal) \\ \pi_i P_{ij} \cancel= \pi_j P_{ji}
∀P(Proposal)πiPij=πjPji
不过对P稍加调整,就能够满足细致平衡条件
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\widetilde{P}_{ij} = P_{ij} min(1,\frac{\pi_j P_{ji}}{\pi_i P_{ij}}) \\ \pi_i \widetilde{P_{ij}} = \pi_j \widetilde{P_{ji}}
Pij=Pijmin(1,πiPijπjPji)πiPij=πjPji
我们进行证明
π
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{
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\pi_i \widetilde{P}_{ij} = \begin{cases} \pi_j P_{ji} & \pi_i P_{ij} \geq \pi_j P_{ji}\\ \pi_iP_{ij} & \pi_i P_{ij} < \pi_j P_{ji} \end{cases}
πiPij={πjPjiπiPijπiPij≥πjPjiπiPij<πjPji
π
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=
{
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\pi_j \widetilde{P}_{ji} = \begin{cases} \pi_j P_{ji} & \pi_i P_{ij} \geq \pi_j P_{ji}\\ \pi_iP_{ij} & \pi_i P_{ij} < \pi_j P_{ji} \end{cases}
πjPji={πjPjiπiPijπiPij≥πjPjiπiPij<πjPji
则
π
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P
i
j
~
=
π
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P
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~
\pi_i \widetilde{P_{ij}} = \pi_j \widetilde{P_{ji}}
πiPij=πjPji
States
P
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Observations
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\text{States} \\ P_{ij} = P(S_n = j|S_{n-1} = i) \\ \text{Observations} B_{ij}(n) = P(O_n=j|S_n = i)
StatesPij=P(Sn=j∣Sn−1=i)ObservationsBij(n)=P(On=j∣Sn=i)
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P(O_1,...,O_n)
P(O1,...,On)
用全概率公式展开
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P(O_1,...,O_n) = \sum_{S_1,..,S_n} P(O_1,...,O_n|S_1,...,S_n)P(S_1,...,S_n) \\ = \sum_{S_1,..,S_n} (\prod_{k=1}^n P(O_k|S_k)) \prod_{k=1}^n P(S_k|S_{k-1})
P(O1,...,On)=S1,..,Sn∑P(O1,...,On∣S1,...,Sn)P(S1,...,Sn)=S1,..,Sn∑(k=1∏nP(Ok∣Sk))k=1∏nP(Sk∣Sk−1)
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\lambda_i(k) = P(O_1,O_2,...,O_k,S_k = i)\\ \lambda_j(k+1) = P(O_1,O_2,...,O_k,O_{k+1},S_{k+1} = j)
λi(k)=P(O1,O2,...,Ok,Sk=i)λj(k+1)=P(O1,O2,...,Ok,Ok+1,Sk+1=j)
找一下递推关系
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\lambda_j(k+1) = P(O_1,O_2,...,O_k,O_{k+1},S_{k+1} = j) \\ = \sum_iP(O_1,O_2,...,O_k,O_{k+1},S_{k+1} = j,S_k=i) \\ = \sum_i P(O_{k+1},S_{k+1}=j|O_1,...O_k,S_k = i)P(O_1,...,O_k,S_k=i) \\ = \sum_i P(O_{k+1},S_{k+1}=j|S_k=i) \lambda_i(k)
λj(k+1)=P(O1,O2,...,Ok,Ok+1,Sk+1=j)=i∑P(O1,O2,...,Ok,Ok+1,Sk+1=j,Sk=i)=i∑P(Ok+1,Sk+1=j∣O1,...Ok,Sk=i)P(O1,...,Ok,Sk=i)=i∑P(Ok+1,Sk+1=j∣Sk=i)λi(k)
可以根据马尔科夫性忘记前面的状态。再用全概率公式变形
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\lambda_j(k+1) =\sum_i P(O_{k+1},S_{k+1}=j|S_k=i) \lambda_i(k) \\ = \sum_i P(O_{k+1}|S_{k+1}=j,S_k=i)P(S_{k+1}=j|S_k=i) \lambda_i(k) \\ = P(O_{k+1}|S_{k+1}=j)\sum_i \lambda_i(k) P(S_{k+1}=j|S_k=i) \\ = B_{jO_{k+1}}(k+1)(\sum_i \lambda_i(k)P_{ij})
λj(k+1)=i∑P(Ok+1,Sk+1=j∣Sk=i)λi(k)=i∑P(Ok+1∣Sk+1=j,Sk=i)P(Sk+1=j∣Sk=i)λi(k)=P(Ok+1∣Sk+1=j)i∑λi(k)P(Sk+1=j∣Sk=i)=BjOk+1(k+1)(i∑λi(k)Pij)
我们要求的概率可以表示为
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P(O_1,...,O_n) = \sum_i P(O_1,...,O_n,S_n=i) = \sum_{i}\lambda_i(n)
P(O1,...,On)=i∑P(O1,...,On,Sn=i)=i∑λi(n)
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max_{S_1,...,S_n} P(S_1,...,S_n|O_1,...,O_n)
maxS1,...,SnP(S1,...,Sn∣O1,...,On)
就是在已知某个观测的前提下,计算这个状态后验概率的最大值。
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max_{S_1,...,S_n} P(S_1,...,S_n|O_1,...,O_n) \Leftrightarrow max_{S_1,...,S_n} P(S_1,...,S_n,O_1,...,O_n)
maxS1,...,SnP(S1,...,Sn∣O1,...,On)⇔maxS1,...,SnP(S1,...,Sn,O1,...,On) 如果用贝叶斯公式的话,P(O1,…On)会出现在分母上,这是个常数,不需要关心,因此可以转化为求右边式子的最大值。
有一个标准解法,就是维特比迭代
Viterbi
\text{Viterbi}
Viterbi
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\mu_i(k) = max_{S_1,...,S_{k-1}}P(O_1,...,O_k,S_k=i,...,S_1) \\ \mu_j(k+1) = max_{S_1,...,S_{k}} P(O_1,...,O_{k+1},S_{k+1}=j,...,S_1) \\ =max_{S_1,...,S_{k}} P(O_{k+1},S_{k+1}|O_k,...,O_1,S_k,...,S_1) P(O_k,...,O_1,S_k,...S_1) \\ = max_{S_1,...,S_{k}} P(O_{k+1}|S_{k+1})P(S_{k+1}=j|S_k=i) P(O_k,...,O_1,S_k=i,...S_1) \\ = max_{i} P(O_{k+1}|S_{k+1}=j)max_{S_1,...,S_{k-1}}P(S_{k+1}=j|S_k=i) P(O_k,...,O_1,S_k=i,...S_1) \\ = max_i( B_{j,O_{k+1}(k+1)}P_{ij} \mu_i(k))
μi(k)=maxS1,...,Sk−1P(O1,...,Ok,Sk=i,...,S1)μj(k+1)=maxS1,...,SkP(O1,...,Ok+1,Sk+1=j,...,S1)=maxS1,...,SkP(Ok+1,Sk+1∣Ok,...,O1,Sk,...,S1)P(Ok,...,O1,Sk,...S1)=maxS1,...,SkP(Ok+1∣Sk+1)P(Sk+1=j∣Sk=i)P(Ok,...,O1,Sk=i,...S1)=maxiP(Ok+1∣Sk+1=j)maxS1,...,Sk−1P(Sk+1=j∣Sk=i)P(Ok,...,O1,Sk=i,...S1)=maxi(Bj,Ok+1(k+1)Pijμi(k))
维特比的递推和前向递推具有相似的结构
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\mu_j(k+1) = max_i B_{j,O_{k+1}(k+1)}(P_{ij} \mu_i(k)) \\
μj(k+1)=maxiBj,Ok+1(k+1)(Pijμi(k))
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\lambda_j(k+1)=\sum_i B_{jO_{k+1}}(k+1)( \lambda_i(k)P_{ij}) \\
λj(k+1)=i∑BjOk+1(k+1)(λi(k)Pij)
其中
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\mu_i(k) = max_{S_1,...,S_{k-1}}P(O_1,...,O_k,S_k=i,...,S_1) \\ \lambda_i(k) = P(O_1,O_2,...,O_k,S_k = i)
μi(k)=maxS1,...,Sk−1P(O1,...,Ok,Sk=i,...,S1)λi(k)=P(O1,O2,...,Ok,Sk=i)