# 广义投入产出分析方法

1. 如果某个部门，例如第$$N$$部门，增加了对其他某个$$j$$部门的投入或者需求——需求的意思就是反过来$$j$$部门到$$N$$部门的投入，整个系统的各个部门的产出会发生什么变化
2. 各个部门对整体系统的重要程度或者说有影响力
3. $$j$$部门对其他各个部门的影响力

$$N$$封闭系统各个部门之间的投入产出关系有以下矩阵代表
$x = \left(x^{i}_{j}\right)_{N\times N},$

$X^{i} = \sum_{j}x^{i}_{j}, X_{i} = \sum_{j}x^{j}_{i}.$

$B^{i}_{j} = \frac{x^{i}_{j}}{X^{j}}.$

$F^{i}_{j} = \frac{x^{i}_{j}}{X_{i}}.$

$MB^{i}_{j} = \frac{x^{i}_{j}}{X^{i}}.$

$MF^{i}_{j} = \frac{x^{i}_{j}}{X_{j}}.$

\begin{align}
\left\langle \lambda_{B} \right| B = \lambda_{B} \left\langle \lambda_{B} \right| \notag \\
\Rightarrow \lambda_{B} \left\langle \lambda_{B} \right| \left. i \right\rangle = \sum_{j} \left\langle \lambda_{B} \right| \left. j \right\rangle \left\langle j \right| B | \left. i \right\rangle = \sum_{j} \left\langle \lambda_{B} \right| \left. j \right\rangle B^{j}_{i} \notag \\
\Rightarrow \lambda_{B} \left\langle \lambda_{B} \right| \left. i \right\rangle = \sum_{j} \left\langle \lambda_{B} \right| \left. j \right\rangle \frac{x^{j}_{i}}{X^{i}} = \sum_{j} \left\langle \lambda_{B} \right| \left. j \right\rangle \frac{x^{j}_{i}}{X^{j}}\frac{X^{j}}{X^{i}} \notag \\
\Rightarrow \lambda_{B} \left\langle \lambda_{B} \right| \left. i \right\rangle X^{i}=\sum_{j} \left\langle \lambda_{B} \right| \left. j \right\rangle X^{j} \frac{x^{j}_{i}}{X^{j}}.
\Rightarrow \lambda_{B} \left\langle \lambda_{B} \right| \left. i \right\rangle X^{i}=\sum_{j} \left\langle \lambda_{B} \right| \left. j \right\rangle X^{j} MB^{j}_{i}.
\end{align}

\begin{align}
\sum_{j} B^{i}_{j} X^{j} = \sum_{j} \frac{x^{i}_{j}}{X^{j}}X^{j} = X^{i}.
\end{align}

$X^{i}\neq X^{i}$

\begin{align}
X^i=\sum^{N-1}_{j=1}x^{i}_{j}+x^i_{N} = \sum^{N-1}_{j=1} x^{i}_{j}+Y^i = \sum^{N-1}_{j=1}\frac{x^i_j}{X^{j}}X^{j}+Y^i \notag \\
\Rightarrow X^{\left(-N\right)} = B^{\left(-N\right)}X + Y^{\left(-N\right)} \notag \\
\Rightarrow X^{\left(-N\right)} = \left(1-B^{\left(-N\right)}\right)^{-1} Y^{\left(-N\right)} \\
\Rightarrow \Delta X^{\left(-N\right)} = \left(1-B^{\left(-N\right)}\right)^{-1} \Delta Y^{\left(-N\right)} = L^{\left(-N\right)}_{B} \Delta Y^{\left(-N\right)}
\end{align}

\begin{align}
X_i=\sum^{N-1}_{j=1}x^{j}_{i}+x^{N}_{i} = \sum^{N-1}_{j=1} x^{j}_{i}+V_i = \sum^{N-1}_{j=1}\frac{x^{j}_{i}}{X_{j}}X_{j}+V_i \notag \\
\Rightarrow X^{\left(-N\right)} = XF^{\left(-N\right)} + V^{\left(-N\right)} \notag \\
\Rightarrow X^{\left(-N\right)} = V\left(1-F^{\left(-N\right)}\right)^{-1} \\
\Rightarrow \Delta X^{\left(-N\right)} = \Delta V^{\left(-N\right)} \left(1-F^{\left(-N\right)}\right)^{-1} = \Delta V^{\left(-N\right)} L^{\left(-N\right)}_{F}
\end{align}

Hypothetical Extraction Method (HEM)的意思是假想地从系统中去掉一个部门，然后看一看，在这个新的系统中，如果我们还要实现同样的需求（或者提供同样的投入，针对$$F$$），各个部门的总产出的变化。具体计算如下。

\begin{align}
L^{\left(-N-j\right)}_{B} = \left(1-B^{\left(-N-j\right)}\right)^{-1}.
\end{align}

\begin{align}
L^{\left(-N-j\right)}_{B} Y^{\left(-j\right)}, \left(L^{\left(-N\right)}_{B} Y\right)^{\left(-j\right)}.
\end{align}

\begin{align}
X^{\left(-j\right)}\left(m+1\right)= B^{\left(-N-j\right)} X^{\left(-j\right)}\left(m\right) + Y^{\left(-j\right)}
\end{align}

$$F$$的问题可以做类似分析。

\begin{align}
L^{\left(-k\right)}_{B} = \left(1-B^{\left(-k\right)}\right)^{-1}.
\end{align}

\begin{align}
Z_{B} = \left(Z^{k}_{j}\right)_{N\times N}.
\end{align}

$$F$$的问题可以做类似的分析。

\begin{align}
B \left| 1 \right\rangle_{B} = \left| 1 \right\rangle_{B},
\end{align}

\begin{align}
\left| 1 \right\rangle_{B} = \left(X^{1}, X^{2}, \cdots, X^{N}\right)^{T}.
\end{align}

\begin{align}
\left\langle 1 \right|_{B} B = \left\langle 1 \right|_{B},
\end{align}

\begin{align}
B^{\left(-k\right)}\left| \lambda^{\left(-k\right)}_{Max} \right\rangle_{B^{\left(-k\right)}} = \lambda^{\left(-k\right)}_{Max} \left| \lambda^{\left(-k\right)}_{Max} \right\rangle_{B^{\left(-k\right)}}.
\end{align}

\begin{align}
IOF^{k} = 1-\lambda^{\left(-k\right)}_{Max} ,
\end{align}

\begin{align}
IOMI^{k}_{j} = \left\langle j \left| \right. \lambda^{\left(-k\right)}_{Max} \right\rangle_{B^{\left(-k\right)}} – \left\langle j \left| \right. 1 \right\rangle_{B}.
\end{align}

PagerRank和PageRank的HEM

\begin{align}
\left\langle 1 \right|_{MB} MB = \left\langle 1 \right|_{MB}.
\end{align}

\begin{align}
\left\langle \lambda^{\left(-k\right)}_{Max} \right|_{MB^{\left(-k\right)}} MB^{\left(-k\right)}= \lambda^{\left(-k\right)}_{Max} \left\langle \lambda^{\left(-k\right)}_{Max} \right|_{MB^{\left(-k\right)}}.
\end{align}

\begin{align}
\left\langle 1 \right|_{\hat{MB}^{\left(-k\right)}} \hat{MB}^{\left(-k\right)}= \left\langle 1 \right|_{\hat{MB}^{\left(-k\right)}}.
\end{align}

\begin{align}
PRF^{k} = 1- \left\langle 1 \right|_{\hat{MB}^{\left(-k\right)}} \left(\left| 1 \right\rangle_{MB}\right)^{\left(-k\right)} .
\end{align}

\begin{align}
PRMI^{k}_{j} = \left\langle 1 \right|_{\hat{MB}^{\left(-k\right)}} \left| j \right\rangle – \left\langle 1 \right|_{MB} \left| j \right\rangle .
\end{align}

1. 小思科学学的工作提出和运用了本征向量HEM
2. 小勇的工作运用了传统HEM并且正在考察目标外界HEM
3. 秦磊的工作考虑运用目标外界HEM
4. 崔浩川城市的工作先运用本征向量HEM方法，将来再对比多个方法
5. 小思方法对比的工作需要做各个方法的对比
6. 李梦辉关于专利和文献合起来的领域相互影响的问题，考虑用封闭系统方法（本征向量HEM和目标外界HEM）
7. 张江在贸易网络上的工作就相当于用了传统的当作开放系统的HEM（并且进一步讨论了体量和影响力的关系，以及这个关系的幂律指数的含义）
8. 当作封闭系统的国家之间贸易网络的工作还没有人做
9. Dyson方程的工作解决的是以上各个技术中计算简化的问题