GES Model :Combining Pearson Correlation Coefficient Analysis with Multilayer Perceptron

$\divideontimes$ Consistency Check

After getting the eigenvalues of the weight coefficient matrix, we perform a consistency check, the equations are as follow

$$CI=\frac{\lambda_{max}-n}{n-1}=\frac{7.72-7}{7-1}=0.12.$$

(1)

Substituting the data obtained before, we get

$$CR=\frac{CI}{RI}=\frac{0.12}{1.32}=0.0909<0.1.$$

(2)

Finally, we get the formula to measure the national development score($Eq_{k}$):

	$\displaystyle Eq_{k}$	$\displaystyle=0.187\times EI+0.387\times IDG+0.097\times CEA$		(3)
		$\displaystyle+0.0436\times MA+0.086\times HR+0.0831\times ER$
		$\displaystyle+0.117\times SA.$

$\divideontimes$ Model Validation

Take Income Distribution Gap(IDG) as an example, according to the factors affecting IDG obtained from the previous analysis, the income inequality score of each country is finally shown in figure 6 .

For a certain country, first we de-self the country by comparing its development score($Eq_{k}$) with the average of other countries’ development scores($Eq_{m}$)($k\neq m$). Then we subtract it from the mean of all selected countries($\overline{Eq_{k}}$) and take the mean value. That is, using the following formula

$$GE=\frac{1}{10n}\sum{(\frac{Eq_{k}}{\sum\limits_{i\neq k}Eq_{i}/(n-1)}-\overline{Eq_{k}})^{2}}.$$

(4)

Considering the different contributions of different countries in asteroid mining, the countries that contribute more to science and technology deserve to share more resources.

After defining how these variables are determined, we use the TOPSIS method to continue our analysis of the problem[17].

$\divideontimes$ The basic Steps of TOPSIS

$\bullet$ Normalize the Original Matrix

Suppose ${x_{i}}$ is a set of intermediate type indicator series and the optimal value is $x_{best}$, then equation 5 , 6 are the forwarding equations.

$$M=\max{|x_{i}-x_{best}|}$$

(5)

$$\tilde{x_{i}}=1-\frac{|x_{i}-x_{best}|}{M}$$

(6)

$\bullet$ Normalize the Normalization Matrix

If the normalized matrix is noted as Z, then we use equation 7 to normalize the matrix X.

$$z_{ij}=x_{ij}/\sqrt{\sum\limits_{i=1}^{n}x_{ij}^{2}}.$$

(7)

Assume that there are n objects to be evaluated and a standardized matrix of m evaluation indicators.

	$\displaystyle\boldsymbol{Z}$	$\displaystyle=\begin{bmatrix}z_{11}&z_{12}&\cdots&z_{1m}\\ z_{21}&z_{22}&\cdots&z_{2m}\\ \vdots&\vdots&\ddots&\vdots\\ z_{n1}&z_{n2}&\cdots&z_{nm}\end{bmatrix}$		(8)
		$\displaystyle=\begin{bmatrix}0.4056&0.2154&\cdots&0.2433\\ 0.3187&0.1031&\cdots&0.2305\\ \vdots&\vdots&\ddots&\vdots\\ 0.0290&0.1511&\cdots&0.2122\\ \end{bmatrix}$

. Then we define maximum value($Z^{+}$):

	$\displaystyle Z^{+}$	$\displaystyle=(Z_{1}^{+},Z_{2}^{+},\cdots Z_{m}^{+})$		(9)
		$\displaystyle=(\max{\{z_{11},z_{21},\cdots,z_{n1}\}},\max{\{z_{12},z_{22},\cdots,z_{n2}\}},$
		$\displaystyle\cdots,\max{\{z_{1m},z_{2m},\cdots,z_{nm}\}}),$

minimun value($Z^{-}$):

	$\displaystyle Z^{-}$	$\displaystyle=(Z_{1}^{-},Z_{2}^{-},\cdots Z_{m}^{-})$		(10)
		$\displaystyle=(\min{\{z_{11},z_{21},\cdots,z_{n1}\}},\min{\{z_{12},z_{22},\cdots,z_{n2}\}},$
		$\displaystyle\cdots,\min{\{z_{1m},z_{2m},\cdots,z_{nm}\}}),$

Distance of the $i^{th}(i=1,2,\cdots,n)$ evaluation object from the maximum value($D^{+}_{i}$):

$$D^{+}_{i}=\sqrt{\sum\limits_{j=1}^{m}(Z^{+}_{j}-z_{ij})^{2}},$$

(11)

Distance of the $i^{th}(i=1,2,\cdots,n)$ evaluation object from the minimum value($D^{-}_{i}$):

$$D^{-}_{i}=\sqrt{\sum\limits_{j=1}^{m}(Z^{-}_{j}-z_{ij})^{2}}.$$

(12)

Then, we can calculate the un-normalized score of the $i^{th}$ evaluation object:

$$S_{i}=\frac{D^{-}_{i}}{D^{+}_{i}+D^{-}_{i}}.$$

(13)

Next, we normalize the score using equation 14:

$$\tilde{S_{i}}=S_{i}/\sum\limits_{i=1}^{n}S_{i}.$$

(14)

Considering that the technology is not fully mature in the early stage, the mining rate should rise first and then fall. We use the t-distribution probability density function as our mining curve[20]. The function is defined as follows.

$$p(t)=\frac{\Gamma(\frac{n+1}{2})}{\sqrt{n\pi}\Gamma(\frac{n}{2})}(1+\frac{y^{2}}{n})^{-\frac{n+1}{2}},0<t<+\infty$$

(15)

With this definition of the mining curve, we then calculate the income.

$$income=\int_{t_{1}}^{t_{2}}p^{\prime}(t)\cdot Vdt$$

(16)

where V denotes total mineral value and we take V = 70 trillion. Then,

$$Profit=income-cost$$

(17)

Let $\gamma$ be the poverty index of the kth country and we specify the value of $\gamma$ could be calculated by equation 18 .

$$\gamma=\left\{\begin{aligned} &\ 1.2\text{, if the country is one of the 20 countries with the lowest GDP},\\ &\ 1.0\text{, otherwise}.\end{aligned}\right.$$

(18)

Then, the formula for total profit is:

$$pro_{c}=\gamma[(income-cost)\cdot\frac{Eq_{k}}{\sum{Eq_{k}}}]\\$$

(19)

where $pro_{c}$ represents the total profit of Country C.

Finally, we calculated the Global Inequity Index for 2030-2039 and the results are shown in figure 11 .

The backpropagation algorithm is a supervised learning method used in conjunction with optimization algorithms such as gradient descent[12]. It is a generalization of the Delta rule for multilayer feedforward networks, which allows the gradient to be computed using a chain rule for each layer iteration.

In a word, the following equation holds:

$$\left\{\begin{aligned} z_{1}^{(l)}=&\ w^{(l)}_{11}a^{(l-1)}_{1}+w^{(l)}_{12}a^{(l-1)}_{2}+\cdots\\ &+w^{(l)}_{1N^{(l-1)}}a^{(l-1)}_{N^{(l-1)}}+b^{(l)}_{1}\\ z_{2}^{(l)}=&\ w^{(l)}_{21}a^{(l-1)}_{1}+w^{(l)}_{22}a^{(l-1)}_{2}+\cdots\\ &+w^{(l)}_{2N^{(l-1)}}a^{(l-1)}_{N^{(l-1)}}+b^{(l)}_{2}\\ \vdots&\\ z^{(l)}_{N^{(l)}}=&\ w^{(l)}_{N^{(l)}1}a^{(l-1)}_{1}+w^{(l)}_{N^{(l)}2}a^{(l-1)}_{2}+\cdots\\ &+w^{(l)}_{N^{(l)}N^{(l-1)}}a^{(l-1)}_{N^{(l-1)}}+b^{(l)}_{N^{(l)}}\\ \end{aligned}\right.,$$

(20)

written in the form of matrix multiplication:

	$\displaystyle\begin{bmatrix}z^{(l)}_{1}\\ z^{(l)}_{2}\\ \vdots\\ z^{(l)}_{N^{(l)}}\\ \end{bmatrix}$	$\displaystyle=\begin{bmatrix}w^{(l)}_{11}&w^{(l)}_{12}\cdots&w^{(l)}_{1N^{(l-1)}}\\ w^{(l)}_{21}&w^{(l)}_{22}\cdots&w^{(l)}_{2N^{(l-1)}}\\ \vdots\\ w^{(l)}_{N^{(l)}1}&w^{(l)}_{N^{(l)}2}\cdots&w^{(l)}_{N^{(l)}N^{(l-1)}}\\ \end{bmatrix}\begin{bmatrix}a^{(l-1)}_{1}\\ a^{(l-1)}_{2}\\ \vdots\\ a^{(l-1)}_{N^{(l-1)}}\\ \end{bmatrix}$		(21)
		$\displaystyle+\begin{bmatrix}b^{(l)}_{1}\\ b^{(l)}_{2}\\ \vdots\\ b^{(l)}_{N^{(l)}}\\ \end{bmatrix},$

that is:

$$\boldsymbol{z^{(l)}}=\boldsymbol{w^{(l)}}\boldsymbol{a^{(l-1)}}+\boldsymbol{b^{(l)}}.$$

(22)

Since $\boldsymbol{a^{(l)}}=\sigma(\boldsymbol{z^{(l)}})$, we can conclude that

$$\boldsymbol{a^{(l)}}=\sigma(\boldsymbol{w^{(l)}}\boldsymbol{a^{(l-1)}}+\boldsymbol{b^{(l)}})$$

(23)

where $\sigma(x)$ denotes Activation Function, ususally take $\sigma(x)=\frac{1}{1+e^{-x}}$.

$\divideontimes$ Errors in the output layer

According to the transmissibility of error propagation $\Delta z^{(L)}_{j}\rightarrow\Delta a^{(L)}_{j}\rightarrow\Delta C(\theta)$ and combined with the chain derivative rule, we found the error of the loss function on the output layer neurons $\delta^{(L)}_{j}$.

	$\displaystyle\delta^{(L)}_{j}$	$\displaystyle=\frac{\partial C(\theta)}{\partial z^{(L)}_{j}}=\frac{\partial C(\theta)}{\partial a^{(L)}_{j}}\frac{\partial a^{(L)}_{j}}{\partial z^{(L)}_{j}}$		(24)
		$\displaystyle=\frac{\partial C(\theta)}{\partial a^{(L)}_{j}}\frac{\partial\sigma(z^{(L)}_{j})}{\partial z^{(L)}_{j}}$
		$\displaystyle=\frac{\partial C(\theta)}{\partial a^{(L)}_{j}}\sigma^{{}^{\prime}}(z^{(L)}_{j})$

For all neurons on the output layer, this can be represented as a vector form.

	$\displaystyle\boldsymbol{\sigma^{(L)}}$	$\displaystyle=\begin{bmatrix}\sigma^{(L)}_{1}\\ \sigma^{(L)}_{2}\\ \vdots\\ \sigma^{(L)}_{N^{(L)}}\\ \end{bmatrix}=\begin{bmatrix}\frac{\partial C(\theta)}{\partial a^{(L)}_{1}}\sigma^{{}^{\prime}}(z^{(L)}_{1})\\ \frac{\partial C(\theta)}{\partial a^{(L)}_{2}}\sigma^{{}^{\prime}}(z^{(L)}_{2})\\ \vdots\\ \frac{\partial C(\theta)}{\partial a^{(L)}_{N^{L}}}\sigma^{{}^{\prime}}(z^{(L)}_{N^{L}})\\ \end{bmatrix}$		(25)
		$\displaystyle=\begin{bmatrix}\frac{\partial C(\theta)}{\partial a^{(L)}_{1}}\\ \frac{\partial C(\theta)}{\partial a^{(L)}_{2}}\\ \vdots\\ \frac{\partial C(\theta)}{\partial a^{(L)}_{N^{L}}}\\ \end{bmatrix}\odot\begin{bmatrix}\sigma^{{}^{\prime}}(z^{(L)}_{1})\\ \sigma^{{}^{\prime}}(z^{(L)}_{2})\\ \vdots\\ \sigma^{{}^{\prime}}(z^{(L)}_{N^{L}})\\ \end{bmatrix}$
		$\displaystyle=\nabla_{\boldsymbol{a^{(L)}}}C(\theta)\odot\sigma^{{}^{\prime}}(\boldsymbol{z^{(L)}})$

where $\odot$ is the Hadamard Product(the product of the corresponding elements of the two matrices).

$\divideontimes$ Error in the Hidden Layer

Because the error of the output layer has been found above, according to the principle of error back propagation, the error of the current layer can be understood as a composite function of the error of all neurons in the previous layer(the error of the previous layer is used to represent the error of the current layer, and so on).

$$\begin{aligned} \sigma^{(l)}_{j}&=\frac{\partial C(\theta)}{\partial z^{(l)}_{j}}\\ &=\sum\limits_{k=1}^{N^{(l+1)}}\frac{\partial C(\theta)}{\partial z^{(l+1)_{k}}}\frac{\partial z^{(l+1)}_{k}}{\partial a^{(l)}_{j}}\frac{\partial a^{(l)}_{j}}{\partial z^{(l)}_{j}}\\ &=\sum\limits_{k=1}^{N^{(l+1)}}\delta^{(l+1)}_{k}\frac{\partial(\sum\limits_{s=1}^{N^{(l)}}w^{(l+1)}_{ks}a^{(l)}_{s}+b^{(l+1)}_{k})}{\partial a^{(l)}_{j}}\frac{\partial a^{(l)}_{j}}{\partial z^{(l)}_{j}}\\ &=\sum\limits_{k=1}^{N^{(l+1)}}\delta^{(l+1)}_{k}w^{(l+1)}_{kj}\sigma^{{}^{\prime}}(z^{(l)}_{j})\\ \end{aligned}.$$

(26)

Similarly for the error of all neurons in the hidden layer, it can be written in vector form.

	$\displaystyle\boldsymbol{\delta^{(l)}}$	$\displaystyle=\begin{bmatrix}\delta^{(l)}_{1}\\ \delta^{(l)}_{2}\\ \vdots\\ \delta^{(l)}_{N^{(l)}}\\ \end{bmatrix}=\begin{bmatrix}\sum\limits_{k=1}^{N^{(l+1)}}\delta^{(l+1)}_{k}w^{(l+1)}_{k1}\sigma^{{}^{\prime}}(z^{(l)}_{1})\\ \sum\limits_{k=1}^{N^{(l+1)}}\delta^{(l+1)}_{k}w^{(l+1)}_{k2}\sigma^{{}^{\prime}}(z^{(l)}_{2})\\ \vdots\\ \sum\limits_{k=1}^{N^{(l+1)}}\delta^{(l+1)}_{k}w^{(l+1)}_{kN^{(l)}}\sigma^{{}^{\prime}}(z^{(l)}_{N^{(l)}})\\ \end{bmatrix}$		(27)
		$\displaystyle=\begin{bmatrix}\sum\limits_{k=1}^{N^{(l+1)}}\delta^{(l+1)}_{k}w^{(l+1)}_{k1}\\ \sum\limits_{k=1}^{N^{(l+1)}}\delta^{(l+1)}_{k}w^{(l+1)}_{k2}\\ \vdots\\ \sum\limits_{k=1}^{N^{(l+1)}}\delta^{(l+1)}_{k}w^{(l+1)}_{kN^{(l)}}\\ \end{bmatrix}\odot\begin{bmatrix}\sigma^{{}^{\prime}}(z^{(l)}_{1})\\ \sigma^{{}^{\prime}}(z^{(l)}_{2})\\ \vdots\\ \sigma^{{}^{\prime}}(z^{(l)}_{N^{(l)}})\\ \end{bmatrix}$
		$\displaystyle=(\boldsymbol{w^{(l+1)}})^{T}\boldsymbol{\delta^{(l+1)}}\odot\sigma^{{}^{\prime}}(\boldsymbol{z^{(l)}})$

Pearson correlation analysis is used to explore the correlation between world equity scores and each representative indicator of each country. The Pearson correlation coefficient between the two variables is defined by the equation 28:

$$\rho_{(Eq_{k},GDP)}=\frac{cov(Eq_{k},GDP)}{\sigma_{Eq_{k}}\sigma_{GDP}}=\frac{E[(Eq_{k}-\overline{Eq_{k}})(GDP-\overline{GDP})]}{\sigma_{Eq_{k}}\sigma_{GDP}}$$

(28)

Then we use equation 29 to calculate the correlation coefficient r.

$$r=\frac{\sum\limits_{i=1}^{n}(Eq_{k}-\overline{Eq_{k}})(GDP-\overline{GDP})}{\sqrt{\sum\limits_{i=1}^{n}(Eq_{k}-\overline{Eq_{k}})^{2}}\sqrt{\sum\limits_{i=1}^{n}(GDP-\overline{GDP})^{2}}}$$

(29)

As can be seen from figure 13 , the closer the absolute value of r is to 1, the stronger the correlation between the two variables. Meanwhile, the positive or negative of r determines the positive or negative of the correlation.

$\divideontimes$ Calculation of Correlation Coefficient

Take analysis of the correlation between GDP and countries’ equity scores as an example, through the analysis of the first question we know that the mean value of $Eq_{k}$ is 34.7227. Then, we use equation 30 .

	$\displaystyle\sigma_{Eq_{k}}$	$\displaystyle=\sqrt{\sum\limits(Eq_{k}-\overline{Eq_{k}})^{2}}$		(30)
		$\displaystyle=\sqrt{\sum\limits(Eq_{k}-34.7227)^{2}}$
		$\displaystyle=5.7147$

to got the variance is 32.6574 and the standard deviation is 5.7147.

Similarly, since the average value of GDP is 11540.8620, we substitute it into equation 31 .

	$\displaystyle\sigma_{GDP}$	$\displaystyle=\sqrt{\sum(GDP-\overline{GDP})^{2}}$		(31)
		$\displaystyle=\sqrt{\sum(GDP-11540.8620)^{2}}$
		$\displaystyle=11.9809$

that is, ehile variance is 143.5418, the standard deviation is 11.9809.

Finally, we calculate the sum of the outlying product of the score and GDP by equation 32 .

	$\displaystyle cov(Eq_{k},GDP)$	$\displaystyle=\sqrt{\sum\limits(Eq_{k}-\overline{Eq_{k}})(GDP-\overline{GDP})}$		(32)
		$\displaystyle=\sqrt{\sum\limits(Eq_{k}-34.7227)(GDP-11540.8620)}$
		$\displaystyle=6.6509.$

The sum of the outlying product of the score and GDP is 6.6509. Substitute into equation 33.

	$\displaystyle\rho_{(Eq_{k},GDP)}$	$\displaystyle=\frac{cov(Eq_{k},GDP)}{\sigma_{Eq_{k}}\sigma_{GDP}}=\frac{\sum\limits_{i=1}^{n}\frac{(Eq_{k}-\overline{Eq_{k}})}{\sigma_{Eq_{k}}}\frac{(GDP-\overline{GDP})}{\sigma_{GDP}}}{n}$		(33)
		$\displaystyle=\frac{\sum\limits_{i=1}^{n}\frac{(Eq_{k}-34.7227)}{\sigma_{Eq_{k}}}\frac{(GDP-11540.8620)}{\sigma_{GDP}}}{n}$
		$\displaystyle=0.871$

$\divideontimes$ Test of Pearson’s Correlation Coefficient

After obtaining the correlation coefficient, we use the Pearson Correlation Coefficient Method to test it which using equation 34 .

	$\displaystyle r$	$\displaystyle=\frac{\sum\limits_{i=1}^{n}(Eq_{k}-34.7227)(GDP-11540.8620)}{\sqrt{\sum\limits_{i=1}^{n}(Eq_{k}-34.7227)^{2}}\sqrt{\sum\limits_{i=1}^{n}(GDP-11540.8620)^{2}}}$		(34)
		$\displaystyle=0.78.$

Next, we propose the hypothesis:

$\bullet$ $H_{0}:P=0$, score is not related to GDP;

$\bullet$ $H_{1}:P\neq 0$, score is not related to GDP;

Meanwhile, we determine the corresponding significant level of 0.05.

Using equation 35 , we obtain the value of r.

	$\displaystyle t_{r}$	$\displaystyle=\frac{|r-0|}{\sqrt{(1-r^{2})/(n-2)}}$		(35)
		$\displaystyle=\frac{0.78}{\sqrt{(1-0.78^{2})/(7-2)}}$
		$\displaystyle=19.5894$