## 5 Simulation and Experiment
### Simulation Examples  
Four types of structural similarity indices were simulated to evaluate node pairs with and without links. The pdfs of the structural similarity indices are also provided. We construct 3 groups of known distributions for the similarity indices pdfs. One thousand samples extracted from 10,000 existing links and 100,000 samples of nonexistent links were generated following the appropriate pdfs. The 1000 samples serve as probe set; the 100,000 samples with 1000 probe links serve as unknown links for training; and the remaining 9000 samples serve as train set of existing links. Each sample had 4 dimensions to simulate 4 similarity scores. We first compute AUC and Precision for each dimension, then use proposed TLF method to obtain the synthetic score and calculate the AUC and Precision, compared with other combination methods such as Naive Bayes and logistic regression. Finally, we calculate AUC and Precision using the theoretical limit theorem and compare with the above methods. Let random vectors \(\mathbf{X}=\left(X_{1},X_{2},X_{3},X_{4}\right)^{T}\) and \(\mathbf{Y}=\left(Y_{1},Y_{2},Y_{3},Y_{4}\right)^{T}\) be the scores of existing and nonexistent links, which follow \(f(\mathbf{x})=f(x_{1},x_{2},\cdots,x_{n})\) and \(g(\mathbf{x})=g(x_{1},x_{2},\cdots,x_{n})\) pdfs, respectively. Let \(f(\mathbf{x}),\ g(\mathbf{x})\) are 4-dimensional normal distributions,  
\[f(\mathbf{x})=\frac{1}{\left(2\pi\right)^{p/2}\left|\mathbf{\Sigma}\right|^{ 1/2}}\exp\biggl{\{}-\frac{1}{2}(\mathbf{x}-\boldsymbol{\mu})^{T}\mathbf{ \Sigma}^{-1}(\mathbf{x}-\boldsymbol{\mu})\biggr{\}}, \tag{19}\]  
where \(\text{diag}(\mathbf{\Sigma})\mathbf{1}=\left(\sigma_{1}^{2},\sigma_{2}^{2}, \sigma_{3}^{2},\sigma_{4}^{2}\right)^{T}\), and \(\mathbf{\Sigma}_{ij}=\mathbf{r}_{ij}\sigma_{i}\sigma_{j}\). The parameter sets for the 2 groups of simulation examples are as follows. Group 1: \(\Theta_{1f}=\left\{\boldsymbol{\mu}_{1f},\mathbf{\Sigma}_{1f}\right\}\), and \(\Theta_{1g}=\left\{\boldsymbol{\mu}_{1g},\mathbf{\Sigma}_{1g}\right\}\);  
Group 2: \(\Theta_{2f}=\left\{\boldsymbol{\mu}_{2f},\mathbf{\Sigma}_{2f}\right\}\) and \(\Theta_{2g}=\left\{\boldsymbol{\mu}_{2g},\mathbf{\Sigma}_{2g}\right\}\). In each group, \(\boldsymbol{\mu}_{1f}=\left(1,2,1.7,2.1\right)^{T}\), \(\boldsymbol{\mu}_{1g}=\left(1.3,2.5,2.1,2.8\right)^{T}\), \(\boldsymbol{\mu}_{2f}=\left(1,2,1.7,2.1\right)^{T}\), \(\boldsymbol{\mu}_{2g}=\left(1.5,3.5,2.8,3\right)^{T}\), \(\text{diag}\left(\mathbf{\Sigma}_{1f}\right)\mathbf{1}=\left(1.5^{2},2.2^{2}, 3^{2},2.5^{2}\right)^{T}\), \(\text{diag}\left(\mathbf{\Sigma}_{1g}\right)\mathbf{1}=\left(2^{2},2.2^{2},3^{ 2},2.5^{2}\right)^{T}\), \(\text{diag}\left(\mathbf{\Sigma}_{2g}\right)\mathbf{1}=\left(2.5^{2},2.2^{2}, 3^{2},2.5^{2}\right)^{T}\), \(\text{diag}\left(\mathbf{\Sigma}_{2g}\right)\mathbf{1}=\left(2.5^{2},3.5^{2},4 ^{2},2.5^{2}\right)^{T}\), \(\boldsymbol{r}_{1f}=\boldsymbol{r}_{1g}=\left[ \begin{array}{cccc}1&0.8&0.76&0. 56\\ 0.8&1&0.85&0.74\\ 0.76&0.85&1&0.93\\ 0.56&0.74&0.93&1\end{array} \right]\), and \(\boldsymbol{r}_{2f}=\boldsymbol{r}_{2g}=\left[ \begin{array}{cccc}1&0.62&0.45&0. 34\\ 0.62&1&0.28&0.47\\ 0.45&0.28&1&0.65\\ 0.34&0.47&0.65&1\end{array} \right]\). The window width \(h\) of TLF method in the group 1 and 2 is \(h=0.1\). Group 3: Let  
\[f_{3}(\mathbf{x})=x_{1}x_{2}x_{3}x_{4}+x_{1}x_{4}+x_{3}\exp(x_{1})\log(x_{2})\]  
\[(0\leq x_{1}\leq 3,1\leq x_{2}\leq 3,3\leq x_{3}\leq 5,2\leq x_{4}\leq 3.5)\]  
and  
\[g_{3}(\mathbf{x})=x_{1}x_{2}x_{3}x_{4}+x_{3}\exp(x_{1})\log(x_{2})\]  
\[(0\leq x_{1}\leq 4,1\leq x_{2}\leq 3,3\leq x_{3}\leq 5,2.5\leq x_{4}\leq 5)\]  
We ignore the constant that makes the integral of \(f(\mathbf{x}),~{}g(\mathbf{x})\) equal to 1. The simulation results of group 3 are shown as Table 2.  
The window width \(h\) of TLF method in the group 3 is \(h=0.1\).