Are extreme value estimation methods useful for network data?

Abstract

Preferential attachment is an appealing edge generating mechanism for modeling social networks. It provides both an intuitive description of network growth and an explanation for the observed power laws in degree distributions. However, there are often difficulties fitting parametric network models to data due to either model error or data corruption. In this paper, we consider semi-parametric estimation based on an extreme value approach that begins by estimating tail indices of the power laws of in- and out-degree for the nodes of the network using nodes with large in- and out-degree. This method uses tail behavior of both the marginal and joint degree distributions. We compare the extreme value method with the existing parametric approaches and demonstrate how it can provide more robust estimates of parameters associated with the network when the data are corrupted or when the model is misspecified.

Appendices

Appendix A: Parameter estimation for linear PA model

Parameter estimation for the linear PA model was studied in Wan et al. (2017). If the complete history of the network evolution is available (i.e., timestamps of edge creation are known), then MLE estimates exist and are computable. On the other hand, if only a snapshot of the network is given at a single point in time (i.e., timestamp information for the creation of the edges is unavailable), an approximate MLE was proposed. This procedure combined elements of method of moments with an approximation to the likelihood. In the following we provide a brief summary of these two estimation methods. Asymptotic properties of these estimators can be found in Wan et al. (2017).

1.1 A.1 MLE

Given the full evolution of the network G(n), assuming the graph began with n₀ initial edges, the MLE estimator of 𝜃 = (α,β,γ,δ_in,δ_out),

$$ \widehat{\boldsymbol{\theta}}_{n}^{MLE}:=(\hat{\alpha}^{MLE}, \hat{\beta}^{MLE}, \hat{\gamma}^{MLE}, \hat{\delta}_{\text{in}}^{MLE},\hat{\delta}_{\text{out}}^{MLE}), $$

is obtained by setting

$$ \begin{array}{@{}rcl@{}} \hat{\alpha}^{MLE} &=& \frac{1}{n-n_{0}} \sum\limits_{t=n_{0}+1}^{n} \textbf{1}_{\{J_{t}=1\}},\\ \hat{\beta}^{MLE} &=& \frac{1}{n-n_{0}} \sum\limits_{t=n_{0}+1}^{n} \textbf{1}_{\{J_{t}=2\}},\\ \hat{\gamma}^{MLE} &=& 1- \hat{\alpha}^{MLE} - \hat{\beta}^{MLE}, \end{array} $$

and solving for $(\hat {\delta }_{\text {in}}^{MLE}, \hat {\delta }_{\text {out}}^{MLE})$ from

$$ \begin{array}{@{}rcl@{}} \sum\limits_{i=0}^{\infty} \frac{N^{\text{in}}_{>i}(n) - N^{\text{in}}_{>i}(n_{0})}{i+\hat{\delta}_{\text{in}}^{MLE}} &= \frac{n-n_{0}}{\hat{\delta}_{\text{in}}^{MLE}}\hat{\gamma}^{MLE} +\sum\limits_{t=n_{0}+1}^{n} \frac{N(t-1)}{t-1+\hat{\delta}_{\text{in}}^{MLE} N(t-1)}\textbf{1}_{\{J_{t}\in\lbrace 1, 2\rbrace\}},\\ \sum\limits_{j=0}^{\infty} \frac{N^{\text{out}}_{>j}(n) - N^{\text{out}}_{>j}(n_{0})}{j+\hat{\delta}_{\text{out}}^{MLE}} &= \frac{n-n_{0}}{\hat{\delta}_{\text{out}}^{MLE}}\hat{\alpha}^{MLE} +\sum\limits_{t=n_{0}+1}^{n} \frac{N(t-1)}{t-1+\hat{\delta}_{\text{out}}^{MLE} N(t-1)}\textbf{1}_{\{J_{t}\in\lbrace 2, 3\rbrace\}}, \end{array} $$

where

$$ N^{\text{in}}_{>i}(n) := \sum\limits_{i^{\prime}>i} N^{\text{in}}_{i^{\prime}}(n),\qquad N^{\text{out}}_{>j}(n) := \sum\limits_{j^{\prime}>j} N^{\text{out}}_{j^{\prime}}(n). $$

By Wan et al. (2017, Theorem 3.3), $\widehat {\boldsymbol {\theta }}_{n}^{MLE}$ is strongly consistent, asymptotically normal and efficient.

1.2 A.2 Snapshot

The estimation method for 𝜃 from the snapshot G(n) is summarized in the following 7-step procedure:

• 1. Estimate β by $\hat {\beta }^{SN}=1-N(n)/n$.

• 2. Obtain $\hat {\delta }_{\text {in}}^{0}$ by solving

  $$ \sum\limits_{i=1}^{\infty} \frac{N^{\text{in}}_{>i}(n)}{n}\frac{i}{i+\hat{\delta}_{\text{in}}^{0}}(1+\hat{\delta}_{\text{in}}^{0}(1-\hat{\beta}^{SN})) =\frac{\frac{N^{\text{in}}_{0}(n)}{n} + \hat{\beta}^{SN} }{1-\frac{N^{\text{in}}_{0}(n)}{n} \frac{\hat{\delta}_{\text{in}}^{0}}{1+(1-\hat{\beta}^{SN})\hat{\delta}_{\text{in}}^{0}}}, $$

  where $N^{\text {in}}_{0}(n)$ denotes the number of nodes with in-degree 0 in G(n).

• 3. Estimate α by

  $$ \hat\alpha^{0} = \frac{\frac{N^{\text{in}}_{0}(n)}{n} +\hat{\beta}^{SN}}{1-\frac{N^{\text{in}}_{0}(n)}{n} \frac{\hat{\delta}_{\text{in}}^{0}}{1+(1-\hat{\beta}^{SN})\hat{\delta}_{\text{in}}^{0}}} - \hat{\beta}^{SN}. $$

• 4. Obtain $\hat {\delta }_{\text {out}}^{0}$ by solving

  $$ \sum\limits_{j=1}^{\infty} \frac{N^{\text{out}}_{>j}(n)}{n}\frac{j}{j+\hat{\delta}_{\text{out}}^{0}}(1+\hat{\delta}_{\text{out}}^{0}(1-\hat{\beta}^{SN})) = \frac{\frac{N^{\text{out}}_{0}(n)}{n} + \hat{\beta}^{SN} }{1-\frac{N^{\text{out}}_{0}(n)}{n} \frac{\hat{\delta}_{\text{out}}^{0}}{1+(1-\hat{\beta}^{SN})\hat{\delta}_{\text{out}}^{0}}}, $$

  where $N^{\text {out}}_{0}(n)$ denotes the number of nodes with out-degree 0 in G(n).

• 5. Estimate γ by

  $$ \hat\gamma^{0} = \frac{\frac{N^{\text{out}}_{0}(n)}{n} + \hat{\beta}^{SN}}{1-\frac{N^{\text{out}}_{0}(n)}{n} \frac{\hat{\delta}_{\text{out}}^{0}}{1+(1-\hat{\beta}^{SN})\hat{\delta}_{\text{out}}^{0}}} - \hat{\beta}^{SN}. $$

• 6. Re-normalize the probabilities

  $$ (\hat\alpha^{SN},\hat{\beta}^{SN},\hat\gamma^{SN}) \leftarrow \left( \frac{\hat\alpha^{0}(1-\hat{\beta}^{SN})}{\hat\alpha^{0}+\hat\gamma^{0}},\hat{\beta}^{SN},\frac{\hat\gamma^{0}(1-\hat{\beta}^{SN})}{\hat\alpha^{0}+\hat\gamma^{0}}\right). $$

• 7. Solve for $\hat {\delta }_{\text {in}}^{SN}$ from

  $$ \sum\limits_{i=0}^{\infty} \frac{N^{\text{in}}_{>i}(n)/n}{i+\hat{\delta}_{\text{in}}^{SN}}-\frac{1-\hat\alpha^{SN}-\hat{\beta}^{SN}}{\hat{\delta}_{\text{in}}^{SN}} -\frac{(\hat\alpha^{SN}+\hat{\beta}^{SN})(1-\hat{\beta}^{SN})}{1+(1-\hat{\beta}^{SN})\hat{\delta}_{\text{in}}^{SN}} = 0. $$

  Similarly, solve for $\hat {\delta }_{\text {out}}^{SN}$ from

  $$ \sum\limits_{j=0}^{\infty} \frac{N^{\text{out}}_{>j}(n)/n}{j+\hat{\delta}_{\text{out}}^{SN}}-\frac{1-\hat\gamma^{SN}-\hat{\beta}^{SN}}{\hat{\delta}_{\text{out}}^{SN}} -\frac{(\hat\gamma^{SN}+\hat{\beta}^{SN})(1-\hat{\beta}^{SN})}{1+(1-\hat{\beta}^{SN})\hat{\delta}_{\text{out}}^{SN}} = 0. $$

Note that Step 6 ensures that

$$ \hat\alpha^{SN}+\hat{\beta}^{SN}+\hat\gamma^{SN}=1. $$

It is shown in Wan et al.(2017, Theorem 4.1) that $\widehat {\boldsymbol \theta }^{SN}_{n} := (\hat \alpha ^{SN},\hat \beta ^{SN},\hat \gamma ^{SN}, \hat {\delta }_{\text {in}}^{SN}, \hat {\delta }_{\text {out}}^{SN})\overset {\text {a.s.}}{\longrightarrow } \boldsymbol \theta$. Its asymptotic normality and efficiency are analyzed through simulation studies in the same paper.

Appendix B: Proof of theorem 2.1

Proof

We first prove the out-degree part of Theorem 2.1. Note that

$$ \begin{array}{@{}rcl@{}} \textbf{E}\left( N^{\text{out}}_{j}(n+1)| G(n)\right)& =& N^{\text{out}}_{j}(n)+\gamma\boldsymbol{1}_{\{j=0\}}+\alpha\boldsymbol{1}_{\{j=1\}}\\ &&+ (\beta+\gamma) \left( N^{\text{out}}_{j-1}(n)\frac{j-1+\delta_{\text{out}}}{n+\delta_{\text{out}}|V^{0}(n)|} - N^{\text{out}}_{j}(n)\frac{j+\delta_{\text{out}}}{n+\delta_{\text{out}}|V^{0}(n)|}\right).\\ \end{array} $$

(B.1)

Meanwhile, by the definition of V⁰(n), we have

$$ |V^0(n)|+1 = N(n) \sim \text{Binomial}(n, 1-\beta). $$

(B.2)

Applying the arguments in the proof of Theorem 3.1 of Bollobás et al. (2003), it follows that the out-degree distribution of a linear superstar model coincides with that of a standard linear preferential attachment network with parameters (α,β,γ,δ_in,δ_out). Moreover,

$$ \frac{N^{\text{out}}_{j}(n)}{n} \overset{\text{a.s.}}{\longrightarrow} q^{\text{out}}_{j}, \quad j>0,\quad n\to\infty, $$

where {qj out} := {pj out} is the limiting out-degree distribution of PA(α,β,γ,δ_in,δ_out). In particular,

$$ q^{\text{out}}_{j} \sim C_{\text{out}}^{\prime} j^{-(1+\iota_{\text{out}})}\qquad\text{ as}j\to\infty, $$

for $C_{\text {out}}^{\prime }$ positive and

$$ \iota_{\text{out}}^{-1} = \frac{\beta+\gamma}{1+\delta_{\text{out}}(\alpha+\gamma)}. $$

Next we consider the in-degree counts of non-superstar nodes. Observe also from the construction of the superstar model that

$$ |E^0(n)| \sim \text{Binomial}(n, 1-(\alpha+\beta)p). $$

(B.3)

Applying the Chernoff bound to both Eqs. B.2 and B.3 gives

$$ \begin{array}{@{}rcl@{}} \left|V^{0}(n)\right| &=& (1-\beta)n + O(n^{1/2}\log n),\\ \left|E^{0}(n)\right| &=& (1-(\alpha+\beta)p)n + O(n^{1/2}\log n). \end{array} $$

Taking expectation on both sides of Eq. B.1 then gives

$$ \begin{array}{@{}rcl@{}} &&\textbf{E} \left( (\alpha+\beta)(1-p)N^{\text{in}}_{i}(n)\frac{i+\delta_{\text{in}}}{|E^{0}(n)|+\delta_{\text{in}} |V^{0}(n)|}\right)\\ &&\quad= (\alpha+\beta)(1 - p) \frac{i+\delta_{\text{in}}}{n(1 - (\alpha+\beta)p)+\delta_{\text{in}} \cdot n(1 - \beta)} \textbf{E}(N^{\text{in}}_{i}(n)) + O(n^{-1/2}\log n).\\ \end{array} $$

(B.4)

By the rule of the superstar model, given G(n), $N^{\text {in}}_{i}(n)$ will increase by 1 if either scenario (1b) or (2b) happens and a node with $I_{n}^{(n)}(v) = i-1$ is chosen as the ending point of the edge. Also, it will decrease by 1 if either scenario (1b) or (2b) happens, but a node with $I_{n}^{(n)}(v) = i$ is chosen as the ending point of the edge. Moreover, with probability α a new node with in-degree 0 will be added to the graph, and with probability γ a new node with in-degree 1 is created in the next step. Hence, $\{N^{\text {in}}_{i}(n)\}_{n\ge 1}$ satisfies the following:

$$ \begin{array}{@{}rcl@{}} \textbf{E}\left( N^{\text{in}}_{i}(n+1)|G(n)\right) &= & N^{\text{in}}_{i}(n) + \alpha\boldsymbol{1}_{\{i=0\}}+\gamma\boldsymbol{1}_{\{i=1\}}\\ &&+ (\alpha+\beta)(1-p)N^{\text{in}}_{i-1}(n)\frac{i-1+\delta_{\text{in}}}{|E^{0}(n)|+\delta_{\text{in}}|V^{0}(n)|}\\ &&- (\alpha+\beta)(1-p)N^{\text{in}}_{i}(n)\frac{i+\delta_{\text{in}}}{|E^{0}(n)|+\delta_{\text{in}} |V^{0}(n)|}. \end{array} $$

Now let $q^{\text {in}}_{-1} = 0$, and define $\{q^{\text {in}}_{i}\}_{i\ge 0}$ by

$$ \begin{array}{@{}rcl@{}} q^{\text{in}}_i &=& \frac{(\alpha+\beta)(1-p)}{1-(\alpha+\beta)p+\delta_{\text{in}}(\alpha+\gamma)} \left( (i-1+\delta_{\text{in}})q^{\text{in}}_{i-1} - (i+\delta_{\text{in}})q^{\text{in}}_i\right)\\ &&+ \alpha\boldsymbol{1}_{\{i=0\}}+\gamma\boldsymbol{1}_{\{i=1\}}. \end{array} $$

(B.5)

According to the approximation in Eq. B.4, we use the same proof technique as in Bollobás et al. (2003, Theorem 3.1) to obtain

$$ \frac{N^{\text{in}}_{i}(n)}{n}\overset{\text{a.s.}}{\longrightarrow} q^{\text{in}}_{i},\qquad\text{as}\quad n\to\infty. $$

Also, solving the recursion in Eq. B.5 yields

$$ \begin{array}{@{}rcl@{}} q^{\text{in}}_{0} &=& \frac{\alpha}{1+\iota_{\text{in}}^{-1} \delta_{\text{in}}}, \end{array} $$

(B.6)

$$ \begin{array}{@{}rcl@{}} q^{\text{in}}_{1} &=& (1+\delta_{\text{in}}+\iota_{\text{in}})^{-1}\left( \frac{\alpha\delta_{\text{in}}}{1+\iota_{\text{in}}^{-1}\delta_{\text{in}}}+\frac{\gamma}{\iota_{\text{in}}^{-1}}\right),\\ q^{\text{in}}_{i} &=& \frac{{\Gamma}(i+\delta_{\text{in}})}{{\Gamma}(i+\delta_{\text{in}}+\iota_{\text{in}}+1)} \frac{{\Gamma}(2+\delta_{\text{in}}+\iota_{\text{in}})}{{\Gamma}(1+\delta_{\text{in}})} q^{\text{in}}_{1}, \quad i\ge2, \end{array} $$

(B.7)

where

$$ \iota_{\text{in}}^{-1} := \frac{(\alpha+\beta)(1-p)}{1-(\alpha+\beta)p+\delta_{\text{in}}(\alpha+\gamma)}. $$

Therefore, applying Stirling’s approximation to Eqs. B.6–B.7 gives

$$ q^{\text{in}}_{i} \sim C^{\prime}_{\text{in}} i^{-(1+\iota_{\text{in}})}, \qquad \text{as}n\to\infty, $$

for some positive constant $C^{\prime }_{\text {in}}$. This completes the proof. □