Constrained expectation-maximisation for inference of social graphs explaining online user–user interactions

Abstract

Current network inference algorithms fail to generate graphs with edges that can explain whole sequences of node interactions in a given dataset or trace. To quantify how well an inferred graph can explain a trace, we introduce feasibility, a novel quality criterion, and suggest that it is linked to the result’s accuracy. In addition, we propose CEM-*, a network inference method that guarantees 100% feasibility given online social media traces, which are a non-trivial extension of the Expectation-Maximization algorithm developed by Newman (Nature Phys 14:67–75, 2018). We propose a set of linear optimization updates that incorporate a set of auxiliary variables and a set of feasibility constraints; the latter takes into consideration all the hidden paths that are possible between users based on their timestamps of interaction and guides the inference toward feasibility. We provide two CEM-* variations, that assume either an Erdős–Rényi (ER) or a Stochastic Block Model (SBM) prior for the underlying graph’s unknown distribution. Extensive experiments on one synthetic and one real-world Twitter dataset show that for both priors CEM-* can generate a posterior distribution of graphs that explains the whole trace while being closer to the ground truth. As an additional benefit, the use of the SBM prior infers and clusters users simultaneously during optimization. CEM-* outperforms baseline and state-of-the-art methods in terms of feasibility, run-time, and precision of the inferred graph and communities. Finally, we propose a heuristic to adapt the inference to lower feasibility requirements and show how it can affect the precision of the result.

Appendices

Appendix A: CEM-er

For the E-step, we modify the Newman algorithm by taking the expectation over the set of random variables $Y_{ij}$ at both sides of (10):

$$\begin{aligned}&{\mathbb {E}}[\log {P}(\theta \text { }|\text { } \mathcal {T})] \ge {\mathbb {E}} [\sum _{{{\textbf {A}}}} q({{\textbf {A}}}) \log \frac{{P}({{\textbf {A}}}, \theta \text { }|\text { } \mathcal {T})}{q({{\textbf {A}}}}] \nonumber \\&\quad = \sum _{{{\textbf {A}}}} q({{\textbf {A}}})\big ({\mathbb {E}}[\log {P}({{\textbf {A}}}, \theta \text { }|\text { } \mathcal {T})] - \log q({{\textbf {A}}} \big )). \end{aligned}$$

(28)

To find ${\mathbb {E}}[\log {P}({{\textbf {A}}},\theta \text { }|\text { }\mathcal {T})]$, we replace (9) into (8). Setting $\Gamma ={P}(\theta )/{P}(\mathcal {T})$, the expectation of the log of (8) becomes:

$$\begin{aligned}&{\mathbb {E}}[\log {P}({{\textbf {A}}},\theta \text { }|\text { }\mathcal {T})] = log \Gamma + \sum _{i \ne j} \Big [{A_{ij}} \Big (\log \rho + {{\mathbb {E}}[Y_{ij}]}\log \alpha \nonumber \\&\quad +(M_{ij}-{\mathbb {E}}[Y_{ij}])\log {(1 - \alpha )}\Big ) \nonumber +(1-A_{ij}\Big (\log (1-\rho ) + \nonumber \\&\quad + {{\mathbb {E}}[Y_{ij}]}\log \beta +(M_{ij} - {\mathbb {E}}[Y_{ij}])\log {(1 - \beta )\Big )\Big ]}. \end{aligned}$$

(29)

Then, by replacing (7) into (29), and then (29) into (28), we get:

$$\begin{aligned}&{\mathbb {E}}[\log {P}(\theta \text { }|\text { } \mathcal {T})] \ge \sum _{{{\textbf {A}}}} q({{\textbf {A}}})\log \frac{D_{ij}}{q({{\textbf {A}}})} \end{aligned}$$

(30)

$$\begin{aligned}&\text {where, } D_{ij} = \Gamma \prod _{i \ne j}{\left[ \rho \alpha ^{M_{ij}\sigma _{ij}}{(1 - \alpha )}^{M_{ij}(1-\sigma _{ij})}\right] }^{A_{ij}} \nonumber \\&\quad \times {\left[ (1-\rho )\beta ^{M_{ij}\sigma _{ij}}{(1 - \beta )}^{M_{ij}(1-\sigma _{ij})}\right] }^{1-A_{ij} }. \end{aligned}$$

(31)

For the M-step of the EM algorithm, the function that we want to maximize is ${\mathbb {E}}[\log {P}(\theta \text { }|\text { } \mathcal {T})]$. To do so, we need to find the unknown values, $q({{\textbf {A}}})$ and $\theta =${$\alpha , \beta , \rho , \varvec{\sigma }$}, that maximize the expectation on the left-hand side of (11), under the feasibility constraints on the parameters set $\theta$. From these, only the $\sigma _{ij}$ have an important constraint set, specified in (5) and (6).

Solution with respect to ${\varvec{q}}({{\textbf {A}}})$. We notice that the choice of $q({{\textbf {A}}})$ that achieves equality (i.e. maximizes the right-hand side) in (11) is:

$$\begin{aligned} q({{\textbf {A}}}) = \dfrac{D_{ij}}{\sum _{{{\textbf {A}}}}D_{ij}}, \end{aligned}$$

(32)

which leads us to Eq. (13).

Solution with respect to $\varvec{\alpha }, \varvec{\beta }, \varvec{\rho }$. To maximize the right-hand side of (11) in terms of parameter $\alpha$ we differentiate it with respect to $\alpha$ and then setting it equal to zero (while holding $\sigma _{ij}$, q constant):

$$\begin{aligned} \sum _{i \ne j} Q_{ij} M_{ij} \left( \frac{\sigma _{ij}}{\alpha } - \frac{1-\sigma _{ij}}{1-\alpha }\right) = 0. \end{aligned}$$

(33)

After rearranging, we get the updates shown in Eq. (15), and we repeat likewise for $\beta$ and $\rho$.

Solution with respect to $\varvec{\sigma }_{ij}$. If we take into account that $Q_{ij} = \sum _{{{\textbf {A}}}}q({{\textbf {A}}})A_{ij}$ and also that $\sum _{{{\textbf {A}}}} q({{\textbf {A}}}) = 1$, by rearranging the right-hand side of (11), the problem becomes equivalent to maximizing:

$$\begin{aligned}&\sum _{{{\textbf {A}}}} q({{\textbf {A}}}) \sum _{i \ne j} \sigma _{ij} M_{ij}\left( A_{ij}\log \frac{\alpha }{1-\alpha } + (1-A_{ij})\log \frac{\beta }{1-\beta }\right) \ \nonumber \\&\quad = \sum _{i \ne j} \sigma _{ij} M_{ij}\left( Q_{ij}\log \frac{\alpha }{1-\alpha }+ (1-Q_{ij})\log \frac{\beta }{1-\beta }\right) . \end{aligned}$$

(34)

This leads us to the constrained optimization problem of Eq. (17).

Appendix B: CEM-sbm

For the E-step of the EM algorithm, we modify the Newman algorithm by taking the expectation over the set of random variables $Y_{ij}$ at both sides of (20):

$$\begin{aligned}&{\mathbb {E}}[\log {P}(\theta \text { }| \text { } \mathcal {T})] \ge {\mathbb {E}} [\sum _{{{\textbf {A}}}} q({{\textbf {A}}}, {{\textbf {g}}}) \log \frac{{P}({{\textbf {A}}}, {{\textbf {g}}}, \theta \text { }|\text { }\mathcal {T})}{q({{\textbf {A}}}, {{\textbf {g}}})}] \nonumber \\&\quad = \sum _{{{\textbf {A}}}} q({{\textbf {A}}}, {{\textbf {g}}})\big ({\mathbb {E}}[\log {P}({{\textbf {A}}}, {{\textbf {g}}}, \theta \text { }|\text { }\mathcal {T})] - \log q({{\textbf {A}}}, {{\textbf {g}}} \big )). \end{aligned}$$

(35)

To find ${\mathbb {E}}[\log {P}({{\textbf {A}}}, {{\textbf {g}}}, \theta \text { }| \text { } \mathcal {T})]$, we replace (19) into (18). Setting $\Gamma ={P}(\theta )/{P}(\mathcal {T})$, the expectation of the log of (18) becomes:

$$\begin{aligned}{} & {} {\mathbb {E}}[\log {P}({{\textbf {A}}}, {{\textbf {g}}}, \theta \text {} \text { }|\text { } \text {} \mathcal {T})] = log \Gamma + \sum _{\begin{array}{c} {i \ne j}\\ {g_{i} = g{j}} \end{array}} \Big [{A_{ij}} \Big (\log p + {{\mathbb {E}}[Y_{ij}]}\log \alpha \nonumber \\{} & {} \quad +(M_{ij}-{\mathbb {E}}[Y_{ij}])\log {(1 - \alpha )}\Big ) \nonumber +(1-A_{ij})\Big (\log (1-p) \nonumber \\{} & {} \quad + {{\mathbb {E}}[Y_{ij}]}\log \beta +(M_{ij} - {\mathbb {E}}[Y_{ij}])\log {(1 - \beta )\Big )\Big ]} \nonumber \\{} & {} \quad + \sum _{\begin{array}{c} {i \ne j}\\ {g_{i} \ne g{j}} \end{array}} \Big [{A_{ij}} \Big (\log q \nonumber \\{} & {} \quad + {{\mathbb {E}}[Y_{ij}]}\log \alpha +(M_{ij}-{\mathbb {E}}[Y_{ij}])\log {(1 - \alpha )}\Big ) \nonumber \\{} & {} \quad +(1-A_{ij})\Big (\log (1-q) \nonumber \\{} & {} \quad + {{\mathbb {E}}[Y_{ij}]}\log \beta +(M_{ij} - {\mathbb {E}}[Y_{ij}])\log {(1 - \beta )\Big )\Big ]}. \end{aligned}$$

(36)

By replacing (7) into (36), and then (36) into (35), we get:

$$\begin{aligned} {\mathbb {E}}[\log {P}(\theta \text { }|\text { } \mathcal {T})] \ge \sum _{{{\textbf {A}}}} q({{\textbf {A}}}, {{\textbf {g}}})\log \frac{D({{\textbf {A}}}, {{\textbf {g}}})}{q({{\textbf {A}}}, {{\textbf {g}}})}, \end{aligned}$$

(37)

where,

$$\begin{aligned}&{ D({{\textbf {A}}}, {{\textbf {g}}})= \Gamma \prod _{\begin{array}{c} {i \ne j}\\ {g_{i} = g{j}} \end{array}}{\left[ p \alpha ^{M_{ij}\sigma _{ij}}{(1 - \alpha )}^{M_{ij}(1-\sigma _{ij})}\right] }^{A_{ij}} } \nonumber \\&{\left[ (1-p)\beta ^{M_{ij}\sigma _{ij}}{(1 - \beta )}^{M_{ij}(1-\sigma _{ij})}\right] }^{1-A_{ij} }\nonumber \\&\quad \prod _{\begin{array}{c} {i \ne j} \\ {g_{i} \ne g{j}} \end{array}}{\left[ q \alpha ^{M_{ij}\sigma _{ij}}{(1 - \alpha )}^{M_{ij}(1-\sigma _{ij})}\right] }^{A_{ij}} \nonumber \\&{\left[ (1-q)\beta ^{M_{ij}\sigma _{ij}}{(1 - \beta )}^{M_{ij}(1-\sigma _{ij})}\right] }^{1-A_{ij} }. \end{aligned}$$

(38)

For the M-step of EM, we maximize the expectation ${\mathbb {E}}[\log {P}(\theta \text { } |\text { } \mathcal {T})]$ as we did in the CEM-er prior.

Solution with respect to ${\varvec{q}}({{\textbf {A}}}, {{\textbf {g}}}$). We notice that the choice of $q({{\textbf {A}}}, {{\textbf {g}}})$ that achieves equality (i.e. maximizes the right-hand side) in (37) is:

$$\begin{aligned} q({{\textbf {A}}}, {{\textbf {g}}}) = \dfrac{D({{\textbf {A}}}, {{\textbf {g}}})}{\sum _{{{\textbf {A}}}}D({{\textbf {A}}}, {{\textbf {g}}})}. \end{aligned}$$

(39)

From (39), in a similar fashion to Newman’s method [Eq. (13), 20], and because $\Gamma$ cancels out, we get:

$$\begin{aligned}&q({{\textbf {A}}}, {{\textbf {g}}}) = \prod _{i \ne j, (g_{i} = g_{j})}Q_{ij}(g_{i},g_{j})^{A_{ij}}(1-Q_{ij}(g_{i},g_{j}))^{1-A_{ij}} \nonumber \\&\quad \prod _{i \ne j, (g_{i} \ne g_{j})}Q_{ij}(g_{i},g_{j})^{A_{ij}}(1-Q_{ij}(g_{i},g_{j}))^{1-A_{ij}}. \end{aligned}$$

(40)

Hence, given Eq. (38), the values of $Q_{ij}$ are found to be the ones in Eq. (21) and (22).

Our goal is to find the unknown parameters $\theta =${$\alpha , \beta , p, q, \varvec{\sigma }$} that maximize the right-hand size of (37), given the maximising distribution for $q({{\textbf {A}}}, {{\textbf {g}}})$ in (39), hence given the values of $Q_{ij}(g_{i},g_{j})$ in (40).

Solution with respect to $\varvec{\alpha }, \varvec{\beta }, {\varvec{p}}, {\varvec{q}}$. To maximize the right-hand side of (37) in terms of parameter $\alpha$, we differentiate the equation with respect to $\alpha$ and we set it equal to zero (while holding the rest of the parameters $\theta$ constant):

$$\begin{aligned} \sum _{i \ne j} Q_{ij}(g_{i},g_{j}) M_{ij} \left( \frac{\sigma _{ij}}{\alpha } - \frac{1-\sigma _{ij}}{1-\alpha }\right) = 0. \end{aligned}$$

(41)

After rearranging, we get the value in Eq. (23). By repeating the same procedure for $\beta$, we get Eq. (24). Likewise, differentiating the r.h.s. of (37) with respect to p and then setting it equal to zero we get:

$$\begin{aligned} \sum _{{{\textbf {A}}}} q({\textbf {A, g}}) \sum _ {\begin{array}{c} {i \ne j}\\ {g_{i} = g{j}} \end{array}} \left(\frac{A_{ij}}{p} - \frac{1-A_{ij}}{1-p} \right) = 0. \end{aligned}$$

(42)

This is how we get the updates for p in Eq. (25), and, likewise, for q in Eq. (26).

Solution with respect to $\varvec{\sigma }_{ij}$. If we take into account that $Q_{ij}(g_{i},g_{j}) =\sum _{{{\textbf {A}}}}q({{\textbf {A}}}, {{\textbf {g}}})A_{ij}$ and also that $\sum _{{{\textbf {A}}}} q({{\textbf {A}}}, {{\textbf {g}}}) = 1$, by rearranging the right-hand side of (37), the problem becomes equivalent to maximizing:

$$\begin{aligned}&\sum _{{{\textbf {A}}}} q({{\textbf {A}}}, {{\textbf {g}}}) \sum _{i \ne j} \sigma _{ij} M_{ij}\left( A_{ij}\log \frac{\alpha }{1-\alpha } \right. \nonumber \\&\qquad \left. + (1-A_{ij})\log \frac{\beta }{1-\beta }\right) \nonumber \\&\quad = \sum _{i \ne j} \sigma _{ij} M_{ij}\left( Q_{ij}(g_{i},g_{j})\log \frac{\alpha }{1-\alpha }\right. \nonumber \\&\qquad \left. + (1-Q_{ij}(g_{i},g_{j}))\log \frac{\beta }{1-\beta }\right) . \end{aligned}$$

(43)

This leads us to the optimization problem of Eq. (27) through which we can find the $\sigma _{ij}$ values.