Influence learning for cascade diffusion models: focus on partial orders of infections

Abstract

Probabilistic cascade models consider information diffusion as an iterative process in which information transits between users of a network. The problem of diffusion modeling then comes down to learning transmission probability distributions, depending on hidden influence relationships between users, in order to discover the main diffusion channels of the network. Various learning models have been proposed in the literature, but we argue that the diffusion mechanisms defined in most of these models are not well-adapted to deal with noisy diffusion events observed from real social networks, where transmissions of content occur between humans. Classical models usually have some difficulties for extracting the main regularities in such real-world settings. In this paper, we propose a relaxed learning process of the well-known independent cascade model that, rather than attempting to explain exact timestamps of users’ infections, focus on infection probabilities knowing sets of previously infected users. Furthermore, we propose a regularized learning scheme that allows the model to extract more generalizable transmission probabilities from training social data. Experiments show the effectiveness of our proposals, by considering the learned models for real-world prediction tasks.

Appendix

1.1 Proof of Proposition 1

Let us denote $\theta _{u,v}^{(i)}$ the transmission probability from user u to user v at the i-th iteration of the learning process. Let also denote $P^{D^{(i)}}_{v}$ the estimation of the infection probability of v in the episode D (computed using formula 1 using current transmission probabilities) at the i-th iteration of the learning process.

First, with $A_{u,v}=\frac{|D^+_{u,v}|}{|D^+_{u,v}|+|D^-_{u,v}|}$, let us consider the following Lemma:

Lemma 1

$$\begin{aligned} \forall i \in \mathbb {N}, \quad \forall I_{u,v} \in \mathscr {I} : \left( \theta ^{(i)}_{u,v} \le A_{u,v} \right) \end{aligned}$$

Proof

Lemma 1 can be easily deduced from the update formula applied at each step of the learning process (Eq. 7), since we know from (Eq. 1) that $\frac{\theta ^{(i)}_{u,v}}{P_v^{D^{(i)}}}\le 1$ for all $I_{u,v} \in \mathscr {I}$ at every iteration $i>0$ of the process. Note that, without loss of generality, for getting the lemma valid for $i=0$, we assume that the probabilities $\theta$ are all initialized such that for all $I_{u,v} \in \mathscr {I}: \theta ^{(0)}_{u,v} \in [0,A_{u,v}]$. $\square$

Let’s now consider the following lemma:

Lemma 2

$$\begin{aligned} \forall I_{u,v} \in \mathscr {I} : (|\mathscr {D}_{u,v}^-|=0 \implies \forall i \in \mathbb {N} : (\theta ^{(i+1)}_{u,v} \ge \theta ^{(i)}_{u,v})) \end{aligned}$$

Proof

If $|\mathscr {D}^-_{u,v}|=0$, we get, from formula 7:

$$\begin{aligned} \frac{\theta ^{(i+1)}_{u,v}}{\theta ^{(i)}_{u,v}} = \frac{1}{|\mathscr {D}^+_{u,v}|} {\sum _{D\in \mathscr {D}^+_{u,v}} \frac{1}{P_j^{D^{(i)}}} } \ge \frac{1}{|\mathscr {D}^+_{u,v}|} {\sum _{D\in \mathscr {D}^+_{u,v}} 1} = 1 \end{aligned}$$

where we used the fact that $P_j^{D^{(i)}}$ is included in ]0; 1[. $\square$

For simplicity, let us now state $I_v^D =(U^D_{v} \cap Preds_v)$. For every episode $D \in \mathscr {D}$ and every user $v \in U^D_{\infty }$, we have at any iteration i of the process:

$$\begin{aligned} P_v^{D^{(i)}}= & {} 1- \prod _{u \in I_v^D} (1-\theta ^{(i)}_{u,v}) \\= & {} 1- \prod _{u \in I_v^D, |\mathscr {D}_{u,v}^-|>0} (1-\theta ^{(i)}_{u,v}) \prod _{u \in I_v^D, |\mathscr {D}_{u,v}^-|=0} (1-\theta ^{(i)}_{u,v}) \\\le & {} 1- \prod _{u \in I_v^D, |\mathscr {D}_{u,v}^-|>0} (1-A_{u,v}) \prod _{u \in I_v^D, |\mathscr {D}_{u,v}^-|=0} (1-\theta ^{(i)}_{u,v}) \end{aligned}$$

Let state $B_v^D = \prod \limits _{u \in I_v^D, |\mathscr {D}_{u,v}^-|>0} (1-A_{u,v})$. Note that $B_v^D$ is a constant over the whole learning process. Now, let’s consider the case of the proposition, where it exists at least one user $u \in I_j^D$ such that $|\mathscr {D}_{u,v}^-|=0$. In that case, we can rewrite the inequality as :

$$\begin{aligned} P_v^{D^{(i)}}\le & {} 1 - B_v^D (1-\theta ^{(i)}_{u,v}) \prod _{u' \in I_v^D\setminus \{u\}, |\mathscr {D}_{u',v}^-|=0} (1-\theta ^{(i)}_{u',v}) \nonumber \\\le & {} 1 - B_v^D (1-\theta ^{(i)}_{u,v}) \left(1 - \max \limits _{u' \in I_v^D\setminus \{u\}, |\mathscr {D}_{u',v}^-|=0} \theta ^{(i)}_{u',v}\right)^{|\{u' \in I_v^D\setminus \{u\}, |\mathscr {D}_{u',v}^-|=0\}|} \end{aligned}$$

(19)

Now, let us consider the sequence V defined as:

$$\begin{aligned} V_n=\left(1 - \max \limits _{u' \in I_v^D\setminus \{u\}, |\mathscr {D}_{u',v}^-|=0} \theta ^{(n)}_{u',v}\right)^{|\{u' \in I_v^D\setminus \{u\}, |\mathscr {D}_{u',v}^-|=0\}|} \end{aligned}$$

From Lemma 2, we know that V is decreasing, since any component of the max function does not own any counter-example in the training set. Moreover, This sequence is lower-bounded by 0. Then, V converges toward its fixed point, which we denote as l. From this, two possibilities: either l equals 0 or is strictly ${>}0$.

If $l=0$, then we know that:

$$\begin{aligned} \lim \limits _{n \rightarrow \infty } \max \limits _{u' \in I_v^D\setminus \{u\}, |\mathscr {D}_{u',v}^-|=0} \theta ^{(n)}_{u',v} = 1 \end{aligned}$$

Now, the formula 1 leads to know that, at every iteration i, $\forall u' \in I_v^D: P_v^{D^{(i)}} \ge \theta ^{(i)}_{u',v}$. Therefore, at every iteration i, we have: $P_v^{D^{(i)}} \ge \max \limits _{u' \in I_v^D\setminus \{u\}, |\mathscr {D}_{u',v}^-|=0} \theta ^{(i)}_{u',v}$. Since we know that $P_v^{D^{(i)}}$ is also upper-bounded by 1 at every iteration i, we can state that, in that case, $\lim \limits _{n \rightarrow \infty } P_v^{D^{(n)}}=1$.

Else, we have at every iteration i:

$$\begin{aligned} \left(1-\max \limits _{u' \in I_v^D\setminus \{u\}, |\mathscr {D}_{u',v}^-|=0} \theta _{u',v}^{(i)}\right)^{|\{u' \in I_v^D\setminus \{u\}, |\mathscr {D}_{u',v}^-|=0\}|} \ge l \end{aligned}$$

Plugging this in inequality (19), we get for every i:

$$\begin{aligned} P_v^{D^{(i)}} \le 1- l B_j^D (1-\theta ^{(i)}_{u,v}) \le 1- \lambda + \lambda \theta ^{(i)}_{u,v} \end{aligned}$$

with $\lambda = l B_v^D$. Then, we can rewrite the update formula 7 as:

$$\begin{aligned} \theta ^{(i+1)}_{u,v} = \frac{\sum _{D'\in \mathscr {D}^+_{u,v} \setminus D} \frac{\theta ^{(i)}_{u,v}}{P_v^{D'^{(i)}}} + \frac{\theta ^{(i)}_{u,v}}{P_v^{D^{(n)}}}}{|\mathscr {D}^+_{u,v}|} \ge \frac{(|\mathscr {D}^+_{u,v}|-1) \theta ^{(i)}_{u,v} + \frac{\theta ^{(i)}_{u,v}}{1- \lambda + \lambda \theta _{u,v}}}{|\mathscr {D}^+_{u,v}|} \end{aligned}$$

(20)

Let us consider now the sequence W such that:

$$\begin{aligned} {\left\{ \begin{array}{l} W_0 = \theta _{u,v}^{(0)} \\ W_{n+1} = \frac{(|\mathscr {D}^+_{u,v}|-1) W_{n} + \frac{W_{n}}{1-\lambda + \lambda W_{n}}}{|\mathscr {D}^+_{u,v}|} \end{array}\right. } \end{aligned}$$

Then, since W takes its values in ]0; 1[, and that $\lambda$ is also in ]0; 1[, we can state that:

$$\begin{aligned} \frac{W_{n+1}}{W_{n}} = \frac{|\mathscr {D}^+_{u,v}|-1+\frac{1}{1-\lambda + \lambda W_{n}}}{|\mathscr {D}^+_{u,v}|} > 1 \end{aligned}$$

The sequence is thus strictly increasing. Since it is upper-bounded by its fixed point 1, we know that it converges to 1. Now, since we know that, from inequality (20), $\forall n : \theta _{u,v}^{(n)} \ge W_n$, we can get that $\lim \limits _{n \rightarrow \infty }\theta ^{(n)}_{u,v}=1$. This concludes the proof since therefore: $\lim \limits _{n \rightarrow \infty } P_v^{D^{(n)}}=1$.

1.2 Proof of Proposition 2

If, for a given relationship $I_{u,v} \in \mathscr {I}$ such that $|\mathscr {D}^-_{u,v}|>0$, it exists in each $D \in \mathscr {D}^+_{u,v}$ at least one user $u' \in U^D_v \cap Preds_v$ such that $|\mathscr {D}^-_{u',v}|=0$, we can deduce from Proposition 1 that:

$$\begin{aligned} \forall D \in \mathscr {D}^+_{u,v} : \lim _{n \rightarrow +\infty } P^{D^{(n)}}_v = 1 \end{aligned}$$

In that case, we can state that, after a given iteration m, it exists a value $x \in ]A_{u,v};1[$ such that $\forall D \in \mathscr {D}^+_{u,v}: P^{D^{(n)}}_v > x$. Then, we know that: $\forall n>m, \theta ^{(n+1)}_{u,v} < \theta ^{(n)}_{u,v}\frac{A_{u,v}}{x} = \gamma \theta ^{(n)}_{u,v}$, with $\gamma = \frac{A_{u,v}}{x}$. Note that $\gamma \in ]0;1[$ since $x>A_{u,v}$. Let us consider now the following sequence V:

$$\begin{aligned} {\left\{ \begin{array}{ll} V_0 = \theta _{u,v}^{(0)} &{} \\ V_{n+1} = \gamma V_n \end{array}\right. } \end{aligned}$$

This sequence converges to its unique fixed point 0 since $\gamma \in ]0;1[$. Since we know that: $\forall n>m, \theta ^{(n)}_{u,v} \le V_n$ and that $\theta ^{(n)}_{u,v}$ is lower-bounded by 0, then we get: $\lim \limits _{n \rightarrow +\infty } \theta ^{(n)}_{u,v} = 0$.

1.3 Proof of Proposition 3

Proving that the solution given by (14), denoted hereafter $\theta ^*_{u,v}$, is nonnegative is straightforward. Inequality $\theta ^*_{u,v} \ge 0$ can indeed be transformed into $\upbeta \ge \sqrt{\varDelta }$ whose both sides are nonnegative terms and which can thus be verified by considering its square: as $\varDelta -\upbeta ^2 = - 4\lambda \gamma \le 0$, $\upbeta ^2 \ge \varDelta$ is always true.

Proving that $\theta ^*_{u,v} \le 1$ requires showing that $\upbeta - \sqrt{\varDelta } \le 2\lambda$, which is equivalent to $\upbeta -2\lambda \le \sqrt{\varDelta }$. If $\lambda \ge (|\mathscr {D}_{u,v}^-|+|\mathscr {D}_{u,v}^+|)$, the verification of the latter is direct since in that case $\upbeta -2\lambda \le 0$ (and we know that $\sqrt{\varDelta } \ge 0$). In the opposite case, both sides of the inequality are nonnegative. It is then possible to consider the square of the inequality: $(\upbeta - 2\lambda )^2 \le \varDelta$ is equivalent to $|\mathscr {D}_{u,v}^-|+|\mathscr {D}_{u,v}^+|-\gamma \ge 0$, that is always true since we know that $|\mathscr {D}_{u,v}^+| \ge \gamma$. Then, $\theta ^*_{u,v}$ always lies in [0, 1].

Proving that the solution given by (14) can be used as an update rule at each maximization step for solving the estimator of formula (12) implies to show that it maximizes, for any pair (u, v), the quantity $Q=\mathscr {Q}(\theta |\hat{\theta }) - \lambda \sum _{\theta _{u,v} \in \theta } \theta _{u,v}$. Since we already know that $\theta ^*_{u,v}$ corresponds to one of the two possible solutions of the cancellation of the derivative of Q from Eq. (13), it suffices to show that it corresponds to a maximum. This can be easily verified by considering the second derivative of Q w.r.t. $\theta _{u,v}$, which equals:

$$\begin{aligned} \frac{\partial Q}{\partial \theta _{u,v}^2}=-\sum \limits _{D \in \mathscr {D}_{u,v}^+} \left(\frac{\hat{\theta }^D_{u\rightarrow v}}{\theta _{u,v}^2} + \frac{(1-\hat{\theta }^D_{u\rightarrow v})}{(1-\theta _{u,v})^2}\right) - \sum \limits _{D \in \mathscr {D}_{u,v}^-} \frac{1}{(1-\theta _{u,v})^2} \end{aligned}$$

where $\hat{\theta }^D_{u\rightarrow v}$ is a shortcut for $\frac{\hat{\theta }_{u,v}}{\hat{P}^D_v}$. From this formulation, it is easy to see that the second derivative of Q w.r.t. $\theta _{u,v}$ is always negative on ]0; 1[, which concludes the proof: taking $\theta ^*_{u,v}$ as an update of $\theta _{u,v}$ allows us to maximize Q at each step of the EM algorithm.