I NFORMATION C ASCADE Priyanka Garg. OUTLINE Information Propagation Virus Propagation Model How to model infection? Inferring Latent Social Networks.

INFORMATION CASCADEPriyanka Garg

OUTLINE

Information Propagation Virus Propagation Model How to model infection?

Inferring Latent Social Networks Inferring edge influence Inferring influence volume

INFORMATION PROPAGATION

How information/infection/influence flows in the network?

Epidemiology: Question: Will a virus take over the network? Type of virus:

Susceptible Infected Susceptible (SIS) Example: Flu

Susceptible Infected Removed (SIR) Example: Chicken-pox , deadly disease

Viral Marketing: Once a node is infected, it remains infected. Question: How to select a subset of persons such

that maximum number of persons can be influenced?

HOW TO MODEL INFECTION?

Simple model: Each infected node infects its neighbor with a

fixed probability. SIS:

A node infects its neighbor with probability b (how infectious is the virus?)

Node recovers with probability a (how easy is it to get cured?)

Strength of virus = b/a Result: If virus strength < t then virus will

instinct eventually. t = 1/largest eigen value of adjacency matrix A.

HOW TO MODEL INFECTION?

Independent Contagion Model Each infected node infects its neighboring node

with probability pij.

Threshold Model Each infected node i infect its neighboring node j

with weight wij.

The node j becomes active if ∑j=neigh(i)wij > thi.

thi is the threshold of node i.

HOW TO MODEL INFECTION?: GENERAL CONTAGION MODEL

General language to describe information diffusion.

Model: S infected nodes tried but failed to infect node v. New node u becomes infected. Probability of node u successfully influencing node v

also depends on S. pv(u, S)

Example Node becomes active if k of its neighbors are active.

ie. if |S + 1| > k then pv(u, S) = 1 else 0

Independent Cascade: pv(u,S) = p(u,v)

Threshold model: if (p(S,v) + p(u,v)) > t then pv(u,S) = 1 else 0

HOW TO MODEL INFECTION?: GENERAL CONTAGION MODEL

Can also model the diminishing returns property S>T then Gain(S + u) < Gain (T + u) Gain = Probability of infecting neighbor j

CHALLENGES IN USING THESE MODELS

Problem under consideration Viral marketing: How to select a subset of persons such

that maximum number of persons can be influenced?

How to find the infection probability/weights of every edge?

INFERRING INFECTION PROBABILITIES

We know the time of infections over a lots of cascades.

Train: Maximize the likelihood of node infections over

all the nodes in all the cascades. Likelihood = ∏c∏iPi,c

Pi = P(i gets infected at time ti| infected nodes)

Independent Contagion Model Pi=At least one of the already infected node

infects node i Pi= 1 - ∏j(1-(probability of infection from node j

to node i at time ti))

INFERRING INFECTION PROBABILITIES

Variability with time: Infection probabilities vary with time. Let w(t) is

the distribution which captures the variability with time.

Probability of node j infecting node i at time t is w(t-tj)*Aji. Here tj is the infection time of node j.

Thus: Pi= 1 - ∏j(1- w(ti-tj)Aji)

The log-likelihood maximization problem can be shown to be a convex optimization problem

ANOTHER APPROACH: MORE DIRECT

Find number of infected nodes at any time t?

Number of infected nodes at time t depends only on number of already infected nodes.

Model: V(t) is the number of nodes infected at time t

V(t+1) = ∑u=1,N ∑l=0,L-1 Mu(t-l) Iu(l+1) Mu(t) = 1 if node u is infected at time t Iu(t) = Infection variability with time

Minimize the difference between V(t) and observed volume at every time t.

Accounting for novelty: V(t+1) = α(t)∑u=1,N ∑l=0,L-1 Mu(t-l) Iu(l+1)

THANK YOU

SIS

Let pit = P(i is infected at time t)

tit = P(i doesn’t receive infection from its neighbor)

tit = ∏j=neigh(i) (pj(t-1) (1-b) + 1 – pj(t-1))

1-pit=P(i is healthy at t-1 and didn’t receive infection) + P(i is infected at t-1 and got recovered and didn’t receive infection) + P(i is not infected at t-1 but got cured after infection at t).

1 – pit = (1-pi(t-1)) tit + pi(t-1)a tit + (1-pi(t-1))tita 0.5