CS 284r: Incentives and Information in Networks Fall 2013 Lecture 13 — October 21, 2013 Prof. Yaron Singer Scribe: Bo Waggoner

1 Overview

We now have some nice models of influence and algorithms for maximizing influence in a network, but we need to confront some challenges with implementing these models in the real world. Namely, the problem of maximizing influence when we only have access to a sample of the graph.

2 Motivation

Suppose we work at, say, the Gap. We can think of the users who enter our store as a random sample of some large . Suppose we want to maximize influence on this network; then we need to decide whether this sample will be useful as our “seed set”. Question: What is the liklihood of seeing an influential user in our store?

−α Well, think of a power law distribution: P[d(v) ≥ d] ≈ d , for α ≈ 2. For example, the probability √ 1 of seeing a node with d(n) ≥ n is n . Answer: It’s tiny! A small random sample of users is very unlikely to contain high-degree nodes.

3 The Friendship Paradox

The friendship paradox informally states, “Your friends have more friends than you.”

Theorem 1 (Feld 91). : In any graph the average degree of neighbors is µ + σ2/µ. Here µ is expected degree of a node and σ2 is variance.

Proof. Let G = (V,E) and, for shorthand, number the vertices V = {1, . . . , n} and let di be the degree of vertex i. We want the quantity E [di] where i is selected by picking a random edge, then picking a random endpoint of that edge. This is n   X di d . 2|E| i i=1

di 1 (i is selected with probability 2|E| because each of i’s edges is selected with probability |E| and i is 1 selected as the endpoint of that edge with probability 2 .)

1 We can rewrite the sum (note the denominator is the same because each edge is counted twice): Pn 2 i=1 di Pn . i=1 di Pn i=1 di Let µ = n , the average degree. Add and subtract the mean: Pn 2 Pn i=1 di i=1 di Pn + µ − i=1 di n Pn d2 i=1 i − µ2 = µ + n µ σ2 = µ + . µ 2 2 2 (Recall that σ = E[di ] − (E[di]) .)

This may be a useful fact for us, but there’s a problem: The fact that this holds on average doesn’t mean that it holds “with constant probability” over, for instance, our sample (which is what we want!). Example: Let’s construct a graph as follows. Sample a value d from a power law distribution, then create a (fully connected component) of size d + 1 and add it to your graph. Every vertex in this clique has degree d. Repeat as long as necessary. (We can also connect the to each other with a few extra edges.) This graph has a power law , but the ratio between an average degree of a node and its neighbor is 1! To deal with this sort of bad example, let’s recall the configuration model of random graphs. We start with all of our nodes. For each one, draw a degree from our preferred distribution and create that many “stub” edges poking out of that node. Then randomly and uniformly connect all the stubs. (We may get some occasional self loops, but let’s not worry about them!)

There’s a minor issue: how do we adapt power laws to a finite graph? Easy: Just fix P[d(v) = 1 ∆] = N , where ∆ is the highest degree of a node. Theorem 2. : For random finite power law graphs in the configuration model with 1 < α < 2,  α−1 − ∀ > 0 the ratio between the degree of a random node and its random neighbor is Ω N α .

Proof: The number of nodes of degree d is c · d−α (for some constant c), by definition of power law graphs.

−α 1 What is the highest degree? c∆ = 1 ⇐⇒ ∆ = c α . P∆ −α P∞ −α What is N (the number of nodes)? N = i=1 c · i = c i=1 i = Θ(c). 1 P∆ −α c Pc1/α 1−α Pt −β What is the number of edges? M = 2 i=1 i · c · i = 2 i=1 i . Use the fact that i=1 i = 1−β  2−α  2/α Θ(t ) for β > 1. So we get Θ c · c α = Θ(c ).

 2/α  2−α M c α Average degree of a node is N = Θ c = Θ(c ). So with constant probability, we sample 2−α [X] α E a node of degree no more than Θ(c ). (This is from Markov’s inequality: P[X ≥ a] ≤ a . So 2−α 1 α P[d(node) ≥ 2µ] ≤ 2 , where µ = Θ(c ).)

2 Now, for a small detour, what is the total number of endpoints of edges adjacent to nodes of degree Pc1/α −α   2 −1 2−α −α ≥ k? i=k i · c · i = Θ c · c α − k . (Recall that c · i is the number of nodes of degree i.)

1 − This means that, if we take k = c α , the total number of edge endpoints adjacent to nodes of  2 2 −  2  degree ≥ k is Θ c α − c α = Θ c α = Θ(M). So every edge is incident to some high-degree node with constant probability.

2−α Ok, so if we take a random vertex, it has degree Θ(c α ) and it has an edge to a node of degree 1 − 1 − 2−α − 1− 1 − ≥ k = c α (both things happen with constant probability). The ratio is c α α = c α . Since c = Θ(n), this completes the proof. 

3