q is fraction of nodes in largest component
look at any node: chance it is in the giant component is q
chance that this node is outside of the giant component is the chance that all of its neighbors are outside of the giant component
Probability that a node is outside of the giant component = 1-q
Probability 1-q that a node is outside of the giant component is:
Solve fo q...
It's a heuristic calculation because I'm not worrying about correlations and some of these variables I'm treating this as if they're independent. They're not quite independent. You can do a more exact calculation using generating functions. There's some of that in the book, but this will give us a fairly good approximation and work fairly well especially on large networks
Solve $1-q = \sum_d(1-q)^dP(d)$
plug in a poisson distribution for PFD: $P(d) = [(n-1)^d/d!]p^d e^{-(n-1)p}$
so $1-q = e^{-(n-1)p} \sum_d[(1-q)(n-1)p]^d/d!$
Taylor series approximation:
$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}...$
$= \sum \frac{x^d}{d!}$
$[f(x) = f(a) + \frac{f'(a)(x-a)}{1!} + \frac{f''(a)(x-a)^2}{2!}...]$
so $1-q = e^{-(n-1)p} \sum_d[(1-q)(n-1)p]^d/d!$
$= e^{-(n-1)p}e^{(n-1)p(1-q)}$
$= e^{-q(n-1)p}$
$(n-1)p$ is the expected degree, p is the probability of any link
so $1-q = e^{-qE(d)}$
or $-\frac{log(1-q)}{q} = (n-1)p = E[d]$
We get these Tyche phase transitions, so if expected neighbors less than "1" expected interactions that would transmit a diffusion of less than "1", we don't get diffusion. Once you hit one we get a diffusion and then it grows fairly rapidly and by the time you have three interactions then we're already to the point where you're very likely to hit most of the population. So if you remember in our promo video I said tell at least two friends. Well, if you tell just one friend about this then we don't get much diffusion. Two friends is at least enough to start pushing us fairly far along this. And you get to three friends and you're most of the way there.
So just in terms of giant component size here we are. We see that there's a fairly tight transition from having no giant component to having almost everybody being connected and so most of the play is in this one region. And indeed when people study epidemiology there's this idea that having at least one contact that's going to transmit something if people are doing more than one you're going to get diffusion or contagions and if people are having less than one transmission then things are going to die out. And that makes perfect sense. You need at least, you know, more than one and it's growing, fewer than one and it's shrinking. That's the critical point. And then here we find that it actually transitions very rapidly so that by the time you've got three interactions that are likely to lead to contagion then our diffusion is quite extensive.
Probability of being in the giant component
$1-(1-q)^d$ increasing in d
More connected, more likely to be infected
(more likely to infected at any point in time...)
Immunity: delete a fraction of nodes and study the giant component on remaining nodes
Probabilistic infection
Some node is initially exposed to infection
$\pi$ of the nodes are immune naturally
only some links result in contagion - fraction f
What is the extent of the infection?
Consider a random network on n nodes
Delete fraction $\pi$ of the nodes
Delete fraction f of the links
If starts at a node in giant component of the remaining network, then the giant component of that network is the extent of the infection; otherwise negligible
Let q be the fraction of nodes of the remaining network in its giant component
$q(1-\pi)$ is the probability of a nontrivial contagion
Conditional on a contagion it infects $q(1-\pi)$ of the original nodes
q solves $-log(1-q)/q = (n-1)p(1-\pi)f$
Infection can fail if $\pi$ is high enough or f or p are low enough
High $\pi$ - immunization, low virulence
Low f - low contagiousness
Low p - low contact among population