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In a problem it's stated the following:

Consider a simplified version of the Internet, which has only two distinct websites: Friendface, and Metube. Metube has 1 link to Friendface, and 1 link to itself. Friendface contains 2 links to Metube and 1 link to itself. We model a user’s behavior on the web as follows: at every time step k, the user follows a link on the current page; we assume that this link is chosen uniformly at random. Let x(k) ∈ {Friendface, Metube} represent the website that the user is on after k clicks.

You are given that x(0) = Friendface with probability 1 . Calculate the probability that x(k) = Metube for k → ∞.

Now in the solution basically it's stated that as time goes to infinity the probability of x at time k+1 is equal to the probability of x at time k.

My question is why? How to we know that this limit converge to this value?

Tommaso Bendinelli
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    Are you familiar with markov chains ? – nicomezi Aug 31 '18 at 06:57
  • I have never directly studied it, but I realized that you can model this system as Markov Chain. Is there a Theorem that when k → ∞ the probability values converge? – Tommaso Bendinelli Aug 31 '18 at 06:59
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    Perron-Frobenius theorem may be what you are looking for. – nicomezi Aug 31 '18 at 07:02
  • I didn't know the theorem and I understand that might be used in some way but how? – Tommaso Bendinelli Aug 31 '18 at 07:07
  • Btw I found online two different version of the theorem, probably they means the same but I can't figure out how to go from one version to the other. Version 1 https://textbooks.math.gatech.edu/ila/stochastic-matrices.html Version 2 https://math.stackexchange.com/questions/1166068/what-is-the-implication-of-perron-frobenius-theorem – Tommaso Bendinelli Aug 31 '18 at 07:21

1 Answers1

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OK let's try this so the probability that $x(0)=Friend Face$ is $1$. I will write this as $PF(0)=1$ and so $PM(0)=0$.

What is $PF(1)$? Well there is a $2/3$ chance the person will head on over to Metube and a $1/3$ chance they will stay so $PF(1)=1/3$ and $PM(1)=2/3$. What next? For $PF(2)$ we have $1/3.PF(1)+1/2.PM(1)$ which is $1/9+1/3=4/9$. $PM(2)=2/3.PF(1)+1/2.PM(1)=2/9+1/3=5/9$. Phew, they add to $1$ :).

$PF(3)=1/3.PF(2)+1/2.PM(2)$ and $PM(3)=2/3.PF(2)+1/2.PM(2)$.

$PF(3)=4/27+5/18=23/54$. $PM(3)=8/27+5/18=31/54$.

OK let us just assume that the sequence does converge (I know this is assuming what you want to prove but let's just take a look).

This means that $PF(n)=PF(n=1)$ and $PM(n)=PM(n=1)$.

Well generalising the above we have.

$PF(n+1)=1/3.PF(n)+1/2.PM(n)$ and $PM(n+1)=2/3.PF(n)+1/2.PM(n)$.

Substituting in $PF(n)=PF(n+1)$ and $PM(n)=PM(n+1)$ and just calling them $F$ and $M$ respectively we have.

$F=1/3.F+1/2.M$ and $M=2/3.F+1/2M$.

These both simplify to $2/3.F=1/2M$. So since $F+M=1$ we know that $F=3/7$ and $M=4/7$.

Now to answer your question. How do we know it coverges? There is a common sense answer. It doesn't look like it will oscillate and it can't shoot off to infinity. The most extreme position would me $F$ or $M$ equal to $1$ or $0$ but that will never happen as there are links back to the other site.

Our algebra has shown the existence of a (unique) equilibrium position which will act as an upper bound and the early analysis shows that we are moving there (strongly monotonically) so, by the monotone convergence theorem, the sequence will converge. https://en.wikipedia.org/wiki/Monotone_convergence_theorem

Simon Terrington
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  • Yeah, Intuitively makes complete sense. But rigorously showing the existence of a unique equilibrium position doesn't imply that the system will converge to that equilibrium, does it? I think with Perron-Frobenius theorem https://math.stackexchange.com/questions/1166068/what-is-the-implication-of-perron-frobenius-theorem you can prove it, but I can't see how – Tommaso Bendinelli Aug 31 '18 at 07:31
  • Yeah, I completely agree. I divided in but it turned out to be a little more tricky than I expected. I have got a few more ideas so I may come back to this soon. I'll have a look at the link that you provided too. Thanks. – Simon Terrington Sep 01 '18 at 11:59
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    I have had a think about this and I actually think that the montone convergence theorem finishes the proof. I have made an edit to reflect this. – Simon Terrington Sep 01 '18 at 19:18