# Transition rates of the number of infected vertices for the contact process on the triangle

Consider the contact process with parameter $\lambda \in (0,1)$ on the triangle. Let $X(t)$ denote the total number of infections at time $t$. Compute the transition rates of this process.

Our teacher gave us the solutions: The transition rates $r(i, j)$ from $i$ to $j$ are: $r(3; 2) = 3$, $r(2; 1) = 2$, $r(1; 0) = 1$, $r(1; 2) = 2 \lambda$, $r(2; 3) = 2 \lambda$, and all other rates are zero.

The definition of the rates is the following:

$$c(x,n) = 1 \text { if } n(x)=1$$ $$c(x,n) = \lambda \sum_{y~x} n(y) \text{ if } n(x)=0$$

By calculating the rates with the definition I get for $r(2,1) = 1$, because it goes from a state were two people are injected to a state were just one is.. so $n(x)=1$ and therefore the rate is equal to 1?

And so on... $r(1,0) =1$, $r(1,2) = \lambda$, $r(2,3)= 2 \lambda$.

What am I doing wrong? How am I supposed to compute the rates?

• To get a constructive question, you should explain in details why you think that $r(2,1) = 1$ and $r(1,2) = \lambda$. – Did Dec 12 '16 at 14:40
• ((The tag (contact-topology) was wildly inaccurate (but it made me smile). Try to read the content of the tags you are not familiar with before using them.)) – Did Dec 12 '16 at 14:41
• A added something to my question, can you maybe help me now? @Did – Yuhe Dec 13 '16 at 10:12
• The rate $r(1,2)$ corresponds to a transition from the state $(n(0),n(1),n(2))=(1,0,0)$, say, to $(n(0),n(1),n(2))=(1,1,0)$ or $(n(0),n(1),n(2))=(1,0,1)$. Each of these transitions happens at rate $\lambda$ hence $r(1,2)=2\lambda$. – Did Dec 13 '16 at 18:04
• Ahh thanks a lot .) – Yuhe Dec 14 '16 at 13:22