transition rates

[tatos]

Consider 2 machines
Time to break down follows an exponential distribution with rate s
Time to repair also follows an exponential distribution but with rate r

Only one machine can be repaired at a time

the generator matrix is given by

-r r 0
s -(s+r) r
0 2s -2s

My question is: why isn't there a transition rate to describe a movement from 2 machines in working order to 0 machines in working order (entry 3,1 in the matrix above, which is 0)? i.e. Why does it seem like it has to pass through "state 1" en route to zero machines in working order?

I understand how you can go from 0 to 1 only and not from 0 to 2 (because only one machine can be repaired at a time) but I don't understand why you can't go from 2 to 0 immediately (since surely the machines can simultaneously break down?)

 

[didster]

It's a continuous time model and the transition rates are the rates of transition in a really really short period of time.
If two drops down (or up) appear to happen simultaneously, you could always conceptually chop up the "instant" into two where in each a jump occurs.

Academically if you go back to first principles of the poisson process, the multiple jumps are o(dt^2) and higher and thus "disappear".

 

[tatos]

thanks didster