Hi Madhur,
For a small time period, h
pHS(h) = h*mu(HS) + o(h)
Notice how the mu works, it's a rate, so if we multiply by h, it gives us the probability of making the transition over the time period h. Check out this post too...
http://www.acted.co.uk/forums/showthread.php?t=9108
Now, I'm going to subtract 0 from the LHS
pHS(h) - pHS(0) = h*mu(HS) + o(h)
I can subtract 0 from something whenever I like ;-)
Divide through by h
[pHS(h) - pHS(0)] / h = mu(HS) + o(h) /h
The LHS is a simplified version of [pHS(t+h) - pHS(t)] / h, the limit of this as h tends to 0 is d/dtpHS(t) by definition.
So, we can see that mu(HS) is the limit of d/dtpHS(t) as t tends to 0. You are correct to point this out. This is only as t tends to 0 though. For all other values of t, I would try and think about them in the way that I've said
IMPORTANT:
I didn't say "derivative of pij(t) is the rate of change over t years"
I said "d/dt pHS(t) is the RATE OF CHANGE of the PROBABILITY that we go from Healthy to Sick over the next t years"
I will put brackets into this sentence...
d/dt pHS(t) is the RATE OF CHANGE of {the PROBABILITY that we go from Healthy to Sick over the next t years}
Good luck!
John