doubt in kolmogorov's forward equation v/s the transition rate

[Madhur]

Can any one tell me the difference between the transition rate \( \mu_{ij} \) and this \( \frac{d}{dx} p_{ij}(t) \) used in the forward komogorov's forward equation.
The rate of change of the transition probability from state i to state j should be the same as the transition rate.

Regards,
Madhur

 

[John Potter]

They are very different things. In fact, the easiest way to answer your question is to say why would you ever think they were the same?

Take the Helathy-Sick-Dead model

1) mu(HS) is the force of transition from Healthy to Sick

It is ALWAYS a positive number.

It shows the RATE at which we move from Healthy to Sick, so if mu(HS) = 0.1, this means 10% of a person goes from healthy to sick every year (or maybe nicer to say 1 person every 10 years!)

2) d/dt pHS(t) is the RATE OF CHANGE of the PROBABILITY that we go from Healthy to Sick over the next t years.

Let's think about what happens to this probability:

For small values of t, the probability that we go from healthy to sick should be going up, so d/dt pHS(t) is positive

For larger values of t, the probability that we go from healthy to sick should be going down, because much more likely that the person is already dead so d/dt pHS(t) is negative.

So, one is a rate, the other is a rate of change of a probability,

Good luck!
John

 

[Madhur]

But \(\mu_{ij}\) is defined to be the derivative of \(p_{ij}(t)\) evaluated at t=0. And in time homogeneous case, for it does not depend upon 't' so derivative of \(p_{ij}(t)\) should equal.

I understand the example you gave, but I am not able to understand this concept which I mentioned above.

And also in the second point you mentioned that derivative of pij(t) is the rate of change over t years, but if we are differentiating a function then we should get the rate of change in that instance or in a very small amount of time.

 

[John Potter]

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

 

[Madhur]

thank you sir.
Now I am comfortable with the idea.

 

[sahildh]

I didnt get the idea of why the transition rate is found out as a rate of change of probablility of moving from one state to another as h -> 0 i.e. in a very short interval ?

It would be very helpful if someone could please explain me the idea behind transition rate and the probability of moving from one state to another in simple terms. I am confused as the transition rate makes use of the probability and the probability involves multiplying the transition rate with the time period !!!! Why and how ?

I am very confused and a response would be a great help. Thank you.

 

[Hemant Rupani]

I didn't get the idea of why the transition rate is found out as a rate of change of probability of moving from one state to another as \( h \to 0 \) i.e. in a very short time interval ?
its just the definition of transition intensity, proved helpful in continuous time process... suppose, if you want to calculate the \( P_{ij}(0,1) \) and change of state can be done in discrete time then its okay, you can find it directly(presumably you know Probability), But if the change of State can occur anytime then you'll gonna need transition rates.

the idea behind transition rate and the probability of moving from one state to another in simple terms.
suppose time measured in years...
Transition rate_{ij}: number of lives change from i to j, from life-year... suppose 3 lives change state from i to j, where 50 life-year totally spent in i(say, from 100 lives, 40 lives spent half year, 40 lives spent quarter year, 20 lives spent full year, then 40*0.5+40*0.25+20=50), then Transition rate_{ij}=3/50
Probability_{ij}: number of lives in state j at time 1 from number of lives in state at time 0...(it may be lower or higher than transition rates)

the transition rate makes use of the probability and the probability involves multiplying the transition rate with the time period, Why and how ?
\(P_{ij}(h)=\mu_{ij} h \) it only good for infinitesimally small time interval... hence, we add o(h) term to make it \(P_{ij}(h)=\mu_{ij} h +o(h) \)

 

[sahildh]

Transition rate is the rate of change of probability of moving from state i to j w.r t to t at t=0. What does this imply?I mean why at t = 0?

 

[Hemant Rupani]

Transition rate is the rate of change of probability of moving from state i to j w.r t to t at t=0. What does this imply?I mean why at t = 0?

it imply that transition rate is instance[time interval from now to then=h where \( h\to 0\)](t=0) change from i to j.