View Full Version : Two-state vs. Poisson Model

06-04-2009, 01:58 AM
When comparing these two models, a point given was:

The two-state model can be specified to allow for increments, whereas the Poisson model can not.

What does it mean by increments? Is it the increments in exposed to risk? Or more than one decrement?

On a separate note, what are the similarities between a two-state and Poisson model, besides 'both estimates the force of mortality'?

06-04-2009, 03:13 PM
An increment denotes the transition back from a 'lower' state to a 'higher state' e.g. from sick to healthy. Decrement is the opposite i.e. from a 'higher' state to 'lower' state e.g. healthy to sick or healthy to dead. The Poisson model doesn't allow for increments, while the 2 state Markov does.

Intuitively, the Poisson model is used for calculating forces of mortality mu-x and can be used for multiple decrements, unlike the Binomial model which can allow only one decrement viz. death. Poisson can be used for multiple decrements but not for increments. 2 state can model multiple increments as well as decrements.

Similarities between 2 state and Poisson model:
1. MLE approach is possible to determine the force of decrement in 2 state and Poisson; the binomial model can estimate q-x by method of moments, not MLE.
2. Exact calculation of MLE possible in Poisson and 2 state (subtle difference between pt 1 and 2)
3. mu-x calculated using 2 state and Poisson has lower variance around the force of mortality as compared to the binomial model, where q-x can be estimated from (1 - exp {-mu})

The important difference between Poisson and 2 state is that Poisson assumes waiting times are fixed in advance, while 2 state doesn't make any such assumption.

I trust this helps. Regards

06-04-2009, 03:14 PM
One more point of similarity:

Poisson and 2 state consider information from 'time of death' and not just 'number of deaths', which is a flaw of the binomial model

07-04-2009, 02:02 AM
Thanks Ankmola

In your earlier point 3, I assume you are talking about estimating mu from the formula q_x = 1 - exp (mu).


08-04-2009, 03:37 AM
My point was that mu_x of 2-state and Poisson have lower variance (that's the similarity between 2-state and Poisson) than the Binomial model which estimates q_x from the 1 - exp (mu_x).

I had tried to draw out the similarities between 2-state and Poisson, and only by a passing reference compared this to the variance of q_x of the binomial model.


09-04-2009, 05:05 AM
I thought in the Binomial model we estimate q_x, and in the Poisson model we estimate mu?

10-04-2009, 12:22 PM
You are right :). I had mentioned about the binomial model just to contrast it against the commonality of Poisson and 2-state.


11-04-2009, 11:46 AM
Oh i see. I thought i was getting it all wrong.

Thanks Ankmola!

20-04-2009, 10:02 AM
If Binomial model cannot handle multiple decrements, does this means, Multinomial models can?

21-09-2010, 10:47 PM
I have a question on this thread (I realise the previous responses are a bit old).

I don't understand this assumption about the central exposed to risk (or "waiting time") being fixed in advance. Surely to use the Poisson model you take data where you have observed lives for a certain amount of time, calculate the total waiting time and then work out what the force of mortality is based on the number of deaths witnessed.

Why do you need to make an assumption about the total waiting time when you can just calculate it?


Mark Mitchell
27-09-2010, 11:13 AM
Usually we're interested in using the Poisson model to estimate the (unknown) true force of mortality. Calculating, or knowing, the central exposed to risk is really only useful to us in the sense that it helps us get to an estimate of mu.

The assumption that the central exposed to risk is known (or fixed) is so that the parameter of the Poisson random variable for the number of deaths is fixed. So D (the number of deaths) ~ Poi (Ecx * mu). The parameter being fixed and not a random variable makes manipulation of the model easier.