Poisson Model vs Two state Model

Discussion in 'CS2' started by Darshan Mody, Mar 19, 2019.

  1. Darshan Mody

    Darshan Mody Member

    Hi,

    There are two points I am confused about with regards to the comparison of both these models (Poisson & Two State):

    1) "The two-state model can be specified so as to allow for increments (lives entering the poppulation), which is not possible for the poisson model"

    I dont understand how does the two state model allow for it ?

    2) "The Poisson model assumes that exposed to risk remains constant and estimation of transition rates in the model only involves measurement of the observed no of decrements
    WHEREAS the estimation of transition rates in the two-state model involves the measurement of two random variables - the observed no of decrements & the exposed to risk that gave rise to these decrements."

    I dont understand how its assumed in poisson model that central exposed to risk is constant?
    We actually calculate the central exposed to risk so how is the assumption made that it is constant?

    If possible please explain with an example wherever possible.


    Thanks!
     
  2. Calm

    Calm Ton up Member

    1) Central exposed to risk is used in the Binomial model. This means that each person contributes from the time of joining (whether at the beginning or midway through) to the time of leaving (death/withdrawal etc.)

    2) The central exposed to risk that is calculated already takes that assumption into account. Usually, the µ-rate is based on that of the middle of the interval of data measured; since the true mortality rate only undergo small changes within that time, it is considered to be a good approximation.
     
  3. Darshan Mody

    Darshan Mody Member

    Thanks.

    Just further query:
    1) I understood how it works for Binomial Model but how does the two-state model allow for increments and how does the poisson model not allow for it?

    2) So does this mean that central exposed to risk calculated for a particular year is then assumed constant for the next year?
    Sorry i didn't get the second point at all.
     
  4. Hi Darshan

    This is how I understand it.

    The Poisson models only allows decrements [e.g. alive to dead] and cannot model increments [dead to alive].

    In the example I have given the Poisson model makes sense when you have a model with only these states [Alive, Dead]. Note it is just a fancy way of saying the transitions only happen in one direction.

    When you look at a Poisson model, by definition we fully define a Poisson process as having mean lambda * t.

    This lambda parameter by definition is constant -> by assumption. It is not even a good assumption unless the time period is small enough to make this reasonable!

    With a two state model [example with Healthy and Sick], of course we can go from Healthy to Sick [Decrement] and Sick to Healthy [increment]. But I do not think it is necessary to have increments for the model to be two-state. In fact, surely the Alive -> Dead model is poisson and two-state with the right assumptions but will still be two-state if you remove the assumption of constant mu, etc.

    With the second point, in a Poisson model, because the exposed to risk has not been accounted for in the model, it is surely an assumption that it does not matter what the exposed to risk is, it will not effect the mortality. So if mu = no of deaths / exposed to risk. Both of these are either constant or change proportionately to give the same constant parameter of mu, and therefore, lambda.

    mu = 1/lambda => lambda = exposed to risk / number of deaths

    With a two state model, of course we can have a situation where both of these need to be estimated since we do not actually know in advance the number of decrements or the central exposed to risk.

    Regards
     
    Last edited by a moderator: Mar 20, 2019
  5. Darshan Mody

    Darshan Mody Member


    Hi,

    So with regards to the 1st point, I don't think if we remove the assumption of constant mu the model will be a two-state model because constant mu is an assumption required for a model to be two-state.

    With regards to the second point, the poisson model does account for the central exposed to risk, isn't it? By taking (mu*central exposed to risk) as the parameter, so number of deaths and hence the mortality will definitely be affected by the value of central exposed to risk.
    Please correct me where I am wrong.

    I am pretty confused about both the points still.

    Though I understood the fact that "The Poisson models only allows decrements [e.g. alive to dead] and cannot model increments [dead to alive]."

    Thanks!
     
  6. Hi Darshan

    I think I was wrong about some of the assumptions.

    The force of mortality is assumed constant between integer ages in both models. I think the only difference really is how that force of mortality is calculated.

    With the two state model we try to estimate the number of deaths and central exposed to risk.

    With Poisson we take the number of lives alive at beginning of observation and assume this is fixed. Then just take number of deaths in the year and calculate mu.

    With the two state model we do that thing where we calculate the number of months each life was actually exposed to the risk and measure this. I think this answers your question.

    I think the confusion comes in with probability mass function. This uses the central exposed to risk. However even this is generally assumed constant in the Poisson model. Why is this. Well if we are at the beginning of the period, we do not actually know how the lives will be exposed to the risk in that period. It may not be necessary to try and estimate the months etc. with people leaving the study for reasons other than death so it simplifies our assumption and provides an approximation to the more accurate two state model.
     
    Last edited by a moderator: Mar 21, 2019

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