Your inclinations are basically correct. Setting f=0 where there is no effort makes sense. Based on what you've described, the fact that you got a reasonable result under a constant S model, and some that did not make sense under the St model also is not surprising.
Because I do not have your recovery matrix in front of me I cannot comment on which parameters I believe would not be estimable, but here is a general guideline, based on the probability structure for the recovery matrix below. (Note that the structure of the matrix will be left-justified by the time you see this on the bulletin. Therefore, to follow what I say below you will want to re-align the rows so that it is an upper triangular matrix with the cells where f is alone are on the diagonal).
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f1 S1f2 S1S2f3 S1S2S3f4
f2 S2f3 S2S3f4
f3 S3f4
f4
When there is not recovery effort in a given year you basically set an entire column of the recovery matrix to 0. If the rest of the recovery matrix is "rich", i.e., the cells are filled, you should still be able to estimate all S's. You can see this from the above matrix. For example, the most direct way to get an estimate of S1 is algebraically from the two cells in column 2. You have direct information on f2 from direct recoveries, and direct information on the product of S1f2. This constitutes two equations and two unknowns, which is easily solved. If f2=0, then there is no information on S1 in column 2. However, looking at column 3, you can derive an estimate of S1 from the combination of data from the top two cells in this column, and derive an estimate of S2 from the combination data from the bottom two cells in that column. Normally you would not do these things, because S1 and S2 are also found in other cells in the model. The Brownie et al. model uses all information from the cells, to maximize precision. However, this at least tells you whether the minimum data necessary for estimation is available.
The bigger problem lies in years where you do not release any animals. In this case you are taking out an entire row of the recovery matrix. For example, if you did not release any animals in year 2. In this case, if you play the same algebraic game, you will see that you lose your ability to estimate S1 and S2. You don't have direct information on f2 and therefore cannot use it to extract S1 from the product S1f2. In addition, you lose information on the product S2f3, and therefore again cannot use it to extract S2. The best you can do in this case is estimate the product S1S2. In general, then, when you do not release animals in year t, you cannot get separate estimates of St-1 and St.
Cavell Brownie taught me this approach a long time ago, and I have found it very useful. This same algebraic game can be played fairly easily with the CJS model, but gets a lot tougher when you move to more complex models.
This was a long answer to a short question, but it might help you tease apart what can be estimated. It can get more complex when you have years with neither releases nor recovery effort.
Cheers,
Bill
