Problem computing expected values under POPAN?

[bromaghin]

I am trying to compare observed and expected frequencies as a aid in assessing fit across data types (POPAN versus closed, no covariates). Expected frequencies under the closed models seem reasonable, summing to roughly half of the sample size. However, the sum of the expected frequencies based on a POPAN model are over 10 times the sample size, even though the parameter estimates seem reasonable. Actually, the expected value of the last history listed is equal to the sample size.

Does anyone have any suggestions for assessing fit across data types?

Thanks!
Jeff

[sbonner]

Hi all,

I hate to revive such an old topic, but I've run into the same confusing result as Jeff and found this post when searching the forums.

Like Jeff was doing in 2006, I've been fitting a POPAN model and hoped to compare the observed and expected values to diagnose possible sources of heterogeneity. It looks like a transient problem, but I am confused by the expected values computed in MARK. As Jeff notes, the expected value for the number of individuals captured only on the last occasion is exactly equal to the sample size for all models fit.

Further, the expected values for the remaining histories don't seem to jive with the parameter estimates. E.g., for the phi(.),p(.) model the expected count for history 1000000000 is given as 167. However, the estimate of N is 644, of pent[1] is .10, and of p is .10 so I would have expected the expected count for this history to be less than 6 (I'd have guessed that the sum of expected counts for all histories with a leading 1 should be 6.4).

Can anyone explain where I'm going wrong with this?

Thanks in advance!

Simon

[jlaake]

Sure you have the correct pent? Can't remember now which pent is computed by subtraction for POPAN. In some models it is last pent and in others the first pent. Maybe what you want is pent(0) - proportion in population before sampling started. But not certain.

--jeff

[gwhite]

For the POPAN model (data type 19), the probability of being present prior to the first sample is computed by subtraction.

Gary

[sbonner]

You guys enjoyed that didn't you!

OK, so I was looking at the wrong value, but I'm still confused... about life in general but this point in particular.

Parameter estimates for the phi(.), p(.), pent(t) model are copied below. By subtraction pent[1]=.28. So, I would compute the expected number of individuals captured on the first occasion to be 644*.28*.19=34.26, but the value given in the residuals file is 167.30. The sum of the expected counts in the residual file yields 2289.

I don't see how the expected number of individuals captured in a similar experiment can be 2289 if the total population size is estimated at 644 with 95%CI (553,765). I also noticed that the expected counts for the histories with one captured run in increasing order from 1000000000 through 0000000001 and sum to 1985. Removing these the expected number of individuals captured twice or more would be 304 -- which seems reasonable given a population size of 644 and capture probability of .28.

Am I misinterpreting these values?

Thanks Jeff and Gary for your previous replies,

Simon

Code: Select all

             
               estimate    se    lcl    ucl
Phi g1 a0 t1       0.79  0.03   0.73   0.85
p g1 a0 t1         0.19  0.02   0.15   0.24
N g1 a0 t1       644.05 53.68 553.35 765.63
pent g1 a1 t2      0.10  0.06   0.03   0.28
pent g1 a2 t3      0.05  0.05   0.01   0.33
pent g1 a3 t4      0.04  0.05   0.00   0.36
pent g1 a4 t5      0.03  0.04   0.00   0.40
pent g1 a5 t6      0.03  0.04   0.00   0.31
pent g1 a6 t7      0.08  0.04   0.03   0.20
pent g1 a7 t8      0.28  0.05   0.20   0.39
pent g1 a8 t9      0.00  0.00   0.00   1.00
pent g1 a9 t10     0.11  0.05   0.05   0.24


[gwhite]

Simon:

The expected values are just the CJS portion of the likelihood times the total number of animals captured [M(t+1)]. The last encounter history of all zeros but with a release on the last occasion has a probability of 1 under the CJS model, so hence gets the value of M(t+1).

The reason for this is that all lack of fit is only due to the recaptures in the CJS portion of the likelihood. There is no lack of fit from the new animals recruited under the pent(g*t) model. The same applies to all of the Jolly-Seber models: Pradel, Link-Barker, etc. Only the recaptures provide information on lack of fit.

To get what I think you are after, you will have to compute the encounter history probability manually using the parameter estimates from the model.

Gary

[sbonner]

Hey Gary,

Thanks for clarifying -- the calculations make sense now and I get the point about there not being lack of fit due to recruitment.

Cheers,

Simon