schnabel method

questions concerning analysis/theory using program MARK

schnabel method

Postby eronje » Thu May 23, 2019 1:35 pm

Hi,

I used the Huggin's closed capture models in program MARK to estimate abundance for a primary with 3 sampling occasions. My preferred model is the Mt model. As a check for myself, I also estimated abundance in Excel using a simple matrix and the Schnabel method, which as I understand it (please correct me if wrong) most closely resembles the closed capture Mt model. The Schnabel method I used was this: Image
upload

The estimates were quite close so that was reassuring, but with some of the the smaller sample sizes approaching 30, the Schnabel results were negatively biased so that the estimate was less than the input. I don't understand WHY they are a little different from the program MARK estimates. Can you enlighten me?
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Re: schnabel method

Postby cooch » Thu May 23, 2019 2:35 pm

Several reasons, potentially, but short of seeing the data, try adding +1 to the denominator:

$\hat{N} = \frac{\sum_{t=1}^{S} (C_t M_t)}{\left( \sum_{t=1}^{S} R_t\right) + 1}$

Often, for low sampling fractions (say, <15% of the population size),the '+1' adjustment tends to help.
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Re: schnabel method

Postby eronje » Thu May 23, 2019 4:24 pm

Thanks. I tried the +1 in the denominator but it decreased the estimate beneath the input even further. Here are the capture probabilities as estimated in MARK. Would the large difference between probabilities be a likely culprit? Even the program MARK estimates are barely over the "n" .

Sample size for the estimates below is n=32. MARK estimates slightly over 32 and Schnabel estimates 29-30.

Huggins p and c

Real Function Parameters of {Mt}
95% Confidence Interval
Parameter Estimate Standard Error Lower Upper
-------------------------- -------------- -------------- -------------- --------------
1:p 0.8250379 0.0709599 0.6427594 0.9251423
2:p 0.8555949 0.0666850 0.6728934 0.9446453
3:p 0.1222278 0.0573769 0.0465441 0.2842844
4:c 0.8555949 0.0666850 0.6728934 0.9446453
5:c 0.1222278 0.0573769 0.0465441 0.2842844
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Re: schnabel method

Postby gwhite » Thu May 23, 2019 4:46 pm

The Schnabel estimator is a completely different estimator from the Huggins estimator -- it is not a maximum likelihood estimator, and hence does not share all the good characteristics of an MLE. Further, I'm not sure that the Schnabel estimator conditions on just the encounter histories where animals are captured at least once like the Huggins estimator does.

You can also run the full likelihood estimator, and see you will again get a very similar estimate but not identical to the Huggins estimator. The full likelihood estimator does not condition on encounter histories with the probability that animals are captured >0. The all zero encounter history is part of the likelihood.

The estimator from the literature most similar to the M(t) Huggins estimator is the Darroch estimator, although it is exactly the full likelihood M(t) estimator. These subtleties escape users until you actually write out the likelihoods.

You might ask why the Huggins estimator is used and recommended instead of the full likelihood estimator -- because 1) individual covariates can be used with the Huggins estimator as all individual in the sample were captured to measure the covariate value (which is not the case for uncaptured individuals with the all zero encounter history), and 2) the slight loss in efficiency is negligible compared to the full likelihood estimator. Further, there are fewer numerical problems with optimizing the Huggins estimator compared to the full likelihood estimator (although this issue is pretty much not an issue now days). The Otis et al. monograph and Program CAPTURE used the full likelihood version for all likelihood estimators in the program. The idea of using individual covariates was not discussed at the time. Huggins and Alho brought this idea to the literature.

Gary


An appropriate quote:
“The most misleading assumptions are the ones you don't even know you're making...”
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Re: schnabel method

Postby eronje » Mon May 27, 2019 1:32 pm

Thanks Dr. White,

Can you clarify how the full-likelihood model conditions upon the animals never seen? I have used the full-likelihood model a number of times, but have struggled to understand how an estimate is derived with an unknown "n" of unseen animals?
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Re: schnabel method

Postby gwhite » Mon May 27, 2019 1:37 pm

The full likelihood parameterization does not "condition" on the number of animals not seen. Rather, the number of animals not seen in the likelihood is the current value being optimized of N minus M(t+1), the number of animals encountered 1 or more times. So the probability of the all zero encounter history is raised to the power N - M(t+1) in the likelihood.

There are also some additional terms involving N and M(t+1) in the multinomial coefficient, but I think the above explanation is enough for your to understand what "n" is.

Gary
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Re: schnabel method

Postby cooch » Mon May 27, 2019 1:54 pm

All of which is covered in Chapter 14. Section 14.2, specifically.
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Re: schnabel method

Postby eronje » Tue May 28, 2019 9:40 am

Thanks to you both! I will re-read chapter 14...again...
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Re: schnabel method

Postby eronje » Sat Jun 15, 2019 5:26 pm

After re-reading Ch. 14, I think I have a better handle on how the full-likelihood approach works but I still struggle to understand why my full-likelihood estimate has a negative SE. The estimate is exactly the same as the input, and I understand that it cannot be lower than that because the estimate of N has been constrained to be greater than or equal to Mt+1.

I noted in Chapter 14.7 (first paragraph) that negatively biased estimates should be expected if there is individual capture heterogeneity. I am leaning towards that conclusion, would you agree?

Real Function Parameters of {Mt}
95% Confidence Interval
Parameter Estimate Standard Error Lower Upper
-------------------------- -------------- -------------- -------------- --------------
1:p 0.8437495 0.0641863 0.6752752 0.9334323
2:p 0.8750020 0.0584630 0.7105912 0.9522841
3:p 0.4374995 0.0876951 0.2789332 0.6099558
4:c 0.8750020 0.0584630 0.7105912 0.9522841
5:c 0.4374995 0.0876951 0.2789332 0.6099558
6:f0 0.1115002E-07 0.8284761E-04 0.2839883E-11 0.4377748E-04

/* Begin input file */
/* 1000 */ 110 1;
/* 2000 */ 111 1;
/* 2001 */ 110 1;
/* 2002 */ 011 1;
/* 2004 */ 110 1;
/* 2007 */ 110 1;
/* 2016 */ 111 1;
/* 6000 */ 110 1;
/* 6001 */ 110 1;
/* 6002 */ 111 1;
/* 6003 */ 110 1;
/* 6004 */ 110 1;
/* 6005 */ 100 1;
/* 6006 */ 110 1;
/* 6007 */ 010 1;
/* 6008 */ 111 1;
/* 6014 */ 110 1;
/* 6031 */ 111 1;
/* 6032 */ 010 1;
/* 6033 */ 011 1;
/* 6565 */ 111 1;
/* 6566 */ 100 1;
/* 7000 */ 111 1;
/* 7001 */ 111 1;
/* 7004 */ 110 1;
/* 7007 */ 101 1;
/* 7036 */ 001 1;
/* 7320 */ 110 1;
/* 7397 */ 110 1;
/* 8000 */ 110 1;
/* 8001 */ 111 1;
/* 8002 */ 111 1;
/* End Input File */
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Re: schnabel method

Postby cooch » Sat Jun 15, 2019 6:00 pm

eronje wrote:After re-reading Ch. 14, I think I have a better handle on how the full-likelihood approach works but I still struggle to understand why my full-likelihood estimate has a negative SE.


I suggest a refresher on how to read scientific notation. A value like

Code: Select all
0.4353121E-03


is not negative. It corresponds to

Code: Select all
 0.0004353121


Basically, you have a zero SE for f0. And the reason for that has only to do with the fact that you need to apply a constraint to final p and c parameters in M(t) to get anything sensible (if you don't know why, you need to re-read Chapter 14, section 14.3.1). Typically, you set p=c. If you fit M0, which is perhaps reasonable with only 3 occasions, then

Code: Select all
                           Real Function Parameters of {M0}
                                                               95% Confidence Interval
  Parameter                  Estimate       Standard Error      Lower           Upper
 --------------------------  --------------  --------------  --------------  --------------
     1:p                      0.7138792       0.0511398       0.6043375       0.8029806
     2:p                      0.7138792       0.0511398       0.6043375       0.8029806
     3:p                      0.7138792       0.0511398       0.6043375       0.8029806
     4:c                      0.7138792       0.0511398       0.6043375       0.8029806
     5:c                      0.7138792       0.0511398       0.6043375       0.8029806
     6:f0                     0.2183338       1.0112231       0.0068811       6.9276097     


and estimate of N is

Code: Select all
                     Estimates of Derived Parameters
                       Population Estimates of {M0}
                                                95% Confidence Interval
 Grp. Sess.     N-hat        Standard Error      Lower           Upper
 ---- -----  --------------  --------------  --------------  --------------
   1     1    32.218334       1.0112231       32.006881       38.927610   
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