Lambda estimates output ?

[bp]

Hi Folks,

I understand that lambda is not estimable in the first and last occassions, but is MARK reporting occasion K1 through Kx-1 or is it reporting K2 through Kx? I have 14 encounter histories and 13 estimates of Lambda. Both 1 and 13 don't have SE estimates, so either K2 or K13 also isn't estimable with this data set- not sure which it is.

Thanks for the help.
bp

[jhines]

Assuming you are talking about the full time-specific model, the first and last Lambda are not identifiable. If you have 14 years, you'll get 13 Lambda's (I call them Lambda(1)..Lambda(13)), but the first and last ones are products of Lambda and phi or p. If you use a model where the first and last p are estimable, then you should be able to estimate all of the Lambda's.

JHines

[bp]

Thanks for the reply Jim. Yes I'm using a fully time dependent model (phi(t)p(t)lambda(t)). My first and last p (p(1) and p(14)) estimates are reasonable, but SE's came out to zero. In my case lambda(1) and lambda (13) have reasonable estimates, but zero's for SE, and there is no estimate for lambda (14) listed in the output. So in this case the lambda's are not estimable in the first and last two occasions? Your message states that if the first and last p are estimable then I should be able to estimate all the lambda's, shouldn't that be all but the last one?

Thanks again for the help and the lesson.


bp

Assuming you are talking about the full time-specific model, the first and last Lambda are not identifiable. If you have 14 years, you'll get 13 Lambda's (I call them Lambda(1)..Lambda(13)), but the first and last ones are products of Lambda and phi or p. If you use a model where the first and last p are estimable, then you should be able to estimate all of the Lambda's.

JHines

[jhines]

If you have 14 years, there can only be 13 Lambda's because Lambda(i)=N(i+1)/N(i). So, the last Lambda is Lambda(13)=N(14)/N(13). With time-specific phi and p, the last N (N(14)) is inestimable, so Lambda(13) is inestimable. Similarly, the first N (N(1)) is inestimable meaning Lambda(1) is also inestimable. If p is constant over time, then all 14 N's are estimable, so all 13 Lambda's are estimable.

[bp]

Got it, THANKS!

If you have 14 years, there can only be 13 Lambda's because Lambda(i)=N(i+1)/N(i). So, the last Lambda is Lambda(13)=N(14)/N(13). With time-specific phi and p, the last N (N(14)) is inestimable, so Lambda(13) is inestimable. Similarly, the first N (N(1)) is inestimable meaning Lambda(1) is also inestimable. If p is constant over time, then all 14 N's are estimable, so all 13 Lambda's are estimable.