by cschwarz@stat.sfu.ca » Thu Jan 21, 2010 1:12 am
Hmmm...
Model 2: phi(stock:time) p(release:time)
My understanding is that the stocks are geographically separated and that for each stock, there are 2 release groups (e.g. week 1 and week 2). The capture occasions are "receivers" at dams downstream so there is no "time" per se (time being the receivers). In fact, the arrivals of each stock at the same receiver could be temporally separated as well, i.e. stock 1 arrives at receiver 5 in july while stock 2 arrives at receiver 5 in august.
SO there a total of 6 groups of fish released, 2 from each stock.
Likely a "better" notation would be
phi(stock:time) p(release(stock):time)
i.e. releases are nested within stocks so there is NO connection between release 1 of stock 1 and release 1 of stock 2 etc.
But, if the release groups are "connected", e.g. males and females so that a sex effect is being estimated for recovery probabilities, then we get the six products as being:
1. phi(s1, k-1) p(r1, k)
2. phi(s1, k-1) p(r2, k)
3. phi(s2, k-1) p(r1, k)
4. phi(s2, k-1) p(r2, k)
5. phi(s3, k-1) p(r1, k)
6. phi(s3, k-1) p(r2, k)
with 3 phis and 2 p's for 5 "raw" parameters.
There is still confounding with estimable parameters being:
(a) phi(s1, k-1) p(r1, k)
(b) p(r2,k)/p(r1,k) => hence (2 above) = (a)*(b)
(c) phi(s2,k-1)/phi(s1,k-1) => hence (3 above) = (a)*(c)
and (4 above) = (a)*(c)*(b)
(d) phi(s3,k-1)/phi(s1,k-1) => hence (5 above) = (a)*(d) and
(6 above) = (a)*(d)*(b).
Rather than being 5 raw parameters there are 4 estimable functions.
"Regardless of the previous issue, did you mean estimable parameters 3...6 here?" - oops - yes.