858 



Fishery Bulletin 100(4) 



because the residuals corresponding to the "never seen 

 again" cells are zero, it is not possible for a row pattern to 

 be expressed in the residuals matrices of either model 1 or 

 model 1. This constraint renders it very difficult to detect 

 assumption violations that are cohort-specific — the most 

 common being tag-induced and handling mortality and 

 short-term tag loss. 



Results 



To verify that the aforementioned constraints about the 

 structure of the residuals matrices associated with model 

 1 and model 1' are true, we offer the following mathemati- 

 cal arguments. The proofs simply involve algebraic manip- 

 ulation of equations involving the analytical formulae for 

 the maximum likelihood parameter estimates (MLE) of 

 model 1. The formulae for the MLEs were originally devel- 

 oped by Seber (1970) and can be applied to both to model 

 1 and model 1 (because r, can be expressed as a function 

 of S, and/",, the invariance property of MLEs implies that 

 an MLE of r, can be obtained by the transform). Hence, 

 the proofs are developed for model 1, and we note that 

 similar arguments could be constructed for the residuals 

 of model 1 . 



Recall that the analytical solutions for the maximum 

 likelihood estimates of/] and S, from Seber (1970) and 

 Brownie et al. ( 1985) are given by 



NT 



and S, = — ^   -^ . 



where R^ and C, are the row and column totals of the 

 observed data in year /; and 



T,=R,^T__,-C,_, , = 2,... J 



Tuj = T,^j_,-C,^j_, j=l,...,J-nfJ>I. 



The (1,1) cell 



To show that the residual associated with the (1,1) cell 

 is always zero, we must demonstrate that the difference 

 between the obsei-ved and estimated expected value in the 

 first cell is always zero. Hence, we have 



1R C ^ 

 — ^ — - (substituting for/", ) 

 7V,T, j ^ " 



= '"ii ~ C, (since T, = i?, ) 



= /jj - Tjj = (because the column total 



in the first recovery year is 1-^). 



The {/,/) cell when / = J 



To show that the residual associated with the (/,/) cell 

 is always zero when / = J. again we must demonstrate 



that the difference between the observed and estimated 

 expected values in that cell is always zero. Hence, we 

 have 



'n-E,i =r„-N,fi 



( R C \ 

 = r,, - N, — - — - (substituting for/",) 



" '\N,T,j ^ " 



= rjj - R, ( because 7) = C; ) 

 = fji - ;•,/ = ( because the row total in the 

 final recovery year is r^;). 



Column sums when I = J 



To show that the column sums of the residuals matrix 

 equal zero, we must demonstrate the column sum of the 

 observed data equals that of the expected values. Consider 

 the sum of the expected values associated with the /"^ 

 column of the recovery matrix, that is 



N.,(S,-S,_,)f,+ -+N,_,S,J,+N,f]. 

 Now substitute for /" and S, on the right hand side: 



' Ri (Ti-C^) N., 

 . N, T, R, 



^/-i T,^, R, J J 



N., 



R, (7?, -C) N., 



R,^ ^T,_, 



Ml] 



C,_,)N, 



N.-, T, 



R, 



N, 



R, I 



R£l 



\ N,T, J 



 + Af,_, 



Rj-i iT,_j -C,_i) R, 

 ^/-i 7)-! N, j 



N,T,] '[n,T, 



Cancel terms and factor out the term 



C/. 



Q 



(£l' 

 [t, 



(R,(T,^C,)](T,-C,] (T,_,-C,A 



[ T., 



Tj., 



R^iT^-C.,)' 

 T. 



(T,-C, 



7^7-1 -Q_i 



£l 



( R,_,{T,_,-C,_,) 



7)-! 



+ R, 



iT -C ) 

 Systematically factor out terms of the form — '■ '— ire- 



call thatT, =/?i): ^' 



