A corporate-crime perspective on fisheries: liability rules and non-compliance

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1 A corporae-crime perspecive on fiseries: liabiliy rules and non-compliance FRANK JENSEN, Corresponding auor Universiy of Copenagen, Deparmen of Food and Resource Economics, Roligedsvej 3, 1958 Frederiksberg C., Copenagen, Denmark. Tel: Pone: LINDA NØSTBAKKEN Norwegian Scool of Economics, Deparmen of Economics, Bergen, Norway. ONLINE APPENDIX

2 Appendix A: Te basic model A.1. Reacion funcions In is secion, we derive and caracerize e reacion funcions presened and analyzed in e paper. From e main ex, e cos funcion saisfies e following properies: c L c 0, 0 L c c, 0, I 0 I c, L I c 0, 0, (A1) x c x I c 0, x L 0. In addiion, e properies of e penaly funcions are: G'( I ) 0, G''( I ) 0, F'( I ) 0, F''( I ) 0. (A) From secion.1, we ave e following firs-order condiions for e privae opimum: c p 0 L (A3) c p F( I ) G( I ) 0 I (A4) L Q. (A5) We can express e equaion sysem (A3)-(A5) as e reacion funcion presened in e main ex: ( Q, x, ). (A6) I I Toal differeniaing (A3)-(A5) yields e following: c c c d d d dx x L I L L I L (A7)

3 c c d [ ( F ( ) G ( ))] d L I I I L I I c dx [ F ( I ) G ( I )] d x I (A8) d L dq. (A9) Insering equaion (A9) ino (A7) and (A8) yields: c c c d d dx dq x I L I L L (A10) c c [ ( F ''( ) G ''( ))] d dx x I I I I I c [ F '( I ) G '( I )] d dq. (A11) L I Noe a equaion (A11) only depends on d I. Using is equaion, we can now find d dx I by seing d dq 0 : c d x. (A1) dx ( F ''( ) G ''( )) I I I c I I In (A1), e denominaor is posiive because 0 1, c 0 I, F''( I ) 0, and G''( I ) (cf. equaions (A1) and (A)). From (A1) we also ave a di 0 dx. c x I 0, wic implies a Turning o di d, we se dx dq 0 in (A11) and obain:

4 di [ F '( I ) G '( I )]. (A13) d c ( F ''( I ) G ''( I )) I Te denominaor is idenical o e one in equaion (A1), and is us posiive, and from (A) we ave a G'( I ) > 0 and F'( I ) >0. Consequenly, we find a 0 d. Le us finally deermine e effec of quoa on illegal arves. We le d dx 0 in (A11), and arrive a: d I c d. (A14) dq ( F ''( ) G ''( )) I I L I c I I From before, we know a bo e denominaor and e numeraor are posiive, since c L I di 0. Tis implies a 0. dq A. Enforcemen coss In is secion, we derive and caracerize e enforcemen cos funcion used in e paper. We sar ou by invering e reacion funcion in (A6), wic yields: ( Q, x, ). (A15) I Toal differeniaing (A15) produces: dq dx di 0. (A16) Q x I Nex, we define e probabiliy of being deeced as a funcion of enforcemen effor, ( e ), and we assume a: 0. (A17) e Noe a we can inver ( e ) o yield e ( ), and because of (A17) we obain e following:

5 e 1 0 e. (A18) Subsiuing e invered reacion funcion ino e ( ) gives e e ( ( Q, x, I )) ( Q, x, ). Now, we wan o find e sign of e derivaives of ( Q, x, ). Firs, we I invesigae e sign of by using: I I e. (A19) I I e From (A18), we ave a 0 and we noe a: 1 I I. (A0) From (A13) we ave a I 0, and consequenly, from (A0) we ge a 0. Using (A19) now implies a 0. Concerning e sign of I we ave a: x I e x x. (A1) e As above 0. By seing dq = 0 in (A16) and solving for we reac: x I x x I. (A)

6 I From (A1), we know a 0, and from (A0) a I 0. Combining is wi x (A) gives us a 0, wic in urn implies a 0. x Finally, we find e sign of by using a: Q x e. (A3) Q Q Seing dx = 0 in (A16) and solving for Q we ge: I Q Q I. (A4) We ave esablised a I I 0, and from (A14) we learned a 0. Terefore, Q (A4) implies a 0, and using is in (A3) gives us a 0. Q Le us nex urn o e enforcemen cos funcion, Ke ( ). We assume a: Q K e 0 and K 0 e. (A5) From above e e ( ( Q, x, )) ( Q, x, ) and insering is ino e enforcemen cos I I funcion gives K( e ( ( Q, x, ))) = F( ( Q, x, )) E( Q, x, ). We now wan o I I I deermine e signs of e derivaives of e enforcemen cos funcion, and we sar by considering E Q : E F K e. (A6) Q Q e Q

7 e K From (A3), 0, and from (A5) we ave a 0, wic implies a Q Q e E Q 0. Nex, for E x we ave a: E F K e x x e x. (A7) e K Using (A1), we ave a 0, and according o (A5), 0, from wic i x x e E follows a 0. x Finally, for E I we ge: E F K e. (A8) e I I I e K From (A19) we know a 0, and using (A3) we ave a 0. Tis e E implies a 0. I I I

8 Appendix B: Sare of profi B.1. Reacion funcions From secion 3 we ave e following firs-order condiions: W L W I c u 0 L c G I 0 I (B1) (B) L Q. (B3) We also ave e following wage sceme from secion 3:, 1,, W L I p L I c L I x F I. (B4) From e wage sceme in (B4) we may obain: W L c ( p (1 ) ) L (B5) W I c ( p (1 ) F ( I )). (B6) I (B5) can be subsiued ino (B1) and (B6) ino (B). Tis gives e following rewrien firsorder condiions: c c ( p ) (1 ) u 0 L L c c ( p ) (1 ) ( F I G I ) 0 I I (B7) (B8) L Q. (B9) (B7) - (B9) may be oal differeniaed wic gives:

9 c c c [ (1 ) ] d [ L L L I L c c c (1 ) ] di du [ (1 ) ] dx x x I L L L (B10) c c c c [ (1 ) ] d [ (1 ) L I L I L I I ( F ( ) G ( ))] d ( F ( ) G ( )) d I I I I I c c [ (1 ) ] dx x x I I (B11) d L dq. (B1) (B1) can be subsiued ino (B10) and (B11) wic gives: c c [ (1 ) ] di du I L I L c c c c [ (1 ) ] dx [ (1 ) ] dq x x L L L L (B13) c c [ (1 ) ( F ( I ) G ( I ))] di I I c c ( F ( I ) G ( I )) d [ (1 ) ] dx x x c c [ (1 ) ] dq I L I L I I (B14) d I is e only variable a eners in (B14) and, erefore, (B14) can be used o caracerize e reacion funcion. In (B14) we may se d dx 0 and reac: d I dq c ( (1 ) ) I L. (B15) c ( (1 ) ) ( F ( I ) G ( I )) I

10 We ave a 0 1, 0 1, 0 1, c I F >0 and G ( ) 0 and is >0, ( ) I I imply a e denominaor in (B15) is posiive. Wi respec o e nominaor c I L >0 so di e nominaor is also posiive. In oal, we, erefore, reac e conclusion a 0 dq. Concerning di d we se dq dx 0 in (B14) and arrive a: di F ( I ) G ( I ). (B16) d c ( (1 ) ) ( F ( I ) G ( I )) I From (B15) we ave a e denominaor is posiive and, in addiion, e nominaor in (B16) is posiive because G ( I ) 0 and F ( I ) 0. Terefore, we obain a 0 d. Las, by seing dq d 0 we reac: d I c ( (1 ) ) x. (B17) x ( (1 ) ) ( F ( ) G ( )) I I c I I I From above e denominaor is posiive. In addiion, we ave a c x I 0 so e I nominaor is negaive. In oal, is implies a 0. x B.. Enforcemen coss Te invered reacion funcion is: From (B18) we ge: ( Q, x, ). (B18) I

11 dq dx di 0. (B19) Q x I Now ( e ) is e probabiliy of being deeced as a funcion of enforcemen effor and we ave: 0. (B0) e We inver ( e ) o ge e ( ) and due o (B0) we obain: e 1 0 e. (B1) ( Q, x, ) can be used in e ( ) and is gives e e ( ( Q, x, )) ( Q, x, ). Now I we can find e sign of by using: I I I e. (B) I I e From (B1) 0 and furermore we ave a: 1 I I. (B3) From (B16) I 0 and by using is in (B3) we obain 0. Now (B) now imply I a 0. I For e sign of we ave: x e x x. (B4)

12 e In (B1) i was saed a 0 and using by dq = 0 in (B19) i is obained a: I x x I. (B5) I From (B17) 0, and in (B16) we reaced a I 0. Combining is in (B5) x 0, wic by using (B4) gives 0. x x Lasly, we urn aenion o e sign of were we ave: Q e. (B6) Q Q Using dx = 0 in (B19) and solving for Q we ge: I Q Q I. (B7) From (B16) I 0 I and using (B15) implies a 0. Terefore, 0 Q Q and by using is in (B6) i follows a 0. Q Now e enforcemen cos funcion is given as Ke ( ) and we assume a: K e 0 and K 0 e. (B8)

13 From before e e ( ( Q, x, )) ( Q, x, ) and insering is in e enforcemen cos I I funcion gives K( e ( ( Q, x, ))) = F( ( Q, x, )) E( Q, x, ). Now we can find e I I I sign of e derivaives and we sar by E Q were we ave: E F K e. (B9) Q Q e Q e In (B6) 0 Q Q K E and from (B8) 0, implying a 0. e Q Nex for e sign of E x we ave a: E F K e x x e x. (B30) e Using (B4) we ave a 0 x x K and from (B8) 0 e wic implies a E x 0. Las for e sign of E I we ge: E F K e. (B31) e I I I e From (B) 0 I I K E and using (B8) 0. In oal, is implies a 0. e I

14 Appendix C: Sare of revenue C.1. Reacion funcions Wi e sare of revenue rule e wage funcion is:, L I L I W p. (C1) Te general firs-order condiions for e employee are given by (B1)-(B3) in appendix B.1. Insering e derivaives of (C1) in e firs-order condiions gives: c p u 0 L c p G I 0 I (C) (C3) L Q. (C4) By oal differeniaing (C) - (C4) we ge a: c c c d d du dx x L I L L I L (C5) c c d [ G ( )] d L I I I L I c dx G ( I ) d x I (C6) d L dq. (C7) (C7) can be insered ino (C5) and (C6) wic yields: c c c d du dx dq x I L I L L (C8) c c c [ G ( )] d dx G ( ) d dq. (C9) x I I I I I I L

15 Since (C9) only depends on properies of e reacion funcion. d I, is equaion is e one we will consider o derive e Firs, we invesigae e sign of I dq and by seing d dx 0 in (C9) we reac: c I I L. (C10) dq c G ( ) I I Concerning (C10) c 0 I and G ( I ) 0 so e denominaor is posiive. Te nominaor is also posiive because I a 0. dq Seing d dq 0 in (C9) gives: c I L 0. In oal, (C10) erefore imply c I Ix. (C11) dx c G ( ) I I As in (C10) e denominaor is posiive. However, now c x I 0 so e nominaor is I negaive and is imply a 0. dx Las, we evaluae e sign of I d by seing dx dq 0 in (C9). Tis gives: I G ( I ). (C1) d c G ( I ) I

16 Te denominaor is posiive from (C10) and e nominaor is also posiive because G ( I ) 0. Tis implies a I 0. d C.. Enforcemen coss As before we ave an invered e reacion funcion given by: ( Q, x, ). (C13) I (C13) can be oal differeniaing: dq dx di 0. (C14) Q x I Now e probabiliy of being deeced is defined as ( e ) and we ave a: 0. (C15) e From ( e ) we ge e ( ) and because of (C15) we ave a: e 1 0 e. (C16) (C13) can be subsiued ino e ( ) o obain e e ( ( Q, x, I )) ( Q, x, I ). Now we wan o find e sign of e derivaives of ( Q, x, ). Firs, we consider e sign of I : I e. (C17) I I e From (C16) i is obained a 0. Furermore, we ave: 1 I I. (C18)

17 (C1) imply a I 0, and erefore we ave a 0 I by using (C18). Now (C17) implies a 0. Concerning e sign of I we ge a: x e x x. (C19) e (C16) express a 0 and by using dq = 0 in (C14) we reac: I x x I. (C0) I In (C11) we ave a 0, and from (C18) we reaced a I 0. Combining is x informaion implies a 0 x and using (C19) gives 0. x Lasly, we find e sign of by using a: Q e. (C1) Q Q By seing dx = 0 in (C14) we ge: I Q Q I. (C) From (C18) I 0 I and furermore we ave a 0 Q in (C10). Terefore, (C) implies a 0 Q and using is in (C1) gives 0. Q Now e enforcemen cos funcion is given as Ke ( ) and we assume a:

18 K e 0 and K 0 e. (C3) Now we ave a e e ( ( Q, x, )) ( Q, x, ) and insering is in e enforcemen I I cos funcion gives K( e ( ( Q, x, ))) = F( ( Q, x, )) E( Q, x, ). Now we can find I I I e sign of e derivaives of e enforcemen cos funcion and we sar by e sign of E Q were we ave: E F K e. (C4) Q Q e Q e From (C1) we ge a 0 Q Q K E and from (C3) 0, implying a 0. e Q Nex for e sign of E x we ave a: E F K e x x e x. (C5) e K Using (C19) we ave a 0 and using a 0 x x e in (C3) is implies a E x 0. Las for E I we ge: E F K e. (C6) e I I I e K (C17) gives 0 and using (C3) we ave a 0. In oal, is implies e E a 0. I I I