Consider Formula (10), which shows dα_i/dα_j. If the second-order condition is satisfied for the maximum of EU_i^N, the sign of the numerator is positive: that is, – ∂2EUjN/∂αj2>0. Therefore, the sign of dαi/dαj is identical with that of the denominator, ∂2EUjN/∂αj∂αi. Using the first-order condition represented by Formula (5), we obtain ∂2EUjN/∂αj∂αi as follows.

Then, Formula (23) can be rearranged:

which implies that dαi/dαj>0.

Appendix 2: Proof of Proposition 2 for the concave quadratic utility function

Consider now the effects on the contingent fees of w and v using the quadratic utility function. Using Lemma 3, we obtain

From Formulae (25) and (26), we find ∂αi∗/∂w>∂αi∗/∂v, which implies that if we show ∂αi∗/∂v>0, then we obtain ∂αi∗/∂w>∂αi∗/∂v>0. Formula (26) can be rearranged:

where 1>δw and 3(δv+w−1)+4δvδv−21−δw+91−δw21/2>0 for α1∗=α2∗>0.

Appendix 3: Proof of Proposition 2 for the negative exponential utility function

In Section 3, we obtain Formulae (8) and (9) by differentiating the first-order conditions represented by Formula (5). Formula (5) can be described by Formulae (15) and (16). First, consider the effect on the contingent fee fraction of w with the negative exponential utility function. From Formulae (15) and (16), we know that the equilibrium contingent fee fraction does not rely on w, which implies ∂αi∗/∂w=0. Next, consider the effect on the contingent fee fraction of v with the negative exponential function. Without loss of generality, we assume litigant 1 as litigant i and litigant 2 as litigant j, and we only consider litigant i’s behavior. Formula (9) can be rearranged:

Putting Formula (28) into Formula (8), we obtain

where the second-order conditions must be satisfied for ∂2EUi/∂αi2 <0 and∂2EUj/∂αj2<0. ^[14] With partial differentiation of Formula (15) with respect to α_j, we obtain ∂2EUi/∂αi∂αj and with partial differentiation of Formula (16) with respect to α_i, we obtain ∂2EUj/∂αj∂αi. The sign of ∂2EUi/∂αi∂αj is as follows.

With partial differentiation of Formulae (15) and (16) with respect to v, we obtain ∂2EUi/∂αi∂v and ∂2EUj/∂αj∂v. The sign of ∂2EUi/∂αi∂v is as follows. First, from Formulae (15) and (16), we obtain αj/αi+αj=αirv⋅exp[−r(1−αi)v]/1−exp[−r(1−αi)v] and αi/αi+αj=αjrv⋅exp[−r(1−αj)v]/1−exp[−r(1−αj)v]. Using the equations, we obtain ∂2EUi/∂αi∂v as follows.

since −1+exp[−r(1−αi)v]+r(1−αi)v>0. We can obtain ∂2EUj/∂αj∂v in the same way. ^[15] Considering ∂2EUi/∂αi2<0and,∂2EUj/∂αj2<0, we find that the sign of the numerator of Formula (29) is positive. Therefore, the sign of ∂αi∗/∂v is the same as the sign of the denominator of Formula (29). For concise exposition, we let [αj/αi+αj21−exp[−r(1−αi)v] as A,rvαj/αi+αj⋅exp[−r(1−αi)v)] as B,rvαj/αi+αj⋅exp[−r(1−αi)v)] as C,αi/αi+αj21−exp[−r(1−αj)v] as D,rvαi/αi+αj⋅exp[−r(1−αj)v] as E,andrv(1+αjrv)⋅exp[−r(1−αj)v] as F, where A, B, C, D, E, and F>0.

which implies ∂αi∗/∂v>0.

Appendix 4: Proof of Proposition 2 for the power utility function

Consider now the effects on the contingent fees of w and v with the power utility function. First, consider ∂αi∗/∂w. Putting Formula (28) into Formula (8), we obtain

where the second-order conditions must be satisfied for ∂2EUi/∂αi2<0and ∂2EUj/∂αj2<0. ^[16] With partial differentiation of Formula (19) with respect to α_j, we obtain ∂2EUi/∂αi∂αj and with partial differentiation of Formula (20) with respect to α_i, we obtain ∂2EUj/∂αj∂αi. The sign of ∂2EUi/∂αi∂αj is as follows.

With partial differentiation of Formulae (19) and (20) with respect to w, we obtain ∂2EUi/∂αi∂w. First, from Formulae (19) and (20), we obtain αj/αi+αj=αin−1/nw+1−αiv−1/n/w+1−αiv(n−1)/n−w(n−1)/n and αi/αi+αj=αjn−1/nw+1−αjv−1/n/w+1−αjv(n−1)/n−w(n−1)/n. Using the equations, we obtain ∂2EUi/∂αi∂w as follows.

We can obtain ∂2EUj/∂αj∂w in the same way. ^[17]

Considering ∂2EUi/∂αi2<0and ∂2EUj/∂αj2<0, we find that the sign of the numerator of Formula (33) is negative. Therefore, ∂αi∗/∂w has the opposite sign of the denominator of Formula (33). For concise exposition, we let −αj/αi+αj2 w+1−αivn−1/n−wn−1/n as G,−n−1/n −αjv/αi+αjw+1−αiv−1/nasH,−n−1/nw+1−αiv−1/n+αivw+1−αiv−1+n/n/n as I,−αi/αi+αj2w+1−αjvn−1/n−wn−1/n as J,−n−1/n−αiv/αi+αjw+1−αjv−1/n as K, and −n−1/nw+1−αjv−1/n+αjvw+1−αjv−1+n/n/n as L: G, H, I, J, K, and L<0.

which implies ∂αi∗/∂w<0.

Next, consider the sign of ∂αi∗/∂v which is shown in Formula (29). We know that the sign of the denominator of Formula (36) is positive. Herein, the sign of ∂αi∗/∂v is the same as that of the numerator of Formula (29). Since ∂2EUi/∂αi20,∂2EUj/∂αj20,∂2EUi/∂αi∂αj0 and ∂2EUj/∂αj∂αi0, the sign of ∂αi∗/∂v depends on the signs of ∂2EUi/∂αi∂v and ∂2EUj/∂αj∂v. Using Formulae (19) and (20), we obtain

which implies ∂αi∗/∂v>0.