Equilibrium with Observable Scores

(* Run this set of commands first!

This Section derives the equations characterizing the equilibrium, leading up to locus A.19 in Appendix A
Recall that we will have M=m0+m1 Y with m1=χ. 
We also let sx:=(Subscript[σ, ξ]) and st:=(Subscript[σ, θ]) *)

Clear["Global`*"]
(*Guess for the policy in terms of primitives, i.e., including the current price p, drives monopoly price ps*)
Q=q0+q1  θ+q2 M-p;
u=Q(θ-p)-(Q^2)/2;
ps=((q0+q1  M+q2 M)/2);
(*Change variables to α,β,δ*)
sol0={q0,q1,q2}/.Solve[{δ==q0/2,α==q1,β==(q2-q1)/2},{q0,q1,q2}][[1]];
q0=sol0[[1]];
q1=sol0[[2]];
q2=sol0[[3]];

(*New expression for quantity Q*)
Simplify[Collect[Q/.p->ps,{θ,M}]];

(*Guess for the value function*)
V=v0+v1 θ+v2 M+v3 (M^2)+v4 (θ^2)+v5 θ M;

(*The consumer’s problem is to max Q given ps*)
OBJ=q(θ-ps)-(q^2)/2-κ(θ-μ)D[V,θ]+D[V,M](m1 q-ϕ (M-m0))+(1/2)(st^2)D[V,{θ,2}]+(1/2)(m1^2)(sx^2)D[V,{M,2}];

(*First-order condition*)
FOC=Simplify[D[OBJ,q]//.{q->Q,p->ps}];

(*Match coefficients*)
v5s=v5/.Solve[(Coefficient[FOC,θ,1])==0,v5][[1]];
v3s=v3/.Solve[(Coefficient[FOC,M,1])==0,v3][[1]];
v2s=v2/.Solve[(FOC/.{v3->v3s,v5->v5s})==0,v2][[1]];

(*Now compute the equilibrium payoff*)
HJB=Q(θ-p)-(Q^2)/2-κ(θ-μ)D[V,θ]+D[V,M](m1 Q-ϕ (M-m0))+(1/2)(st^2)D[V,{θ,2}]+(1/2)(m1^2)(sx^2)D[V,{M,2}]/.p->ps;
MA=-r V+HJB;

(*The envelope is*)
dHJB=D[q(θ-ps)-(q^2)/2-κ(θ-μ)D[V,θ]+D[V,M](m1 q-ϕ (M-m0))+(1/2)(st^2)D[V,{θ,2}]+(1/2)(m1^2)(sx^2)D[V,{M,2}],M]//.{q->Q,p->ps};
(*Therefore, this is a relationship that must hold =0 for all θ and y*)
ENV=(-r D[V,M]+dHJB);

(*We now simplify the envelope conditions’ coefficients wrt θ and M, and the constant, using the solutions from the FOC.
At this stage, we have a system of 4 eqns in 4 unκnowns, including the initial equation for m1*)
sys=FullSimplify[{m1 Coefficient[ENV,θ],m1 Coefficient[ENV,M], m1 ENV/.{θ->0,M->0},χ-m1}//.{v2->v2s,v3->v3s,v5->v5s}];

(*Solve the first two for β and m1*)
{m1s,βs}=FullSimplify[{m1,β}/.Solve[{sys[[2]]==0,sys[[1]]==0},{m1,β}][[1]]];

(*and also derive the constant m0 for M= m0+m1 Y *)
m0s=((μ(ϕ-β m1)-(δ+α μ)m1)/ϕ)//.{m1->m1s,β->βs};

(*Then use the third equation to solve for δ, which is isolated*)
δs=FullSimplify[δ/.Solve[(sys[[3]]//.{m1->m1s,β->βs,m0->m0s})==0,δ][[1]],Assumptions->ϕ>0&&κ>0&&r≥0&&1>α>0];

(*Now recall that we can write Cov(θ,Y)/Var(Y) as... (also display 12) *)
χ=α (st^2)(((ϕ-β m1))/((α^2)(st^2)+κ (sx^2)(ϕ+κ-β m1)));
m1c=Simplify[m1/.Solve[m1==χ,m1][[1]]];
(*Finally, obtain an equation for (α,ϕ)==0, which we solve for ϕ (it is a cubic)*)
fin=FullSimplify[χ-m1//.{m1->m1s,β->βs}];

(* We now have the coefficients (α,β,δ) up to an equation ("fin") that ties α & ϕ. *)
(*Now compute the steady-state variance in equilibrium and compares it to the case of public signal histories*)
γf=FullSimplify[((st^2)/(2κ))((4(q1^2)(st^2)κ+κ (sx^2)(((2(ϕ+κ)-(q2-q1)m1))^2))/((2(q1^2)(st^2)+κ (sx^2)(2(ϕ+κ)-(q2-q1)m1))(2(ϕ+κ)-(q2-q1)m1)))//.{m1->m1s,β->βs}];
(*Now compare with the public histories benchmarκ*)
γs=γ/.Solve[-2κ γ+(st^2)-(((α γ/sx))^2)==0,γ][[2]];
finb=FullSimplify[ϕ-(κ+q1 ((γs)/(sx^2))(q1+((q2-q1)/2)))//.{β->βs}];
fin/.{α->1,ϕ->0};
af=(-4 (((16 (κ^4) (sx^4) (((-1+α))^2)-12 (κ^3) r (sx^4) (((-1+α))^3)+(st^4) (α^4) (((1-3 (α^2)))^2)+(κ^2) (sx^2) (-1+α) (3 (r^2) (sx^2) (-1+α) (1+3 (α^2))+2 (st^2) (α^2) (1+15 (α^2)))-3 κ r (st^2) (sx^2) (α^2) (1+α (2+3 α (1+2 (-2+α) α)))))^3)+(((-128 (κ^6) (sx^6) (((-1+α))^3)+144 (κ^5) r (sx^6) (((-1+α))^4)+2 (st^6) (α^6) (((-1+3 (α^2)))^3)-9 κ r (st^4) (sx^2) (-1+α) (α^4) (-1+3 (α^2)) (-1+3 α (-1+2 (-1+α) α))-9 (κ^3) r (sx^4) (((-1+α))^2) α (6 (r^2) (sx^2) (((-1+α))^2) (1+α)+(st^2) α (1+7 α) (-7-3 α+6 (α^2)))+6 (κ^4) (sx^4) (((-1+α))^2) (6 (r^2) (sx^2) (-1+α) (-1+3 α (2+α))+(st^2) (α^2) (5+3 α (18+7 α)))+6 (κ^2) (st^2) (sx^2) (-1+α) (α^2) ((st^2) (α^2) (-1-12 (α^2)+45 (α^4))+3 (r^2) (sx^2) (-1+α) (1+3 α (2+α (-1+3 (-1+α) α))))))^2));
daf=Simplify[-D[fin,ϕ]/D[fin,α],Assumptions->κ>0&&r>0&&0<α<1&&st>0&&sx>0];
F1=FullSimplify[fin/.{sx->(Sqrt[s])st},Assumptions->κ>0&&r>0&&1>α>1/2&&st>0&&s>0&&ϕ>0];
xa=FullSimplify[s/.Solve[F1==0,s][[1]]]; (*this solves for the relative variance of the noise terms in an equilibrium with (α,ϕ) *)
P=FullSimplify[α+βs+δs/μ];
G=FullSimplify[(((α λ)/(ϕ+κ-βs λ)))//.{λ->m1s}];
(* End of the equilibrium computation*)

Proposition 7 (Optimal Persistence)

(* We use the expressions for profits and surplus from Supplementary Appendix S.3*)

CS0=μ ((r+ϕ)/(2r+3ϕ))(μ-(3/2)μ ((r+ϕ)/(2r+3ϕ)))+FullSimplify[varθ(α(1-α/2)+((α^2)/2+βs-(3/2)(((α+βs))^2))((α λ)/(ϕ+κ-βs λ)))//.{λ->(ϕ/(α+βs)),α->((2(r+κ+ϕ))/(r+2κ+(Sqrt[(r+2ϕ)(r+6ϕ)])))}];

PS0=(((μ ((r+ϕ)/(2r+3ϕ))))^2)+FullSimplify[varθ(((α+βs))^2)((α λ)/(ϕ+κ-βs λ))//.{λ->(ϕ/(α+βs)),α->((2(r+κ+ϕ))/(r+2κ+(Sqrt[(r+2ϕ)(r+6ϕ)])))}];

(* For part (iii), differentiate both expressions and take limits as ϕ->∞ *)

FullSimplify[Limit[(ϕ^2)(D[PS0,ϕ]),ϕ->∞],Assumptions->κ>0&&ϕ>0&&r>0]

FullSimplify[Limit[(ϕ^2)(D[CS0,ϕ]),ϕ->∞],Assumptions->κ>0&&ϕ>0&&r>0]

(* For part (iv), we look for critical points ϕ of CS0 and PS0 and solve for the inverse functions μf(ϕ) and μc(ϕ) *)

m1=Simplify[m/.Solve[(D[PS0,ϕ]/.{varθ->1,μ->(Sqrt[m])})==0,m][[1]],Assumptions->κ>0&&ϕ>0&&r>0];

m2=Simplify[m/.Solve[(D[CS0,ϕ]/.{varθ->1,μ->(Sqrt[m])})==0,m][[1]],Assumptions->r>0&&κ>0&&ϕ>0];

(* CLAIM: Now find conditions under which m1 is downward sloping and m2 is upward sloping at ϕ=0 *)

FullSimplify[(D[m1,ϕ]//.{r->ρ κ,ϕ->0}),Assumptions->κ>0&&ρ>0]

FullSimplify[Solve[FullSimplify[(D[m1,ϕ]//.{r->ρ κ,ϕ->0})]==0,ρ]][[2]]

FullSimplify[(D[m2,ϕ]//.{r->ρ κ,ϕ->0}),Assumptions->κ>0&&ρ>0]

NSolve[FullSimplify[D[m2,ϕ]//.{r->ρ κ,ϕ->0},Assumptions->κ>0&&r>0]==0,ρ,Reals][[3]]

(* CLAIM: m1 is quasiconcave and m2 is quasiconvex 
For both loci, we consider the second derivative at a critical point. 
We show there is no combination of (ϕ,κ,r) for which the second derivative is positive (negative) at any such point. *)

Proof that (μ^f)(ϕ) is quasiconcave

Simplify[(r (((ϕ+r))^2) (((12 (ϕ^2)+8 ϕ r+(r^2)))^2) (((2 κ+r+(Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)])))^4) (((12 (ϕ^2)+2 κ (r+(Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+ϕ (6 κ+5 r+3 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))))^3))D[m1,ϕ],Assumptions->κ>0&&ϕ>0&&r>0]/.(Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)])->x;

Solve for the square root term

x0=FullSimplify[x/.Solve[%==0,x][[1]],Assumptions->κ>0&&ϕ>0&&r>0];

Plug in the square root term in the second derivative (i.e., evaluate at a critical point)

m11=FullSimplify[Collect[Simplify[(Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)])r (((ϕ+r))^3) (((12 (ϕ^2)+8 ϕ r+(r^2)))^3) (((2 κ+r+(Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)])))^5) (((12 (ϕ^2)+2 κ (r+(Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+ϕ (6 κ+5 r+3 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))))^4)D[m1,{ϕ,2}],Assumptions->κ>0&&ϕ>0&&r>0]/.(Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)])->x,x]/.x->x0];

FindInstance[m11>0&&x0>0&&κ==1&&r>0&&ϕ>0,{r,ϕ,κ}]

Proof that (μ^c)(ϕ) is quasiconvex

Simplify[(r (((2268 (ϕ^8)+4 (κ^2) (r^2) (((κ+r))^3) (r+(Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+216 (ϕ^7) (30 κ+31 r+3 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+27 (ϕ^6) (280 (κ^2)+616 κ r+295 (r^2)+70 κ (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)])+63 r (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+4 ϕ κ r (((κ+r))^2) (4 (r^2) (r+(Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+(κ^2) (11 r+7 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+κ r (21 r+17 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)])))+12 (ϕ^5) (360 (κ^3)+36 (κ^2) (38 r+5 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+4 (r^2) (101 r+36 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+9 κ r (157 r+39 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)])))+2 (ϕ^4) (630 (κ^4)+18 (κ^3) (214 r+35 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+4 (r^3) (197 r+105 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+9 (κ^2) r (769 r+221 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+3 κ (r^2) (1451 r+591 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)])))+4 (ϕ^3) (36 (κ^5)+16 (r^4) (4 r+3 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+6 (κ^4) (73 r+15 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+6 (κ^3) r (215 r+76 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+κ (r^3) (587 r+347 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+(κ^2) (r^2) (1427 r+669 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)])))+(ϕ^2) (16 (r^5) (r+(Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+6 (κ^5) (24 r+7 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+8 κ (r^4) (39 r+31 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+(κ^4) r (857 r+381 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+2 (κ^3) (r^2) (801 r+454 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+(κ^2) (r^3) (1189 r+797 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)])))))^2))((D[m2,ϕ])/((1/(r (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)])))(((3 ϕ+2 r))^2) )),Assumptions->κ>0&&ϕ>0&&r>0]/.(Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)])->y;

x2=FullSimplify[y/.Solve[%==0,y][[1]]];

Now the second derivative

Simplify[(r^2) (((12 (ϕ^2)+8 ϕ r+(r^2)))^(3/2)) (((2268 (ϕ^8)+4 (κ^2) (r^2) (((κ+r))^3) (r+(Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+216 (ϕ^7) (30 κ+31 r+3 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+27 (ϕ^6) (280 (κ^2)+616 κ r+295 (r^2)+70 κ (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)])+63 r (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+4 ϕ κ r (((κ+r))^2) (4 (r^2) (r+(Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+(κ^2) (11 r+7 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+κ r (21 r+17 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)])))+12 (ϕ^5) (360 (κ^3)+36 (κ^2) (38 r+5 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+4 (r^2) (101 r+36 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+9 κ r (157 r+39 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)])))+2 (ϕ^4) (630 (κ^4)+18 (κ^3) (214 r+35 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+4 (r^3) (197 r+105 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+9 (κ^2) r (769 r+221 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+3 κ (r^2) (1451 r+591 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)])))+4 (ϕ^3) (36 (κ^5)+16 (r^4) (4 r+3 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+6 (κ^4) (73 r+15 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+6 (κ^3) r (215 r+76 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+κ (r^3) (587 r+347 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+(κ^2) (r^2) (1427 r+669 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)])))+(ϕ^2) (16 (r^5) (r+(Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+6 (κ^5) (24 r+7 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+8 κ (r^4) (39 r+31 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+(κ^4) r (857 r+381 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+2 (κ^3) (r^2) (801 r+454 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)]))+(κ^2) (r^3) (1189 r+797 (Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)])))))^3)((D[m2,{ϕ,2}])/(2 (3 ϕ+2 r))),Assumptions->κ>0&&r>0&&ϕ>0]/.(Sqrt[12 (ϕ^2)+8 ϕ r+(r^2)])->y;

m22=FullSimplify[Collect[%,y]/.y->x2,Assumptions->κ>0&&r>0&&ϕ>0];

Plug in the square root term in the second derivative (i.e., evaluate at a critical point)

FindInstance[m22<0&&x2>0&&κ==1&&r>0&&ϕ>0,{r,ϕ,κ}]

Proposition 8 (Naive vs. Strategic)

Here are the  expressions for consumer surplus in the observable, naive, and hidden case. In all these expressions, μ denotes  μ (Sqrt[κ])/(Subscript[σ, θ])

CSH=FullSimplify[-z  (((r+ϕ)(r+3ϕ))/(2(r+ϕ)(r+3ϕ)+2(ϕ^2)))μ(μ+((1/2)-(1/z))z(((r+ϕ)(r+3ϕ))/(2(r+ϕ)(r+3ϕ)+2(ϕ^2)))μ)+(1/(2 ))((α-((α^2)/2))+α((α/4)((3/2)+(1/z))-(1/2))(ϕ/(ϕ +κ/2)))//.{α->((r+κ+ϕ)/(r+κ+2ϕ)),z->-((r+2ϕ)/(r+3ϕ)),λ->(ϕ/(α+β)),β->-α/2,r->ρ κ,ϕ->f κ}];CSN=Collect[FullSimplify[(((μ )/2))(μ-(3/4)μ)+(1/(2 ))((α-((α^2)/2))+(((α^2)/2)+β-(3/2)(((α+β))^2))((α λ)/(ϕ+κ-β λ)))//.{α->1,λ->(ϕ/(α+β)),β->-1/2,r->ρ κ,ϕ->f κ},Assumptions->ρ>0&&f>0&&κ>0&&μ>0],μ];

CSO=Simplify[(μ ((r+ϕ)/(2r+3ϕ)))(μ-(3/2)(μ ((r+ϕ)/(2r+3ϕ))))+(1/(2 ))((α-((α^2)/2))+(((α^2)/2)+βs-(3/2)(((α+βs))^2))((α λ)/(ϕ+κ-βs λ)))//.{α->((2(r+κ+ϕ))/(r+2κ+(Sqrt[(r+2ϕ)(r+6ϕ)]))),λ->(ϕ/(α+βs)),r->ρ κ,ϕ->f κ},Assumptions->ρ>0&&f>0&&κ>0&&μ>0];

We obtain the following difference (when μ=0 and wlog κ=1)

diffON=CSO-CSN/.{κ->1,μ->0}

FindInstance[diffON<0&&ρ>0&&f>0,{ρ,f}]

Therefore, CS with Observable scores dominates the Naive case

Supplementary Appendix S.2.3

Now we compute the variance of the price and consider the intersection of the inequalities below, which is empty. This proves that Var[P] is decreasing in ϕ when α is increasing in ϕ

varP=FullSimplify[(((α+βs))^2)G/.{ϕ->f κ,r->ρ κ},Assumptions->f>0&&κ>0&&r≥0&&1>α>0];

xaa=Simplify[(κ^2)xa/.{ϕ->f κ,r->ρ κ}];

FindInstance[D[varP,α]>0&&1>α>0&&f>0&&ρ>0&&xaa>0,{α,f,ρ}]
FindInstance[D[varP,ϕ]>0&&1>α>0&&f>0&&ρ>0,{α,f,ρ}]

Proposition 10 (Observable vs Hidden Scores--Expected Price)

Now compare the expected price under observable scores to the hidden case.
Recall the (α,ϕ) equilibrium locus under observable scores and define its analogue for the hidden case.

FP=((α (r+κ-α (r+κ)+ϕ))/((α^3)+s κ (r-r α+κ+ϕ)))+(((-1+α) (r+κ+ϕ) (r+κ+ϕ+α (r-κ+3 ϕ)))/((α^3) (r+2 ϕ)));
FH=((2 (r+κ+ϕ) (-1+(Subscript[α, h])))/((((Subscript[α, h])))^2))+(((Subscript[α, h]) (r+κ+ϕ-(r+κ)(Subscript[α, h])))/(((((Subscript[α, h])))^3)+s κ (r+κ+ϕ-r (Subscript[α, h]))));
s1=FullSimplify[s/.Solve[FP==0,s][[1]]];;
s2=FullSimplify[s/.Solve[FH==0,s][[1]]];
(*Recall the sufficient condition from Section 7*)
cond=α(α-(Subscript[α, h]))(r+2ϕ)-(((1-α))^2)(Subscript[α, h])(r+κ+ϕ);
(* and cond<0 if αH>(((2 ϕ+r) (α^2))/(ϕ+κ+r-(2 κ+r) α+(ϕ+κ+r) (α^2)))*)
αH=FullSimplify[(Subscript[α, h])/.Solve[cond==0,(Subscript[α, h])][[1]],Assumptions->1>α>0&&r>0&&ϕ>0&&κ>0&&s1>0]

(* Now compare the s solutions*)
FullSimplify[(s1/s2)/.(Subscript[α, h])->αH,Assumptions->1>α>0&&r>0&&ϕ>0&&κ>0]-1

(*Further simplify and ignore positive constants.
Reduce the inequalities below to show their intersection is empty*)FindInstance[1>α>1/2&&r>0&&ϕ>0&&κ==1&&(α^3) (r-r α+κ+ϕ) (((r+2 ϕ))^3) (r+κ+ϕ+α (r-κ+3 ϕ))-2 (r+κ-α κ+ϕ+α ϕ) (((r+κ-α (r+2 κ)+ϕ+(α^2) (r+κ+ϕ)))^2) (-α (r+2 κ) (r+κ+ϕ)+(((r+κ+ϕ))^2)+(α^2) (2 r κ+(((κ+ϕ))^2)))>0,{α,ϕ,r,κ}]
(*Now we know the implied s2<s1. Because we know both α and (Subscript[α, h]) are increasing in s, this means for the same true ratio s we need (Subscript[α, h]) to increase,and hence to satisfy cond<0*)

Proposition 11 (Observable vs. Hidden--Consumer Surplus)

(* We let μ0= μ (Sqrt[κ])/(Subscript[σ, θ]) *)

CSH=FullSimplify[-z  (((r+ϕ)(r+3ϕ))/(2(r+ϕ)(r+3ϕ)+2(ϕ^2)))μ(μ+((1/2)-(1/z))z(((r+ϕ)(r+3ϕ))/(2(r+ϕ)(r+3ϕ)+2(ϕ^2)))μ)+(1/(2 ))((α-((α^2)/2))+α((α/4)((3/2)+(1/z))-(1/2))(ϕ/(ϕ +κ/2)))//.{α->((r+κ+ϕ)/(r+κ+2ϕ)),z->-((r+2ϕ)/(r+3ϕ)),λ->(ϕ/(α+β)),β->-α/2,μ->μ0}];
CSO=Simplify[(μ ((r+ϕ)/(2r+3ϕ)))(μ-(3/2)(μ ((r+ϕ)/(2r+3ϕ))))+(1/(2 ))((α-((α^2)/2))+(((α^2)/2)+βs-(3/2)(((α+βs))^2))((α λ)/(ϕ+κ-βs λ)))//.{α->((2(r+κ+ϕ))/(r+2κ+(Sqrt[(r+2ϕ)(r+6ϕ)]))),λ->(ϕ/(α+βs)),μ->μ0},Assumptions->r>0&&ϕ>0&&κ>0&&μ0>0];

(*Verify that the difference when μ0=0 satisfies

diffOH=(ϕ (r+κ+ϕ) (96 (ϕ^6)-2 r (r-κ) (κ^2) (r+κ) (-r+(Sqrt[(r+2 ϕ) (r+6 ϕ)]))+4 (ϕ^5) (41 r+72 κ+6 (Sqrt[(r+2 ϕ) (r+6 ϕ)]))+(ϕ^4) (111 (r^2)+474 r κ+264 (κ^2)+35 r (Sqrt[(r+2 ϕ) (r+6 ϕ)])+52 κ (Sqrt[(r+2 ϕ) (r+6 ϕ)]))+κ ϕ (2 (r^2) κ (20 r-(Sqrt[(r+2 ϕ) (r+6 ϕ)]))+(r^3) (9 r+(Sqrt[(r+2 ϕ) (r+6 ϕ)]))+(κ^3) (-6 r+8 (Sqrt[(r+2 ϕ) (r+6 ϕ)]))+r (κ^2) (13 r+17 (Sqrt[(r+2 ϕ) (r+6 ϕ)])))+(ϕ^2) (4 (r^3) (r+(Sqrt[(r+2 ϕ) (r+6 ϕ)]))+(r^2) κ (81 r+25 (Sqrt[(r+2 ϕ) (r+6 ϕ)]))+(κ^3) (66 r+32 (Sqrt[(r+2 ϕ) (r+6 ϕ)]))+r (κ^2) (209 r+35 (Sqrt[(r+2 ϕ) (r+6 ϕ)])))+(ϕ^3) (72 (κ^3)+(r^2) (35 r+19 (Sqrt[(r+2 ϕ) (r+6 ϕ)]))+(κ^2) (406 r+52 (Sqrt[(r+2 ϕ) (r+6 ϕ)]))+r κ (293 r+67 (Sqrt[(r+2 ϕ) (r+6 ϕ)])))))/(2 (r+2 ϕ) (κ+2 ϕ) (((r+κ+2 ϕ))^2) (((r+2 κ+(Sqrt[(r+2 ϕ) (r+6 ϕ)])))^2) (12 (ϕ^2)+2 κ (r+(Sqrt[(r+2 ϕ) (r+6 ϕ)]))+ϕ (5 r+6 κ+3 (Sqrt[(r+2 ϕ) (r+6 ϕ)]))));

Simplify[(CSO-CSH)-diffOH/.{μ0->0},Assumptions->κ>0&&ϕ>0&&r>0]

(* Isolating the functions f1 and f2 from Supplementary Appendix S.2.4 *)

f1=((ϕ (k+r+ϕ))/((k+2 ϕ) (r+2 ϕ) (((k+r+2 ϕ))^2) (((2 k+r+(Sqrt[(r+2 ϕ) (r+6 ϕ)])))^2) (12 (ϕ^2)+2 k (r+(Sqrt[(r+2 ϕ) (r+6 ϕ)]))+ϕ (6 k+5 r+3 (Sqrt[(r+2 ϕ) (r+6 ϕ)])))));
f2=96 (ϕ^6)+4 (ϕ^5) (72 k+41 r+6 (Sqrt[(r+2 ϕ) (r+6 ϕ)]))+(ϕ^4) (264 (k^2)+474 k r+111 (r^2)+52 k (Sqrt[(r+2 ϕ) (r+6 ϕ)])+35 r (Sqrt[(r+2 ϕ) (r+6 ϕ)]))+(ϕ^2) (4 (r^3) (r+(Sqrt[(r+2 ϕ) (r+6 ϕ)]))+k (r^2) (81 r+25 (Sqrt[(r+2 ϕ) (r+6 ϕ)]))+(k^3) (66 r+32 (Sqrt[(r+2 ϕ) (r+6 ϕ)]))+(k^2) r (209 r+35 (Sqrt[(r+2 ϕ) (r+6 ϕ)])))+(ϕ^3) (72 (k^3)+(r^2) (35 r+19 (Sqrt[(r+2 ϕ) (r+6 ϕ)]))+(k^2) (406 r+52 (Sqrt[(r+2 ϕ) (r+6 ϕ)]))+k r (293 r+67 (Sqrt[(r+2 ϕ) (r+6 ϕ)])));

(* Clearly f1>0 and f2>0 

The proof of Proposition 11 in the Supplementary Appendix contains the remaining steps*)

Proposition 12 (Naive vs. Hidden--Consumer Surplus)

Now compare consumer surplus with  hidden scores and with naive consumers. In  these expressions, μ denotes  μ (Sqrt[κ])/(Subscript[σ, θ])

CSH=FullSimplify[-z  (((r+ϕ)(r+3ϕ))/(2(r+ϕ)(r+3ϕ)+2(ϕ^2)))μ(μ+((1/2)-(1/z))z(((r+ϕ)(r+3ϕ))/(2(r+ϕ)(r+3ϕ)+2(ϕ^2)))μ)+(1/(2 ))((α-((α^2)/2))+α((α/4)((3/2)+(1/z))-(1/2))(ϕ/(ϕ +κ/2)))//.{α->((r+κ+ϕ)/(r+κ+2ϕ)),z->-((r+2ϕ)/(r+3ϕ)),λ->(ϕ/(α+β)),β->-α/2,r->ρ κ,ϕ->f κ}];CSN=Collect[FullSimplify[(((μ )/2))(μ-(3/4)μ)+(1/(2 ))((α-((α^2)/2))+(((α^2)/2)+β-(3/2)(((α+β))^2))((α λ)/(ϕ+κ-β λ)))//.{α->1,λ->(ϕ/(α+β)),β->-1/2,r->ρ κ,ϕ->f κ},Assumptions->ρ>0&&f>0&&κ>0&&μ>0],μ];

Collect[FullSimplify[CSH-CSN],μ]

FullSimplify[%-(((f^2) ((μ^2) ρ+(((-2+ρ (-4+f)-4 f) (((ρ+2 f))^2))/((1+2 f) (((1+ρ+2 f))^2)))))/(8 (((ρ+2 f))^3)))]

Therefore, for our comparison, we use the term in parentheses in the numerator.

Solve for the critical μ at ϕ=0

FullSimplify[μ/.Solve[(FullSimplify[Limit[(μ^2) ρ+(((-2+ρ (-4+f)-4 f) (((ρ+2 f))^2))/((1+2 f) (((1+ρ+2 f))^2))),f->0],Assumptions->ρ>0])==0,μ][[2]]]

Solve at ϕ=∞

FullSimplify[μ/.Solve[(FullSimplify[Limit[(μ^2) ρ+(((-2+ρ (-4+f)-4 f) (((ρ+2 f))^2))/((1+2 f) (((1+ρ+2 f))^2))),f->∞]])==0,μ][[2]]]

Check that ranking at ϕ=0 and ϕ=∞ suffices

FindInstance[(μ^2) ρ+(((-2+ρ (-4+f)-4 f) (((ρ+2 f))^2))/((1+2 f) (((1+ρ+2 f))^2)))>0&&4>ρ>0&&f>0&&0<μ<((Sqrt[4-ρ])/(Sqrt[2 ρ]))&&μ<(( (Sqrt[2ρ (1+2 ρ)]))/(1+ρ)),{μ,f,ρ}]

So we have a region over which naive consumers are better off than strategic consumers with hidden scores. It is given by the two conditions μ<min{((Sqrt[4-ρ])/(Sqrt[2 ρ])),(( (Sqrt[2ρ (1+2 ρ)]))/(1+ρ))} and ρ<4