Moral Hazard in Search:
Technical Appendix

S. Nuray Akin and Brennan C. Platt

In our model, the equilibrium conditions involve integrals, and a number of proofs require non-trivial algebraic manipulation and/or differentiation.  We provide this technical appendix to elucidate the intermediate steps of such manipulation.

Also, since the model has no closed-form solution, we employed a numerical example in the paper.  We include the numerical solution method below.

Section 2 — Proof details

Service Firm Price Distribution:

The price distribution is pinned down by the requirement that any price be equally profitable.  We equate profit at the highest price, M, to profit at any other price, p:

Solve[(M-r)(1-S)==(p-r)(1-S+2(1-F[p])S),F[p]]

F[p]->1-(((M-p)(1-S))/(2 (p-r) S))

The density is simply the first derivative:

Simplify[D[F[p]->1-(((M-p)(1-S))/(2 (p-r) S)),p]]

F'[p]->(((M-r) (1-S))/(2 (((p-r))^2) S))

And the support’s lower bound is determined by where F[p] = 0:

Solve[0==1-(((M-p)(1-S))/(2 (p-r) S)),p]

p->r+(((M-r)(1-S))/(1+S))

Lemma 2

The ex-interim utility simplifies as follows:

Simplify[(∫_(r+(((M-r)(1-S))/(1+S)))^M)u[w-θ-γ p](1-s+2((((M-p)(1-S))/(2 (p-r) S)))s)(((M-r) (1-S))/(2 (((p-r))^2) S))\[DifferentialD]p-c (s^2)]

The first order condition for s is:

Simplify[D[%,s]]

-2 c s+(∫_((M-M S+2 r S)/(1+S))^M)(((M-r) (1-S) (-p+M (1-S)+r S-r +r) u[w-p γ-θ])/(2 (((p-r))^3) (S^2)))\[DifferentialD]p

-2 c s+(∫_((M-M S+2 r S)/(1+S))^M)(((M-r) (1-S) ((M-r)(1-S)-p+r) u[w-p γ-θ])/(2 (((p-r))^3) (S^2)))\[DifferentialD]p

Proposition 1, Claim 2

First derivative of (Subscript[β, 0]):

FullSimplify[D[((S^2)/(1-S))(-2 c S+(∫_((M-M S+2 r S)/(1+S))^M)(((M-r) (-1+S) (p+M (-1+S)-r S) u[w-p γ-θ])/(2 (((p-r))^3) (S^2)))\[DifferentialD]p),S]]

FullSimplify[(1/(((-1+S))^2))S (2 c S (-3+2 S)- (∫_((M-M S+2 r S)/(1+S))^M)((((-2+S)(M-r) (-1+S) (p+M (-1+S)-r S) u[w-p γ-θ])/(2 (((p-r))^3) (S^2)))+(((-1+S) S(M-r) (2 p+2 M (-1+S)-(p+r) S) u[w-p γ-θ])/(2 (((p-r))^3) (S^3))))\[DifferentialD]p+u[w+(((M (-1+S)-2 r S) γ)/(1+S))-θ])]

((S )/(((1-S))^2))(u[w+(((M (-1+S)-2 r S) γ)/(1+S))-θ]-2 c S (3-2 S)-(∫_((M-M S+2 r S)/(1+S))^M)(((((M-r))^2) (((1-S))^2) u[w-p γ-θ])/(2 (((p-r))^3) S))\[DifferentialD]p)

Second derivative of (Subscript[β, 0]):

FullSimplify[D[((S )/(((1-S))^2))(u[w+(((M (-1+S)-2 r S) γ)/(1+S))-θ]-2 c S (3-2 S)-(∫_((M-M S+2 r S)/(1+S))^M)(((((M-r))^2) (((1-S))^2) u[w-p γ-θ])/(2 (((p-r))^3) S))\[DifferentialD]p),S]]

FullSimplify[-(1/((((1-S))^3) (((1+S))^2)))( (∫_((M-M S+2 r S)/(1+S))^M)((((((1+S))^3)(((M-r))^2) (((-1+S))^2) u[w-p γ-θ])/(2 (((p-r))^3) S))-(((((1+S))^2) (-1+S) S (((M-r))^2) (-1+(S^2)) u[w-p γ-θ])/(2 (((p-r))^3) (S^2))))\[DifferentialD]p+(((1+S))^2) (4 c S (3+(-3+S) S))+2 (M-r) (-1+S) S γ u'[w+(((M (-1+S)-2 r S) γ)/(1+S))-θ])]

-((2 S )/(((1-S))^3))(2 c (3-3 S+(S^2))-(((M-r) (1-S) γ )/(((1+S))^2))u'[w+(((M (-1+S)-2 r S) γ)/(1+S))-θ])

First derivative of (Subscript[β, 1]):

FullSimplify[D[(((((1+S))^2)(((1-S))^2))/(2S))(-((2 S )/(((1-S))^3))(2 c (3-3 S+(S^2))-(((M-r) (1-S) γ )/(((1+S))^2))u'[w+(((M (-1+S)-2 r S) γ)/(1+S))-θ])),S]]

The second term is always negative since u''<0.  The first term is as well for any 0<S<1:

Plot[(-2-(4/(((-1+S))^2))+3 (S^2)),{S,0,1}]

Proposition 1, Claim 3

Note that -3-s+2 (s^2) is always less than -2 for 0 < S < 1:

Plot[-3-s+2 (s^2),{s,0,1}]

Proposition 1, Claim 4

The derivative of Δ w.r.t. gamma is:

(Subscript[Δ, γ])==FullSimplify[D[-2 c S+(∫_((M-M S+2 r S)/(1+S))^M)(((M-r) (-1+S) (p+M (-1+S)-r S) u[w-p γ-θ])/(2 (((p-r))^3) (S^2)))\[DifferentialD]p,γ]]

Note that the integrand is negative iff p < M - (M - r)S:

FullSimplify[-((p (M-r) (-1+S) (p+M (-1+S)-r S))/(2 (((p-r))^3) (S^2)))<0]

p<M (1-S)+r S

We split the integral at M-(M-r)S:

(Subscript[Δ, γ])==(∫_((M-M S+2 r S)/(1+S))^(M-(M-r)S))-((p (M-r) (-1+S) (p+M (-1+S)-r S) u'[w-p γ-θ])/(2 (((p-r))^3) (S^2)))\[DifferentialD]p+(∫_(M-(M-r)S)^M)-((p (M-r) (-1+S) (p+M (-1+S)-r S) u'[w-p γ-θ])/(2 (((p-r))^3) (S^2)))\[DifferentialD]p

Then, to create a lower bound, we replace u[w-p γ-θ] with its largest value in the first term, and its highest value in the second term.  In both cases, that is u[w- (M - (M - r)S) γ-θ].

(Subscript[Δ, γ])>u'[w-(M-(M-r)S) γ-θ](∫_((M-M S+2 r S)/(1+S))^(M-(M-r)S))-((p (M-r) (-1+S) (p+M (-1+S)-r S) )/(2 (((p-r))^3) (S^2)))\[DifferentialD]p+u'[w-(M-(M-r)S) γ-θ](∫_(M-(M-r)S)^M)-((p (M-r) (-1+S) (p+M (-1+S)-r S) )/(2 (((p-r))^3) (S^2)))\[DifferentialD]p==u'[w-(M-(M-r)S) γ-θ](∫_((M-M S+2 r S)/(1+S))^M)-((p (M-r) (-1+S) (p+M (-1+S)-r S) )/(2 (((p-r))^3) (S^2)))\[DifferentialD]p

The integral now evaluates to:

Assuming[0<S<1&&M>r,(∫_((M-M S+2 r S)/(1+S))^M)-((p (M-r) (-1+S) (p+M (-1+S)-r S) )/(2 (((p-r))^3) (S^2)))\[DifferentialD]p]

This value is positive for all 0 < S < 1, so (Subscript[Δ, γ])>0

Plot[(((-1+S) (2 S+Log[((1-S)/(1+S))]))/(2 (S^2))),{S,0,1}]

Section 3 — Proof details

Lemma 3 and Eq. 8:

The expected expenditure is:

Assuming[0<S<1&&M>r,FullSimplify[ρ (∫_(r+(((M-r)(1-S))/(1+S)))^M)p(1-s+2((((M-p)(1-S))/(2 (p-r) S)))s)(((M-r) (1-S))/(2 (((p-r))^2) S))\[DifferentialD]p]]

Note that ArcTanh[S]==(1/2)Log[((1+S)/(1-S))], so this is:

ρ (r+(((1-S)(M-r))/S)(s+(((S-s) ArcTanh[S])/S)))

ρ (r+(((1-S)(M-r))/S)(s+(((S-s))/(2S))Log[((1+S)/(1-S))]))

And when s = S, it is:

Simplify[ρ (r+(((1-S)(M-r))/S)(s+(((S-s))/(2S))Log[((1+S)/(1-S))]))/.s->S]

The expected payout is obviously multiplied by (1 - γ)

Proposition 2

Setup

The consumer surplus from insurance, as defined in Appendix A.4, is:

CS==ρ (∫_((M-M S+2 r S)/(1+S))^M)(((((M-r))^2) (((1-S))^2) (u[w-p γ-θ] -u[w-p]))/(2 (((p-r))^3) S))\[DifferentialD]p+(1-ρ) (u[w-θ]-u[w])

where

θ->ρ(1-γ)(M-M S+r S)+Π

dCS/dΠ

Insurer profit clearly reduces consumer surplus:

FullSimplify[D[ρ (∫_((M-M S+2 r S)/(1+S))^M)(((((M-r))^2) (((1-S))^2) (u[w-p γ-θ] -u[w-p]))/(2 (((p-r))^3) S))\[DifferentialD]p+(1-ρ) (u[w-θ]-u[w])/.θ->ρ(1-γ)(M-M S+r S)+Π,Π]]

The integrand is negative:

-(((((M-r))^2) (((1-S))^2) u'[w-p γ-Π-(M (-1+S)-r S) (-1+γ) ρ])/(2 (((p-r))^3) S))<0

dCS/dγ

Higher coinsurance also reduces surplus:

FullSimplify[D[ρ (∫_((M-M S+2 r S)/(1+S))^M)(((((M-r))^2) (((1-S))^2) (u[w-p γ-θ] -u[w-p]))/(2 (((p-r))^3) S))\[DifferentialD]p+(1-ρ) (u[w-θ]-u[w])/.θ->ρ(1-γ)(M-M S+r S)+Π,γ]]

ρ ((∫_((M-M S+2 r S)/(1+S))^M)(((((M-r))^2) (((1-S))^2) ((M (1-S)+r S) ρ-p) u'[w-p γ-θ])/(2 (((p-r))^3) S))\[DifferentialD]p+(M (-1+S)-r S) (-1+ρ) u'[w-θ])

Assuming ρ < 1/2, integrand is always negative.

FullSimplify[(M-M S+r S) /2<((M-M S+2 r S)/(1+S))]

So we can substitute u'[w-p γ-θ] with the smaller u'[w-θ] to get an upper bound:

(Subscript[CS, γ])<ρ (u'[w-θ](∫_((M-M S+2 r S)/(1+S))^M)(((((M-r))^2) (((1-S))^2) ((M (1-S)+r S) ρ-p))/(2 (((p-r))^3) S))\[DifferentialD]p+(M (-1+S)-r S) (-1+ρ) u'[w-θ])

The integral evaluates to:

Assuming[0<S<1&&M>r,(∫_((M-M S+2 r S)/(1+S))^M)(((((M-r))^2) (((1-S))^2) ((M (1-S)+r S) ρ-p))/(2 (((p-r))^3) S))\[DifferentialD]p]

So:

Simplify[(Subscript[CS, γ])<ρ (-(M (-1+S)-r S) (-1+ρ)u'[w-θ]+(M (-1+S)-r S) (-1+ρ) u'[w-θ])]

This also holds true when S = 0.

FullSimplify[D[ρ (u[w-M γ-Π-M (1-γ) ρ]-u[-M+w])+(-1+ρ) (u[w]-u[w-Π-(M) (1-γ) ρ]),γ]]

M (1-ρ) ρ (u'[w-Π+M (-1+γ) ρ]-u'[w-M γ-Π+M (-1+γ) ρ])<0

dCS/dS

Since dΠ/dγ < 0 for all s, set γ = 0

FullSimplify[D[u[w-Π-(M-M S+r S) ρ]-(1-ρ) u[w]-ρ (∫_((M-M S+2 r S)/(1+S))^M)(((((M-r))^2) (((1-S))^2) (u[-p+w]))/(2 (((p-r))^3) S))\[DifferentialD]p,S]]

(Subscript[CS, S])==ρ ((M-r)  u'[w-θ]-((1+S)/((1-S) S)) (u[((M (-1+S)-2 r S)/(1+S))+w]-(∫_((M-M S+2 r S)/(1+S))^M)(((((M-r))^2) (((1-S))^2) u[-p+w])/(2 (((p-r))^3) S))\[DifferentialD]p))

Within the integral, (((((M-r))^2) (((1-S))^2))/(2 (((p-r))^3) S))is a density function that integrates to 1.

Assuming[0<S<1,(∫_((M-M S+2 r S)/(1+S))^M)(((((M-r))^2) (((1-S))^2))/(2 (((p-r))^3) S))\[DifferentialD]p]

Therefore, we can move u[((M (-1+S)-2 r S)/(1+S))+w]inside the integral:

(Subscript[CS, S])==ρ ((M-r)  u'[w-θ]-((1+S)/((1-S) S)) ((∫_((M-M S+2 r S)/(1+S))^M)(((((M-r))^2) (((1-S))^2) )/(2 (((p-r))^3) S))(u[((M (-1+S)-2 r S)/(1+S))+w]-u[-p+w])\[DifferentialD]p))

Due to diminising marginal utility,

u[((M (-1+S)-2 r S)/(1+S))+w]-u[-p+w]>(((M (-1+S)-2 r S)/(1+S))+w-(-p+w))u'[((M (-1+S)-2 r S)/(1+S))+w]==(((M (-1+S)-2 r S)/(1+S))+p)u'[((M (-1+S)-2 r S)/(1+S))+w]

Therefore:

(Subscript[CS, S])<ρ ((M-r)  u'[w-θ]-((1+S)/((1-S) S)) ((∫_((M-M S+2 r S)/(1+S))^M)(((((M-r))^2) (((1-S))^2) )/(2 (((p-r))^3) S))(((M (-1+S)-2 r S)/(1+S))+p)u'[((M (-1+S)-2 r S)/(1+S))+w]\[DifferentialD]p))==ρ ((M-r)  u'[w-θ]-((1+S)/((1-S) S)) u'[((M (-1+S)-2 r S)/(1+S))+w]((∫_((M-M S+2 r S)/(1+S))^M)(((((M-r))^2) (((1-S))^2) )/(2 (((p-r))^3) S))(((M (-1+S)-2 r S)/(1+S))+p)\[DifferentialD]p))

The integral then evaluates to:

Assuming[0<S<1,FullSimplify[(∫_((M-M S+2 r S)/(1+S))^M)(((((M-r))^2) (((1-S))^2) )/(2 (((p-r))^3) S))(((M (-1+S)-2 r S)/(1+S))+p)\[DifferentialD]p]]

Thus:

(Subscript[CS, S])<ρ ((M-r)  u'[w-θ]-((1+S)/((1-S) S)) u'[((M (-1+S)-2 r S)/(1+S))+w](-(((M-r) (-1+S) S)/(1+S))))==ρ (M-r)( u'[w-θ]- u'[((M (-1+S)-2 r S)/(1+S))+w])

This will be less than 0 so long as  w-θ>w-((M (1-S)+2 r S)/(1+S))

((M (1-S)+2 r S)/(1+S))>Π+(M-M S+r S) ρ

((M (1-S)+2 r S)/(1+S))-(M-M S+r S) ρ>Π

This is strictly decreasing in S

FullSimplify[D[((M (1-S)+2 r S)/(1+S))-(M-M S+r S) ρ,S]]

Since 0 ≤ S ≤ 1 and ρ < 1/2, this is always negative.  Thus, we evaluate the inequality at S = 1 to get:

((M (1-S)+2 r S)/(1+S))-(M-M S+r S) ρ>Π/.S->1

r(1-ρ)>Π

Thus, by assumption, (Subscript[CS, S])<0

Lemma 4

We only need to show that the ζ function is decreasing in s, once Θ[s,γ,S] is substituted in.

FullSimplify[D[(1/(2c))(∫_((M-M S+2 r S)/(1+S))^M)(((M-r) (1-S) ((M-r)(1-S)-p+r) u[w-p γ-θ])/(2 (((p-r))^3) (S^2)))\[DifferentialD]p/.θ->ρ(1-γ) (r+(((1-S)(M-r))/S)(s+(((S-s))/(2S))Log[((1+S)/(1-S))])),s]]

(((1-γ) ρ)/(2 c))( (1/2)Log[((1+S)/(1-S))]-S)(∫_((M-M S+2 r S)/(1+S))^M)(((((M-r))^2) (((1-S))^2) (M (1-S)+r S-p))/(2 (((p-r))^3) (S^4)))  u'[w-p γ-θ]\[DifferentialD]p

Note that (1/2)Log[((1+S)/(1-S))]-S>0:

Plot[(1/2)Log[((1+S)/(1-S))]-S,{S,0,1}]

The integrand is positive iff p < (M (1-S)+r S):

Simplify[(((((M-r))^2) (((1-S))^2) (M (1-S)+r S-p))/(2 (((p-r))^3) (S^4)))>0]

M (1-S)+r S>p

We split the integral and use the upper bound in each half:

(∫_((M-M S+2 r S)/(1+S))^M)(((((M-r))^2) (((1-S))^2) (M (1-S)+r S-p))/(2 (((p-r))^3) (S^4)))  u'[w-p γ-θ]\[DifferentialD]p== (∫_((M-M S+2 r S)/(1+S))^(M (1-S)+r S))(((((M-r))^2) (((1-S))^2) (M (1-S)+r S-p))/(2 (((p-r))^3) (S^4)))  u'[w-p γ-θ]\[DifferentialD]p+(∫_(M (1-S)+r S)^M)(((((M-r))^2) (((1-S))^2) (M (1-S)+r S-p))/(2 (((p-r))^3) (S^4)))  u'[w-p γ-θ]\[DifferentialD]p<u'[w-(M (1-S)+r S) γ-θ](∫_((M-M S+2 r S)/(1+S))^(M (1-S)+r S))(((((M-r))^2) (((1-S))^2) (M (1-S)+r S-p))/(2 (((p-r))^3) (S^4))) \[DifferentialD]p+u'[w-(M (1-S)+r S)  γ-θ](∫_(M (1-S)+r S)^M)(((((M-r))^2) (((1-S))^2) (M (1-S)+r S-p))/(2 (((p-r))^3) (S^4))) \[DifferentialD]p== u'[w-(M (1-S)+r S) γ-θ](∫_((M-M S+2 r S)/(1+S))^M)(((((M-r))^2) (((1-S))^2) (M (1-S)+r S-p))/(2 (((p-r))^3) (S^4))) \[DifferentialD]p

The integral evaluates to 0:

Assuming[0<S<1,(∫_((M-M S+2 r S)/(1+S))^M)(((((M-r))^2) (((1-S))^2) (M (1-S)+r S-p))/(2 (((p-r))^3) (S^4))) \[DifferentialD]p]

So the ζ function is decreasing in s.

Proposition 3

We only have to show how dEU/dγ differs when s is included in the total derivative.  Thus, treating all other variables as constants, the consider the following partial derivatives.

dEU/dS

Unambiguously improving.  Lowers the price distribution, so increases utility (no extra search cost z).

FullSimplify[D[(∫_((M-M S+2 r S)/(1+S))^M)(((M-r) (-1+S) (p s+M s (-1+S)-p S+r S-r s S) u[w-p γ-θ])/(2 (((p-r))^3) (S^2)))\[DifferentialD]p-c (s^2),S]/.s->S]

dEU/dθ

Unambiguously worse.  Reduces wealth in every state.

FullSimplify[D[(∫_((M-M S+2 r S)/(1+S))^M)(((M-r) (-1+S) (p s+M s (-1+S)-p S+r S-r s S) u[w-p γ-θ])/(2 (((p-r))^3) (S^2)))\[DifferentialD]p-c (s^2),θ]/.s->S]

dθ/ds

Unambiguously decreasing.  Reduces prices, so therefore reduces the average payout.

FullSimplify[D[ρ (r+(((1-S)(M-r))/S)(s+(((S-s))/(2S))Log[((1+S)/(1-S))])),S]/.s->S]

This is always negative:

Plot[-(((2 S+(-1+S) Log[((1+S)/(1-S))]))/(2 (S^2))),{S,0,1}]

Section 4 — Proof details

Measuring Moral Hazard

The expected total cost comes from Lemma 3

ρ(M(1-S)+r S)

In comparing the expected cost of different insured search equilibria (with S and (OverBar[S]), for instance), the difference between them will be:

FullSimplify[ρ(M(1-(OverBar[S]))+r (OverBar[S]))-ρ(M(1-S)+r S)]

Decomposing the direct effect

If s differs from S, the expected cost comes from Eq. 8, calculated above.

ρ (r+(((1-S)(M-r))/S)(s+(((S-s))/(2S))Log[((1+S)/(1-S))]))

We compare what the expected cost would be, holding aggregate search at (OverBar[S]) while allowing individual search to be at S.

Simplify[ρ(M(1-(OverBar[S]))+r (OverBar[S]))-ρ (r+(((1-(OverBar[S]))(M-r))/(OverBar[S]))(S+((((OverBar[S])-S))/(2(OverBar[S])))Log[((1+(OverBar[S]))/(1-(OverBar[S])))]))]

Then we ask what fraction of the total cost change is due to this direct effect:

Simplify[((((-M+r) ρ (-1+(OverBar[S])) (-S+(OverBar[S])) (-Log[((1+(OverBar[S]))/(1-(OverBar[S])))]+2 (OverBar[S])))/(2 ((OverBar[S])^2)))/((M-r) ρ (S-(OverBar[S]))))]

This fraction varies depending on the reference point (OverBar[S]), but never is greater than 0.104

Plot[(((-1+(OverBar[S])) (-Log[((1+(OverBar[S]))/(1-(OverBar[S])))]+2 (OverBar[S])))/(2 ((OverBar[S])^2))),{(OverBar[S]),0,1}]

We take the first order condition:

Simplify[D[(((-1+S) (-Log[((1+S)/(1-S))]+2 S))/(2 (S^2))),S]==0]

And use FindRoot to numerically find the maximizer and compute the maximal value.

{S,(((-1+S) (-Log[((1+S)/(1-S))]+2 S))/(2 (S^2)))}/.FindRoot[2 S (2+S)+(-2-S+(S^2)) Log[((1+S)/(1-S))]==0,{S,0.6,0,1}]

Section 5 — Numerical Solution

Expected Utility

We use the CRRA functional form of utility: U[w]==((w^(1-a))/(1-a)).  For integer values of a, expected utility can be analytically computed by Mathematica.  This is helpful, because numerical integration performs rather poorly for this distribution when S is close to 1.  Below we evaluate and simplify EU for a = 1, 2, 3, and 4.  The simplest expression comes from a = 2.

We also get the first-order condition on s for each.

In each case, the initial evaluation of the integral produces an analytical result, subject to a list of conditions about the parameters.  These conditions are easily satisfied by the fact that the parameters are real numbers (so its imaginary part Im[.] = 0), 0 < s < 1, and w - θ - γ M > 0.

a = 1 (i.e. Log)

(∫_((((1-S)(M-r))/(1+S))+r)^M)Log[w-θ-γ p](((M-r) (-1+S) (M (-1+S) s+p (-S+s)+r (S-S s)))/(2 (((p-r))^3) (S^2)))\[DifferentialD]p

FullSimplify[(1/(2 (S^2)))(M-r) (-1+S) (-1/(2 (M-r) (((-w+r γ+θ))^2))((M-r) (-1+S) s γ (-w+r γ+θ)+(w-M γ-θ) (s (w-M γ-θ)+S (w (-2+s)-2 r (-1+s) γ+M s γ+2 θ-s θ)) Log[w-M γ-θ])+(1/2) (-((γ (S (2 w+r (-2+s) γ-M s γ-2 θ)+s (-2 w+M γ+r γ+2 θ)) Log[((1-S)/(1+S))])/(((-w+r γ+θ))^2))-((-2 S (w-M γ-θ) (w+r (-1+s) γ-M s γ-θ)+s (((-w+M γ+θ))^2)+(S^2) (-2 (w^2)-4 (r^2) (γ^2)+(M^2) s (γ^2)+(r^2) s (γ^2)-6 r γ θ-2 (θ^2)+2 M γ (r γ-r s γ+θ)+w (-2 M γ+6 r γ+4 θ))) Log[(((1+S) w+M (-1+S) γ-2 r S γ-θ-S θ)/(1+S))])/((M-r) (-1+S) (((-w+r γ+θ))^2))-s ((((1+S) γ)/(-w+r γ+θ))+(((S^2) Log[((w+S w-M γ+M S γ-2 r S γ-θ-S θ)/(1+S))])/(M-r-M S+r S)))))]

Ex-interim utility is then inserted into ex-ante utility:

(1-ρ)Log[w-θ]+ρ((1/(4 (S^2) (((-w+r γ+θ))^2)))(M-r) (-1+S) (-γ (S (2 w+r (-2+s) γ-M s γ-2 θ)+s (-2 w+(M+r) γ+2 θ)) Log[-1+(2/(1+S))]+1/((M-r) (-1+S))(-2 (M-r) (-1+S) S s γ (-w+r γ+θ)-(-1+S) (w-M γ-θ) (s (w-M γ-θ)+S (w (-2+s)-2 r (-1+s) γ+M s γ+2 θ-s θ)) Log[w-M γ-θ]+((1+S) w+M (-1+S) γ-2 r S γ-(1+S) θ) (s (-w+M γ+θ)+S (w (2+s)-s (M γ+θ)-2 (r γ+θ))) Log[w+(((M (-1+S)-2 r S) γ)/(1+S))-θ]))-c (s^2))

The first order condition on s:

FullSimplify[D[(1-ρ)Log[w-θ]+ρ((1/(4 (S^2) (((-w+r γ+θ))^2)))(M-r) (-1+S) (-γ (S (2 w+r (-2+s) γ-M s γ-2 θ)+s (-2 w+(M+r) γ+2 θ)) Log[-1+(2/(1+S))]+1/((M-r) (-1+S))(-2 (M-r) (-1+S) S s γ (-w+r γ+θ)-(-1+S) (w-M γ-θ) (s (w-M γ-θ)+S (w (-2+s)-2 r (-1+s) γ+M s γ+2 θ-s θ)) Log[w-M γ-θ]+((1+S) w+M (-1+S) γ-2 r S γ-(1+S) θ) (s (-w+M γ+θ)+S (w (2+s)-s (M γ+θ)-2 (r γ+θ))) Log[w+(((M (-1+S)-2 r S) γ)/(1+S))-θ]))-c (s^2)),s]==0]

a = 2

(∫_((((1-S)(M-r))/(1+S))+r)^M)((((w-θ-γ p))^(-1))/(-1))(((M-r) (-1+S) (M (-1+S) s+p (-S+s)+r (S-S s)))/(2 (((p-r))^3) (S^2)))\[DifferentialD]p

-(1/(2 (S^2)))(((M-r) (-1+S) )/(2 (M-r)(((-w+r γ+θ))^3)))FullSimplify[((w-r γ-θ) (s (w-2 M γ+r γ-θ)+S (w (-2+s)+r (2-3 s) γ+2 M s γ+2 θ-s θ))-2 (M-r) γ (s (w-M γ-θ)+S (-w+r γ+M s γ-r s γ+θ)) Log[(((1+S) w+M (-1+S) γ-2 r S γ-θ-S θ)/((1-S)(w-M γ-θ)))]-((1+S) (w-r γ-θ) (s (-w+2 M γ-r γ+θ)+S (w (2+s)+r (-2+s) γ-2 M s γ-2 θ-s θ)))/(-1+S))]

(1/(4 (S^2) (((w-r γ-θ))^3)))(-4 S (-w+r γ+θ) ((-M+r) s γ+S (-w+(r+M s-r s) γ+θ))-2 (M-r)(-1+S)  γ (s (w-M γ-θ)+S (-w+(r+M s-r s) γ+θ)) Log[-((1+S+((2 (-M+r) S γ)/(-w+M γ+θ)))/(-1+S))])

Ex-interim utility is then inserted into ex-ante utility, and the first order condition on s::

FullSimplify[D[(1-ρ)((((w-θ))^(-1))/(-1))+ρ((1/(4 (S^2) (((w-r γ-θ))^3)))(-4 S (-w+r γ+θ) ((-M+r) s γ+S (-w+(r+M s-r s) γ+θ))-2 (M-r)(-1+S)  γ (s (w-M γ-θ)+S (-w+(r+M s-r s) γ+θ)) Log[-((1+S+((2 (-M+r) S γ)/(-w+M γ+θ)))/(-1+S))])-c (s^2)),s]==0]

2 c s== (((M-r) (-1+S) γ (-2 S (-w+r γ+θ)+(-w+(M-M S+r S) γ+θ) Log[-((1+S+((2 (-M+r) S γ)/(-w+M γ+θ)))/(-1+S))]))/(2 (S^2) (((w-r γ-θ))^3)))

a = 3

(∫_((((1-S)(M-r))/(1+S))+r)^M)((((w-θ-γ p))^(-2))/(-2))(((M-r) (-1+S) (M (-1+S) s+p (-S+s)+r (S-S s)))/(2 (((p-r))^3) (S^2)))\[DifferentialD]p

(1/(4 (S^2)))(((M-r) (-1+S))/(2(((-w+r γ+θ))^4))) FullSimplify[(-((w-r γ-θ) (s (w-M γ-θ) (w-6 M γ+5 r γ-θ)+S ((w^2) (-2+s)+2 (r^2) (γ^2)-6 (M^2) s (γ^2)-2 (r^2) s (γ^2)+5 r s γ θ-2 (θ^2)+s (θ^2)+w (-5 r s γ+M (4+3 s) γ+4 θ-2 s θ)+M γ (r (-4+9 s) γ-(4+3 s) θ))))/((M-r) (-w+M γ+θ) )-2 γ (s (2 w-3 M γ+r γ-2 θ)+S (-2 w+r (2-3 s) γ+3 M s γ+2 θ)) Log[(((1+S) w+M (-1+S) γ-2 r S γ-θ-S θ)/((w-M γ-θ)(1-S)))]-(((1+S) (w-r γ-θ) (-s (w-M γ-θ) (w-6 M γ+5 r γ-θ)+2 S ((w^2)-(r^2) (γ^2)+6 (M^2) s (γ^2)+4 (r^2) s (γ^2)-2 r s γ θ+(θ^2)-2 w (-r s γ+M (1+s) γ+θ)+2 M γ (r γ-5 r s γ+θ+s θ))+(S^2) ((w^2) (2+s)+6 (r^2) (γ^2)-6 (M^2) s (γ^2)-8 (r^2) s (γ^2)+8 r γ θ-r s γ θ+2 (θ^2)+s (θ^2)+w (M (4-3 s) γ+r (-8+s) γ-2 (2+s) θ)+M γ (r (-4+15 s) γ+(-4+3 s) θ))))/((M-r) (-1+S) ((1+S) w+M (-1+S) γ-2 r S γ-θ-S θ))))]

Ex-interim utility is then inserted into ex-ante utility, and the first order condition on s::

FullSimplify[D[(1-ρ)((((w-θ))^(-2))/(-2))+ρ((1/(8 (S^2) (((-w+r γ+θ))^4)))(M-r) (-1+S) (-1/((M-r) (-w+M γ+θ))(w-r γ-θ) (s (w-M γ-θ) (w-6 M γ+5 r γ-θ)+S ((w^2) (-2+s)+(-2 (r^2) (-1+s)-6 (M^2) s+M r (-4+9 s)) (γ^2)+(5 r s-M (4+3 s)) γ θ+(-2+s) (θ^2)+w (-5 r s γ+M (4+3 s) γ+4 θ-2 s θ)))-((1+S) (w-r γ-θ) (-s (w-M γ-θ) (w-6 M γ+5 r γ-θ)+2 S ((w^2)+(2 M r (1-5 s)+6 (M^2) s+(r^2) (-1+4 s)) (γ^2)+2 (M+M s-r s) γ θ+(θ^2)-2 w (-r s γ+M (1+s) γ+θ))+(S^2) ((w^2) (2+s)+(2 (r^2) (3-4 s)-6 (M^2) s+M r (-4+15 s)) (γ^2)+(-r (-8+s)+M (-4+3 s)) γ θ+(2+s) (θ^2)+w (M (4-3 s) γ+r (-8+s) γ-2 (2+s) θ))))/((M-r) (-1+S) ((1+S) w+M (-1+S) γ-2 r S γ-(1+S) θ))-2 γ (s (2 w-3 M γ+r γ-2 θ)+S (-2 w+r (2-3 s) γ+3 M s γ+2 θ)) Log[-((1+S+((2 (-M+r) S γ)/(-w+M γ+θ)))/(-1+S))])-c (s^2)),s]==0]

a = 4

(∫_((((1-S)(M-r))/(1+S))+r)^M)((((w-θ-γ p))^(-3))/(-3))(((M-r) (-1+S) (M (-1+S) s+p (-S+s)+r (S-S s)))/(2 (((p-r))^3) (S^2)))\[DifferentialD]p

(1/(12 (S^2)(((w-r γ-θ))^5)))(M-r) (-1+S) (FullSimplify[((w-r γ-θ) (s (w-M γ-θ) ((w^2)+12 (M^2) (γ^2)+(r^2) (γ^2)-12 M γ (r γ-θ)-10 r γ θ+(θ^2)-2 w (6 M γ-5 r γ+θ))+S ((w^3) (-2+s)-(r^3) (γ^3)+12 (M^3) s (γ^3)+(r^3) s (γ^3)-6 (r^2) (γ^2) θ+7 (r^2) s (γ^2) θ-3 r γ (θ^2)-7 r s γ (θ^2)+2 (θ^3)-s (θ^3)+(w^2) (M (9+4 s) γ-r (3+7 s) γ-3 (-2+s) θ)+6 (M^2) (γ^2) (r (γ-3 s γ)+θ+3 s θ)+M γ ((r^2) (-3+4 s) (γ^2)+2 r (3-14 s) γ θ+(9+4 s) (θ^2))-w (6 (M^2) (1+3 s) (γ^2)+(r^2) (-6+7 s) (γ^2)-2 r (3+7 s) γ θ-3 (-2+s) (θ^2)+2 M γ (r (3-14 s) γ+(9+4 s) θ)))))/((M-r) (((-w+M γ+θ))^2) )-(((1+S) (w-r γ-θ) (-s (w-M γ-θ) ((w^2)+12 (M^2) (γ^2)+(r^2) (γ^2)-12 M γ (r γ-θ)-10 r γ θ+(θ^2)-2 w (6 M γ-5 r γ+θ))+S (-(w^3) (-2+s)+(r^3) (γ^3)-36 (M^3) s (γ^3)+3 (r^3) s (γ^3)+6 (r^2) (γ^2) θ-35 (r^2) s (γ^2) θ+3 r γ (θ^2)-r s γ (θ^2)-2 (θ^3)+s (θ^3)+(w^2) (-r (-3+s) γ+M (-9+4 s) γ+3 (-2+s) θ)+6 (M^2) (γ^2) (r (-1+13 s) γ-(1+5 s) θ)+M γ ((r^2) (3-44 s) (γ^2)+2 r (-3+34 s) γ θ+(-9+4 s) (θ^2))+w (6 (M^2) (1+5 s) (γ^2)+(r^2) (-6+35 s) (γ^2)+2 r (-3+s) γ θ-3 (-2+s) (θ^2)+2 M γ (r (3-34 s) γ+(9-4 s) θ)))+(S^2) ((w^3) (4+s)+8 (r^3) (γ^3)+36 (M^3) s (γ^3)-36 (r^3) s (γ^3)+11 (r^2) s (γ^2) θ-12 r γ (θ^2)+10 r s γ (θ^2)-4 (θ^3)-s (θ^3)-12 (M^2) (γ^2) (r (-1+10 s) γ+(-1+s) θ)+(w^2) (-13 M s γ+2 r (-6+5 s) γ-3 (4+s) θ)+M γ ((r^2) (-24+119 s) (γ^2)-2 r (12+s) γ θ-13 s (θ^2))+w (12 (M^2) (-1+s) (γ^2)-11 (r^2) s (γ^2)-4 r (-6+5 s) γ θ+3 (4+s) (θ^2)+2 M γ (r (12+s) γ+13 s θ)))+(S^3) ((w^3) (2+s)-17 (r^3) (γ^3)-12 (M^3) s (γ^3)+33 (r^3) s (γ^3)-30 (r^2) (γ^2) θ+23 (r^2) s (γ^2) θ-15 r γ (θ^2)+r s γ (θ^2)-2 (θ^3)-s (θ^3)+(w^2) (M (9-4 s) γ+r (-15+s) γ-3 (2+s) θ)+6 (M^2) (γ^2) (r (-1+9 s) γ+(-1+3 s) θ)+M γ ((r^2) (21-76 s) (γ^2)+2 r (15-22 s) γ θ+(9-4 s) (θ^2))+w ((r^2) (30-23 s) (γ^2)+6 (M^2) (1-3 s) (γ^2)-2 r (-15+s) γ θ+3 (2+s) (θ^2)+2 M γ (r (-15+22 s) γ+(-9+4 s) θ)))))/((M-r) (-1+S) (((-(1+S) w+2 r S γ+M (γ-S γ)+θ+S θ))^2)))]+6 γ (S (w-2 M s γ+r (-1+2 s) γ-θ)+s (-w+2 M γ-r γ+θ)) Log[(((1+S) w+M (-1+S) γ-2 r S γ-θ-S θ)/((1-S)(w-M γ-θ)))])

Ex-interim utility is then inserted into ex-ante utility, and the first order condition on s::

FullSimplify[D[(1-ρ)((((w-θ))^(-3))/(-3))+ρ((1/(12 (S^2) (((w-r γ-θ))^5)))(M-r) (-1+S) (1/(M-r)(w-r γ-θ) (((4 S (3 (-M+r) s γ+S (-w+(r+3 M s-3 r s) γ)))/(-1+S))+((4 (S^2) θ)/(-1+S))+(((((-M+r))^3) S (-1+s) (γ^3))/(((-w+M γ+θ))^2))+(((((M-r))^2) (-s+S (-6+7 s)) (γ^2))/(-w+M γ+θ))+(((((M-r))^2) (-1+S) (s+S (6+7 s)) (γ^2))/((1+S) w+M (-1+S) γ-2 r S γ-(1+S) θ))+(((((-M+r))^3) (((-1+S))^2) S (1+s) (γ^3))/(((-(1+S) w+M γ+θ+S (-M γ+2 r γ+θ)))^2)))+6 γ (S (w-2 M s γ+r (-1+2 s) γ-θ)+s (-w+2 M γ-r γ+θ)) Log[(((1+S) w+M (-1+S) γ-2 r S γ-θ-S θ)/((1-S) (w-M γ-θ)))])-c (s^2)),s]==0]

Numerical Example

Assignment of Parameters

To compute a numerical solution, we must assign values to each parameter.  We include the following command so as to clear those assignments if needed.

Clear[w,M,r,c,ρ,γ]

We also assign a = 4 throughout; for other levels of risk aversion, one only needs to replace the EU and Δ equations from above.

w=40000;r=2000;M =4000;ρ=0.05;

We compute c by assigning an approximate dollar value, and converting it to a utility cost:

c=N[x/.Flatten[Solve[w-(((((-1)/(3*(((((w))^(-3))/(-3))-x)))))^(1/3))==40,x]]]

As we will need to examine γ over a grid from 0 to 1, we set the step size here:

step=.005;

Solution for all Insured Search Equilibria

We begin by finding the insured search equilibrium associated with each coinsurance rate γ, assuming that an actuarially-fair insurance premium will be charged.

In practice, this is done by fixing a value for γ and solving for the S where Δ = 0.  This is repeated for 200 different values of γ, which are stored in the “eqt” table.

eqt=Table[{γ,S/.FindRoot[-2 c S+(1/(12 (S^2) (((w-r γ-θ))^5)))(M-r) (-1+S) (γ (w-r γ-θ) (12  S+((-(((-M+r))^2) S (γ^2))/(((-w+M γ+θ))^2))+(((M-r) (-1+7 S) γ)/(-w+M γ+θ))+(((M-r) (-1+S) (1+7 S) γ)/((1+S) w+M (-1+S) γ-2 r S γ-(1+S) θ))+((-(((-M+r))^2) (((-1+S))^2) S (γ^2))/(((-(1+S) w+M γ+θ+S (-M γ+2 r γ+θ)))^2)))-6 γ (w+(r+2 M (-1+S)-2 r S) γ-θ) Log[-((1+S+((2 (-M+r) S γ)/(-w+M γ+θ)))/(-1+S))])/.θ->ρ(1-γ)(M(1-S)+r S),{S,.8,0.000000001,0.9999999}]},{γ,0,0.99,step}];
ListPlot[eqt]

We can then compute the expected utility associated with each (γ,S) pair.  This is stored in the table ExpUtil.

ExpUtil = Table[{γ,(1-ρ)((((w-θ))^(-3))/(-3))+ρ((1/(12 (S^2) (((w-r γ-θ))^5)))(M-r) (-1+S) (1/(M-r)(w-r γ-θ) (((4 S (3 (-M+r) s γ+S (-w+(r+3 M s-3 r s) γ)))/(-1+S))+((4 (S^2) θ)/(-1+S))+(((((-M+r))^3) S (-1+s) (γ^3))/(((-w+M γ+θ))^2))+(((((M-r))^2) (-s+S (-6+7 s)) (γ^2))/(-w+M γ+θ))+(((((M-r))^2) (-1+S) (s+S (6+7 s)) (γ^2))/((1+S) w+M (-1+S) γ-2 r S γ-(1+S) θ))+(((((-M+r))^3) (((-1+S))^2) S (1+s) (γ^3))/(((-(1+S) w+M γ+θ+S (-M γ+2 r γ+θ)))^2)))+6 γ (S (w-2 M s γ+r (-1+2 s) γ-θ)+s (-w+2 M γ-r γ+θ)) Log[(((1+S) w+M (-1+S) γ-2 r S γ-θ-S θ)/((1-S) (w-M γ-θ)))])-c (s^2))/.θ->ρ(1-γ)(M(1-S)+r S)/.s->S}/.S->eqt[[i]][[2]]/.γ->eqt[[i]][[1]],{i,1,Length[eqt],1}];
ListPlot[ExpUtil]

The γ which maximizes expected utility in ExpUtil will be the second-best, utility-maximizing contract, stored in EUGamma.  We also record the position of that γ on the grid.

UMGamma =First[Sort[ExpUtil,#1[[2]]>#2[[2]]&]][[1]]
UM =Floor[ UMGamma/step]+1

Solution for Competitively-Insured Search Equilibrium

The remaining step is to find the competitively-insured search equilibrium.  This requires search across two grids, over γ and S.  That is, for each S and each γ, we compute the resulting individual search effort s (i.e. the unique solution as indicated in Lemma 4).

We then compute the expected utility associated with that s, γ, and S.  In the next step, we find the EU-maximizing coinsurance rate γ for each S.  Then in the final step, we search for the S such that utility-maximizing γ will produce search effort s = S.

The table “deveqt” records for all (S,γ) pairs the individual search effort s and associated EU.  Note that the actuarially-fair profit from Eq. 8 is used.

The table “maxeqt” records the maximal EU for each S, along with the maximizer γ and its associated s, while “Sdev” records, for each γ, the deviation of individual effort s from aggregate effort S.  The plot output shows this difference; the equilibrium occurs where this crosses the x axis (s = S).

deveqt=Table[{S,γ,s,(1-ρ)((((w-θ))^(-3))/(-3))+ρ((1/(12 (S^2) (((w-r γ-θ))^5)))(M-r) (-1+S) (1/(M-r)(w-r γ-θ) (((4 S (3 (-M+r) s γ+S (-w+(r+3 M s-3 r s) γ)))/(-1+S))+((4 (S^2) θ)/(-1+S))+(((((-M+r))^3) S (-1+s) (γ^3))/(((-w+M γ+θ))^2))+(((((M-r))^2) (-s+S (-6+7 s)) (γ^2))/(-w+M γ+θ))+(((((M-r))^2) (-1+S) (s+S (6+7 s)) (γ^2))/((1+S) w+M (-1+S) γ-2 r S γ-(1+S) θ))+(((((-M+r))^3) (((-1+S))^2) S (1+s) (γ^3))/(((-(1+S) w+M γ+θ+S (-M γ+2 r γ+θ)))^2)))+6 γ (S (w-2 M s γ+r (-1+2 s) γ-θ)+s (-w+2 M γ-r γ+θ)) Log[(((1+S) w+M (-1+S) γ-2 r S γ-θ-S θ)/((1-S) (w-M γ-θ)))])-c (s^2))}/.θ->(((-1+γ) ρ (-S (r S+(-M+r) (-1+S) s)+(M-r) (-1+S) (S-s) ArcTanh[S]))/(S^2))/.First[FindRoot[-2 c s+(1/(12 (S^2) (((w-r γ-θ))^5)))(M-r) (-1+S) (γ (w-r γ-θ) (12  S+((-(((-M+r))^2) S (γ^2))/(((-w+M γ+θ))^2))+(((M-r) (-1+7 S) γ)/(-w+M γ+θ))+(((M-r) (-1+S) (1+7 S) γ)/((1+S) w+M (-1+S) γ-2 r S γ-(1+S) θ))+((-(((-M+r))^2) (((-1+S))^2) S (γ^2))/(((-(1+S) w+M γ+θ+S (-M γ+2 r γ+θ)))^2)))-6 γ (w+(r+2 M (-1+S)-2 r S) γ-θ) Log[-((1+S+((2 (-M+r) S γ)/(-w+M γ+θ)))/(-1+S))])/.θ->(((-1+γ) ρ (-S (r S+(-M+r) (-1+S) s)+(M-r) (-1+S) (S-s) ArcTanh[S]))/(S^2)),{s,.99,0,1}]],{S,0.01,0.5,step},{γ,0.01,0.3,step}];

maxeqt=Table[{First[Sort[deveqt[[i]],#1[[4]]>#2[[4]]&]]},{i,1,Length[deveqt],1}];

Sdev =Table[{Flatten[maxeqt[[i]]][[2]],Flatten[maxeqt[[i]]][[1]]-Flatten[maxeqt[[i]]][[3]]},{i,1,Length[deveqt],1}];

ListPlot[Sdev,Joined->True]

To speed up computation, it is useful to limit the search grid in a non-binding way.  For instance, it is not necessary to consider γ greater than the UMgamma, since these will never be the equilibrium (Prop. 3).  As one decreases the step size to get a more accurate solution, one can also reduce the range of S considered similarly.

We then record the perfectly-competitive contract as PCgamma, and its position on the grid as PC.

PCGamma =First[Sort[Sdev,Abs[#1[[2]]]<Abs[#2[[2]]]&]][[1]]
PC =Floor[ PCGamma/step]+1

Graphs

To construct the graphs in the paper, we use the following commands.

First, we transform utils into their certainty-equivalent to be expressed in dollar units.

CEquiv = Table[{Extract[ExpUtil,{i,1}],(((((-1)/(3*Extract[ExpUtil,{i,2}]))))^(1/3))},{i,1,Length[ExpUtil],1}];
EquilUtil=Graphics[{PointSize[Large],Darker[Black],Point[{CEquiv[[UM]]}],Point[{CEquiv[[PC]]}],Point[{CEquiv[[1]]}]}];Show[ListPlot[CEquiv,PlotStyle->{Directive[Thick,Black]},Joined->True,AxesLabel->{"Coinsurance \nRate","Certainty \nEquivalent\nWealth"}],EquilUtil]

Second, we demonstrate how search effort varies with coinsurance in equilibrium:

EquilPairs=Graphics[{PointSize[Large],Darker[Black],Point[{eqt[[UM]]}],Point[{eqt[[PC]]}],Point[{eqt[[1]]}]}];
Show[ListPlot[eqt,PlotStyle->{Directive[Thick,Black]},Joined->True,AxesLabel->{"Coinsurance \nRate","S"}],EquilPairs,Plot[0,{x,0,1},PlotStyle->{Directive[Thick,Black]}]]

Third, we demonstrate how this search effort affects the price distribution:

dist1=Table[{p,Max[0,1-(((M-p)(1- Extract[eqt,{UM,2}]))/(2 Extract[eqt,{UM,2}] (p-r)))]},{p,r+1,M,(M-r)/100}];dist2=Table[{p,Max[0,1-(((M-p) (1-Extract[eqt,{PC,2}]))/(2 Extract[eqt,{PC,2}] (p-r)))]},{p,r+1,M,(M-r)/100}];dist3=Table[{p,If[p<M,0,1]},{p,r,M,(M-r)/100}];ListPlot[{dist3,dist2,dist1},PlotStyle->{Directive[Thick,GrayLevel[.5]],Directive[Thick,Dashed,GrayLevel[.3]],Directive[Thick,Dotted,Black]},Joined->True,AxesOrigin->{r,0},AxesLabel->{"Price","CDF"}]

Finally, we compute the ex-ante total cost and its decomposition into out-of-pocket expenses and insurance premium:

InsPremium=Table[{Extract[eqt,{i,1}],ρ(1-Extract[eqt,{i,1}])(M(1-Extract[eqt,{i,2}])+r Extract[eqt,{i,2}])},{i,1,Length[eqt],1}];OOP=Table[{Extract[eqt,{i,1}],ρ Extract[eqt,{i,1}](M(1-Extract[eqt,{i,2}])+r Extract[eqt,{i,2}])},{i,1,Length[eqt],1}];TotalCost=Table[{Extract[eqt,{i,1}],Extract[OOP,{i,2}]+Extract[InsPremium,{i,2}]},{i,1,Length[eqt],1}];ListPlot[{InsPremium,OOP,TotalCost},PlotStyle->{Directive[Thick,GrayLevel[.5]],Directive[Thick,Dashed,GrayLevel[.3]],Directive[Thick,Dotted,Black]},Joined->True,AxesOrigin->{0,0},AxesLabel->{"Coinsurance Rate","Ex-Ante Cost"}]

The following cost values are mentioned in the text:

TotalCost[[UM]]
TotalCost[[PC]]
TotalCost[[1]]

The decomposition of costs reveals the direct effect to be responsible for 5.8% of the cost change:

(((-1+S) (-Log[((1+S)/(1-S))]+2 S))/(2 (S^2)))/.S->eqt[[PC]][[2]]

(TotalCost[[PC]][[2]]-TotalCost[[UM]][[2]])*%

We also mention the certainty-equivalent utility when a monopolist fully extracts the consumer surplus under the monopolist outcome.

w-θ/.First[Solve[(1-ρ)((((w))^(-3))/(-3))+ρ((((w-4000))^(-3))/(-3))==((((w-θ))^(-3))/(-3)),θ]]

Comparative statics are computed by adjusting the parameter values above.