Asymmetric Information in Economic Policy Noah Williams University of Wisconsin - Madison Williams Econ 899
Asymmetric Information Risk-neutral moneylender. Borrow and lend at rate R = 1/β. Strictly risk-averse household Maximize E T t τ=t βτ t u(c τ ). u has decreasing absolute risk aversion (DARA). u is differentiable. i.i.d. unobservable stochastic endowment. No storage technology. Finite horizon game. In each period, household reports endowment. moneylender makes transfer. 1
Notation: States s = 1, 2,..., S. Endowment in state s is e s, e 1 < < e S. Probability of state s is π s. Transfer if individual reports states is s is b s. States are i.i.d. 2
Simplest Case: 2 states, 1 period Value for planner as a function of value v for household: P (v) = max b 1,b 2 π 1 b 1 π 2 b 2 s.t. π 1 u(e 1 + b 1 ) + π 2 u(e 2 + b 2 ) = v u(e 1 + b 1 ) u(e 1 + b 2 ) u(e 2 + b 2 ) u(e 2 + b 1 ) (PK) (IC1) (IC2) IC1 implies b 1 b 2. IC2 implies b 2 b 1. So b 1 = b 2. So the static problem is not very interesting. 3
Note P (v) is decreasing and strictly concave. P (v) = max b b s.t. π 1 u(e 1 + b) + π 2 u(e 2 + b) = v Lagrangian: L(b, λ) = b + λ(eu(e + b) v). To show P (v) is decreasing: Envelope condition: P (v) = λ. First order condition: λeu (e + b) = 1. So P 1 (v) = Eu (e + b). To show P (v) is strictly concave: Higher v requires higher b, hence lower u (e + b), lower P (v). Or it can be done directly: P (v) = Eu (e + b) ( Eu (e + b) ) 3. 4
Simplest Interesting Case: 2 states, 2 periods The second period looks just like the one period model. P 2 (v) = max b 2 b 2 s.t. Eu(e + b 2 ) = v P 2 (v) is decreasing and strictly concave, with P 2 (v) = 1 Eu (e + b 2 ). In the first period, things get more interesting: P 1 (v) = max π ( 1 b1 + βp 2 (w 1 ) ) ( + π 2 b2 + βp 2 (w 2 ) ) {b,w} subject to π 1 ( u(e1 + b 1 ) + βw 1 ) + π2 ( u(e2 + b 2 ) + βw 2 ) = v u(e 1 + b 1 ) + βw 1 u(e 1 + b 2 ) + βw 2 u(e 2 + b 2 ) + βw 2 u(e 2 + b 1 ) + βw 1 5
Claim: b 1 b 2 and w 1 w 2. Proof: Add the incentive compatibility constraints together: u(e 2 + b 2 ) u(e 1 + b 2 ) u(e 2 + b 1 ) u(e 1 + b 1 ) u concave implies u(e 2 + b) u(e 1 + b) is decreasing, so b 2 b 1. (IC2) then requires w 2 w 1. Note that b 1 = b 2 w 1 = w 2. 6
Claim: (IC2) binds. Proof: Suppose not, i.e. u(e 2 + b 2 ) + βw 2 > u(e 2 + b 1 ) + βw 1. b 1 b 2 implies w 2 > w 1. Reduce w 2 to w 2 and raise w 1 to w 1, keeping constant Note that still w 2 w 1. π 1 w 1 + π 2 w 2 = π 1 w 1 + π 2 w 2, until u(e 2 + b 2 ) + βw 2 = u(e 2 + b 1 ) + βw 1. w is a mean preserving spread of w. This makes (IC1) easier to satisfy. Since P 2 (v) is concave, this raises the moneylender s payoff. 7
Claim: The household and moneylender prefer high endowments. Household: From (IC2), u(e 2 + b 2 ) + βw 2 = u(e 2 + b 1 ) + βw 1 > u(e 1 + b 1 ) + βw 1 Note that we don t really need to use (IC2) binding. Moneylender: Suppose not, b 1 + βp 2 (w 1 ) > b 2 + βp 2 (w 2 ). Switch from {b 1, b 2, w 1, w 2 } to {b 1, b 1, w 1, w 1 }. This trivially satisfies the IC constraints. Household utility is unchanged since (IC2) binds: u(e 2 + b 2 ) + βw 2 = u(e 2 + b 1 ) + βw 1 But the moneylender does strictly better, a contradiction. 8
Claim: The household s expected marginal utility rises over time. Proof: Write the Lagrangian, ignoring (IC1): ( P 1 (v) = π 1 b1 + βp 2 (w 1 ) ) ( + π 2 b2 + βp 2 (w 2 ) ) ( ) ( ) ) + λ (π 1 u(e1 + b 1 ) + βw 1 + π2 u(e2 + b 2 ) + βw 2 v ) + µ (u(e 2 + b 2 ) + βw 2 u(e 2 + b 1 ) βw 1 The other first order conditions are b 1 : π 1 + λπ 1 u (e 1 + b 1 ) µu (e 2 + b 1 ) = 0 (1) b 2 : π 2 + λπ 2 u (e 2 + b 2 ) + µu (e 2 + b 2 ) = 0 (2) w 1 : π 1 P 2(w 1 ) + λπ 1 µ = 0 (3) w 2 : π 2 P 2(w 2 ) + λπ 2 + µ = 0 (4) 9
Kuhn-Tucker implies µ 0. If µ = 0: Equations (1) and (2) imply e 1 + b 1 = e 2 + b 2 or b 1 > b 2. Equations (3) and (4) imply w 1 = w 2. This contradicts earlier argument that b 1 = b 2 w 1 = w 2. So µ > 0, i.e. (IC2) binds. Equation (3) implies λπ 1 > µ, since P 2 (w 1) < 0. Equations (2) and (4) imply u (e 2 + b 2 ) = π 2 λπ 2 + µ = 1 P 2 (w 2) = Eu (e + b 2 (w 2 )) i.e. when the endowment is high, the Euler equation holds. 10
Equations (1) and (3) imply λπ 1 u (e 1 + b 1 ) µu (e 2 + b 1 ) λπ 1 µ = π 1 λπ 1 µ = 1 P 2 (w 1) = Eu (e+b 2 (w 1 )) e 1 < e 2 implies u (e 1 + b 1 ) > u (e 2 + b 1 ). Since λπ 1 > µ > 0, u (e 1 + b 1 ) < λπ 1u (e 1 + b 1 ) µu (e 2 + b 1 ) λπ 1 µ Combining these gives u (e 1 + b 1 ) < Eu (e + b 2 (w 1 )). i.e. when the endowment is low, b 1 is low relative to b 2 (w 1 ). Putting this together, E 1 u (e + b 1 ) < E 1 u (e + b 2 ). Marginal utility grows over time with βr = 1. Contrast this with exogenous incomplete markets. 11
What happens with more than 2 periods? The recursive structure suggests an inductive argument. For this to work, P 1 (v) must be strictly concave and differentiable. Differentiability is easy. The envelope theorem implies P 1 (v) = λ. Add together (3) and (4) to get π 1 P 2 (w 1) + π 2 P 2 (w 2) + λ = 0. So P 1 (v) = EP 2 (w), which exists. Strict concavity is harder... 12
T period model It is possible to work through all the main arguments inductively. Conclusions: Eu (e + b) is a super-martingale. Since u > 0, super-martingale convergence theorem applies. Expected marginal utility converges to infinity. EP t(v) is a martingale. Note that over long-time horizons, P t (v) P (v). P (v) (strictly?) concave implies v is a sub-martingale. The household s expected utility falls (without bound?). 16
Lower bound on the Pareto frontier P (v): A constant transfer b in every period is feasible: P (v) b 1 β P a(v) s.t. π 1u(e 1 + b)+π 2 u(e 2 + b) 1 β = v For example, if u(c) =logc and π 1 = π 2 = 1 2, this becomes b = 1 ( ) (e2 e 1 ) 2 2 +4exp(2(1 β)v) e 1 e 2 So a lower bound on the moneylender s profit is (e2 e 1 ) P a (v) = 2 +4exp(2(1 β)v) e 1 e 2. 2(1 β) In this case, P a ( ) =e 1 /(1 β) andp a ( ) =. 1
Upper bound on Pareto frontier P (v): Constant consumption e + b in every state is cheapest: P (v) π 1b 1 π 2 b 2 1 β P c (v) s.t. u(e 1 + b 1 ) 1 β = u(e 2 + b 2 ) 1 β = v For example, if u(c) =logc and π 1 = π 2 = 1 2, this becomes b 1 = e 1 + exp((1 β)v) andb 2 = e 2 + exp((1 β)v). So an upper bound on the moneylender s profit is P c (v) = e 1 + e 2 2 exp((1 β)v). 2(1 β) In this case, P c ( ) = e 1 + e 2 2(1 β) and P c( ) =. 2
Upper and lower bound functions: u(c) =logc, β =0.9, e 1 =0,e 2 =1,π 1 = π 2 = 1 2. -20-15 -10-5 5 10-5 -10-15 -20 3
Pareto Frontier: u(c) =logc, β =0.9, e 1 =0,e 2 =1,π 1 = π 2 = 1 2. -20-15 -10-5 5 10-5 -10-15 -20 4
Does v? We know P (v) is a nonpositive Martingale, and hence converges. We hope P (v) isstrictly concave, so there is a one-to-one mapping between v and P (v). We know any finite v cannot be a limit, since w 1 (v) <w 2 (v). We conclude that v or v. L-S assume utility is bounded above, hence preclude v. Atkeson and Lucas (1992) prove v in some special cases. 8
Optimal Unemployment Insurance Preferences over consumption c and job search effort a: E β t (u(c t ) a t ) t=0 with c t 0anda t 0. Standard assumptions on u: increasing, concave, twice differentiable. All jobs pay a wage of w and last forever. Probability of finding a job: p(a), increasing, concave, differentiable. p(0) = 0. lim a p(a) =1. p (0) = 9
Autarky: Consumes w once employed: V e = u(w)+βv e = u(w) 1 β Consumes 0 and searches a while unemployed V aut =max a u(0) a + β ( p(a)v e +(1 p(a))v aut ) Note that search effort a is time-invariant. The first order condition is βp (a)(v e V aut )=1 Combining with the Bellman equations gives u(w) u(0) = 10 1 β(1 p(a)) βp (a) a
Full Information: How much does it cost to deliver utility V>V aut to unemployed? Insurance agency controls c and a for unemployed. Formulate cost minimization problem recursively: ( ) C(V )= min c + β(1 p(a))c(w ) c,a,w where V = u(c) a + β ( p(a)v e +(1 p(a))w ) Note V e is unchanged. No taxes once employed. Claim: C(V ) is strictly convex. 11
Write this problem as a Lagrangian: C(V )=maxl = c + β(1 p(a))c(w ) θ ( u(c) a + β ( p(a)v e +(1 p(a))w ) V ) Envelope condition: C (V )=θ. F.O.C. w.r.t. W : C (W )=θ. So C (V )=C (W )orequivalentlyv = W. Induction establishes c and a are constant during unemployment. 12
Restate the dual problem: subject to V =max c,a u(c) a + β( p(a)v e +(1 p(a))v ) C = c + β(1 p(a))c The first order conditions from the Lagrangian are u (c) =λ. 1+βp (a)(v e V )= λβp (a)c. Using Bellman equations, we can rewrite this as ( ) 1 u(w) u(c) =(1 β(1 p(a))) βp (a) u (c)c or u(w) = 1 β(1 p(a)) βp (a) +(u(c) u (c)c) 13
u(w) = 1 β(1 p(a)) βp (a) +(u(c) u (c)c) This describes a decreasing relationship between c and a. The level of c and a is chosen to satisfy promise-keeping. Higher V requires higher c and lower a. So c is higher and a is lower than under autarky. But suppose the worker could change a taking c as given: Replicating the earlier argument, she sets u(w) = 1 β(1 p(a)) βp (a) + u(c). 1 β(1 p(a)) This reduces βp, or equivalently reduces a. (a) So there is an incentive problem. 14
Asymmetric Information The insurer chooses a time path for c while unemployed. The worker chooses a, taking the consumption path as given. Formulate cost minimization problem recursively: ( ) C(V )= min c + β(1 p(a))c(w ) c,a,w where and V = u(c) a + β ( p(a)v e +(1 p(a))w ) 1=βp (a)(v e W ). 15
Write as a Lagrangian: C(V )=L = c + β(1 p(a))c(w ) θ ( u(c) a + β ( p(a)v e +(1 p(a))w ) V ) Envelope condition is C (V )=θ. η ( βp (a)(v e W ) 1 ) First order condition for W is C (W )=θ ηp (a) 1 p(a). So C (V ) >C (W )orv>w(assuming C is convex). Worker s utility must decline while unemployed. The optimal choice of c implies θ =1/u (c). Since θ declines over time, c must fall as well. Worker s first order condition implies a increases. 16