Moral Hazard with risk neutral agent

Question

We have a principal-agent model with hidden actions in which the principal is risk averse and the agent is risk neutral; Assume also there are two levels of output, $x$ and $x'$ (with $x'>x$) and two actions $a,a'$. Define $p(a),p(a')$ the probabilities of $x'$ under actions $a,a'$ respectively. Also, the agent disutility from action $a'$ is $-1$. The wages associated to $x,x'$ are $w,w'$ respectively.

My problem is that I am not sure how to show that the optimal contract requires $x'-w' =x-w$, i.e. that the agent, being risk neutral, takes on all the variability associated with the project.

I formalize the problem (assume the principal wants to induce $a'$, otherwise my question is trivial)

$\max\limits_{\{w,w'\}} u(x'-w')p(a') + u(x-w)(1-p(a'))$

st

$w'p(a') + w(1-p(a')) - 1 \geq 0 $

$w'p(a') + w(1-p(a')) - 1 \geq w'p(a) + w(1-p(a))$

In particular, when I try to solve the problem by maximizing the principal expected payoff subject to the "standard" Individual rationality (with $\lambda$ multiplier) and incentive compatibility (with $\mu$ multiplier) constraints (I assume the principal is interested in the more costly action $a'$) I end up with two equations which are not consistent with the aforementioned result. In particular:

$ u'(x-w) = \lambda + \mu [1- \frac{(1-p(a))}{(1-p(a'))}]$

$ u'(x'-w') = \lambda + \mu [1- \frac{p(a)}{p(a')}]$

It is evident that $x-w = x'-w'$ holds iff $p(a) =p(a')$ which is not the case in this problem (here we have that $p(a') >p(a)$). Another possibility would be to assume that the Incentive compatibility constraint is slack (hence $\mu = 0$); however I cannot understand why that should hold, when the principal wants to induce the most costly action $a'$ (help here)

I have read online that another approach would be to assume that the principal "sells" the project to the agent and the agent, after having chosen which level of effort maximizes its expected utility, pays back a fixed amount to the principal (call it $\beta_{a}, \beta_{a'}$)

So we would have something like:

$w'p(a') + w(1-p(a')) - 1 -\beta_{a'} \geq 0 $ if the agent chose to undertake high effort and $w'p(a) + w(1-p(a)) -\beta_a \geq 0 $ otherwise.

But then how to go from there? How to insure that the agent is going to choose the action $a'$? How are the fixed amounts determined? Why are they optimal?

Answer 1

This answer shows three things:

1. We do not need the Lagrangian approach to solve your maximization problem.
2. We do not need the assumption that $x'-x=\frac{1}{p(a')-p(a)}$ either.
3. The condition $x'-w'=x-w$ is not necessarily satisfied for the optimal contract.

Fix indeed the payment $w$. The problem can be written \begin{equation*} \max_{w'}{u(x'-w')p(a')} \end{equation*} given the constraints \begin{align*} & w'p(a') \geq 1 - w[1-p(a')] \\ & w'[p(a')-p(a)] \geq 1+w[p(a')-p(a)] \end{align*} It is clear that the principal has interest to set the lowest possible value for $w'$ given this set of constraint, since the objective function is decreasing in $w'$. Therefore he will set \begin{equation} w' = \max\{\frac{1-w[1-p(a')]}{p(a')},\frac{1+w[p(a')-p(a)]}{p(a')-p(a)}\} \end{equation}

As @Alecos_Papadopoulos did, it makes sense to assume that the agent is protected by limited liability, i.e. that his payments are nonnegative. Otherwise the problem does not necessarily have a solution: the principal could always benefit from decreasing $w$ and increasing $w'$ so as to keep the individual rationality constraint satisfied. But the contract $(w=-\infty,w'=+\infty)$ is obviously not a satisfactory solution. Therefore I restrict attention to the case where $w \geq 0$ and $w' \geq 0$.

The condition $w \geq 0$ implies \begin{equation*} \dfrac{1+w[p(a')-p(a)]}{p(a')-p(a)} \geq \dfrac{1-w[1-p(a')]}{p(a')} \end{equation*} and therefore \begin{equation*} w' = \dfrac{1+w[p(a')-p(a)]}{p(a')-p(a)} \end{equation*}

Plugging this equation into the objective function, the principal's problem becomes

\begin{equation*} \max_{w \geq 0}{u(x'-\frac{1}{p(a')-p(a)}-w)p(a')+u(x-w)(1-p(a'))} \end{equation*} This objective function is decreasing in $w$. Therefore he simply sets $w=0$ and $w'=\dfrac{1}{p(a')-p(a)}$. As a conclusion, the equality $x'-w'=x-w$ has no reason to be satisfied unless one assumes that $x'-x=\dfrac{1}{p(a')-p(a)}$, i.e. that \begin{equation*} p(a') x' + (1-p(a'))x -1= p(a)x' + (1-p(a))x \end{equation*} This latter equation means that the social surplus resulting from $a'$ equals the surplus resulting from $a$: it is a very particular case where the cost of effort for the agent is exactly compensated by the increase in expected output for the principal. In all other cases, we have $x'-w' \ne x-w$.

I think the reason why the agent does not take on all the risk is because his actions are not observable, and therefore not contractible upon. This property would be true in a risk-sharing economy with unconstrained allocations. But the allocation is here distorted by the need to incentivize the agent to exert a high effort.

Answer 2

A thing that bothers me here is the following: the Incentive Compatibility constraint is

$$IC: w'p(a') + w(1-p(a')) - 1 \geq w'p(a) + w(1-p(a))$$

$$\implies w'-w \geq \frac {1}{p(a')-p(a)} \tag{1}$$

...since by assumption $p(a')-p(a) >0$. We are told that we should find that at the optimum, $$x'-w' = x-w \implies x'-x = w'-w \tag{2}$$

Combining $(1)$ and $(2)$, if indeed this is the optimum under the given constraints, we must also have

$$x'-x \geq \frac {1}{p(a')-p(a)} \tag{3}$$

But this is an additional, necessary constraint on a priori magnitudes, that must hold if the postulated optimal solution is to be admissible. Even if indeed such a constraint is assumed, in any case, it visibly reduces the generality of the problem (which purports to show something general, i.e. how the risk-neutrality of the agent affects the solution).

Nevertheless, let's work this a bit more formally. I will assume that $w, w'$ can be zero, but not negative. This is a maximization problem in normal form with inequality constraints, non-negative decision variables and non-negative multipliers. The full Lagrangean of the problem therefore is (I will compact notation in an obvious way),

$$\Lambda = u(x'-w')p' + u(x-w)(1-p') + \lambda\cdot [w'p' + w(1-p') - 1 ]\\ + \mu \cdot [ w'p' + w(1-p') - 1 - w'p - w(1-p)] + \xi w + \xi' w'$$

The essential first order conditions are

$$\frac {\partial \Lambda}{\partial w} \leq 0, \;\;\frac {\partial \Lambda}{\partial w} \cdot w = 0$$

and analogously for $w'$. These result in

$$\frac {\partial \Lambda}{\partial w} = -u'(x-w)(1-p') +\lambda (1-p') - \mu (p'-p) + \xi \leq 0$$

$$\implies u'(x-w)(1-p') \geq \lambda (1-p') - \mu (p'-p) + \xi $$

$$\implies u'(x-w) \geq \lambda - \mu \frac {p'-p}{1-p'} + \frac {\xi}{1-p'} \tag{4}$$

$$\frac {\partial \Lambda}{\partial w'} = -u'(x'-w')p' +\lambda p' + \mu (p'-p) + \xi' \leq 0$$

$$\implies u'(x'-w') \geq \lambda + \mu\frac {p'-p}{1-p'} + \frac {\xi'}{p'} \tag{5}$$

First note that not both wages can be zero, because the constraints would be violated. Given this, consider the possibility that the $IR$ is binding (so $\lambda >0$). If it is binding, then with not both wages zero, the $IC$ constraint will necessarily be violated. So we conclude that

$$\lambda_* = 0$$

and the first-order conditions now become

$$u'(x-w) \geq - \mu \frac {p'-p}{1-p'} + \frac {\xi}{1-p'} \tag {4a}$$

$$ u'(x'-w') \geq \mu \frac{p'-p}{1-p'} + \frac {\xi'}{p'} \tag{5a}$$

Now note that if $\xi =0$ (i.e. $w >0$) then $(4a)$ should hold as an equality and with the last term on the right equal to zero. But this would require negative marginal utility which is inadmissible. We also know that not both wages can be zero. So we conclude that we must have

$$\xi_* > 0 , w_*=0,\;\;\; \xi'_* = 0,\; w'_* >0$$

and the conditions now become

$$u'(x) \geq - \mu \frac {p'-p}{1-p'} + \frac {\xi_*}{1-p'} \tag {4b}$$

$$ u'(x'-w') = \mu \frac{p'-p}{1-p'} \tag{5b}$$

Eq. $(5b)$ implies that $\mu_*>0$ , under a usual utility function specification, which does not give zero marginal utility except at infinity. This in turn means that the $IC$ constraint should hold as an equality. Given that $w_*=0$ this gives

$$IC: w'p' - 1 - w'p =0 \implies = w'_* = \frac 1{p'-p} \tag{6}$$

This should ring a bell, because the right-hand-side of $(6)$ is the same as the right-hand-sides of $(1)$ and $(3)$.

Namely, if we are assuming a priori that $x'-x = \frac 1{p'-p}$, then the solution we have arrived at validates the claim $x'-w'_* = x-w_*$

Under this additional assumption, we also obtain

$$u'(x) \geq - \mu_* \frac {p'-p}{1-p'} + \frac {\xi_*}{1-p'} \tag {4c}$$

$$ u'(x) = \mu_* \frac{p'-p}{1-p'} \tag{5c}$$

Combining, we obtain

$$ \mu \frac{p'-p}{1-p'} \geq - \mu \frac {p'-p}{1-p'} + \frac {\xi_*}{1-p'}$$

$$\implies \mu_* \geq \frac {\xi_*}{2(p'-p)} \tag{7}$$

This is admissible. So under $x'-x = \frac 1{p'-p}$, we obtain the solution

$$\left \{w'_* = x'-x = 1/(p'-p), w_* =0, \lambda_*=0, \mu_* \geq \frac {\xi_*}{2(p'-p)}, \xi_* >0, \xi'_* =0\right\}$$