A risk neutral person has following type of utility function.
It is a
strictly increasing differentiable linear function. It is shown in
figure 1.
Take the same lottery as before.
Suppose a risk averse individual has m0 amount of currency. With
currency this guy can buy a lottery. The lottery is of following
nature:
It gives m1 with probability p
It gives m2 with probability (1 − p).
Suppose m0 = m1 p + m2 (1 − p), 0 < m1 < m0 < m2 .
A risk neutral person is such that
u(m0 = m1 p + m2 (1 − p)) = u(m1 ) × p + u(m2 ) × (1 − p). It is
shown in figure 1.
The risk neutral person is indifferent between buying this lottery
and holding the money m0 .
An example is u(m) = am where a > 0.
A person is risk lover if the utility function is strictly increasing in
nature. It is shown in figure 2.
Take the same lottery as before.
Suppose a risk averse individual has m0 amount of currency. With
currency this guy can buy a lottery. The lottery is of following
nature:
It gives m1 with probability p
It gives m2 with probability (1 − p).
Suppose m0 = m1 p + m2 (1 − p), 0 < m1 < m0 < m2 .
A risk lover person is such that
u(m0 = m1 p + m2 (1 − p)) < u(m1 ) × p + u(m2 ) × (1 − p). It is
shown in figure 2.
The risk lover person prefers to buy this lottery rather than holding
the money m0 .
An example is u(m) = mα where α > 1.
From the feasible set in terms of money or currency
M = {m = (m1 , m2 ) : m1 + m2 ≤ 1}. This set was shown in last
class. It is feasible set.
We define the feasible set in terms of utility of the two individuals.
It is
X = {u = (u1 (m1 ), u2 (m2 )) : m ∈ M}.
We want the feasible set to be a convex set.
A set S is convex set, if for all x , y ∈ S then
λ × x + (1 − λ) × y ∈ S where 0 < λ < 1. It is shown in figure 3.
It is obvious that M is a convex set.
We have to show that X is also a convex set when the individuals
are risk averse.
We want the following properties to be satisfied by the feasible set
X in the bargaining problem.
1) X is a convex set.
2) X is a closed and bounded.
3) Free disposal is allowed.
X is closed and bounded. It means that if there a sequence
belonging to the set X then the limit point should also belong to
that set.
There are numbers ū1 , ū2 > 0 such that for all u ∈ X , we have
u = (u1 , u2 ) < ū = (ū1 , ū2 ). This means that the set X is
bounded. It is obvious that X contains only non-negative number.
Free disposal means that if individual 1 has m1 amount of the cake
and want to consume only m1 − amount. No additional cost is
incurred to dispose off amount. It also means that If u(m) ∈ X
then u(m − ) ∈ X .
We show that X is a convex set.
Take any m and n belonging to M and u(m) ∈ X and also
u(n) ∈ X . We have to show that λ × u(m) + (1 − λ) × u(n) ∈ X ,
where 0 < λ < 1.
We know that ui (mi ), i = 1, 2 is strictly concave in nature because
individuals are risk averse in nature. So we have
ui (mi λ + ni (1 − λ)) > ui (mi )λ + ui (ni )(1 − λ), i = 1, 2
We know that (u1 (m1 λ + n1 (1 − λ)), u2 (mi λ + ni (1 − λ))) ∈ X
since m1 λ + n1 (1 − λ) ∈ M. Because M is a convex set.
(u1 (m1 )λ + u1 (n1 )(1 − λ), u2 (m2 )λ + u2 (n2 )(1 − λ)) ∈ X from the
free disposal condition. Thus, X is a convex set.
A Nash bargaining problem is given by a pair (X , d) where X is the
feasible set and d is the disagreement point. The disagreement
point d = (d1 , d2 ) is such that that d ∈ X and is the payoff
received by each individual when the two players ( person) cannot
come to agreement on a particular division.
It is also called the status-qua point, the payoff received when
there is disagreement. It is shown in figure 4.
Suppose bargaining solution or the payoff on which both the
person ( players) agree upon is u ∗ = (u1∗ , u2∗ ). It should be such
that u1∗ ≥ d1 and u2∗ ≥ d2 and u ∗ ∈ X .
Given the bargaining problem (X , d) objective is to find a feasible
solution u ∗ which is acceptable to both persons (players).
The solution of the Nash bargaining problem is given by
max (u1 (m1 ) − d1 )(u2 (m2 ) − d2 )
subject to (u1 , u2 ) ∈ X .
This is also called as Nash product. The optimal (m1 , m2 ) is
obtained by maximising the Nash product by taking m1 = m and
m2 = 1 − m.
The optimal solution is
differentiate (u1 (m) − d1 )(u2 (1 − m) − d2 ) with respect to m. This
gives
u10 (m)(u2 (1 − m) − d2 ) − (u1 (m) − d1 )u20 (1 − m)
The first order condition gives
u10 (m)(u2 (1 − m) − d2 ) = (u1 (m) − d1 )u20 (1 − m)
We get a unique m satisfying the above equation ( first order
condition).
The second order condition is
u100 (m)(u2 (1 − m) − d2 ) − u10 (m)u20 (1 − m) + (u1 (m) − d1 )u200 (1 −
m) − u20 (1 − m)u10 (m).
It is negative for all m > 0 since ui00 () < 0, i = 1, 2. It is due to
concavity of the utility function.
Thus, the optimal m∗ , we get from the first order condition is
maximizing the Nash product. It is shown in figure 5.
The Nash solution to the bargaining problem (X , d) is given by the
maximization of the Nash product. This optimization is
characterized based on four axioms.
