Hello everyone,
I am working on my thesis and am currently checking whether my calculations are correct by using Mathematica. I extend a classical binary agency model, where a principal lets a agent fulfill tasks for him, by adding a third profit state x3. The principal cannot observe the agent so he must create financial rewards or incentives to motivate the agent to take the desired actions. The principal can only evaluate the outcome x (with x1 < x2 < x3), not the agent's level of effort, which can either be high (H) or low (L). The outcome and effort level are linked via different possibilities Phi, which sum up to 1 and lie between 0 and 1. The principal wants to maximize his expected profit Z and therefore uses different remuneration schemes (bonus or malus).
I set up a Lagrange-Function, which needs to fulfill the Kuhn-Tucker constraints.
- The first derivative of Lagrangea3a after u1a and after u3a need to be 0.
- (u1a([Phi]H1 + [Phi]H2) + u3a[Phi]H3 <= vH + U and u1a([Phi]H1 + [Phi]H2) + u3a[Phi]H3 - vH <= u1a([Phi]L1 + [Phi]L2) + u3a[Phi]L3 - vL
- a (u1a ([Phi]H1 + [Phi]H2) + u3a [Phi]H3 - (vH + U)) and b (-vH + vL + u1a ([Phi]H1 + [Phi]H2) + u3a [Phi]H3 - u1a ([Phi]L1 + [Phi]L2) - u3a [Phi]L3) must equal 0
- And a>=0 and b>=0
So how could I solve my Lagrange-function for the Lagrange variables a and b under the given conditions?
Thank you so much!
Mariana
In[1]:= ClearAll
Out[1]= ClearAll
In[2]:= $Assumptions = 0 < \[Phi]H1 < \[Phi]H2 < \[Phi]H3 < 1
$Assumptions = 0 < \[Phi]L1 > \[Phi]L2 > \[Phi]L3 < 1
$Assumptions = 0 < \[Phi]H1 < \[Phi]L1 < 1
$Assumptions = 0 < \[Phi]L3 < \[Phi]H3 < 1
$Assumptions = 0 < vL < vH
$Assumptions = vL != vH
$Assumptions = a >= 0
$Assumptions = b >= 0
Out[2]= 0 < \[Phi]H1 < \[Phi]H2 < \[Phi]H3 < 1
Out[3]= 0 < \[Phi]L1 && \[Phi]L1 > \[Phi]L2 > \[Phi]L3 && \[Phi]L3 < 1
Out[4]= 0 < \[Phi]H1 < \[Phi]L1 < 1
Out[5]= 0 < \[Phi]L3 < \[Phi]H3 < 1
Out[6]= 0 < vL < vH
Out[7]= vL != vH
Out[8]= a >= 0
Out[9]= b >= 0
In[10]:= Za3a[\[Phi]H1, \[Phi]H2, \[Phi]H3, x1, x2, x3, u1a,
u3a] = \[Phi]H1*(x1 - u1a^2) + \[Phi]H2*(x2 -
u1a^2) + \[Phi]H3*(x3 - u3a^2)
Out[10]= (-u1a^2 + x1) \[Phi]H1 + (-u1a^2 + x2) \[Phi]H2 + (-u3a^2 +
x3) \[Phi]H3
In[11]:= PCa = u1a*(\[Phi]H1 + \[Phi]H2) + u3a*\[Phi]H3 == vH + U
ICa = u1a*(\[Phi]H1 + \[Phi]H2) + u3a*\[Phi]H3 - vH ==
u1a*(\[Phi]L1 + \[Phi]L2) + u3a*\[Phi]L3 - vL
Out[11]= u1a (\[Phi]H1 + \[Phi]H2) + u3a \[Phi]H3 == U + vH
Out[12]= -vH + u1a (\[Phi]H1 + \[Phi]H2) + u3a \[Phi]H3 == -vL +
u1a (\[Phi]L1 + \[Phi]L2) + u3a \[Phi]L3
In[20]:= Lagrangea3a[\[Phi]H1, \[Phi]H2, \[Phi]H3, \[Phi]L1, \
\[Phi]L2, \[Phi]L3, x1, x2, x3, u1a, u3a, a,
b] = (\[Phi]H1*u1a^2 + \[Phi]H2*u1a^2 + \[Phi]H3*u3a^2) -
a (u1a (\[Phi]H1 + \[Phi]H2) + u3a \[Phi]H3 - (vH + U)) -
b (u1a*(\[Phi]H1 + \[Phi]H2) + u3a*\[Phi]H3 -
vH - (u1a*(\[Phi]L1 + \[Phi]L2) + u3a*\[Phi]L3 - vL))
Out[20]= u1a^2 \[Phi]H1 + u1a^2 \[Phi]H2 + u3a^2 \[Phi]H3 -
a (-U - vH + u1a (\[Phi]H1 + \[Phi]H2) + u3a \[Phi]H3) -
b (-vH + vL + u1a (\[Phi]H1 + \[Phi]H2) + u3a \[Phi]H3 -
u1a (\[Phi]L1 + \[Phi]L2) - u3a \[Phi]L3)
In[24]:= Solve[
Lagrangea3a[\[Phi]H1, \[Phi]H2, \[Phi]H3, \[Phi]L1, \[Phi]L2, \
\[Phi]L3, x1, x2, x3, u1a, u3a, a, b], {a, b}]
During evaluation of In[24]:= Solve::naqs: u1a^2 \[Phi]H1+u1a^2 \[Phi]H2+u3a^2 \[Phi]H3-a (-U-vH+u1a (\[Phi]H1+\[Phi]H2)+u3a \[Phi]H3)-b (-vH+vL+u1a (\[Phi]H1+\[Phi]H2)+u3a \[Phi]H3-u1a (\[Phi]L1+\[Phi]L2)-u3a \[Phi]L3) is not a quantified system of equations and inequalities. >>
Out[24]= Solve[
u1a^2 \[Phi]H1 + u1a^2 \[Phi]H2 + u3a^2 \[Phi]H3 -
a (-U - vH + u1a (\[Phi]H1 + \[Phi]H2) + u3a \[Phi]H3) -
b (-vH + vL + u1a (\[Phi]H1 + \[Phi]H2) + u3a \[Phi]H3 -
u1a (\[Phi]L1 + \[Phi]L2) - u3a \[Phi]L3), {a, b}]
