# Game theory - Write a maximization problem (with restriction)

I have a problem with finding a way how to solve this example: We have the company, which has two types of customers indexed as $H$ and $L$. We assume: $θ_H > θ_L$. Quantity$(Q)$ of $H$ indexed customers is in population. Payout of type of the customer$(i)$, which buys products($x$) for a price($p$) is $U_i=θ_iv(x)−p$. Company determines quantity and price for that quantity for every type of a customer ($x_i,p_i$). Costs of the company are $C(x_h,x_l)=x_h+x_l$. Company does not know type of the customer and maximizes its profit.

Find maximization problem which company solves (with restriction). Solve this example with $v(x)=\ln x$.

• The function $v$ does not occur in your description of the model. May 14 '18 at 18:53
• The function $v$ is in $U_i$ payout. I editted it. May 14 '18 at 21:24
• You could start writing the incentive compatibility constraints that guarantee that a type $i$ customer chooses the contract $(x_i,p_i)$.There should also be some individual rationality constraints. Is the outside option $0$? May 14 '18 at 21:30
• I think that outside option should be $0$. But I don't know the meaning of the θ, is it some willingness to pay? May 14 '18 at 21:37
• Pretty much. The $\theta_i$ guarantee that for every $x$, the marginal utility when measured in terms of the numeraire good (m) differs by the same proportion. May 14 '18 at 21:39

Here is the maximization problem solved by the company :

\begin{eqnarray*} \max_{x_h, p_h, x_l, p_l} & p_h + p_l - x_h - x_l \\ \text{s.t} & \theta_hv(x_h) - p_h \geq 0 \tag{IR$_h$} \\ & \theta_lv(x_l) - p_l \geq 0 \tag{IR$_l$} \\ & \theta_hv(x_h) - p_h \geq \theta_hv(x_l) - p_l \tag{IC$_h$} \\ & \theta_lv(x_l) - p_l \geq \theta_lv(x_h) - p_h \tag{IC$_l$}\end{eqnarray*}

Now we solve the above maximization problem for $v(x) = \ln x$. Given that $\theta_l < \theta_h$, it is easy to see that only IC$_h$ and IR$_l$ will be binding in optimum, and we can rewrite the optimization problem as :

\begin{eqnarray*} \max_{x_h, p_h, x_l, p_l} & p_h + p_l - x_h - x_l \\ \text{s.t} & \theta_l\ln x_l - p_l = 0 \tag{IR$_l$} \\ & \theta_h\ln x_h - p_h = \theta_h\ln x_l - p_l \tag{IC$_h$} \end{eqnarray*}

Substituting the constraints in the objective, we can rewrite the objective as follows :

\begin{eqnarray*} \max_{x_h, x_l} & \ \ \theta_h\ln x_h + (2\theta_l-\theta_h)\ln x_l - x_h - x_l \end{eqnarray*}

Further assuming $\theta_h + 1 < 2\theta_l$, and solving the problem yields the following:

\begin{eqnarray*} x_h^* & = & \theta_h \\ x_l^* & = & 2\theta_l-\theta_h \end{eqnarray*} Consequently, \begin{eqnarray*} p_l^* & = & \theta_l \ln(2\theta_l-\theta_h) \\ p_h^* & = & \theta_h\ln \theta_h + (\theta_l-\theta_h)\ln (2\theta_l-\theta_h) \end{eqnarray*}