## Q5
$$
% My definitions
\def\ve{{\varepsilon}}
\def\dd{{\text{ d}}}
\newcommand{\dif}[2]{\frac{d #1}{d #2}} % for derivatives
\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}} % for partial derivatives
\def\R{\text{R}}
\def\E{\mathbb{E}}
$$

Derive the pricing equation under each of the following equilibrium assumptions:

(a) marginal cost pricing (no derivation needed).

(b) single product firms in a Bertrand Nash Equilibrium

(c ) multiproduct firms in a Bertrand Nash Equilibrium

(d) perfect collusion / joint profit maximization

### Answer
#### a
For each product $j$,
$$
p_j = MC_j.
$$

#### b
The firm produce product $j$ aims to
$$
\max_{p_j} \pi_j (\cdot) = (p_j - mc_{j}) s_j(\cdot)
$$

The F.O.C is

$$
s_j(\cdot) + (p_j - mc_j) \pd{s_j}{p_j} = 0
$$

#### c
Suppose firm $f$ produce $j \in J_f$ products and $p$ is the vector of price of all of its products. The firm aims to
$$
\max_{p} \pi_f (\cdot) = \sum_{j \in J_f}(p_j - mc_{j}) s_j(\cdot).
$$

The F.O.C is

$$
s_j(\cdot) + \sum_{r \in J_f}(p_r - mc_r) \pd{s_r}{p_j} = 0, j \in J_f
$$

#### d
Let $J$ be the set of all products, and $p$ be the vector of price of all products, the cartel aims to

$$
\max_{p} \pi_c (\cdot) = \sum_{j \in J}(p_j - mc_{j}) s_j(\cdot).
$$

The F.O.C is

$$
s_j(\cdot) + \sum_{r \in J}(p_r - mc_r) \pd{s_r}{p_j} = 0, j \in J
$$

#### Cross-price derivative
Under the above vertical differenciation model, define $\psi_j = \frac{\delta_j- \delta_{j-1}}{p_j - p_{j-1}}, j \ge 2$. We know

$$
s_1 =F(\frac{\delta_1}{p_1}) - F(\frac{\delta_2- \delta_1}{p_2 - p_1})
$$

Hence,

$$
\pd{s_1}{p_1} = -F'(\frac{\delta_1}{p_1}) \frac{s_1}{p_1^2} - F'(\psi_2) \frac{\delta_2-\delta_1}{(p_2-p_2)^2}.
$$

$$
\pd{s_1}{p_2} = F'(\psi_2)\frac{\delta_2-\delta_1}{(p_2-p_1)^2}
$$

In general,

$$
s_j =F(\frac{\delta_j- \delta_{j-1}}{p_j - p_{j-1}}) - F(\frac{\delta_{j+1}- \delta_j}{p_{j+1} - p_j})
$$

We have

$$
\pd{s_j}{p_j} = -F'(\psi_j) \frac{\delta_j-\delta_{j-1}}{(p_j-p_{j-1})^2} - F'(\psi_{j+1}) \frac{\delta_{j+1}-\delta_{j}}{(p_{j+1}-p_{j})^2}.
$$

and

$$
\pd{s_j}{p_{j-1}} = F'(\psi_j) \frac{\delta_j-\delta_{j-1}}{(p_j-p_{j-1})^2}.
$$

$$
\pd{s_j}{p_{j+1}} = F'(\psi_{j+1}) \frac{\delta_{j+1}-\delta_{j}}{(p_{j+1}-p_{j})^2}.
$$

For product $j$, $\pd{s_j}{p_{k}}=0$ if $k \notin \{j-1, j, j+1\}$.

Based on the distribution,
$$
F(x) = 1 - e^{-\lambda x} \implies F'(x) = \lambda e^{-\lambda x}
$$