# Bertrand game with delivery risk and side payments

Consider three agents $$A_i$$, who engage in a Bertrand game. All agents have perfect knowledge on all parameters and the distribution $$F()$$.

• $$A_1$$ moves first and selects price $$0\leq p_1\in \mathbb{R}$$. $$A_2$$ moves second and selects price $$0\leq p_2\in \mathbb{R}$$. $$A_3$$ moves third and selects $$0\leq q_1\in \mathbb{R}$$ and $$0\leq q_2\in \mathbb{R}$$, that is, the quantities that it buys from $$A_1$$ and $$A_2$$, which it sells at market clearing price.
• $$A_1$$ tries to maximize $$q_1 p_1$$.
• $$A_2$$ tries to maximize $$q_2 p_2$$.
• $$A_3$$ tries to maximize $$((A - q_1 - q_2) (q_1+q_2) - q_1 p_1 - q_2 p_2) (1-F(-q_1 p_1))+ ( (A - q_2) q_2- q_2 p_2+B) F(- q_1 p_1)$$, with $$F$$ being a continuous cumulative distribution function. $$A_3$$'s profit function captures that buying from $$A_2$$ comes at no risk, while $$A_1$$ might be unable to deliver. However, the higher $$A_1$$'s profits the smaller that risk and the higher the chance to capture the benefits/losses $$B$$.

Note, all agents have perfect knowledge on all parameters and the distribution $$F()$$.

Does $$A_3$$'s profit increase in $$B$$? How to prove this?

(If $$F$$ is uniform, $$A_3$$ will only buy from one of the agents and it is easy to show that the answer is yes. If $$p_1$$ and $$p_2$$, then it is easy to show that the answer is yes.)
In the general case, which is outlined above, my numerical studies suggest that the answer is yes, but I struggle to write down the analytical proof.

• What exactly do $A_2$ and $A_3$ know when they make their choices? How does $A_3$'s payoff function come about? May 16 at 15:39
• Is there anything that prevents you from applying the envelope theorem? I could be missing something but can you not simply differentiate profit with respect to $B$ and get $F(-q_1p_1)$ and evaluate at optimal $q_1$? May 16 at 16:30
• @ Grada Gukovic: All agents have perfect knowledge on all parameters and the distribution $F()$. What exactly do you mean with how does it come about?
– Paul
May 16 at 17:46
• @Jesper Hybel: I guess you are missing that this is Game Theory and not straight forward Optimization. $A_1$ and $A_2$ may change their price in response to a different $B$. However, perhaps it is me who underestimates the envelope theorem.
– Paul
May 16 at 17:48
• Well maybe and maybe not. Anyway Caputo, M.R., 1996. "The envelope theorem and comparative statics of Nash equilibria". Games Econ. Behav. 13, 201–224 extends the envelope theorem to cover static games with locally differentiable Nash equilibria. So perhaps the main problem arises where prices $p_1$ and $p_2$ are not differentiable with respect to $B$ which I would expect to happen where the firms "choose" 0 production. May 16 at 18:22

I am not going to give a solution to you question (as I am not able too). However, let me try to indicate the difficulty of obtaining one if you cannot find a closed form solution for $$q_1$$ and $$q_2$$ in the last stage of the game.
Let $$\Pi(q_1, q_2, p_1, p_2, B)$$ be the objective function that we want to maximize. Assuming we can totally differentiate this, we get: $$d\Pi = \Pi_{q_1} dq_1 + \Pi_{q_2} dq_2 + \Pi_{p_1} dp_1 + \Pi_{p_2} dp_2 + \Pi_B dB. \tag{1}$$ Assume the first order conditions hold: $$\Pi_{q_1} = 0\\ \Pi_{q_2} = 0 \tag{foc-stage3}$$ This then simplifies $$(1)$$ to: $$d\Pi = \Pi_{p_1} dp_1 + \Pi_{p_2} dp_2 + \Pi_B dB \tag{1.a}$$ The problem here is that we can not drop the $$dp_1$$ and $$dp_2$$ terms as they are (in general) both going to be a function of $$B$$.
Assume that the first order conditions $$\rm{(foc-stage3)}$$ can be solved to give a solution $$q_2(p_1, p_2, B)$$ and $$q_1(p_1, p_2, B)$$. Then in stage 2 of the game, Firm 2 maximizes: $$p_2 q_2(p_1, p_2, B)$$ Again, assume that the first order conditions hold: $$q_2 + p_2 \frac{\partial q_2}{\partial p_2} = 0 \tag{foc-stage2}$$ We can totally differentiate this with respect to $$p_1$$ and $$p_2$$ and $$B$$ (notice that $$q_2$$ is a function of all three these variables). This gives: $$\left(2\frac{\partial q_2}{\partial p_2} + p_2 \frac{\partial^2 q_2}{\partial p_2\partial p_2}\right) dp_2 = -\left(\frac{\partial q_2}{\partial p_1} + p_2 \frac{\partial^2 q_2}{\partial p_2 \partial p_1}\right) dp_1 - \left(\frac{\partial q_2}{\partial B} + p_2 \frac{\partial^2 q_2}{\partial p_2 \partial B}\right) dB \tag{2}$$
Let $$p_2(p_1, B)$$ be the optimal solution of firm 2 (by solving $$({\rm foc-stage2})$$. Then in stage 1, Firm 1 maximizes $$p_1 q_1(p_1, p_2(p_1, B), B)$$ So: $$q_1 + p_1 \frac{\partial q_1}{\partial p_1} + p_1 \frac{\partial q_1}{\partial p_2} \frac{\partial p_2}{\partial p_1}= 0 \tag{foc-stage1}$$
Assuming again that this can be totally differentiated with respect to $$p_1$$ and $$B$$, we obtain: \begin{align*} \left(\frac{\partial q_1}{\partial p_1} + \frac{\partial q_1}{\partial p_1} + p_1 \frac{\partial^2 q_1}{\partial p_1 \partial p_1} + \frac{\partial^2 q_1}{\partial p_2\partial p_1}\frac{\partial p_2}{\partial p_1} + \frac{\partial q_1}{\partial p_2}\frac{\partial^2 p_2}{\partial p_1\partial p_1}\right) dp_1\\ =- \left(\frac{\partial q_1}{\partial B} + p_1 \frac{\partial^2 q}{\partial p_1 \partial B} + \frac{\partial^2 q_1}{\partial p_2 \partial B} \frac{\partial p_2}{\partial p_1} + \frac{\partial q_1}{\partial p_2} \frac{\partial^2 p_2}{\partial p_1 \partial B}\right) dB \tag{3} \end{align*}
Then we can use $$(3)$$ to substitute for $$d p_1$$ in $$(2)$$. Then $$(2)$$ and $$(3)$$ give $$dp_1$$ and $$dp_2$$ in terms of only $$dB$$. This can be substituted into $$(1.a)$$ to give the total effect of $$B$$ on $$\Pi$$.