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

CC BY-SA 4.0
2021-05-20T14:48:16.640Z
2021-05-20T14:48:16.640Z
tdm
<p>Consider three agents <span class="math-container">$A_i$</span>, who engage in a Bertrand game. All agents have perfect knowledge on all parameters and the distribution <span class="math-container">$F()$</span>.</p>
<ul>
<li><span class="math-container">$A_1$</span> moves first and selects price <span class="math-container">$0\leq p_1\in \mathbb{R}$</span>. <span class="math-container">$A_2$</span> moves second and selects price <span class="math-container">$0\leq p_2\in \mathbb{R}$</span>. <span class="math-container">$A_3$</span> moves third and selects <span class="math-container">$0\leq q_1\in \mathbb{R}$</span> and <span class="math-container">$0\leq q_2\in \mathbb{R}$</span>, that is, the quantities that it buys from <span class="math-container">$A_1$</span> and <span class="math-container">$A_2$</span>, which it sells at market clearing price.</li>
<li><span class="math-container">$A_1$</span> tries to maximize <span class="math-container">$  q_1 p_1$</span>.</li>
<li><span class="math-container">$A_2$</span> tries to maximize <span class="math-container">$ q_2 p_2$</span>.</li>
<li><span class="math-container">$A_3$</span> tries to maximize <span class="math-container">$$((A - q_1 - q_2) (q_1+q_2) - q_1 p_1 - q_2 p_2) (1-F(-q_1 p_1))+
( (A - q_2) q_2- q_2 p_2+B) F(- q_1 p_1) $$</span>, with <span class="math-container">$F$</span> being a continuous cumulative distribution function. <span class="math-container">$A_3$</span>'s profit function captures that buying from <span class="math-container">$A_2$</span> comes at no risk, while <span class="math-container">$A_1$</span> might be unable to deliver. However, the higher <span class="math-container">$A_1$</span>'s profits the smaller that risk and the higher the chance to capture the benefits/losses <span class="math-container">$B$</span>.</li>
</ul>
<p>Note, all agents have perfect knowledge on all parameters and the distribution <span class="math-container">$F()$</span>.</p>
<p><strong>Does <span class="math-container">$A_3$</span>'s profit increase in <span class="math-container">$B$</span>? How to prove this?</strong></p>
<p>(If <span class="math-container">$F$</span> is uniform, <span class="math-container">$A_3$</span> will only buy from one of the agents and it is easy to show that the answer is yes. If <span class="math-container">$p_1$</span> and <span class="math-container">$p_2$</span>, then it is easy to show that the answer is yes.)<br />
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.</p>

CC BY-SA 4.0
2021-05-16T15:39:25.703Z
What exactly do $A_2$ and $A_3$ know when they make their choices? How does $A_3$'s payoff function come about?
Grada Gukovic
CC BY-SA 4.0
2021-05-16T16:30:04.267Z
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 $q_1$?
Jesper Hybel
CC BY-SA 4.0
2021-05-16T17:46:36.487Z
@ 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
CC BY-SA 4.0
2021-05-16T17:48:18.563Z
@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
CC BY-SA 4.0
2021-05-16T18:22:42.037Z
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.
Jesper Hybel
<game-theory><risk>
Bertrand game with delivery risk and side payments
Paul
# Bertrand game with delivery risk and side payments

(Asked by: username_0 on 2021-05-12)

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.

**Tags:** game-theory, risk

> **Comments:**
> **username_1**: What exactly do $A_2$ and $A_3$ know when they make their choices? How does $A_3$'s payoff function come about?  
> **username_2**: 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 $q_1$?  
> **username_0**: @ username_1 username_1: All agents have perfect knowledge on all parameters and the distribution $F()$. What exactly do you mean with how does it come about?  
> **username_0**: @username_2: 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.  
> **username_2**: 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.  

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## 1 answers

### Answer by username_3 on 2021-05-20. Score: 2 | Accepted answer

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 $\Pi$.


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