SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Both equilibrium quantities should be the best responses to the opponent's expected amounts
$$
q _ {i} ^ {*} = \frac {\mathbf {E} (1 - c _ {i} - \gamma \sum_ {j \neq i} q _ {j} ^ {*})}{2}.
$$
$$
\sum_ {i} q _ {i} ^ {*} = \frac {\mathbf {E} (m - \sum_ {i} c _ {i} - (m - 1) \gamma \sum_ {i} q _ {i} ^ {*})}{2},
$$
$$
\mathbf {E} \left(q _ {i} ^ {*}\right) = \frac {\mathbf {E} \mathbf {E} (\dots)}{2} = \frac {\mathbf {E} (\dots)}{2} = q _ {i} ^ {*}.
$$
$$
\sum_ {i} q _ {i} ^ {*} = \frac {m - \sum_ {i} c _ {i} ^ {e}}{2 + \gamma (m - 1)}.
$$
The best response equation gives
$$
q _ {i} ^ {*} = \frac {1 - c _ {i} ^ {e} - \gamma \sum_ {j} q _ {j} ^ {*} + \gamma q _ {i} ^ {*}}{2} \Longrightarrow q _ {i} ^ {*} = \frac {2 - \gamma - (2 + \gamma (m - 2)) c _ {i} ^ {e} + \gamma \sum_ {j \neq i} c _ {j} ^ {e}}{(2 - \gamma) (2 + (m - 1) \gamma)}.
$$
We get some expression for the quantities produced by the firms. However, we need to ensure that these quantities are positive. This property is equivalent to the following inequalities
$$
\forall i (2 - \gamma) \left(1 - c _ {i} ^ {e}\right) \geq \gamma \sum_ {j \neq i} \left(c _ {i} ^ {e} - c _ {j} ^ {e}\right). \tag {10}
$$
These inequalities hold, for example, when  $1 > a$  and  $(1 - a) \gg b$ , where  $a$  and  $b$  are parameters from Equation (7).
Now, notice that the expected profit function is a second-degree polynomial in  $q_{i}$
$$
\Pi_ {i} ^ {e} = - q _ {i} ^ {2} + \beta q _ {i}.
$$
Since the maximum value is achieved at  $q_{i}^{*}(q_{-i})$ , this polynomial looks like
$$
\varPi_ {i} ^ {e} = - \left(q _ {i} - q _ {i} ^ {*} (q _ {- i})\right) ^ {2} + q _ {i} ^ {*} (q _ {- i}) ^ {2}.
$$
And it gives the following equilibrium expected profits
$$
\Pi_ {i} ^ {e} = (q _ {i} ^ {*}) ^ {2}.
$$
A.3 Proof of Lemma 4.4
The expected profit maximization tasks of the firms are
$$
\max  _ {p _ {i}} \mathbf {E} \left(\left(p _ {i} - c _ {i}\right) \left(\frac {1 - \gamma - (1 + \gamma (m - 2)) p _ {i} + \gamma \sum_ {j \neq i} p _ {j}}{(1 - \gamma) (1 + \gamma (m - 1))}\right)\right).
$$
Thus, the best responses are
$$
p _ {i} ^ {*} (p _ {- i}) = \frac {\mathbf {E} (1 - \gamma + (1 + \gamma (m - 2)) c _ {i} + \gamma \sum_ {j \neq i} p _ {j})}{2 (1 + \gamma (m - 2))}.
$$
$$
\mathbf {E} \left(p _ {i} ^ {*}\right) = \frac {\mathbf {E E} (\dots)}{2 (1 + \gamma (m - 2))} = \frac {\mathbf {E} (\dots)}{2 (1 + \gamma (m - 2))} = p _ {i} ^ {*}.
$$
$$
p _ {i} ^ {*} = \frac {1 - \gamma + (1 + \gamma (m - 2)) c _ {i} ^ {e} + \gamma \sum_ {j \neq i} p _ {j} ^ {*}}{2 (1 + \gamma (m - 2))}.
$$
If we sum these equations, we get
$$
\sum_ {i} p _ {i} ^ {*} = \frac {(1 - \gamma) m + (1 + \gamma (m - 2)) \sum_ {i} c _ {i} ^ {e} + \gamma (m - 1) \sum_ {i} p _ {i} ^ {*}}{2 (1 + \gamma (m - 2))} \Rightarrow
$$
$$
\sum_ {i} p _ {i} ^ {*} = \frac {(1 - \gamma) m + (1 + \gamma (m - 2)) \sum_ {i} c _ {i} ^ {e}}{2 + \gamma (m - 3)}.
$$
After substitution of this sum into the best response equation, we have
$$
p _ {i} ^ {*} = \frac {d _ {1} + d _ {2} c _ {i} ^ {e} + d _ {3} \sum_ {j \neq i} c _ {j} ^ {e}}{d _ {4}},
$$
$$
d _ {1} = 2 + \gamma (2 m - 5) - \gamma^ {2} (2 m - 3),
$$
$$
d _ {2} = 2 + 3 \gamma (m - 2) + \gamma^ {2} (m ^ {2} - 4 m + 4),
$$
$$
d _ {3} = \gamma + \gamma^ {2} (m - 2),
$$
$$
d _ {4} = 4 + 6 \gamma (m - 2) + \gamma^ {2} (2 m ^ {2} - 9 m + 9).
$$
Here, we again need to ensure that quantities produced by the firms are positive. This property is equivalent the following inequalities
$$
\forall p _ {i} - c _ {i} ^ {e} \geq 0 \Longleftrightarrow d _ {1} \left(1 - c _ {i} ^ {e}\right) \geq d _ {3} \sum_ {j \neq i} \left(c _ {i} ^ {e} - c _ {j} ^ {e}\right). \tag {11}
$$
Notice that the expected profit of the firm is a second degree polynomial
$$
\varPi_ {i} ^ {e} = - \alpha p _ {i} ^ {2} + 2 \beta p _ {i} - \gamma .
$$
Since best response  $p_i^* (p_{-i})$  is a maximizer of this polynomial, we get
$$
\varPi_ {i} ^ {e} = - \alpha (p _ {i} - p _ {i} ^ {*} (p _ {- i})) ^ {2} + \delta .
$$
Also, notice that
$$
\Pi_ {i} ^ {e} (c _ {i} ^ {e}, p _ {- i}) = \mathbf {E} ((c _ {i} ^ {e} - c _ {i}) q _ {i} (c _ {i} ^ {e}, p _ {- i})) = 0.
$$
$$
\Pi_ {i} ^ {e} = \alpha ((c _ {i} ^ {e} - p _ {i} ^ {*} (p _ {- i})) ^ {2} - (p _ {i} - p _ {i} ^ {*} (p _ {- i})) ^ {2}).
$$
Direct calculations show that
$$
\alpha = \frac {1 + \gamma (m + 2)}{(1 - \gamma) (1 + \gamma (m - 1))}.
$$
These equations imply the following equilibrium profits
$$
\Pi_ {i} ^ {e} (p _ {i} ^ {*}, p _ {- i} ^ {*}) = \frac {(1 + \gamma (m + 2)) (p _ {i} ^ {*} - c _ {i} ^ {e}) ^ {2}}{(1 - \gamma) (1 + \gamma (m - 1))},
$$
A.4 Derivation of Equation (8)
Formally, we assume that the cost of production of one unit of good  $C$  is a random variable that depends on random noise  $\mathbf{X} \in \mathbb{R}^m$  and the decisions of a firm  $s \in \mathbb{R}^n$ . The firm knows that the noise is distributed according to a density function
$$
\boldsymbol {X} \sim f (\cdot , \boldsymbol {\theta}), \boldsymbol {\theta} \in \mathbb {R} ^ {d}.
$$
However, it does not know the parameters of the distribution  $\theta$ .
The firm wants to minimize its marginal cost on average
$$
c (\boldsymbol {s}) = \mathbf {E} _ {\boldsymbol {X} \sim f (\cdot , \boldsymbol {\theta})} (C (\boldsymbol {s}, \boldsymbol {X})).
$$
To achieve it, the firm decides to use the maximum likelihood estimates  $\widehat{\theta}$  of  $\theta$  using its data about the realizations of the noise. After that, the firm chooses an action that minimizes an estimated cost on average
$$
\boldsymbol {s} _ {\mathrm {f i n}} = \underset {\boldsymbol {s}} {\arg \min } \mathbf {E} _ {\boldsymbol {X} \sim f (\cdot , \widehat {\boldsymbol {\theta}})} (C (\boldsymbol {s}, \boldsymbol {X})).
$$
The resulting cost  $c(s_{\mathrm{fin}})$  is a random variable because of randomness of the firm's sample  $S$ . However, the asymptotic normality of MLE allows us to reason about a cost distribution.
We use the delta method to achieve this goal. By the asymptotic normality,
$$
\sqrt {n} (\widehat {\boldsymbol {\theta}} - \boldsymbol {\theta}) \xrightarrow {d} \mathrm {N} (0, \mathbf {I} (\boldsymbol {\theta}) ^ {- 1}).
$$
Further, by the delta method,
$$
\sqrt {n} \left(\boldsymbol {s} _ {\text {f i n}} - \boldsymbol {s} ^ {*}\right) \stackrel {{d}} {{\rightarrow}} \sqrt {n} \left(\frac {\partial \boldsymbol {s}}{\partial \boldsymbol {\theta}}\right) ^ {\top} (\widehat {\boldsymbol {\theta}} - \boldsymbol {\theta}).
$$
Finally, by the second-order delta method,
$$
\begin{array}{l} n \left(c \left(\boldsymbol {s} _ {\text {f i n}}\right) - c \left(\boldsymbol {s} ^ {*}\right)\right) \xrightarrow {d} n \nabla c \left(\boldsymbol {s} ^ {*}\right) ^ {\top} \left(\boldsymbol {s} _ {\text {f i n}} - \boldsymbol {s} ^ {*}\right) + n \left(\boldsymbol {s} _ {\text {f i n}} - \boldsymbol {s} ^ {*}\right) ^ {\top} \frac {\nabla^ {2} c \left(\boldsymbol {s} ^ {*}\right)}{2} \left(\boldsymbol {s} _ {\text {f i n}} - \boldsymbol {s} ^ {*}\right) \\ = n \left(\boldsymbol {s} _ {\mathrm {f i n}} - \boldsymbol {s} ^ {*}\right) ^ {\mathrm {T}} \frac {\nabla^ {2} c \left(\boldsymbol {s} ^ {*}\right)}{2} \left(\boldsymbol {s} _ {\mathrm {f i n}} - \boldsymbol {s} ^ {*}\right). \\ \end{array}
$$
Thus, the marginal cost will be distributed approximately as generalized chi-squared distribution. The expected marginal cost will be approximately
$$
c ^ {e} = \mathbf {E} _ {S} \left(c \left(\boldsymbol {s} _ {\text {f i n}}\right)\right) \approx c \left(\boldsymbol {s} ^ {*}\right) + \frac {1}{2 n} \operatorname {T r} \left(\nabla^ {2} c \left(\boldsymbol {s} ^ {*}\right) \left(\frac {\partial \boldsymbol {s}}{\partial \boldsymbol {\theta}}\right) ^ {\top} \mathbf {I} (\boldsymbol {\theta}) ^ {- 1} \frac {\partial \boldsymbol {s}}{\partial \boldsymbol {\theta}}\right).
$$
Therefore, one of the natural choices for the dependency of expected marginal cost on the number of data points is
$$
c ^ {e} (n) = a + \frac {b}{n}.
$$
A.5 Proof of Theorem 5.2
In the case  $\gamma \leq 0$ , the collaboration is always profitable because the right-hand side of the criterion is less than zero, and the left-hand side of the criterion is not less than zero.
In the case  $\gamma > 0$ , the criteria for the firm  $i$  are the following. In the case of Cournot competition, we have
$$
\begin{array}{l} (2 - \gamma) \left(n _ {i} ^ {- \beta} - \left(n _ {1} + n _ {2}\right) ^ {- \beta}\right) > \gamma \left(n _ {- i} ^ {- \beta} - n _ {i} ^ {- \beta}\right) \\ \Longleftrightarrow f _ {\text {C o u r n o t}} (x, \gamma , \beta) := 2 x ^ {\beta} - \gamma (1 - x) ^ {\beta} - (2 - \gamma) x ^ {\beta} (1 - x) ^ {\beta} > 0, \\ \end{array}
$$
where  $x = n_{-i} / (n_1 + n_2)$ . The right-hand side of the criterion  $f$  is increasing in  $x$ ,  $f_{\mathrm{Cournot}}(0, \gamma, \beta) < 0$ , and  $f_{\mathrm{Cournot}}(1, \gamma, \beta) > 0$ . Thus, there exists a break-even point  $x_{\mathrm{Cournot}}(\gamma, \beta)$  such that
$$
f _ {\text {C o u r n o t}} (x, \gamma , \beta) \left\{ \begin{array}{l l} <   0, & x <   x _ {\text {C o u r n o t}} (\gamma , \beta), \\ = 0, & x = x _ {\text {C o u r n o t}} (\gamma , \beta), \\ > 0, & x > x _ {\text {C o u r n o t}} (\gamma , \beta). \end{array} \right.
$$
Similarly, a break-even point exists in the Bertrand case.
Now, we prove the properties of  $x$ .
The first property follows from the fact that  $f$  is decreasing in  $\gamma$ . The second property follows from the fact that  $f_{\mathrm{Cournot}} > f_{\mathrm{Bertrand}}$ . The proof of the third property is more complicated.
In the case of Cournot, the proof is the following. Let  $x(\beta) = x_{\mathrm{Cournot}}(\gamma, \beta)$ . We want to show that  $x(\beta)$  is increasing in  $\beta$ . To achieve it, we will show that  $f_{\mathrm{Cournot}}(x(\beta), \gamma, \beta + \epsilon) < 0$ . From the definition,
$$
x (\beta) = \left(\frac {\gamma (1 - x (\beta)) ^ {\beta}}{2 - (2 - \gamma) (1 - x (\beta)) ^ {\beta}}\right) ^ {\frac {1}{\beta}}.
$$
$$
\begin{array}{l} f _ {\text {C o u r n o t}} (x (\beta), \gamma , \beta + \epsilon) <   0 \\ \Longleftrightarrow \left(\frac {\gamma (1 - x (\beta)) ^ {\beta}}{2 - (2 - \gamma) (1 - x (\beta)) ^ {\beta}}\right) ^ {\frac {1}{\beta}} <   \left(\frac {\gamma (1 - x (\beta)) ^ {\beta + \epsilon}}{2 - (2 - \gamma) (1 - x (\beta)) ^ {\beta + \epsilon}}\right) ^ {\frac {1}{\beta + \epsilon}} \\ \Longleftrightarrow \frac {\gamma}{2} \Bigg (\frac {2 - (2 - \gamma) (1 - x (\beta)) ^ {\beta + \epsilon}}{\gamma} \Bigg) ^ {\frac {\beta}{\beta + \epsilon}} + \frac {2 - \gamma}{2} (1 - x) ^ {\beta} <   1. \\ \end{array}
$$
The last inequality follows from the concavity of the function  $x^{\frac{\beta}{\beta + \epsilon}}$ . The proof is similar in the Bertrand case.
A.6 Proof of Theorem 6.2
Direct computations show
$$
\begin{array}{l} \varPi_ {i} ^ {e} \left(\lambda_ {1}, \lambda_ {2}\right) - \varPi_ {i} ^ {e} (0, 0) = \\ \left(2 c _ {i} ^ {e} \left(n _ {i}\right) - \gamma c _ {- i} ^ {e} \left(n _ {- i}\right) - 2 c _ {i} ^ {e} \left(n _ {i} + n _ {- i} \lambda_ {- i}\right) + \gamma c _ {- i} ^ {e} \left(n _ {- i} + n _ {1} \lambda_ {i}\right)\right) \times \\ \frac {\left(4 - 2 \gamma - 2 c _ {i} ^ {e} (n _ {i}) + \gamma c _ {- i} ^ {e} (n _ {- i}) - 2 c _ {i} ^ {e} (n _ {i} + n _ {- i} \lambda_ {- i}) + \gamma c _ {- i} ^ {e} (n _ {- i} + n _ {i} \lambda_ {i})\right)}{(4 - \gamma^ {2}) ^ {2}}. \\ \end{array}
$$
$$
u _ {i} = \frac {1}{n _ {i} ^ {\beta}} - \frac {1}{(n _ {i} + \lambda_ {- i} n _ {- i}) ^ {\beta}}.
$$
It gives
$$
c _ {i} ^ {e} (n _ {i}) - c _ {i} ^ {e} (n _ {i} + \lambda_ {- i} n _ {- i}) = a + \frac {b}{n _ {i} ^ {\beta}} - a - \frac {b}{(n _ {i} + n _ {- i} \lambda_ {- i}) ^ {\beta}} = b u _ {i}.
$$
So, the profit gain will be equal to
$$
\Pi_ {i} ^ {e} (\lambda_ {1}, \lambda_ {2}) - \Pi_ {i} ^ {e} (0, 0) = (2 u _ {i} - \gamma u _ {- i}) \left(1 - \frac {4 \xi}{n _ {i} ^ {\beta}} + \frac {2 \gamma \xi}{n _ {- i} ^ {\beta}} + \xi (2 u _ {i} - \gamma u _ {- i})\right) \frac {b (4 - 2 \gamma) (1 - a)}{(4 - \gamma^ {2}) ^ {2}},
$$
where  $\xi = \frac{b}{(4 - 2\gamma)(1 - a)}$  . Denote
$$
h _ {i} = \frac {1}{n _ {i} ^ {\beta}} - \frac {1}{(n _ {1} + n _ {2}) ^ {\beta}},
$$
$$
g _ {i} = \frac {4}{n _ {i} ^ {\beta}} - \frac {2 \gamma}{n _ {- i} ^ {\beta}}.
$$
Then Equation (9) will simplify to
$$
\max _ {u _ {i} \in [ 0, h _ {i} ]} (2 u _ {1} - \gamma u _ {2}) (2 u _ {2} - \gamma u _ {1}) (1 - \xi g _ {1} + \xi (2 u _ {1} - \gamma u _ {2})) (1 - \xi g _ {2} + \xi (2 u _ {2} - \gamma u _ {1})) \mathrm {s . t .} \forall i   2 u _ {i} - \gamma u _ {- i} \geq 0.
$$
Now, we will change variables
$$
v _ {i} := 2 u _ {i} - \gamma u _ {- i} \Rightarrow u _ {i} = \frac {2 v _ {i} + \gamma v _ {- i}}{4 - \gamma^ {2}},
$$
which will give the following problem
$$
\max  _ {v _ {i} \geq 0} v _ {1} v _ {2} (1 - \xi g _ {1} + \xi v _ {1}) (1 - \xi g _ {2} + \xi v _ {2}) \text {s . t .} \forall i   2 v _ {i} + \gamma v _ {- i} \leq (4 - \gamma^ {2}) h _ {i}.
$$
Notice that the restrictions  $v_{i} \geq 0$  are not binding since the point  $(v_{1}, v_{2}) = (\epsilon, \epsilon)$  delivers positive Nash product and satisfies the restrictions. However, one of the restrictions  $2v_{i} + \gamma v_{-i} \leq (4 - \gamma^{2})h_{i}$  should be binding because any internal point  $(v_{1}, v_{2})$  could be transformed to the point  $((1 + \epsilon)v_{1}, (1 + \epsilon)v_{2})$ , which would increase the objective function and satisfy the restrictions.
First, consider the case  $2v_{1}^{*} + \gamma v_{2}^{*} < (4 - \gamma^{2})h_{1}$  at an optimum  $(v_{1}^{*},v_{2}^{*})$ . According to the previous paragraph, we have  $2v_{2}^{*} + \gamma v_{1}^{*} = (4 - \gamma^{2})h_{2}$ , which gives  $v_{2}^{*} - v_{1}^{*} > (2 + \gamma)(h_{2} - h_{1})\geq 0$ . The derivative of the objective along the direction  $(2, - \gamma)$  equals to
$$
\left. \left(2 v _ {2} ^ {*} - \gamma v _ {1} ^ {*}\right) \left(1 - \xi g _ {1} + \xi v _ {1} ^ {*}\right) \left(1 - \xi g _ {2} + \xi v _ {2} ^ {*}\right) + \xi v _ {1} ^ {*} v _ {2} ^ {*} \left(2 - \gamma - \xi \left(2 g _ {2} - \gamma g _ {1} - 2 v _ {2} ^ {*} + \gamma v _ {1} ^ {*}\right)\right). \right.
$$
Since  $2v_{2}^{*} - \gamma v_{1}^{*} > 0$ ,  $|2g_{2} - \gamma g_{1} - 2v_{2}^{*} + \gamma v_{1}^{*}| \leq 10$ , and  $\xi \leq \frac{b}{2(1 - a)} \leq \frac{1}{10}$ , the derivative is positive. Thus, the point  $(v_{1}^{*}, v_{2}^{*})$  can not be optimal because a small shift in the direction  $(2, -\gamma)$  would increase the objective and would not violate the restrictions. Therefore, in the optimum, we will have  $2v_{1}^{*} = (4 - \gamma^{2})h_{1} - \gamma v_{2}^{*}$ .
If we substitute the last expression into the objective function, the objective will become a fourth-degree polynomial of  $v_{2}$ . If the stationary point of the objective function  $v_{2}^{s}$  satisfies the restriction  $2v_{2}^{s} + \gamma v_{1}(v_{2}^{s}) \leq (4 - \gamma^{2})h_{2}$ , it will be a solution to the original problem. Otherwise, the solution will come from the equation  $2v_{2}^{0} + \gamma v_{1}(v_{2}^{0}) = (4 - \gamma^{2})h_{2}$  since all other restrictions are not binding. So, the solution will be equal to  $v_{2}^{*} = \min(v_{2}^{s}, v_{2}^{0})$  (the restriction  $2v_{2} + \gamma v_{1}(v_{2}) \leq (4 - \gamma^{2})h_{2}$  does not hold only when  $v_{2}$  is big).
Consider the following relaxed problem
$$
\max  _ {v _ {i} \geq 0} v _ {1} v _ {2} (1 - \xi g _ {1} + \xi v _ {1}) (1 - \xi g _ {2} + \xi v _ {2}) \text {s . t .} 2 v _ {1} + \gamma v _ {2} = (4 - \gamma^ {2}) h _ {1}.
$$
The stationary point will be the root of the derivative and hence will satisfy the following equation
$$
\begin{array}{l} \left(\gamma v _ {2} ^ {s} - 2 v _ {1} \left(v _ {2} ^ {s}\right)\right) \left(1 - \xi g _ {1} + \xi v _ {1} \left(v _ {2} ^ {s}\right)\right) \left(1 - \xi g _ {2} + \xi v _ {2} ^ {s}\right) = \\ \xi v _ {1} \left(v _ {2} ^ {s}\right) v _ {2} ^ {s} \left(2 \left(1 - \xi g _ {1} + \xi v _ {1} \left(v _ {2} ^ {s}\right)\right) - \gamma \left(1 - \xi g _ {2} + \xi v _ {2} ^ {s}\right)\right) \\ \Longrightarrow \gamma v _ {2} ^ {s} - 2 v _ {1} (v _ {2} ^ {s}) = \xi \frac {v _ {1} (v _ {2} ^ {s}) v _ {2} ^ {s} (2 (1 - \xi g _ {1} + \xi v _ {1} (v _ {2} ^ {s})) - \gamma (1 - \xi g _ {2} + \xi v _ {2} ^ {s}))}{(1 - \xi g _ {1} + \xi v _ {1} (v _ {2} ^ {s})) (1 - \xi g _ {2} + \xi v _ {2} ^ {s})}. \\ \end{array}
$$
Since  $-2\leq g_{i}\leq 4,0\leq v_{i}\leq 2$  , and  $v_{1}(v_{2})v_{2}\leq \frac{(4 - \gamma^{2})^{2}h_{1}^{2}}{8\gamma}$  , we get
$$
| \gamma v _ {2} ^ {s} - 2 v _ {1} (v _ {2} ^ {s}) | \leq \xi \frac {(4 - \gamma^ {2}) h _ {1} ^ {2}}{8 \gamma} \frac {2 - \gamma + 1 2 \xi}{(1 - 4 \xi) ^ {2}} = \mathrm {O} (\xi).
$$
So, the stationary point is approximately  $v_{2}^{s} = \frac{(4 - \gamma^{2})h_{1}}{2\gamma} + \mathrm{O}(\xi)$ .
Therefore, the solution to the original problem is approximately
$$
v _ {2} ^ {*} = \min  \left(\frac {(4 - \gamma^ {2}) h _ {1}}{2 \gamma} + \mathrm {O} (\xi), 2 h _ {2} - \gamma h _ {1}\right) = \min  \left(\frac {(4 - \gamma^ {2}) h _ {1}}{2 \gamma}, 2 h _ {2} - \gamma h _ {1}\right) + \mathrm {O} (\xi).
$$
Substituting everything back, we get the solution presented in the theorem statement.
Now, we will prove that  $\tilde{\lambda}_1$  is non-increasing in  $\gamma$ .
First, the criterion function for choosing non-unit solution
$$
\frac {(2 - \gamma) ^ {2}}{(n _ {1} + n _ {2}) ^ {\beta}} + \frac {4 \gamma}{n _ {2} ^ {\beta}} - \frac {4 + \gamma^ {2}}{n _ {1} ^ {\beta}} = 4 \gamma h _ {2} - (4 + \gamma^ {2}) h _ {1}.
$$
is increasing in  $\gamma$  since its derivative  $4h_{2} - 2\gamma h_{1}\geq 0$  is positive. So, when  $\gamma$  increases non-unit solution becomes more probable.
Second, the non-unit solution is decreasing in  $\gamma$  since the function  $\frac{4 + \gamma^2}{\gamma} = \frac{4}{\gamma} + \gamma$  is decreasing in  $\gamma$  on  $(0,1)$ . (Note that  $\tilde{\lambda}_1 = 1$ , when  $\gamma \leq 0$ .)
Now, we will prove that  $\tilde{\lambda}_1$  is non-increasing in  $\beta$ .
First, notice that the non-unit criterion is equivalent to
$$
f (x, \beta) = 4 \gamma (1 - x) ^ {\beta} (1 - x ^ {\beta}) - (4 + \gamma^ {2}) x ^ {\beta} (1 - (1 - x) ^ {\beta}) \geq 0,
$$
where  $x = \frac{n_2}{n_1 + n_2}$ . The criterion function  $f$  is decreasing in  $x$  and have one root  $x(\beta)$  on the interval [0, 1]. We will prove that the root is shifting to the right, making more size pairs produce non-unit solution. Notice the criterion can be also expressed as
$$
x \leq \left(\frac {4 \gamma (1 - x) ^ {\beta}}{4 + \gamma^ {2} - (2 - \gamma) ^ {2} (1 - x) ^ {\beta}}\right) ^ {\frac {1}{\beta}}.
$$
So, to prove that  $x(\beta) \leq x(\beta + \epsilon)$  it is sufficient to show that
$$
\begin{array}{l} \left(\frac {4 \gamma (1 - x (\beta)) ^ {\beta}}{4 + \gamma^ {2} - (2 - \gamma) ^ {2} (1 - x (\beta)) ^ {\beta}}\right) ^ {\frac {1}{\beta}} = x (\beta) <   \left(\frac {4 \gamma (1 - x (\beta)) ^ {\beta + \epsilon}}{4 + \gamma^ {2} - (2 - \gamma) ^ {2} (1 - x (\beta)) ^ {\beta + \epsilon}}\right) ^ {\frac {1}{\beta + \epsilon}} \\ \Longleftrightarrow \frac {(2 - \gamma) ^ {2}}{4 + \gamma^ {2}} (1 - x) ^ {\beta} + \frac {4 \gamma}{4 + \gamma^ {2}} \bigg (\frac {4 + \gamma^ {2} - (2 - \gamma) ^ {2} (1 - x) ^ {\beta + \epsilon}}{4 \gamma} \bigg) ^ {\frac {\beta}{\beta + \epsilon}} \leq 1. \\ \end{array}
$$
The last inequality follows from the concavity of the function  $x^{\frac{\beta}{\beta + \epsilon}}$
Now, we will prove that the non-unit solution is decreasing in  $\beta$  given that non-unit criterion holds. Notice that the non-unit solution  $\tilde{\lambda}_1(\beta)$  satisfies
$$
\frac {1}{n _ {2} ^ {\beta}} - \frac {1}{(n _ {2} + \tilde {\lambda} _ {1} (\beta) n _ {1}) ^ {\beta}} = \frac {4 + \gamma^ {2}}{4 \gamma} \left(\frac {1}{n _ {1} ^ {\beta}} - \frac {1}{(n _ {1} + n _ {2}) ^ {\beta}}\right).
$$
So, to show that  $\tilde{\lambda}_1(\beta) \geq \tilde{\lambda}_1(\beta + \epsilon)$ , it is sufficient to show that
$$
(x ^ {- \beta} - \delta (1 - x) ^ {- \beta} + \delta) ^ {\frac {1}{\beta}} \leq (x ^ {- \beta - \epsilon} - \delta (1 - x) ^ {- \beta - \epsilon} + \delta) ^ {\frac {1}{\beta + \epsilon}},
$$
where  $x = \frac{n_2}{n_1 + n_2}$  and  $\delta = \frac{4 + \gamma^2}{4\gamma}$ . This inequality is equivalent to
$$
\left((1 - x) ^ {\beta} x ^ {- \beta} - \delta + \delta (1 - x) ^ {\beta}\right) ^ {\frac {\beta + \epsilon}{\beta}} \leq (1 - x) ^ {\beta + \epsilon} x ^ {- \beta - \epsilon} - \delta + \delta (1 - x) ^ {\beta + \epsilon}.
$$
Denote  $u_0 = \left(\frac{1 - x}{x}\right)^\beta$ ,  $v_0 = (1 - x)^\beta$ , and  $\alpha = \frac{\beta + \epsilon}{\beta}$ . We get the following inequality
$$
\left(u _ {0} - \delta + \delta v _ {0}\right) ^ {\alpha} \leq u _ {0} ^ {\alpha} - \delta + \delta v _ {0} ^ {\alpha}.
$$
$$
f (u, v) = u ^ {\alpha} - \delta + \delta v ^ {\alpha} - (u - \delta + \delta v) ^ {\alpha}.
$$
Notice that  $f(u,1) = 0$ . Additionally notice that
$$
\frac {\partial f}{\partial v} = \delta \alpha v ^ {\alpha - 1} - \delta \alpha (u - \delta + \delta v) ^ {\alpha - 1} \leq 0 \iff v \leq u - \delta + \delta v.
$$
If we prove that this derivative is negative for all points  $(u_0, x)$ , where  $x \in (v_0, 1)$ , we will prove the desired result. Notice that the last inequality is the most stringent for the point  $(u_0, v_0)$ . At this point, we have
$$
v _ {0} \leq u _ {0} - \delta + \delta v _ {0} \iff 0 \leq \frac {4 \gamma}{n _ {2} ^ {\beta}} - \frac {4 + \gamma^ {2}}{n _ {1} ^ {\beta}} + \frac {(2 - \gamma) ^ {2}}{(n _ {1} + n _ {2}) ^ {\beta}}.
$$
The last inequality holds because we consider the non-unit solution.
The last property of the theorem follows from the fact that, for small ratio  $\frac{n_2}{n_1}$ , the non-unit criterion holds and the Taylor formula for this ratio results in the expression presented in the statement.
B Examples of data-sharing problem
B.1 Taxi rides
In this section, how our framework works using motivating example from Section 3. In this example, two taxi firms use data to optimize the driver scheduling algorithm. This optimization allows them to use less driver time; thus, making their expected marginal costs lower.
We assume that the only uncertainty arising in scheduling algorithm is the duration of one kilometer ride  $X \sim \mathrm{N}(\mu, 1)$ . The firms do not know the parameter  $\mu$ . However, each firm  $F_{i}$  has  $n_i$  observations of this random variable  $S_{i} = \{X_{i}^{j}\}_{j=1}^{n_{i}}$ .
Their scheduling algorithm requires the estimate  $s$  of ride duration  $X$ . If the estimate is too low  $s < X$ , the company will need to attract more drivers and pay them more. If the estimate is too big  $s > X$ , the drivers will be underutilized. We assume that costs associated with these losses have MSE form  $C(s, X) = a - b + b(s - X)^2$ . This functional form will imply the following costs on average
$$
c (s) = \mathbf {E} _ {X} (C (s, X)) = a - b + b \mathbf {E} ((s - X) ^ {2}) = a + b (s - \mu) ^ {2}.
$$
If the firms do not collaborate,  $F_{i}$  uses an average of its own points
$$
s _ {i, \text {i n d}} = \frac {1}{n _ {i}} \sum_ {j = 1} ^ {n _ {i}} X _ {j} ^ {i} \Rightarrow s _ {i, \text {i n d}} \sim \mathrm {N} \left(\mu , \frac {1}{n _ {i}}\right)
$$
to estimate  $\mu$ . This will give the following expected marginal costs
$$
c _ {i, \text {i n d}} ^ {e} = \mathbf {E} _ {S _ {1}, S _ {2}} c (s _ {i, \text {i n d}}) = a + b \mathbf {E} (s _ {i, \text {i n d}} - \mu) ^ {2} = a + \frac {b}{n _ {i}}.
$$
If the firms collaborate, they use both samples to calculate average
$$
s _ {\text {s h a r e d}} = \frac {1}{n _ {1} + n _ {2}} \sum_ {i, j} X _ {j} ^ {i} \Rightarrow s _ {\text {s h a r e d}} \sim \mathrm {N} \left(\mu , \frac {1}{n _ {1} + n _ {2}}\right)
$$
and the expected marginal costs will be
$$
c _ {\mathrm {s h a r e d}} ^ {e} = \mathbf {E} _ {S _ {1}, S _ {2}} c (s _ {\mathrm {s h a r e d}}) = a + \frac {b}{n _ {1} + n _ {2}}.
$$
To solve the data-sharing problem, we need to describe the competition between firms. For simplicity we will only consider the case of Cournot competition.
As in Section 4.1.2, the firms are interested in increasing their expected profits
$$
\boldsymbol {\Pi} _ {i} ^ {e} = \mathbf {E} _ {S _ {1}, S _ {2}} \left(p _ {i} q _ {i} - c _ {i} q _ {i}\right).
$$