SlowGuess/NeurIPS_2023_OCR huggingface dataset:
A globally optimal mechanism without strategic considerations: The following simple mechanism  $M_{\mathrm{pool}}$ , pools all the data from the other agents and gives it back to an agent. Precisely, it chooses the space of allocation  $\mathcal{A} = \bigcup_{n \geq 0} \mathbb{R}^n$  to be datasets of arbitrary length, and sets agent  $i$ 's allocation to be  $A_i = \bigcup_{j \neq i} Y_i$ . The recommended strategies  $s^{\mathrm{pool}} = \{(n_i^{\mathrm{pool}}, f_i^{\mathrm{pool}}, h_i^{\mathrm{pool}})\}_i$  asks each agent to collect  $n_i^{\mathrm{pool}} = \sigma / \sqrt{cm}$  points², submit it as is  $f_i^{\mathrm{pool}} = \mathbf{I}$ , and use the sample mean of all points
2To avoid rounding effects, henceforth we will treat  $\sigma /\sqrt{cm}$  , and  $\sigma /\sqrt{c}$  as integers.
Algorithm 1  $M_{\mathrm{C3D}}$
1: Mechanism designer publishes:  
2: The allocation space  $\mathcal{A} = \bigcup_{n\geq 0}\mathbb{R}^n\times \bigcup_{n\geq 0}\mathbb{R}^n\times \mathbb{R}_+$  and the procedure in lines 6-15.  
3: Each agent i:  
4: Choose strategy  $s_i = (n_i,f_i,h_i)$ . # See (8) for recommended strategies.  
5: Sample  $n_i$  points  $X_{i} = \{x_{i,j}\}_{j = 1}^{n_{i}}$  and submit  $Y_{i} = f_{i}(X_{i})$  to the mechanism.  
6: Mechanism:  
7: For each agent  $i\in [m]$ : # can be done simultaneously for all agents  
8:  $Y_{-i}\gets \bigcup_{j\neq i}Y_{j}$ .  
9: If  $m\leq 4$ : # Simply pool and return all of the other agents' data to agent i.  
10:  $\bar{A}_i\gets (Y_{-i},\emptyset ,0)$ . Return  $A_{i}$  to agent i.  
11: Else:  
12:  $Z_{i}\gets$  sample min{|Y-|,  $\sigma /\sqrt{cm}$ } points in  $Y_{-i}$  without replacement.  
13:  $\eta_i^2\gets \alpha^2\left(\frac{1}{|Y_i|}\sum_{y\in Y_i}y - \frac{1}{|Z_i|}\sum_{z\in Z_i}z\right)^2$  # See (7) for  $\alpha$ .  
14:  $Z_{i}^{\prime}\gets \{z + \epsilon_{z,i},$  for all  $z\in Y_{-i}\backslash Z_i$  where  $\epsilon_{z,i}\sim \mathcal{N}(0,\eta_i^2)\}$   
15:  $A_{i}\gets (Z_{i},Z_{i}^{\prime},\eta_{i}^{2})$ . Return  $A_{i}$  to agent i.  
16: Each agent i:  
17: Compute estimate  $h_i(X_i,Y_i,A_i)$ . # See (8) for recommended estimator.
as her estimate  $h_i^{\mathrm{pool}}(X_i, X_i, A_i) = \frac{1}{|X_i \cup A_i|} \sum_{z \in X_i \cup A_i} z$ . It is straightforward to show that this minimizes the social penalty if all agents follow  $s^{\mathrm{pool}}$ . After each agent has collected their datasets  $\{X_i\}_i$ , the social penalty is minimized if all agents have access to each other's datasets and they all use a minimax optimal estimator: this justifies using  $M_{\mathrm{pool}}$  with  $f_i^{\mathrm{pool}} = \mathbf{I}$  and setting  $h_i^{\mathrm{pool}}$  to be the sample mean. The following simple calculation justifies the choice of  $\sum_{i} n_i^{\mathrm{pool}}$ :
$$
\inf  _ {s \in \mathcal {S} ^ {m}} \sum_ {i = 1} ^ {m} \left(\sup  _ {\mu} \mathbb {E} \left[ (h _ {i} (X _ {i}, f _ {i}, A _ {i}) - \mu) ^ {2} \mid \mu \right] + c n _ {i}\right) = \min  _ {\{n _ {i} \} _ {i}} \left(\frac {m \sigma^ {2}}{\sum_ {i} n _ {i}} + c \sum_ {i} n _ {i}\right) = 2 \sigma \sqrt {m c}.
$$
However,  $s^{\mathrm{pool}}$  is not a Nash equilibrium of this mechanism, as an agent will find it beneficial to free-ride. If all other agents are submitting  $\sigma / \sqrt{cm}$  points, by collecting no points, an agent's penalty is  $\sigma \sqrt{mc} / (m - 1)$ , as she does not incur any data collection cost. This is strictly smaller than  $2\sigma \sqrt{c / m}$  when  $m \geq 3$ . In fact, it is not hard to show that  $M_{\mathrm{pool}}$  is at a Nash equilibrium only when the total amount of data is  $\sigma / \sqrt{c}$ ; for additional points, the marginal reduction in the estimation error for an individual agent does not offset her data collection costs. The social penalty at these equilibria is  $\sigma \sqrt{c} (m + 1)$  which is significantly larger than the global minimum when there are many agents.
A seemingly simple way to fix this mechanism is to only return the datasets of the other agents if an agent submits at least  $\sigma / \sqrt{cm}$  points. However, as we will see in §4.1, such a mechanism can also be manipulated by an agent who submits a fabricated dataset of  $\sigma / \sqrt{cm}$  points without actually collecting any data and incurring any cost and then discarding the fabricated dataset when estimating. Any naive test to check for the quality of the data can also be sidestepped by agents who sample only a few points, and use that to fabricate a larger dataset (e.g by sampling a large number of points from a Gaussian fitted to the small sample). Next, we will present our mechanism for this problem which satisfies all three desiderata.
3 Method and Results
We have outlined our mechanism  $M_{\mathrm{c3d}}$ , and its interaction with the agents in Algorithm 1 in the natural order of events. We will first describe it procedurally, and then motivate our design choices. Our mechanism uses the following allocation space,  $\mathcal{A} = \bigcup_{n\geq 0}\mathbb{R}^n\times \bigcup_{n\geq 0}\mathbb{R}^n\times \mathbb{R}_+$ . An allocation  $A_{i} = (Z_{i},Z_{i}^{\prime},\eta_{i}^{2})\in \mathcal{A}$  consists of an uncorrupted dataset  $Z_{i}$ , a corrupted dataset  $Z_{i}^{\prime}$ , and the variance  $\eta_i^2$  of the noise added to  $Z_{i}^{\prime}$  for corruption. Once the mechanism and the allocation space are published, agent  $i$  chooses her strategy  $s = (n_{i},f_{i},h_{i})$ . She collects a dataset  $X_{i} = \{x_{i,j}\}_{j = 1}^{n_{i}}$  where  $x_{i,j}\sim \mathcal{N}(\mu ,\sigma^2)$ , and submits  $Y_{i} = f_{i}(X_{i})$  to the mechanism.
Our mechanism determines agent  $i$ 's allocation as follows. Let  $Y_{-i}$  be the union of all datasets submitted by the other agents. If there are at most four agents, we simply return all of the other
agents' data without corruption by setting  $A_{i} \gets (Y_{-i}, \emptyset, 0)$ . If there are more agents, the mechanism first chooses a random subset of size  $\min \{|Y_{-i}|, \sigma / \sqrt{cm}\}$  from  $Y_{-i}$ ; denote this  $Z_{i}$ . In line 13, the mechanism individually adds Gaussian noise to the remaining points  $Y_{-i} \backslash Z_{i}$  to obtain  $Z_{i}'$  (line 14). The variance  $\eta_{i}^{2}$  of the noise depends on the difference between the sample means of the subset  $Z_{i}$  and the agent's submission  $Y_{i}$ . It is modulated by a value  $\alpha$ , which is a function of  $c, m,$  and  $\sigma^2$ . Precisely,  $\alpha$  is the smallest number larger than  $\sqrt{\sigma}(cm)^{-1/4}$  which satisfies  $G(\alpha) = 0$ , where:
$$
G (\alpha) := \left(\frac {m - 4}{m - 2} \frac {4 \alpha^ {2}}{\sigma / \sqrt {c m}} - 1\right) \frac {4 \alpha}{\sqrt {\sigma} (m / c) ^ {1 / 4}} - \left(4 (m + 1) \frac {\alpha^ {2}}{\sigma \sqrt {m / c}} - 1\right) \sqrt {2 \pi} \exp \left(\frac {\sigma \sqrt {m / c}}{8 \alpha^ {2}}\right) \operatorname {E r f c} \left(\frac {\sqrt {\sigma} (m / c) ^ {1 / 4}}{2 \sqrt {2} \alpha}\right) \tag {7}
$$
Finally, the mechanism returns the allocation  $A_{i} = (Z_{i}, Z_{i}^{\prime}, \eta_{i}^{2})$  to agent  $i$  and the agent estimates  $\mu$ .
Recommended strategies: The recommended strategy  $s_i^\star = (n_i^\star, f_i^\star, h_i^\star)$  for agent  $i$  is given in (8). The agent should collect  $n_i^\star = \sigma / (m\sqrt{c})$  samples if there are at most four agents, and  $n_i^\star = \sigma / \sqrt{cm}$  samples otherwise. She should submit it without fabrication or alteration  $f_i = \mathbf{I}$ , and then use a weighted average of the datasets  $(X_i, Z_i, Z_i')$  to estimate  $\mu$ . The weighting is proportional to the inverse variance of the data. For  $X_i$  and  $Z_i$  this is simply  $\sigma^2$ , but for  $Z_i'$ , the variance is  $\sigma^2 + \eta_i^2$  since the mechanism adds Gaussian noise with variance  $\eta_i^2$ . We have:
$$
n _ {i} ^ {\star} = \left\{ \begin{array}{l l} \frac {\sigma}{m \sqrt {c}} & \text {i f} m \leq 4 \\ \frac {\sigma}{\sqrt {c m}} & \text {i f} m > 4 \end{array} \right., \hskip 2 8. 4 5 2 7 5 6 p t f _ {i} ^ {\star} = \mathbf {I},
$$
$$
h _ {i} ^ {\star} \left(X _ {i}, Y _ {i}, \left(Z _ {i}, Z _ {i} ^ {\prime}, \eta_ {i} ^ {2}\right)\right) = \frac {\frac {1}{\sigma^ {2}} \sum_ {u \in X _ {i} \cup Z _ {i}} u + \frac {1}{\sigma^ {2} + \eta_ {i} ^ {2}} \sum_ {u \in Z _ {i} ^ {\prime}} u}{\frac {1}{\sigma^ {2}} \left| X _ {i} \cup Z _ {i} ^ {\prime} \right| + \frac {1}{\sigma^ {2} + \eta_ {i} ^ {2}} \left| Z _ {i} ^ {\prime} \right|} \tag {8}
$$
Design choices: Next, we will describe our design choices and highlight some key challenges. When  $m \leq 4$ , it is straightforward to show that the mechanism satisfies all our desired properties (see beginning of §3.1), so we will focus on the case  $m > 4$ . First, recall that the mechanism needs to incentivize agents to collect a sufficient amount of samples. However, simply counting the number of samples can be easily manipulated by an agent who simply submits a fabricated dataset of a large number of points. Instead, Algorithm 1 attempts to infer the quality of the data submitted by the agents using how well an agent's submission  $Y_{i}$  approximates  $\mu$ . Ideally, we would set the variance  $\eta_{i}^{2}$  of this corruption to be proportional to the difference  $\left(\frac{1}{|Y_i|}\sum_{y\in Y_i}y - \mu\right)^2$ , so that the more data she submits, the less the variance of  $Z_{i}^{\prime}$ , which in turn yields a more accurate estimate for  $\mu$ . However, since  $\mu$  is unknown, we use a subset  $Z_{i}$  obtained from other agents' data as a proxy for  $\mu$ , and set  $\eta_{i}^{2}$  proportional to  $\left(\frac{1}{|Y_i|}\sum_{y\in Y_i}y - \frac{1}{|Z_i|}\sum_{z\in Z_i}z\right)^2$ . If all agents are following  $s^{\star}$ , then  $|Y_{i}| = |Z_{i}| = \sigma / \sqrt{cm} = n_{i}^{\star}$ ; it is sufficient to use only  $n_{i}^{\star}$  points for validating  $Y_{i}$  since both  $\frac{1}{|Y_i|}\sum_{y\in Y_i}y$  and  $\frac{1}{|Z_i|}\sum_{z\in Z_i}z$  will have the same order of error in approximating  $\mu$ .
The second main challenge is the design of the recommended estimator  $h_i^\star$ . In §3.1 we show how splitting  $Y_{-i}$  into a clean and corrupted parts  $Z_i, Z_i'$  allows us to design a minimax optimal estimator. A minimax optimal estimator is crucial to achieving NIC. To explain this, say that the mechanism assumes that agents will use a sub-optimal estimator, e.g. sample mean of  $X_i \cup Z_i \cup Z_i'$ . Then, to account for the larger estimation error, it will need to choose a lower level of corruption  $\eta_i^2$  to minimize the social penalty. However, a smart agent will realize that she can achieve a lower maximum risk by using a better estimator, such as the weighted average, instead of collecting more data in order to reduce the amount of corruption used by the mechanism. She can leverage this insight to collect less data and reduce her overall penalty.
This concludes the description of our mechanism. The following theorem, which is the main theoretical result of this paper, states that  $M_{\mathrm{c3D}}$  achieves the three desiderata outlined in §2.
Theorem 1. Let  $m > 1$ ,  $\alpha$  be as defined in (7), and  $s_i^\star$  be as defined in (8). Then, the following statements are true about the mechanism  $M_{\mathrm{C3D}}$  in Algorithm 1. (i) The strategy profile  $s^\star$  is a Nash equilibrium. (ii) The mechanism is individually rational at  $s^\star$ . (iii) The mechanism is approximately efficient, with  $\mathrm{PR}(M_{\mathrm{C3D}}, s^\star) \leq 2$ .
The mechanism is NIC as, provided that others are following  $s_i^\star$ , there is no reason for any one agent to deviate. Moreover, we achieve low social penalty at  $s_i^\star$ . Other than  $s^\star$ , there is also a set of similar Nash equilibria with the same social penalty: the agents can each add a same constant to the data points they collect and subtract the same value from the final estimate. Before we proceed,
the expression for  $\alpha$  in (7) warrants explanation. If we treat  $\alpha$  is a variable, we find that different choices of  $\alpha$  can lead to other Nash equilibria with corresponding bounds on PR. This specific choice of  $\alpha$  leads to a Nash equilibrium where agents collect  $\sigma / \sqrt{cm}$  points, and a small bound on PR. Throughout this manuscript, we will treat  $\alpha$  as the specific value obtained by solving (7), and not as a variable.
High dimensional non-Gaussian distributions: In Appendix E, we study  $M_{\mathrm{C3D}}$  when applied to  $d$ -dimensional distributions. In Theorem 7, we show that under bounded variance assumptions,  $s^{\star}$  is an  $\varepsilon_{m}$ -approximate Nash equilibrium and that  $\operatorname{PR}(M_{\mathrm{C3D}}, s^{\star}) \leq 2 + \varepsilon_{m}$  where  $\varepsilon_{m} \in \mathcal{O}(1 / m)$ .
3.1 Proof sketch of Theorem 1
When  $m \leq 4$ : First, consider the (easy) case  $m \leq 4$ . At  $s_i^\star$ , the total amount of data collected is  $\sigma / \sqrt{c}$  (see  $n_i^\star$  in (8)), and as there is no corrupted dataset,  $h_i^\star$  simply reduces to the sample mean of  $X_i \cup Y_{-i}$ . The mechanism is IR since an agent's penalty will be  $\sigma \sqrt{c} (1 + 1/m)$  which is smaller than  $p_{\min}^{\mathrm{IR}}$  (6). It is approximately efficient since the social penalty is  $\sigma \sqrt{c} (m + 1)$  which is at most twice the global minimum  $2\sigma \sqrt{mc}$  when  $m \leq 4$ . Finally, NIC is guaranteed by the same argument used in (6); as long as the total amount of data is less than  $\sigma / \sqrt{c}$ , the cost of collecting one more point is offset by the marginal decrease in the estimation error; hence, the agent is incentivized to collect more data. Moreover, as  $A_i$  does not depend on  $f_i$  under these conditions, there is no incentive to fabricate or falsify data.
When  $m > 4$ : We will divide this proof into four parts. We first show that  $G(\alpha) = 0$  in line (6) has a solution  $\alpha$  larger than  $\sqrt{n_i^*} = \sqrt{\sigma} (cm)^{-1/4}$ . This will also be useful when analyzing the efficiency.
1. Equation (7) has a solution. We derive an asymptotic expansion of  $\operatorname{Erfc}(\cdot)$  using integration by parts to analyze the solution to (7). When  $m \geq 5$ , we show that  $G\left(\sqrt{n_i^\star}\right) \times G\left(\sqrt{n_i^\star}(1 + 8/\sqrt{m})\right) < 0$ . By continuity of  $G$ , there exists  $\alpha_m \in \left(\sqrt{n_i^\star}, \sqrt{n_i^\star}(1 + 8/\sqrt{m})\right)$  s.t.  $G(\alpha_m) = 0$ . For  $m$  large enough such that the residual in the asymptotic expansion is negligible, we show  $\alpha_m \in \left(\sqrt{n_i^\star}, \sqrt{n_i^\star}(1 + \log m/m)\right)$  via an identical technique.
2. The strategies  $s^{\star}$  in (8) is a Nash equilibrium: We show this via the following two steps. First (2.1), We show that fixing any  $n_i$ , the maximum risk and thus the penalty  $p_i$  is minimized when agent  $i$  submits the raw data and uses the weighted average as specified in (8), i.e for all  $n_i$ ,
$$
p _ {i} \left(M _ {\mathrm {C 3 D}}, \left(\left(n _ {i}, f _ {i} ^ {\star}, h _ {i} ^ {\star}\right), s _ {- i} ^ {\star}\right)\right) \leq p _ {i} \left(M _ {\mathrm {C 3 D}}, \left(\left(n _ {i}, f _ {i}, h _ {i}\right), s _ {- i} ^ {\star}\right)\right), \forall \left(n _ {i}, f _ {i}, h _ {i}\right) \in \mathbb {N} \times \mathcal {F} \times \mathcal {H}. \tag {9}
$$
Second (2.2), we show that  $p_i$  is minimized when agent  $i$  collects  $n_i^\star$  samples under  $(f_i^\star, h_i^\star)$ , i.e.
$$
p _ {i} \left(M _ {\mathrm {C 3 D}}, \left(\left(n _ {i} ^ {\star}, f _ {i} ^ {\star}, h _ {i} ^ {\star}\right), s _ {- i} ^ {\star}\right)\right) \leq p _ {i} \left(M _ {\mathrm {C 3 D}}, \left(\left(n _ {i}, f _ {i} ^ {\star}, h _ {i} ^ {\star}\right), s _ {- i} ^ {\star}\right)\right), \quad \forall n _ {i} \in \mathbb {N}. \tag {10}
$$
2.1: Proof of (9). As the data collection cost does not change for fixed  $n_i$ , it is sufficient to show that  $(f_i^\star, h_i^\star)$  minimizes the maximum risk. Our proof is inspired by the following well-known recipe for proving minimax optimality of an estimator [24]: design a sequence of priors  $\{\Lambda_\ell\}_\ell$ , compute the minimum Bayes' risk  $\{R_\ell\}_\ell$  for any estimator, and then show that  $R_\ell$  converges to the maximum risk of the proposed estimator as  $\ell \to \infty$ .
To apply this recipe, we use a sequence of normal priors  $\Lambda_{\ell} = \mathcal{N}(0,\ell^{2})$  for  $\mu$ . However, before we proceed, we need to handle two issues. The first of these concerns the posterior for  $\mu$  when conditioned on  $(X_{i},Z_{i},Z_{i}^{\prime})$ . Since the corruption terms  $\epsilon_{z,i}$  added to  $Z_{i}^{\prime}$  depend on  $X_{i}$  and  $Z_{i}$ , this dataset is not independent. Moreover, as the variance  $\eta_i^2$  is the difference between two normal random variables,  $Z_{i}^{\prime}$  is not normal. Despite these, we are able to show that the posterior  $\mu|(X_{i},Z_{i},Z_{i}^{\prime})$  is normal. The second challenge is that the submission  $f_{i}$  also affects the estimation error as it determines the amount of noise  $\eta_i^2$ . We handle this by viewing  $\mathcal{F}\times \mathcal{H}$  as a rich class of estimators and derive the optimal Bayes' estimator  $(f_{i,\ell}^{\mathrm{B}},h_{i,\ell}^{\mathrm{B}})\in \mathcal{F}\times \mathcal{H}$  under the prior  $\Lambda_{\ell}$ . We then show that the minimum Bayes' risk converges to the maximum risk when using  $(f_{i}^{\star},h_{i}^{\star})$ .
Next, under the prior  $\Lambda_{\ell} = \mathcal{N}(0,\ell^{2})$ , we can minimize the Bayes' risk with respect to  $h_i\in \mathcal{H}$  by setting  $h_{i,\ell}^{\mathrm{B}}$  to be the posterior mean. Then, the minimum Bayes' risk  $R_{\ell}$  can be written as,
$$
R _ {\ell} = \inf _ {f _ {i} \in \mathcal {F}} \mathbb {E} \left[ \left(\left| Z _ {i} ^ {\prime} \right| \left(\sigma^ {2} + \alpha^ {2} \left(\frac {1}{\left| Y _ {i} \right|} \sum_ {y \in Y _ {i}} y - \frac {1}{\left| Z _ {i} \right|} \sum_ {z \in Z _ {i}} z\right) ^ {2}\right) ^ {- 1} + \frac {\left| X _ {i} \right| + \left| Z _ {i} \right|}{\sigma^ {2}} + \frac {1}{\ell^ {2}}\right) ^ {- 1} \right]
$$
Note that  $Y_{i} = f_{i}(X_{i})$  depends on  $f_{i}$ . Via the Hardy-Littlewood inequality [10], we can show that the above quantity is minimized when  $f_{i,\ell}^{\mathrm{B}}$  is chosen to be a shrunk version of the agent's initial dataset  $X_{i}$ , i.e.  $f_{i,\ell}^{\mathrm{B}}(X_i) = \left\{\left(1 + \sigma^2 /(|X|\ell^2)\right)^{-1}x, \forall x\in X_i\right\}$ . This gives us an expression for the minimum Bayes' risk  $R_{\ell}$  under prior  $\Lambda_{\ell}$ . To conclude the proof, we note that the minimum Bayes' risk under any prior is a lower bound on the maximum risk, and show that  $R_{\ell}$  approaches the maximum risk of  $(f_{i}^{\star},h_{i}^{\star})$  from below. Hence,  $(f_{i}^{\star},h_{i}^{\star})$  is minimax optimal for any  $n_i$ . (Above, it is worth noting that  $f_{i,\ell}^{\mathrm{B}}\to f_i^\star = \mathbf{I}$  as  $\ell \rightarrow \infty$ . In the Appendix, we also find that  $h_{i,\ell}^{\mathrm{B}}\to h_i^\star$ .)
2.2: Proof of (10). We can now write  $p_i(M_{\mathrm{C3D}},((n_i,f_i^\star ,h_i^\star),s_{-i}^\star)) = R_\infty +cn_i$ , where  $R_{\infty}$  is the maximum risk of  $(f_i^\star ,h_i^\star)$  (and equivalently, the limit of the minimum Bayes' risk):
$$
R _ {\infty} := \mathbb {E} _ {x \sim \mathcal {N} (0, 1)} \bigg [ \left((m - 2) n _ {i} ^ {\star} \big (\sigma^ {2} + \alpha^ {2} \big (\sigma^ {2} / n _ {i} + \sigma^ {2} / n _ {i} ^ {\star} \big) x ^ {2} \big) ^ {- 1} + (n _ {i} + n _ {i} ^ {\star}) \sigma^ {- 2}\right) ^ {- 1} \bigg ]
$$
The term inside the expectation is convex in  $n_i$  for each fixed  $x$ . As expectation preserves convexity, we can conclude that  $p_i$  is a convex function of  $n_i$ . The choice of  $\alpha$  in (7) ensures that the derivative is 0 at  $n^\star$  which implies that  $n^\star$  is a minimum of this function.
3.  $M_{\mathrm{c3D}}$  is individually rational at  $s^{\star}$ : This is a direct consequence of step 2 as we can show that an agent 'working on her own' is a valid strategy in  $M_{\mathrm{c3D}}$ .
4.  $M_{\mathbb{C3D}}$  is approximately efficient at  $s^{\star}$ : By observing that the global minimum penalty is  $2\sigma \sqrt{cm}$ , we use a series of nontrivial algebraic manipulations to show  $\mathrm{PR}(M_{\mathbb{C3D}}, s^{\star}) = \frac{1}{2}\left(\frac{10\alpha^{2}/n_{i}^{\star}-1}{4(m+1)\alpha^{2}/(mn_{i}^{\star})-1}+1\right)$ . As  $\alpha > \sqrt{n_{i}^{\star}}$ , some simple algebra leads to  $\mathrm{PR}(M_{\mathbb{C3D}}, s^{\star}) < 2$ .
4 Special Cases: Restricting the Agents' Strategy Space
In this section, we study two special cases motivated by some natural use cases, where we restrict the agents' strategy space. In addition to providing better guarantees on the efficiency, this will also help us better illustrate the challenges in our original setting.
4.1 Agents cannot fabricate or falsify data
First, we study a setting where agents are not allowed to fabricate data or falsify data. Specifically, in (2),  $\mathcal{F}$  is restricted to functions which map a dataset to any subset. This is applicable when there are regulations preventing such behavior (e.g. government institutions, hospitals)
Mechanism: The discussion at the end of §2 motivates the following modification to the pooling mechanism. We set the allocation space to be  $\mathcal{A} = \bigcup_{n\geq 0}\mathbb{R}^n$ , i.e the space of all datasets. If an agent  $i$  submits at least  $\sigma /\sqrt{cm}$  points, then give her all the other agents' datasets, i.e  $A_{i} = \cup_{j\neq i}Y_{j}$ ; otherwise, set  $A_{i} = \varnothing$ . The recommended strategy  $s_i^\star = (n_i^\star ,f_i^\star ,h_i^\star)$  of each agent is to collect  $\sigma /\sqrt{cm}$  points, submit it as is  $f_{i}^{\star} = \mathbf{I}$ , and then use the sample mean of  $Z_{i}\cup A_{i}$  to estimate  $\mu$ . The theorem below, whose proof is straightforward, states the main properties of this mechanism.
Theorem 2. The following statements about the mechanism and strategy profile  $s^{\star}$  in the paragraph above are true when  $\mathcal{F}$  is restricted to functions which map a dataset to any subset: (i)  $s^{\star}$  is a Nash equilibrium. (ii) The mechanism is individually rational at  $s^{\star}$ . (iii) At  $s^{\star}$ , the mechanism is efficient.
It is not hard to see that this mechanism can be easily manipulated by the agent if there are no restrictions on  $\mathcal{F}$ . As the mechanism only checks for the amount of data submitted, the agent can submit a fabricated dataset of  $\sigma / \sqrt{cm}$  points, and then discard this dataset when computing the estimate, which results in detrimental free-riding.
4.2 Agents accept an estimated value from the mechanism
Our next setting is motivated by use cases where the mechanism may directly deploy the estimated value for  $\mu$  in some downstream application for the agent, i.e. the agents are forced to use this value. This is motivated by federated learning, where agents collect and send data to a server (mechanism), which deploys a model (estimate) directly on the agent's device [9, 23]. This requires modifying the agent's strategy space to  $S = \mathbb{N} \times \mathcal{F}$ . Now, an agent can only choose  $(n_i, f_i)$ , how much data she wishes to collect, and how to fabricate or falsify the dataset. A mechanism is defined as a procedure  $b: (\bigcup_{n \geq 0} \mathbb{R}^n)^m \to \mathbb{R}^m$ , which maps  $m$  datasets to  $m$  estimated mean values.