SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} \mathrm {P R} (M _ {\tt C 3 D}, s ^ {\star}) = \frac {m p (n _ {i} ^ {\star})}{2 \sigma \sqrt {c m}} = \frac {1}{2} \left(\frac {\frac {1 0 \alpha^ {2}}{n _ {i} ^ {\star}} - 1}{\frac {4 \alpha^ {2}}{n _ {i} ^ {\star}} \frac {m + 1}{m} - 1} + 1\right) \\ <   \frac {1}{2} \left(\frac {\frac {1 0 \alpha^ {2}}{n _ {i} ^ {*}} - 1 + \frac {1 0 \alpha^ {2}}{n _ {i} ^ {*}} \frac {1}{m} + \left(\frac {2 \alpha^ {2}}{n _ {i} ^ {*}} \frac {m + 1}{m} - 2\right)}{\frac {4 \alpha^ {2}}{n _ {i} ^ {*}} \frac {m + 1}{m} - 1} + 1\right) = 2. \\ \end{array}
$$
We will use  $M_{\mathrm{PCS}}$  to denote the mechanism in §4.1, as it pools the datasets, but checks for the size of the dataset submitted by each agent. For clarity, we have stated  $M_{\mathrm{PCS}}$  algorithmically in Algorithm 2. We will also re-state the recommended strategies  $s_i^\star = \{(n_i^\star, f_i^\star, h_i^\star)\}_i$  below:
$$
n _ {i} ^ {\star} = \frac {\sigma}{\sqrt {c m}}, \quad f _ {i} ^ {\star} = \mathbf {I}, \quad h _ {i} ^ {\star} \left(X _ {i}, Y _ {i}, A _ {i}\right) = \frac {1}{\left| X _ {i} \cup A _ {i} \right|} \sum_ {u \in X _ {i} \cup A _ {i}} u \tag {26}
$$
Throughout this section,  $s_i^\star$  will refer to (26) (and not (8)).
We will first prove that  $s_i^\star$  is a Nash equilibrium. Because the sample mean achieves minimax error for Normal mean estimation [24], we immediately have, for all  $(n_i, f_i, h_i) \in S$ .
$$
p _ {i} \left(M _ {\mathrm {P C S}}, \left(\left(n _ {i}, f _ {i}, h _ {i} ^ {\star}\right), s _ {- i} ^ {\star}\right)\right) \leq p _ {i} \left(M _ {\mathrm {P C S}}, \left(\left(n _ {i}, f _ {i}, h _ {i}\right), s _ {- i} ^ {\star}\right)\right).
$$
Because the agent can only submit the raw dataset or a subset, and the agent's allocation only depends on the size of the dataset, the size of the dataset she receives can always be maximized by submitting the whole data set she collects, i.e. chooses  $f_{i} = \mathbf{I}$ . Therefore, we have for all  $(n_i, f_i, h_i) \in S$ ,
$$
p _ {i} \left(M _ {\mathrm {P C S}}, \left(\left(n _ {i}, f _ {i} ^ {\star}, h _ {i} ^ {\star}\right), s _ {- i} ^ {\star}\right)\right) \leq p _ {i} \left(M _ {\mathrm {P C S}}, \left(\left(n _ {i}, f _ {i}, h _ {i} ^ {\star}\right), s _ {- i} ^ {\star}\right)\right) \leq p _ {i} \left(M _ {\mathrm {P C S}}, \left(\left(n _ {i}, f _ {i}, h _ {i}\right), s _ {- i} ^ {\star}\right)\right).
$$
Finally, we can use the fact that the maximum risk of the sample mean estimator using  $n$  points is  $\sigma^2 / n$  to show that the penalty is minimized when  $n_i = n_i^\star = \sigma / \sqrt{cm}$ . In particular, we have that if  $n_i < \sigma / \sqrt{cm}$ ,
$$
p _ {i} \left(M _ {\mathrm {P C S}}, \left(\left(n _ {i}, f _ {i} ^ {\star}, h _ {i} ^ {\star}\right), s _ {- i} ^ {\star}\right)\right) = \frac {\sigma^ {2}}{n _ {i}} + c n _ {i} > 2 \sigma \sqrt {c}.
$$
And if  $n_i \geq \sigma / \sqrt{cm}$ ,
$$
p _ {i} (M _ {\mathrm {P C S}}, ((n _ {i}, f _ {i} ^ {\star}, h _ {i} ^ {\star}), s _ {- i} ^ {\star})) = \frac {\sigma^ {2}}{n _ {i} + (m - 1) \sigma / \sqrt {c m}} + c n _ {i} \geq 2 \sigma \sqrt {\frac {c}{m}}
$$
Because  $2\sigma \sqrt{c} \geq 2\sigma \sqrt{c / m}$ ,  $p_i(M_{\mathrm{PCS}},((n_i, f_i^{\star}, h_i^{\star}), s_{-i}^{\star}))$  is minimized when  $n_i = \sigma / \sqrt{cm}$ . We thus conclude that  $s^{\star}$  is a Nash equilibrium. That is, for all  $(n_i, f_i, h_i) \in \mathbb{N} \times \mathcal{F} \times \mathcal{H}$
$$
p _ {i} \left(M _ {\mathrm {P C S}}, s ^ {\star}\right) \leq p _ {i} \left(M _ {\mathrm {P C S}}, \left(\left(n _ {i}, f _ {i}, h _ {i}\right), s _ {- i} ^ {\star}\right)\right).
$$
Next, the IR and efficiency properties follow trivially from the fact that  $p_i(M_{\mathrm{PCS}}, s^{\star}) = 2\sigma \sqrt{c / m}$  for each agent  $i$ . In particular,  $p_i(M_{\mathrm{PCS}}, s^{\star}) < p_{\mathrm{min}}^{\mathrm{IR}}$  and  $P(M_{\mathrm{PCS}}, s^{\star}) = 2\sigma \sqrt{cm}$ .
We will use  $M_{\mathrm{CDED}}$  to denote our mechanism in §4.2, as it corrupts the deployed estimate based on the difference. We have stated this mechanism formally in Algorithm 3. We will also re-state the recommended strategies  $s_i^\star = \{(n_i^\star, f_i^\star)\}_i$  below:
$$
n _ {i} ^ {\star} = \frac {\sigma}{\sqrt {c m}}, \quad f _ {i} ^ {\star} = \mathbf {I}. \tag {27}
$$
Throughout this section,  $s_i^\star$  will refer to (27) (and not (8) or (26)).
We will now present the proof of Theorem 3. First, in §C.1, we show that  $s^{\star}$  is a Nash equilibrium of  $M_{\mathrm{CDED}}$  as the Nash incentive compatibility result. Then, in §C.2, we show individual rationality at  $s_i^{\star}$ . In §C.3, we conclude by showing that  $M_{\mathrm{CDED}}$  is approximately efficient by showing that its social penalty at most a  $(1 + \epsilon)$  factor of the global minimum.
C.1 Algorithm 3 is Nash incentive compatible
Step 1. We will first show that fixing any  $n_i$ , the best strategy is to submit the raw data, i.e. for all  $(n_i, f_i) \in \mathbb{N} \times \mathcal{F}$ .
$$
p _ {i} \left(M _ {\mathrm {C D E D}}, \left(\left(n _ {i}, f _ {i} ^ {\star}\right), s _ {- i} ^ {\star}\right)\right) \leq p _ {i} \left(M _ {\mathrm {C D E D}}, \left(\left(n _ {i}, f _ {i}\right), s _ {- i} ^ {\star}\right)\right). \tag {28}
$$
Let  $e_{z,i} = \epsilon_{z,i} / \eta_i$ , where  $\eta_i$ , and  $\epsilon_{z,i}$  are as given in lines 9 and 10 respectively. We have that  $e_{z,i}$ 's are i.i.d. standard Normal samples. Because the cost term  $cn_i$  is fixed when  $n_i$  is fixed, we only need to consider the risk term. We will first define,
$$
\widehat {\mu} \left(X _ {i}\right) := \frac {1}{\left| X _ {i} \right|} \sum_ {x \in X _ {i}} x, \quad \widehat {\mu} \left(Y _ {i}\right) := \frac {1}{\left| Y _ {i} \right|} \sum_ {x \in Y _ {i}} x, \quad \widehat {\mu} \left(Y _ {- i}\right) := \frac {1}{\left| Y _ {- i} \right|} \sum_ {x \in Y _ {- i}} x. \tag {29}
$$
Via some algebraic manipulations, we can express the maximum risk as:
$$
\begin{array}{l} \sup _ {\mu} \mathbb {E} \left[ \right.\left.\left(\frac {1}{| Y _ {i} | + (m - 1) n _ {i} ^ {\star}} \left(\sum_ {y \in Y _ {i}} (y - \mu) + \sum_ {z \in Y _ {- i}} (z + e _ {z, i} \eta_ {i} - \mu)\right)\right) ^ {2} \middle | \mu \right] \\ = \frac {1}{\left(\left| Y _ {i} \right| + (m - 1) n _ {i} ^ {\star}\right) ^ {2}} \sup  _ {\mu} \mathbb {E} \left[ \left(\sum_ {y \in Y _ {i}} (y - \mu)\right) ^ {2} + \left(\sum_ {z \in Y _ {- i}} (z + e _ {z, i} \eta_ {i} - \mu)\right) ^ {2} \Bigg | \mu \right] \\ = \frac {1}{\left(\left| Y _ {i} \right| + (m - 1) n _ {i} ^ {\star}\right) ^ {2}} \sup  _ {\mu} \mathbb {E} \left[ \left(\left| Y _ {i} \right| (\widehat {\mu} \left(Y _ {i}\right) - \mu)\right) ^ {2} + \left(\sum_ {z \in Y _ {- i}} (z - \mu)\right) ^ {2} + \left(\sum_ {z \in Y _ {- i}} e _ {z, i} \eta_ {i}\right) ^ {2} \Bigg | \mu \right] \\ = \frac {1}{\left(\left| Y _ {i} \right| + (m - 1) n _ {i} ^ {\star}\right) ^ {2}} \sup _ {\mu} \mathbb {E} \Big [ \left(\left| Y _ {i} \right| (\widehat {\mu} (Y _ {i}) - \mu)\right) ^ {2} + (m - 1) n _ {i} ^ {\star} \beta_ {\epsilon} ^ {2} (\widehat {\mu} (Y _ {i}) - \widehat {\mu} (Y _ {- i})) ^ {2 k _ {\epsilon}} \Big | \mu \Big ] \\ + \frac {(m - 1) n _ {i} ^ {\star} \sigma^ {2}}{(| Y _ {i} | + (m - 1) n _ {i} ^ {\star}) ^ {2}} \\ \end{array}
$$
Recall that  $\beta_{\epsilon}$  also involves  $|Y_i|$ . Note that as we have fixed  $n_i$  and  $s_{-i} = s_{-i}^{\star}$ , the maximum risk depends only on  $|Y_i|$  and  $\widehat{\mu}(Y_i)$ , that is, the agent's maximum risk and hence penalty only depends on the number of points she submitted, and their average value. Hence, to find the optimal submission
Algorithm 3  $M_{\mathrm{CDED}}$
Require: Approximation parameter  $\epsilon >0$  # to obtain a  $1 + \epsilon$  bound on PR.
1: Mechanism designer publishes: The procedure in lines 5-11.
2: Each agent  $i$ :
3: Choose strategy  $s_i = (n_i, f_i)$ .
4: Sample  $n_i$  points  $X_i = \{x_{i,j}\}_{j=1}^{n_i}$  and submit  $Y_i = f_i(X_i)$  to the mechanism.
5: Mechanism:
$$
k _ {\epsilon} \leftarrow \lceil \frac {1}{2 \epsilon} \rceil , \quad \beta_ {\epsilon} \leftarrow \sqrt {\frac {\left(\sum_ {i = 1} ^ {m} \left| Y _ {i} \right|\right) ^ {2} (m - 1) ^ {k _ {\epsilon} - 1}}{k _ {\epsilon} (2 k _ {\epsilon} - 1) ! ! \sigma^ {k _ {\epsilon}} c ^ {\frac {k _ {\epsilon} - 2}{2}} m ^ {3 k _ {\epsilon} / 2}}}
$$
7: For each agent  $i\in [m]$  # can be done simultaneously for all agents
$$
8: \quad Y _ {- i} \leftarrow \bigcup_ {j \neq i} Y _ {j}.
$$
$$
9: \quad \eta_ {i} ^ {2} \leftarrow \beta_ {\epsilon} ^ {2} \left(\frac {1}{| Y _ {i} |} \sum_ {y \in Y _ {i}} y - \frac {1}{| Y _ {- i} |} \sum_ {y \in Y _ {- i}} y\right) ^ {2 k _ {\epsilon}}.
$$
10:  $Z_{i}\gets \{z + \epsilon_{z,i},\mathrm{for~all} z\in Y_{-i}\mathrm{~where~}\epsilon_{z,i}\sim \mathcal{N}(0,\eta_{i}^{2})\}$
11: Deploy estimate  $\left(\frac{1}{|Y_i\cup Z_i|}\sum_{u\in Y_i\cup Z_i}u\right)$  for agent  $i$
$Y_{i}$ , we will first fix the size of the agent's submission  $|Y_{i}|$  and optimize for the sample mean  $\widehat{\mu}(Y_{i})$  (step 1.1), and then we will optimize for  $|Y_{i}|$  (step 1.2).
Step 1.1. Since the other agents have each collected  $\sigma / \sqrt{cm} = n_i^\star$  points and submitted it truthfully, we have  $\widehat{\mu}(Y_{-i}) \sim \mathcal{N}\left(\mu, \frac{\sigma^2}{(m - 1)n_i^\star}\right)$ . Via a binomial expansion, we can write,
$$
\begin{array}{l} \mathbb {E} \left[ \left(\widehat {\mu} (Y _ {i}) - \widehat {\mu} (Y _ {- i})\right) ^ {2 k _ {\epsilon}} \right] = \mathbb {E} \left[ \left(\left(\widehat {\mu} (Y _ {i}) - \mu\right) - \left(\widehat {\mu} (Y _ {- i}) - \mu\right)\right) ^ {2 k _ {\epsilon}} \right] \\ = \sum_ {j = 0} ^ {2 k _ {\epsilon}} (- 1) ^ {j} \binom {2 k _ {\epsilon}} {j} \mathbb {E} \left[ (\widehat {\mu} (Y _ {i}) - \mu) ^ {j} \right] \mathbb {E} \left[ (\widehat {\mu} (Y _ {- i}) - \mu) ^ {2 k _ {\epsilon} - j} \right] \\ = \sum_ {j = 0} ^ {k _ {\epsilon}} \binom {2 k _ {\epsilon}} {2 j} \mathbb {E} \left[ (\widehat {\mu} (Y _ {i}) - \mu) ^ {2 j} \right] \mathbb {E} \left[ (\widehat {\mu} (Y _ {- i}) - \mu) ^ {2 k _ {\epsilon} - 2 j} \right] \\ \end{array}
$$
Thus the maximum risk can be written as:
$$
\sup  _ {\mu} \mathbb {E} \left[ \sum_ {j = 0} ^ {k _ {\epsilon}} A _ {j} \left(\widehat {\mu} \left(Y _ {i}\right) - \mu\right) ^ {2 j} \mid \mu \right] \tag {30}
$$
where  $A_0,\ldots ,A_{k_\epsilon}$  is a sequence of positive coefficients.
Similar to the proof of Theorem 1, we construct a lower bound on the maximum risk using a sequence of Bayesian risks. Let  $\Lambda_{\ell} \coloneqq \mathcal{N}(0,\ell^{2}), \ell = 1,2,\ldots$  be a sequence of prior for  $\mu$ . For fixed  $\ell$ , the posterior distribution is:
$$
\begin{array}{l} p (\mu | X _ {i}) \propto p (X _ {i} | \mu) p (\mu) \propto \exp \left(- \frac {1}{2 \sigma^ {2}} \sum_ {x \in X _ {i}} (x - \mu) ^ {2}\right) \exp \left(- \frac {1}{2 \ell^ {2}} \mu^ {2}\right) \\ \propto \exp \biggl (- \frac {1}{2} \biggl (\frac {n _ {i}}{\sigma^ {2}} + \frac {1}{\ell^ {2}} \biggr) \mu^ {2} + \frac {1}{2} 2 \frac {\sum_ {x \in X _ {i}} x}{\sigma^ {2}} \mu \biggr). \\ \end{array}
$$
This means the posterior of  $\mu$  given  $X_{i}$  is Gaussian with:
$$
\mu | X _ {i} \sim \mathcal {N} \left(\frac {n _ {i} \widehat {\mu} (X _ {i}) / \sigma^ {2}}{n _ {i} / \sigma^ {2} + 1 / \ell^ {2}}, \frac {1}{n _ {i} / \sigma^ {2} + 1 / \ell^ {2}}\right) =: \mathcal {N} \left(\mu_ {\ell}, \sigma_ {\ell} ^ {2}\right).
$$
Therefore, the posterior risk is:
$$
\mathbb {E} \left[ \sum_ {j = 0} ^ {k _ {\epsilon}} A _ {j} \left(\widehat {\mu} \left(Y _ {i}\right) - \mu\right) ^ {2 j} \mid X _ {i} \right] = \mathbb {E} \left[ \sum_ {j = 0} ^ {k _ {\epsilon}} A _ {j} \left(\left(\widehat {\mu} \left(Y _ {i}\right) - \mu_ {\ell}\right) - \left(\mu - \mu_ {\ell}\right)\right) ^ {2 j} \mid X _ {i} \right]
$$
$$
= \int_ {- \infty} ^ {\infty} \underbrace {\sum_ {j = 0} ^ {k _ {\epsilon}} A _ {j} (e - (\widehat {\mu} (Y _ {i}) - \mu_ {\ell})) ^ {2 j}} _ {=: F _ {1} (e - (\widehat {\mu} (Y _ {i}) - \mu_ {\ell}))} \underbrace {\frac {1}{\sigma_ {\ell} \sqrt {2 \pi}} \exp \left(- \frac {e ^ {2}}{2 \sigma_ {\ell} ^ {2}}\right)} _ {=: F _ {2} (e)} d e
$$
-  $F_{1}(\cdot)$  is even function and increases on  $[0, \infty)$ ;
-  $F_{2}(\cdot)$  is even function and decreases on  $[0, \infty)$ , and  $\int_{\mathbb{R}} F_{2}(e) de < \infty$
- For any  $a \in \mathbb{R}$ ,  $\int_{\mathbb{R}} F_1(e - a) F_2(e) de < \infty$
By the corollary of Hardy-Littlewood inequality in Lemma 9,
$$
\int_ {\mathbb {R}} F _ {1} (e - a) F _ {2} (e) d e \geq \int_ {\mathbb {R}} F _ {1} (e) F _ {2} (e) d e,
$$
which means the posterior risk is minimized when  $\widehat{\mu}(Y_i) = \mu_\ell$ . In Lemma 11, we have stated expressions for the expected value of the power of a normal random variable. Using this, we can write the Bayes risk as:
$$
R _ {\ell} := \mathbb {E} \left[ \sum_ {j = 0} ^ {k _ {\epsilon}} A _ {j} \mathbb {E} \left[ \left(\mu - \mu_ {\ell}\right) ^ {2 j} \Big | X _ {i} \right] \right] = \sum_ {j = 0} ^ {k _ {\epsilon}} A _ {j} (2 j - 1)!! \sigma_ {\ell} ^ {2 j}
$$
and the limit of Bayesian risk as  $\ell \to \infty$  is
$$
R _ {\infty} := \lim  _ {\ell \rightarrow \infty} \sum_ {j = 0} ^ {k _ {\epsilon}} A _ {j} (2 j - 1)!! \frac {\sigma^ {2 j}}{n _ {i} ^ {j}}
$$
When  $\widehat{\mu}(Y_i) = \widehat{\mu}(X_i)$ , the maximum risk is:
$$
\begin{array}{l} \sup  _ {\mu} \mathbb {E} \left[ \sum_ {j = 0} ^ {k _ {\epsilon}} A _ {j} (\widehat {\mu} (Y _ {i}) - \mu) ^ {2 j} \Bigg | \mu \right] = \sup  _ {\mu} \mathbb {E} \left[ \sum_ {j = 0} ^ {k _ {\epsilon}} A _ {j} (\widehat {\mu} (X _ {i}) - \mu) ^ {2 j} \Bigg | \mu \right] \\ = \sum_ {j = 0} ^ {k _ {c}} A _ {j} (2 j - 1)!! \sigma^ {2 j} n _ {i} ^ {- j} = R _ {\infty}. \\ \end{array}
$$
This means, fixing  $n_i$  and  $|Y_i|$ , agent  $i$  achieves minimax risk when choosing  $\widehat{\mu}(Y_i) = \widehat{\mu}(X_i)$ ; as the maximum is larger than the average, this follows using a similar argument to Step 3 in §A.3.
Step 1.2. Next, we will show that the best size of the submission is  $|Y_i| = |X_i| = n_i$ , assuming  $\widehat{\mu}(Y_i) = \widehat{\mu}(X_i)$ . For this, we will first use  $n_i^\star$  to rewrite  $\beta_\epsilon^2$  as
$$
\beta_ {\epsilon} ^ {2} = \frac {n _ {i} ^ {\star k _ {\epsilon} - 2} (m - 1) ^ {k _ {\epsilon} - 1} (| Y _ {i} | + (m - 1) n _ {i} ^ {\star}) ^ {2}}{k _ {\epsilon} (2 k _ {\epsilon} - 1) ! ! m ^ {k _ {\epsilon} + 1} \sigma^ {2 k _ {\epsilon} - 2}}.
$$
$$
\widehat {\mu} (X _ {i}) - \widehat {\mu} (Y _ {- i}) \sim \mathcal {N} \bigg (0, \left(\frac {1}{n _ {i}} + \frac {1}{(m - 1) n _ {i} ^ {\star}}\right) \sigma^ {2} \bigg),
$$
the risk term in the penalty can be rewritten and lower bounded as follows:
$$
\begin{array}{l} \frac {1}{\left(| Y _ {i} | + (m - 1) n _ {i} ^ {\star}\right) ^ {2}} \bigg (\left| Y _ {i} \right| ^ {2} \sigma^ {2} / n _ {i} + (m - 1) n _ {i} ^ {\star} \beta_ {\epsilon} ^ {2} (2 k _ {\epsilon} - 1)!! \bigg (\frac {1}{n _ {i}} + \frac {1}{(m - 1) n _ {i} ^ {\star}} \bigg) ^ {k _ {\epsilon}} \sigma^ {2 k _ {\epsilon}} \bigg) \\ + \frac {(m - 1) n _ {i} ^ {\star} \sigma^ {2}}{\left(| Y _ {i} | + (m - 1) n _ {i} ^ {\star}\right) ^ {2}} \\ = \frac {| Y _ {i} | ^ {2} \frac {\sigma^ {2}}{n _ {i}} + (m - 1) n _ {i} ^ {\star} \sigma^ {2}}{(| Y _ {i} | + (m - 1) n _ {i} ^ {\star}) ^ {2}} + \frac {n _ {i} ^ {\star k _ {\epsilon} - 1} (m - 1) ^ {k _ {\epsilon}}}{k _ {\epsilon} m ^ {k _ {\epsilon} + 1}} \bigg (\frac {1}{n _ {i}} + \frac {1}{(m - 1) n _ {i} ^ {\star}} \bigg) ^ {k _ {\epsilon}} \sigma^ {2} \\ \end{array}
$$
$$
\geq \frac {\sigma^ {2}}{n _ {i} + (m - 1) n _ {i} ^ {\star}} + \frac {n _ {i} ^ {\star k _ {\epsilon} - 1} (m - 1) ^ {k _ {\epsilon}}}{k _ {\epsilon} m ^ {k _ {\epsilon} + 1}} \bigg (\frac {1}{n _ {i}} + \frac {1}{(m - 1) n _ {i} ^ {\star}} \bigg) ^ {k _ {\epsilon}} \sigma^ {2}.
$$
Here, the last step follows from the fact that
$$
\begin{array}{l} \frac {| Y _ {i} | ^ {2} \frac {\sigma^ {2}}{n _ {i}} + (m - 1) n _ {i} ^ {\star} \sigma^ {2}}{\left(| Y _ {i} | + (m - 1) n _ {i} ^ {\star}\right) ^ {2}} = \frac {| Y _ {i} | ^ {2} \frac {\sigma^ {2}}{n _ {i}} + (m - 1) n _ {i} ^ {\star} \sigma^ {2}}{n _ {i} \frac {| Y _ {i} | ^ {2}}{n _ {i}} + 2 | Y _ {i} | (m - 1) n _ {i} ^ {\star} + (m - 1) ^ {2} n _ {i} ^ {\star 2}} \\ \geq \frac {\left| Y _ {i} \right| ^ {2} \frac {\sigma^ {2}}{n _ {i}} + (m - 1) n _ {i} ^ {\star} \sigma^ {2}}{n _ {i} \frac {\left| Y _ {i} \right| ^ {2}}{n _ {i}} + \left(n _ {i} + \frac {\left| Y _ {i} \right| ^ {2}}{n _ {i}}\right) (m - 1) n _ {i} ^ {\star} + (m - 1) ^ {2} n _ {i} ^ {\star^ {2}}} = \frac {\left| Y _ {i} \right| ^ {2} \frac {\sigma^ {2}}{n _ {i}} + (m - 1) n _ {i} ^ {\star} \sigma^ {2}}{\left(n _ {i} + (m - 1) n _ {i} ^ {\star}\right) \left(\frac {\left| Y _ {i} \right| ^ {2}}{n _ {i}} + (m - 1) n _ {i} ^ {\star}\right)} \\ = \frac {\sigma^ {2}}{n _ {i} + (m - 1) n _ {i} ^ {\star}}. \\ \end{array}
$$
Equality holds in this inequality if and only if  $|Y_{i}| = n_{i}$ .
In conclusion, fixing  $n_i$ , the agent can minimize her penalty by submitting  $n_i$  points with the same sample mean as the dataset  $X_i$  she collected. One way to achieve this is set  $f_i = \mathbf{I}$ . This completes the proof of (28).
Step 2: Our next step is to show that the agent's best strategy is to collect  $n_i^\star$  data points. That is, we will show for all  $n_i \in \mathbb{N}$ .
$$
p _ {i} \left(M _ {\mathrm {C D E D}}, \left(\left(n _ {i} ^ {\star}, f _ {i} ^ {\star}\right), s _ {- i} ^ {\star}\right)\right) \leq p _ {i} \left(M _ {\mathrm {C D E D}}, \left(\left(n _ {i}, f _ {i} ^ {\star}\right), s _ {- i} ^ {\star}\right)\right). \tag {31}
$$
In the following, we will use  $p(n_i)$  as a shorthand for  $p_i(M_{\mathrm{CDED}},((n_i,f_i^\star),s_{-i}^\star))$ . The penalty can be rewritten as:
$$
p (n _ {i}) = \frac {\sigma^ {2}}{n _ {i} + (m - 1) n _ {i} ^ {\star}} + \frac {n _ {i} ^ {\star k _ {\epsilon} - 1} (m - 1) ^ {k _ {\epsilon}}}{k _ {\epsilon} m ^ {k _ {\epsilon} + 1}} \bigg (\frac {1}{n _ {i}} + \frac {1}{(m - 1) n _ {i} ^ {\star}} \bigg) ^ {k _ {\epsilon}} \sigma^ {2} + c n _ {i}
$$
We need to show that  $p_i(n_i)$  achieves minimum at  $n_i = n_i^\star$ . The derivative of  $p_i(\cdot)$  is:
$$
p ^ {\prime} (n _ {i}) = - \frac {\sigma^ {2}}{(n _ {i} + (m - 1) n _ {i} ^ {\star}) ^ {2}} + \frac {n _ {i} ^ {\star k _ {\epsilon} - 1} (m - 1) ^ {k _ {\epsilon}}}{m ^ {k _ {\epsilon} + 1}} \bigg (\frac {1}{n _ {i}} + \frac {1}{(m - 1) n _ {i} ^ {\star}} \bigg) ^ {k _ {\epsilon} - 1} \sigma^ {2} \bigg (- \frac {1}{n _ {i} ^ {2}} \bigg) + c
$$
Because  $p'(n_i)$  increase in  $n_i$ ,  $p(n_i)$  is convex. Moreover, because
$$
\begin{array}{l} p ^ {\prime} (n _ {i} ^ {\star}) = - \frac {\sigma^ {2}}{m ^ {2} n _ {i} ^ {\star 2}} + \frac {n _ {i} ^ {\star k _ {\epsilon} - 1} (m - 1) ^ {k _ {\epsilon}}}{m ^ {k _ {\epsilon} + 1}} \left(\frac {1}{n _ {i} ^ {\star}} + \frac {1}{(m - 1) n _ {i} ^ {\star}}\right) ^ {k _ {\epsilon} - 1} \sigma^ {2} \left(- \frac {1}{n _ {i} ^ {\star 2}}\right) + c \\ = - \frac {\sigma^ {2}}{m ^ {2} n _ {i} ^ {\star 2}} - \frac {(m - 1) \sigma^ {2}}{m ^ {2} n _ {i} ^ {\star 2}} + c = - \frac {\sigma^ {2}}{m n _ {i} ^ {\star 2}} + c = 0, \\ \end{array}
$$
we know  $p(n_{i})$  reaches minimum at  $n_i = n_i^\star$ . This concludes the proof for (31).
C.2 Algorithm 3 is individually rational
The penalty of an agent at the recommended strategies can be expressed as:
$$
\begin{array}{l} p _ {i} \left(M _ {\text {C D E D}}, s _ {i} ^ {\star}\right) = p \left(n _ {i} ^ {\star}\right) = \frac {\sigma^ {2}}{m n _ {i} ^ {\star}} + \frac {n _ {i} ^ {\star k _ {\epsilon} - 1} (m - 1) ^ {k _ {\epsilon}}}{k _ {\epsilon} m ^ {k _ {\epsilon} + 1}} \left(\frac {1}{n _ {i} ^ {\star}} + \frac {1}{(m - 1) n _ {i} ^ {\star}}\right) ^ {k _ {\epsilon}} \sigma^ {2} + c n _ {i} ^ {\star} \\ = \frac {\sigma^ {2}}{m n _ {i} ^ {\star}} + \frac {n _ {i} ^ {\star k _ {\epsilon} - 1} (m - 1) ^ {k _ {\epsilon}}}{k _ {\epsilon} m ^ {k _ {\epsilon} + 1}} \frac {m ^ {k _ {\epsilon}}}{n _ {i} ^ {\star k _ {\epsilon}} (m - 1) ^ {k _ {\epsilon}}} \sigma^ {2} + c n _ {i} ^ {\star} \\ = \frac {\sigma^ {2}}{m n _ {i} ^ {\star}} + \frac {1}{k _ {\epsilon}} \frac {\sigma^ {2}}{m n _ {i} ^ {\star}} + c n _ {i} ^ {\star} = \left(2 + \frac {1}{k _ {\epsilon}}\right) \frac {\sigma \sqrt {c}}{\sqrt {m}}. \tag {32} \\ \end{array}
$$
We have that  $M_{\mathrm{CDED}}$  is IR when  $m \geq 2$ , via the following simple calculation:
$$
\left(2 + \frac {1}{k _ {\epsilon}}\right) \frac {\sigma \sqrt {c}}{\sqrt {m}} \leq \left(2 + \frac {1}{2}\right) \frac {\sigma \sqrt {c}}{\sqrt {2}} <   2 \sigma \sqrt {c} = p _ {\mathrm {m i n}} ^ {\mathrm {I R}}
$$
C.3 Algorithm 3 is approximately efficient
Using the expression for  $p_i(M_{\mathrm{CDED}}, s_i^\star)$  in (32), the penalty ratio can be bounded by:
$$
\Pr \left(M _ {\mathrm {C D E D}}, s ^ {\star}\right) = \frac {\left(2 + \frac {1}{k _ {\epsilon}}\right) \sigma \sqrt {c m}}{2 \sigma \sqrt {c m}} = 1 + \frac {1}{2 k _ {\epsilon}} \leq 1 + \epsilon .
$$
D Additional Materials for Section 4.2
D.1 Mechanism detail
See Algorithm 3.
D.2 Using a weighted average under the original strategy space from §2
In this section, we will consider a variation of  $M_{\mathrm{CDED}}$  when applied to our original strategy space  $\mathbb{N} \times \mathcal{F} \times \mathcal{H}$ . For this, we will assume that  $M_{\mathrm{CDED}}$  will return  $A_i = Z_i$  as the agent's allocation, and then an agent can use  $X_i, Y_i, Z_i$  to estimate  $\mu$ . In this situation, below we show that the agent can achieve a smaller penalty using a weighted average over  $X_i \cup Z_i$  instead of the sample mean used by the mechanism. Here, the weights are proportional to the inverse of the variance of each data point. (Our mechanism purposefully uses the sub-optimal sample mean in the restricted strategy space  $\mathbb{N} \times \mathcal{F}$  as a way to shape the agent's penalty and incentivize good behavior.)
This shows that  $M_{\mathrm{CDED}}$  (with the above modification) is not NIC in this more general strategy space. The agent can obtain a lower penalty using a better estimator (such as the weighted average we show over here) and achieve a lower penalty. More importantly, as the agent knows that she can achieve a lower estimation error via a better estimator instead of more data, she can leverage this insight to collect less data and reduce her penalty even further.
We should emphasize that it is unclear if this weighted average is minimax optimal. It is also unclear if there exists a Nash equilibrium for  $M_{\mathrm{CDED}}$  (or any straightforward modification of  $M_{\mathrm{CDED}}$ ) in the expanded strategy space.
The weighted average estimator: We will now present the weighted average estimator that achieves a lower maximum risk. To show this, first note that for all  $x \in X_i$ ,  $\mathbb{V}[x] = \sigma^2$ ; when  $(n_i, f_i) = (n_i^\star, f_i^\star)$ , for all  $x \in Z_i$ ,
$$
\begin{array}{l} \mathbb {V} [ x ] = \mathbb {E} \left[ (z + \epsilon_ {z, i} - \mu) ^ {2} \right] = \sigma^ {2} + \beta_ {\epsilon} ^ {2} \mathbb {E} \left[ (\widehat {\mu} (X _ {i}) - \widehat {\mu} (Y _ {- i})) ^ {2 k _ {\epsilon}} \right] \\ = \sigma^ {2} + \frac {n _ {i} ^ {\star k _ {\epsilon} - 2} (m - 1) ^ {k _ {\epsilon} - 1} (m n _ {i} ^ {\star}) ^ {2}}{k _ {\epsilon} (2 k _ {\epsilon} - 1) ! ! m ^ {k _ {\epsilon} + 1} \sigma^ {2 k _ {\epsilon} - 2}} (2 k _ {\epsilon} - 1)!! \left(\frac {1}{n _ {i} ^ {\star}} + \frac {1}{(m - 1) n _ {i} ^ {\star}}\right) ^ {k _ {\epsilon}} \sigma^ {2 k _ {\epsilon}} \\ = \sigma^ {2} + \frac {n _ {i} ^ {\star k _ {\epsilon}} (m - 1) ^ {k _ {\epsilon} - 1}}{k _ {\epsilon} m ^ {k _ {\epsilon} - 1}} \frac {m ^ {k _ {\epsilon}}}{(m - 1) ^ {k _ {\epsilon}} n _ {i} ^ {\star k _ {\epsilon}}} \sigma^ {2} \\ = \sigma^ {2} + \frac {1}{k _ {\epsilon}} \frac {m}{m - 1} \sigma^ {2} \\ \end{array}
$$
Consider the following weighted-average estimator: