SlowGuess/NeurIPS_2023_OCR huggingface dataset:
For comparison, Caldas et al. (2018) aim for an accuracy threshold of  $75\%$  using  $5\%$  of the training data.
This research was partially funded by the Ministry of Education and Science of Bulgaria (support for INSAIT, part of the Bulgarian National Roadmap for Research Infrastructure). Florian Dorner is grateful for financial support from the Max Planck ETH Center for Learning Systems (CLS). The authors would like to thank Mark Vero and Nikita Tsoy for their helpful feedback on earlier versions of this manuscript.
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A Ethical considerations
Our theoretical results establish that penalties average out to zero in expectation, but substantial client heterogeneity can still cause large payments for individual honest participants. Despite our experimental results indicating that variance can be manageable even for real-world data, this is problematic for two reasons: The first is about fairness: Paying some honest participants large amounts, while demanding large amounts from others, based on what essentially amounts to luck is problematic, especially when the participants are individuals rather than firms, and when the nature of the problem might make it difficult for participants to gauge the order of magnitude of payments in advance. The second is about diversity: Clients that expect their data to deviate strongly from the mean of the overall data distribution might opt to not participate in FL with our mechanisms, even though underrepresented types of clients can provide data that is crucial to a model's broad performance and generalization. In our formalism, this problem is obscured by the assumption that all participants sample their data independently from the same distribution, and are unable to predict whether or not their data represents outliers.
Correspondingly, it is important to keep penalty weights  $C$  as low as possible to reduce the likelihood of overwhelmingly large penalties, only apply our framework in the context of firms rather than individuals (for whom competitive incentives might often play less of a direct role either way), as well as ensure that the our assumptions about a common data distributions are plausible for the problem at hand. The former can be particularly challenging for non-convex problems, or convex problems with unknown problem parameters, for which no strong candidate for  $C$  can be established theoretically.
Furthermore, our results do not establish collusion-proofness of our mechanism. While we expect our mechanism to be collusion-proof against small coalitions, there is a problem once the colluding coalition is large enough to significantly affect the mean estimate, as the shifted mean would reduce the penalty paid by each member of the coalition.
B.1 Nash equilibria in the mean estimation game under bounded attacks
The conclusion that no Nash equilibria exist in the mean estimation game described in 4 is rather intuitive, since all participants have an incentive to send as modified an update as possible, therefore damaging the other players' estimates. In practice, however, attacking in an unbounded manner, that is, sending updates very far from the true mean, may not be plausible. Indeed, if most players send their true mean, one expects the variance of the estimates that the server receives to be of order  $\mathcal{O}\left(\frac{\sigma^2}{n} +\sigma_\star^2\right)$ . Therefore, players might in practice be reluctant to send estimates that are further than  $\mathrm{A}\sqrt{\frac{\sigma^2}{n} + \sigma_\star^2}$  away from there true local mean, for some constant A.
We therefore consider the same game as before, in the case when an upper bound  $A$  on the parameters  $\alpha^i$  is given. Denote the resulting set of attack strategies by  $\mathcal{A}_A$ . Since  $\mathcal{A}_A \subset \mathcal{A}^m$ , Theorem 4.1 holds for the joint set of strategies  $(\mathcal{A}_A^m \times \mathcal{D})$ . Then we have the following
Corollary B.1. In the setup of Theorem 4.1, if the set of available strategies is  $\mathcal{A}_{\mathrm{A}} \times \mathcal{D}$  for some constant  $\mathrm{A} > 0$ , the only Nash equilibria of the game with  $b^{i}(x^{i}) = 0$  fixed for all players  $i$  are the strategy profiles for which:
$$
\left| \alpha_ {i} \right| = A \quad a n d \quad \beta^ {i} = \frac {A ^ {2}}{\left(\frac {\sigma^ {2}}{n} + \sigma_ {\star} ^ {2}\right) N + A ^ {2}} \quad \forall i \in [ N ]. \tag {9}
$$
Furthermore, at each of these equilibria the value of mean squared error of the estimate of each player  $i$  is
$$
\mathbb {E} \left(\| \theta^ {i} - \mu \| ^ {2}\right) = (\frac {\sigma^ {2}}{n} + \sigma_ {\star} ^ {2}) \frac {(1 + \frac {1}{\frac {\sigma^ {2}}{n} + \sigma_ {\star} ^ {2}} A ^ {2})}{(N + \frac {1}{\frac {\sigma^ {2}}{n} + \sigma_ {\star} ^ {2}} A ^ {2})}
$$
As a result, whenever  $\mathrm{A} = \mathcal{O}(1)$ , each player's estimate at any Nash equilibrium achieves a mean squared error of  $\mathcal{O}\left(\frac{\sigma^2}{Nn} +\frac{\sigma_*^2}{N}\right)$ , which is of the same order as the MSE of the estimates that would have been obtained in a fully collaborative setting.
B.2 Additional details on experiments
The network we train is based on the network used in the LEAF repository  $^{3}$  but implemented in pytorch Paszke et al. (2019). It consists of two convolutional layers with relu activations, kernel size 5, (2,2) padding and 32 and 64 filters, respectively, each followed by max pooling with kernel size and stride 2. After the convolutional layers, there is a single hidden dense layer with 2048 neurons and a relu activation, and a dense output layer. All experiments were conducted using a single GPU each  $^{4}$  per run.
We downloaded the FeMNIST and Twitter datasets using the code provided at https://github.com/TalwalkarLab/leaf/tree/master/data/feminist, opting not to filter writers that only have produced a small amount of samples. Correspondingly, our FeMNIST dataset contains 817851 examples of handwritten digits and characters written by a total of 3597 writers rather than the 805263 samples from 3550 writers reported in Caldas et al. (2018).
Figures 4, 5, 6 show results similar to 2 for  $\alpha_{B}$  fixed to 1 instead of 0 (Figure 4), or  $\alpha_{B}$  varying while  $\alpha_{A}$  is fixed to 0 (Figure 5), or 1 (Figure 6), respectively. As payments are redistributed, the average payments for players in group  $B$  increase slower with  $\alpha_{B}$ , as each individual's increase in payment is partially balanced out by an increase in received payments from encountering other members of group  $B$  (Figures 5, 6). Meanwhile, figures 4 and 6 hint at honest players receiving money when others are adding noise: In both cases, players of one group receive slightly higher reward for larger penalties when they are honest ( $\alpha = 0$ ) while the other group slightly cheats ( $\alpha = 1$ ).
It is worth noting, that we do not perform a projection step in our experiments, such that numerical instabilities become an issue for large values of  $\alpha$ . In particular, for  $\alpha_{A} > 6$  or  $\alpha_{B} > 6$  we regularly observed NaN gradients on FeMNIST for one or more of our 10 runs.
Figure 10 show histograms over the total penalty for  $C = 0.0002$  (FeMNIST) and  $C = 0.002$  (Twitter) paid by each individual client over the whole 10650 steps for the honest case of  $\alpha_{A} = \alpha_{B} = 0$ , averaged over 10 runs. Clearly, most penalties are on the order of 0.01 (FeMNIST) or 0.1 (Twitter), which is substantially smaller (FeMNIST) / comparable (Twitter) to the negative effects of larger noise values on the order of 0.1, which are strongly disincentivized by the considered penalty values.
Figure 13 shows additional results using SGD with Median-based rather than Mean-based aggregation as a baseline that is more robust to noisy gradients. We can see that while using Median-based aggregation helps, players can affect the loss as much as before by adding even more noise (see Figure 11), such that adding noise is still incentivized. At the same time, our mechanism still works: As can be seen in Figure 12, rewards still decrease with increased noise for sufficiently large  $C$ . Correspondingly, players remain disincentivized to add noise. This empirically suggests that our mechanism can also be applied to more advanced federated learning protocols that go beyond simple SGD with Mean-based aggregation.
<sup>3</sup>https://github.com/TalwalkarLab/leaf/blob/master/models/feminist/cnn.py
4We used assigned GPUs from a cluster that employs mostly, but not exclusively Nvidia V100 GPUs
C Proofs on mean estimation
First we prove Theorem 4.1.
Theorem C.1. Let  $\mathcal{D}$  be as described in 4. Then, for any strategy profile  $((\alpha^{1},b^{1},\beta^{1}),\ldots ,(\alpha^{N},b^{N},\beta^{N}))\in (\mathcal{A}\times \mathcal{D})^{N}$  and for any player  $i\in [N]$ , the expected mean squared error of player  $i$  is:
$$
\begin{array}{l} \mathbb {E} \left(\left\| \theta^ {i} - \mu \right\| ^ {2}\right) \\ = \left(1 - \beta^ {i}\right) ^ {2} \left(\frac {\sigma^ {2}}{N n} + \frac {\sigma_ {\star} ^ {2}}{N} + \frac {1}{N ^ {2}} \sum_ {j \neq i} \mathbb {E} \left(a ^ {j} \left(x ^ {j}\right) ^ {2}\right) + \frac {1}{N ^ {2}} \mathbb {E} \| \sum_ {j \neq i} b ^ {j} \left(x ^ {j}\right) \| ^ {2}\right) \\ + (\beta^ {i}) ^ {2} \left(\frac {\sigma^ {2}}{n} + \sigma_ {\star} ^ {2}\right) + 2 (1 - \beta^ {i}) \beta^ {i} \left(\frac {\sigma^ {2}}{N n} + \frac {\sigma_ {\star} ^ {2}}{N}\right) \\ \end{array}
$$
Proof. Recall that  $\theta^i = (1 - \beta^i)\bar{m}^i +\beta^i\bar{x}^i = (1 - \beta^i)\frac{1}{N}\left(N\theta^s -m^i +\bar{x}^i\right) + \beta^i\bar{x}^i$ . Therefore,  $\theta^i -\mu = (1 - \beta^i)\left(\frac{1}{N}\left(N\theta^s -m^i +\bar{x}^i\right) - \mu\right) + \beta^i\left(\bar{x}^i -\mu\right)$ .
$$
Y = \frac {1}{N} \left(N \theta^ {s} - m ^ {i} + \bar {x} ^ {i}\right) - \mu , \quad Z = \bar {x} ^ {i} - \mu
$$
$$
\begin{array}{l} \mathbb {E} \left(\left\| \theta^ {i} - \mu \right\| ^ {2}\right) = \mathbb {E} \left(\left\| \left(1 - \beta^ {i}\right) Y + \beta^ {i} Z \right\| ^ {2}\right) \tag {10} \\ = \left(1 - \beta^ {i}\right) ^ {2} \mathbb {E} \| Y \| ^ {2} + \left(\beta^ {i}\right) ^ {2} \mathbb {E} \| Z ^ {2} \| + 2 \left(1 - \beta^ {i}\right) \beta^ {i} \mathbb {E} \left(Y ^ {t} Z\right) \\ \end{array}
$$
The second term is the mean squared error of the (clean) estimate based on the local data of player  $i$  and so:
$$
\begin{array}{l} \mathbb {E} \left(\| Z \| ^ {2}\right) = \mathbb {E} \left(\left\| \bar {x} ^ {i} - \mu \right\| ^ {2}\right) \\ = \mathbb {E} \left(\left\| \frac {1}{n} \sum_ {j} x _ {j} ^ {i} - \mu \right\| ^ {2}\right) \\ = \mathbb {E} \left(\| \frac {1}{n} \sum_ {j} (x _ {j} ^ {i} - \mu) \| ^ {2}\right) \\ = \frac {1}{n ^ {2}} \mathbb {E} \left(\| \sum_ {j} \left(x _ {j} ^ {i} - \mu\right) \| ^ {2}\right) \\ = \frac {1}{n ^ {2}} \mathbb {E} \left(\| \sum_ {j} \left(x _ {j} ^ {i} - \mu^ {i}\right) \| ^ {2} + \| \sum_ {j} \left(\mu^ {i} - \mu\right) \| ^ {2}\right) \\ = \frac {1}{n ^ {2}} \left(\sum_ {j} \mathbb {E} \left(\left\| \left(x _ {j} ^ {i} - \mu^ {i}\right) \right\| ^ {2}\right) + \mathbb {E} \left(\left\| n \left(\mu^ {i} - \mu\right) \right\| ^ {2}\right)\right) \\ = \frac {n}{n ^ {2}} \mathbb {E} \left(\| (X ^ {i} - \mu^ {i}) \| ^ {2}\right) + \mathbb {E} \| (\mu^ {i} - \mu) \| ^ {2} \\ = \mathbb {E} \left(\frac {\sigma_ {i} ^ {2}}{n}\right) + \sigma_ {\star} ^ {2} \\ = \frac {\sigma^ {2}}{n} + \sigma_ {\star} ^ {2} \tag {11} \\ \end{array}
$$
Where we can split  $\mathbb{E}\left(\| \sum_{j}(x_{j}^{i} - \mu)\|^{2}\right)$  because of the tower law of expectation and using that  $\mathbb{E}(x_j^i) = \mu^i$  conditional on the value of  $\mu^i$ ; and the squared norm of the last sum factors, as all  $x_{j}^{i} - \mu^{i}$  terms are independent and have zero mean such that
$$
\begin{array}{l} \mathbb {E} \| \sum_ {j} (x _ {j} ^ {i} - \mu) \| ^ {2} \\ = \mathbb {E} \| \sum_ {j > 1} \left(x _ {j} ^ {i} - \mu\right) \| ^ {2} + \mathbb {E} \| x _ {1} ^ {i} - \mu \| ^ {2} + 2 \mathbb {E} \left(\left(x _ {1} ^ {i} - \mu\right) ^ {t} \left(\sum_ {j > 1} \left(x _ {j} ^ {i} - \mu\right)\right)\right) \\ = \mathbb {E} \| \sum_ {j > 1} \left(x _ {j} ^ {i} - \mu\right) \| ^ {2} + \mathbb {E} \| x _ {1} ^ {i} - \mu \| ^ {2} + 2 \mathbb {E} \left(\sum_ {d} \left(x _ {1} ^ {i} - \mu\right) _ {d} \left(\sum_ {j > 1} \left(x _ {j} ^ {i} - \mu\right) _ {d}\right)\right) \\ = \mathbb {E} \| \sum_ {j > 1} \left(x _ {j} ^ {i} - \mu\right) \| ^ {2} + \mathbb {E} \| x _ {1} ^ {i} - \mu \| ^ {2} + 2 \sum_ {d} \mathbb {E} \left(\left(x _ {1} ^ {i} - \mu\right) _ {d} \left(\sum_ {j > 1} \left(x _ {j} ^ {i} - \mu\right) _ {d}\right)\right) \\ = \mathbb {E} \| \sum_ {j > 1} \left(x _ {j} ^ {i} - \mu\right) \| ^ {2} + \mathbb {E} \| x _ {1} ^ {i} - \mu \| ^ {2} + 2 \sum_ {d} \mathbb {E} \left(x _ {1} ^ {i} - \mu\right) _ {d} \mathbb {E} \left(\sum_ {j > 1} \left(x _ {j} ^ {i} - \mu\right) _ {d}\right) \\ \end{array}
$$
$$
\begin{array}{l} = \mathbb {E} \| \sum_ {j > 1} \left(x _ {j} ^ {i} - \mu\right) \| ^ {2} + \mathbb {E} \| x _ {1} ^ {i} - \mu \| ^ {2} + 2 \sum_ {d} 0 \mathbb {E} \left(\sum_ {j > 1} \left(x _ {j} ^ {i} - \mu\right) _ {d}\right) \\ = \mathbb {E} \| \sum_ {j > 1} \left(x _ {j} ^ {i} - \mu\right) \| ^ {2} + \mathbb {E} \| x _ {1} ^ {i} - \mu \| ^ {2} \\ \end{array}
$$
and so on, allowing us to inductively factor the sum.
Now, to compute the terms that depend on  $Y$ , note that by definition  $\theta^s = \frac{1}{N} \sum_{i=1}^{N} m^i$  and therefore:
$$
\begin{array}{l} Y = \frac {1}{N} \left(N \theta^ {s} - m ^ {i} + \bar {x} ^ {i}\right) - \mu \\ = \frac {1}{N} \left(\sum_ {j = 1} ^ {N} m ^ {j} - m ^ {i} + \bar {x} ^ {i}\right) - \mu \\ = \frac {1}{N} \sum_ {j \neq i} m ^ {j} + \frac {1}{N} \bar {x} ^ {i} - \mu \\ \end{array}
$$
Define random variables:
$$
Y _ {1} = \frac {1}{N} \sum_ {j \neq i} ^ {N} m ^ {j}, \quad Y _ {2} = \frac {1}{N} \bar {x} ^ {i} - \mu
$$
so that  $Y = Y_{1} + Y_{2}$ . This implies that:
$$
\mathbb {E} \left(Y ^ {t} Z\right) = \mathbb {E} \left(Z ^ {t} \left(Y _ {1} + Y _ {2}\right)\right) = \mathbb {E} \left(Z ^ {t} Y _ {1}\right) + \mathbb {E} \left(Z ^ {t} Y _ {2}\right).
$$
Since the value of  $Y_{1}$  only depends on the data of all players  $j \in \{1, 2, \dots, N\} / \{i\}$  and the value of  $Z$  depends only on the data of player  $i$ , it follows that  $Z$  and  $Y_{1}$  are independent and therefore  $\mathbb{E}(Z^{t}Y_{1}) = \mathbb{E}(Z)^{t}\mathbb{E}(Y_{1})$ .
$$
\begin{array}{l} \mathbb {E} \left(Z ^ {t} Y\right) = \mathbb {E} \left(Z ^ {t} \left(Y _ {1} + Y _ {2}\right)\right) \\ = \mathbb {E} (Z) ^ {t} \mathbb {E} (Y _ {1}) + \mathbb {E} (Z ^ {t} Y _ {2}) \\ = \mathbb {E} \left(\bar {x} ^ {i} - \mu\right) ^ {t} \mathbb {E} \left(\frac {1}{N} \sum_ {j \neq i} m ^ {j}\right) + \mathbb {E} \left(\left(\bar {x} ^ {i} - \mu\right) ^ {t} \left(\frac {1}{N} \bar {x} ^ {i} - \mu\right)\right) \\ = 0 + \frac {1}{N} \mathbb {E} \left(\| \bar {x} ^ {i} \| ^ {2}\right) - \frac {1}{N} \mu^ {t} \mathbb {E} \left(\bar {x} ^ {i}\right) - \mu^ {t} \mathbb {E} \left(\bar {x} ^ {i}\right) + \| \mu \| ^ {2} \\ = \frac {1}{N} \mathbb {E} \left(\| \bar {x} ^ {i} \| ^ {2}\right) - \frac {1}{N} \mu^ {t} \mu - \mu^ {t} \mu + \| \mu \| ^ {2} \\ = \frac {1}{N} \mathbb {E} \left(\left\| \bar {x} ^ {i} \right\| ^ {2}\right) - \frac {1}{N} \| \mu \| ^ {2} \\ = \frac {1}{N} \mathbb {E} \left(\| \frac {1}{n} \sum_ {j} x _ {j} ^ {i} \| ^ {2}\right) - \frac {1}{N} \| \mu \| ^ {2} \\ = \frac {1}{N} \mathbb {E} \left(\| \left(\frac {1}{n} \sum_ {j} x _ {j} ^ {i} - \mu\right) + \mu \| ^ {2}\right) - \frac {1}{N} \| \mu \| ^ {2} \\ = \frac {1}{N} \mathbb {E} \left(\| \left(\frac {1}{n} \sum_ {j} x _ {j} ^ {i} - \mu\right) \| ^ {2} + \| \mu \| ^ {2}\right) - \frac {1}{N} \| \mu \| ^ {2} \\ = \frac {1}{N} \mathbb {E} \left(\| \left(\frac {1}{n} \sum_ {j} x _ {j} ^ {i} - \mu\right) \| ^ {2}\right) \\ \end{array}
$$
$$
= \frac {\sigma^ {2}}{N n} + \frac {\sigma_ {\star} ^ {2}}{N} \tag {12}
$$
Where the squared norm of the sum factors as all  $x_{i}^{j} - \mu$  terms have zero mean, while  $\mu$  is a constant and the last equality follows from the computations in (11).
Now, by definition  $m^j = \bar{x}^j + \alpha^j(x^j)\xi^j + b^j(x^j)$  for every  $j \in [N]$ , where the  $\xi^j$  are random variables with zero mean and unit "variance"  $\mathbb{E}(\|\xi^j\|^2)$ . Note that because  $Y_1$  and  $Y_2$  depend only on the data of all players  $j \in \{1, 2, \dots, N\} / \{i\}$  and on the data of player  $i$  respectively, they are independent and we get:
$$
\begin{array}{l} \mathbb {E} \left(\| Y ^ {2} \|\right) = \mathbb {E} \left(\left\| Y _ {1} + Y _ {2} \right\| ^ {2}\right) \\ = \mathbb {E} \left(\left\| Y _ {1} \right\| ^ {2}\right) + \mathbb {E} \left(\left\| Y _ {2} \right\| ^ {2}\right) + 2 \mathbb {E} \left(Y _ {1}\right) ^ {t} \mathbb {E} \left(Y _ {2}\right) \tag {13} \\ \end{array}
$$
In order to calculate the value of  $\mathbb{E}\left(Y^2\right)$  we have to compute  $\mathbb{E}(Y_1),\mathbb{E}(Y_2),\mathbb{E}\left(\| Y_1\| ^2\right)$  and  $\mathbb{E}\left(\| Y_2\| ^2\right)$ . We have:
$$
\begin{array}{l} \mathbb {E} \left(Y _ {1}\right) = \mathbb {E} \left(\frac {1}{N} \sum_ {j \neq i} \left(\bar {x} ^ {j} + \alpha^ {j} \left(x ^ {j}\right) \xi^ {j} + b ^ {j} \left(x ^ {j}\right)\right)\right) \\ = \frac {1}{N} \sum_ {j \neq i} \mathbb {E} (\bar {x} ^ {j}) + \frac {1}{N} \sum_ {j \neq i} \mathbb {E} \left(\alpha^ {j} \left(x ^ {j}\right)\right) \mathbb {E} (\xi_ {j}) + \frac {1}{N} \sum_ {j \neq i} \mathbb {E} \left(b ^ {j} \left(x ^ {j}\right)\right) \\ = \frac {N - 1}{N} (\mu) + \frac {1}{N} \sum_ {j \neq i} \mathbb {E} b ^ {j} (x ^ {j}), \\ \end{array}
$$
since  $\mathbb{E}(\xi^j) = 0$  and  $\xi^j$  is independent from all other variables. In addition,
$$
\mathbb {E} \left(Y _ {2}\right) = \mathbb {E} \left(\frac {1}{N} \bar {x} ^ {i} - \mu\right) = \frac {1}{N} \mathbb {E} \left(\bar {x} ^ {i}\right) - \mu = \frac {(1 - N)}{N} \mu
$$
$$
\begin{array}{l} \mathbb {E} \left(\| Y _ {1} \| ^ {2}\right) = \mathbb {E} \left(\| \frac {1}{N} \sum_ {j \neq i} \left(\bar {x} ^ {j} + \alpha^ {j} (x ^ {j}) \xi^ {j} + b ^ {j} (x ^ {j})\right) \| ^ {2}\right) \\ = \mathbb {E} \left(\| \frac {1}{N} \sum_ {j \neq i} \bar {x} ^ {j} + b ^ {j} (x ^ {j}) \| ^ {2} + \| \frac {1}{N} \sum_ {j \neq i} \alpha^ {j} (x ^ {j}) \xi^ {j} \| ^ {2}\right) \\ = \frac {1}{N ^ {2}} \mathbb {E} \left(\| \sum_ {j \neq i} \bar {x} ^ {j} + \sum_ {j \neq i} b ^ {j} (x ^ {j}) \| ^ {2}\right) + \frac {1}{N ^ {2}} \sum_ {j \neq i} \mathbb {E} \left(\alpha^ {j} (x ^ {j}) ^ {2}\right) \\ = \frac {1}{N ^ {2}} \mathbb {E} \left(\| \sum_ {j \neq i} (\bar {x} ^ {j} - \mu + b ^ {j} (x ^ {j})) + (N - 1) \mu \| ^ {2}\right) + \frac {1}{N ^ {2}} \sum_ {j \neq i} \mathbb {E} \left(\alpha^ {j} (x ^ {j}) ^ {2}\right) \\ = \frac {1}{N ^ {2}} \mathbb {E} \left(\| \sum_ {j \neq i} \bar {x} ^ {j} - \mu \| ^ {2} + \| \sum_ {j \neq i} b ^ {j} (x ^ {j}) + (N - 1) \mu \| ^ {2}\right) + \frac {1}{N ^ {2}} \sum_ {j \neq i} \mathbb {E} \left(\alpha^ {j} (x ^ {j}) ^ {2}\right) \\ = \frac {1}{N ^ {2}} \mathbb {E} \left(\| \sum_ {j \neq i} \sum_ {k} \left(\frac {1}{n} x _ {k} ^ {j} - \mu\right) \| ^ {2}\right) + \frac {1}{N ^ {2}} \mathbb {E} \| (N - 1) \mu + \sum_ {j \neq i} b ^ {j} \left(x ^ {j}\right) \| ^ {2} + \frac {1}{N ^ {2}} \sum_ {j \neq i} \mathbb {E} \left(\alpha^ {j} \left(x ^ {j}\right) ^ {2}\right) \\ = \frac {1}{N ^ {2}} \mathbb {E} \left(\| \frac {1}{n} \left(\sum_ {j \neq i} \sum_ {k} x _ {k} ^ {j} - \mu\right) \| ^ {2}\right) + \frac {1}{N ^ {2}} \mathbb {E} \| (N - 1) \mu + \sum_ {j \neq i} b ^ {j} (x ^ {j}) \| ^ {2} + \frac {1}{N ^ {2}} \sum_ {j \neq i} \mathbb {E} (\alpha^ {j} (x ^ {j}) ^ {2}) \\ = \frac {1}{N ^ {2}} \mathbb {E} \left(\sum_ {j \neq i} \| \frac {1}{n} \left(\sum_ {k} x _ {k} ^ {j} - \mu\right) \| ^ {2}\right) + \frac {1}{N ^ {2}} \mathbb {E} \| (N - 1) \mu + \sum_ {j \neq i} b ^ {j} \left(x ^ {j}\right) \| ^ {2} + \frac {1}{N ^ {2}} \sum_ {j \neq i} \mathbb {E} \left(\alpha^ {j} \left(x ^ {j}\right) ^ {2}\right) \\ \end{array}
$$
$$
= \frac {(N - 1)}{N ^ {2}} \left(\frac {\sigma^ {2}}{n} + \sigma_ {\star} ^ {2}\right) + \frac {1}{N ^ {2}} \mathbb {E} \| (N - 1) \mu + \sum_ {j \neq i} b ^ {j} (x ^ {j}) \| ^ {2} + \frac {1}{N ^ {2}} \sum_ {j \neq i} \mathbb {E} (\alpha^ {j} (x ^ {j}) ^ {2})
$$
Where the first squared norms factors because
$$
\mathbb {E} [ <   \frac {1}{N} \sum_ {j \neq i} \bar {x} ^ {j} + b ^ {j} (x ^ {j}), \alpha^ {j} (x ^ {j}) \xi^ {j} > ] = \mathbb {E} [ <   \alpha^ {j} (x ^ {j}) \frac {1}{N} \sum_ {j \neq i} (\bar {x} ^ {j} + b ^ {j} (x ^ {j})), \xi^ {j} > ] = 0
$$
as the  $\xi_{j}$  are independent of all other variables and have zero mean, the  $\mathbb{E}(\| \sum_{j\neq i}\alpha^{j}(x^{j})\xi^{j}\|^{2})$  term factors because all  $\alpha^j (x^j)$  and  $\xi_{j}$  are independent (and the  $\xi_{j}$  have zero mean), the  $b^{j}(x^{j})$  terms factor because  $\mathbb{E} < \bar{x}^j -\mu ,b^j (x^j) > = 0$  by assumption and because the  $\bar{x}^j -\mu$  have zero mean and are independent of  $b^{i}(x^{i})$  for  $j\neq i$ . The last squared norm factors because all  $\sum_{k}x_{j}^{k} - \mu$  terms are independent and have zero mean.