SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\Pr \left(\lim  _ {m \longrightarrow \infty} \sup  _ {\log \log (m)) ^ {1 / 2}} \frac {\sqrt {m} | F _ {m} - F |}{(2 \log \log (m)) ^ {1 / 2}} = c \leq \frac {1}{2}\right) = 1.
$$
A consequence of this is that for any  $\delta > 0$ , there exists an  $M_1$  such that for all  $m \geq M_1$ :
$$
\Pr \left(| F _ {m} - F | \leq \frac {(2 \log \log (m)) ^ {1 / 2}}{\sqrt {m}}\right) \geq 1 - \delta . \tag {5}
$$
Since  $F$  is continuously differentiable it is continuous, and since  $\frac{dF}{dx} > 0$  on  $(0,1)$ $F$  is strictly increasing and so its inverse  $F^{-1}$  is continuous and well-defined on  $[0,1]$  also. Since  $F^{-1}$  is
continuous and bounded on a closed interval, it is uniformly continuous; this means that for all  $\varepsilon' > 0$  there is some  $\delta' > 0$  such that for all  $b \in [0,1]$ , if  $|y - b| < \delta'$  then  $|F^{-1}(y) - F^{-1}(b)| < \varepsilon'$ .
Next, for given  $m$ , consider the order statistic  $X_{j}$ . We note that the  $j^{th}$  order statistic is equivalent to  $F_{m}^{-1}\left(\frac{j}{m}\right)$ . For quantiles of the form  $b = \frac{j}{m}$ , we will suppose that  $F_{m}^{-1}$  is well-defined and  $F_{m}^{-1}(F_{m}(b)) = b$ . For such a point  $b$ , let  $a := F^{-1}(b)$  and let  $a_{m} := F_{m}^{-1}(b)$ .
We now introduce another condition on  $m$ . Given  $s_{\varepsilon}$ , choose  $M_2$  such that  $\frac{(2\log\log(M_2))^{1/2}}{\sqrt{M_2}} \leq \delta'$ , where  $\delta'$  is the threshold given by the uniform continuity of  $F^{-1}$  with the choice of  $\varepsilon' = s_{\varepsilon}$ . (Since this is a decreasing function of  $M_2$ , this inequality will hold for all  $m \geq M_2$ .) By Equation (5) applied to the value  $a_m$ , we have that
$$
\left| F \left(a _ {m}\right) - F _ {m} \left(a _ {m}\right) \right| \leq \delta^ {\prime}
$$
with probability at least  $1 - \delta$ . If this is the case, then by uniform continuity,
$$
\left| F ^ {- 1} \left(F \left(a _ {m}\right)\right) - F ^ {- 1} \left(F _ {m} \left(a _ {m}\right)\right) \right| \leq s _ {\varepsilon}
$$
$$
\left| F _ {m} ^ {- 1} (b) - F ^ {- 1} (b) \right| \leq s _ {\varepsilon},
$$
where this last inequality follows by applying the definition of  $a_{m}$ .
Using the fact that  $X_{(j)}$  is supported on  $(0,1]$ , this gives that the variance of  $X_{(j)}$  can be bounded as
$$
\operatorname {V a r} (X _ {(j)}) \leq \left(\frac {(2 \log \log (m)) ^ {1 / 2}}{c \sqrt {m}}\right) ^ {2} + \delta \leq s _ {\varepsilon} ^ {2},
$$
where we take  $\delta = \frac{s_{\varepsilon}^2}{2}$  and sufficiently large  $m \geq M = \max(M_1, M_2)$ .
Proof of Theorem 4.2. Recall that  $\mathrm{ewmr}(\sigma)$  is the expected welfare maximizing alternative. Let  $n \geq n_0$ , where  $n_0$  is the solution to the equation  $\frac{\log(n)}{\sqrt{\mu n}} = \varepsilon_0$ . By Lemma F.1, the expected welfare maximizing alternative satisfies  $\mathbb{E}[\mathrm{dsw}(\mathrm{ewmr}, \sigma)] \geq n\mu$ . Furthermore, note that the social welfare of any alternative  $j$  can be represented as  $\mathrm{dsw}(j, \sigma) = \sum_{i=1}^{n} u_{ij}$  where  $u_{ij}$  are  $n$  independent random variables that are supported on  $(0,1]$ . Therefore, we can apply Hoeffding's theorem to get the following:
$$
\begin{array}{l} \Pr (\mathrm {d s w} (\mathrm {e w m r}, \sigma) \leq (1 - \varepsilon) \mathbb {E} [ \mathrm {d s w} (\mathrm {e w m r}, \sigma) ]) \\ = \Pr (\mathrm {d s w} (\mathrm {e w m r}, \boldsymbol {\sigma}) - \mathbb {E} [ \mathrm {d s w} (\mathrm {e w m r}, \boldsymbol {\sigma}) ] \leq - \varepsilon \mathbb {E} [ \mathrm {d s w} (\mathrm {e w m r}, \boldsymbol {\sigma}) ]) \\ \leq \Pr (\mathrm {d s w} (\mathrm {e w m r}, \boldsymbol {\sigma}) - \mathbb {E} [ \mathrm {d s w} (\mathrm {e w m r}, \boldsymbol {\sigma}) ] \leq - \varepsilon n \mu) \\ \leq e ^ {- 2 (\varepsilon n \mu) ^ {2} / n} \\ \leq e ^ {- 2 \varepsilon^ {2} n \mu^ {2}}. \\ \end{array}
$$
For any  $j\neq \mathrm{ewmr}(\pmb {\sigma})$  , we must have  $\mathbb{E}[\mathrm{dsw}(j,\pmb {\sigma})]\leq \mathbb{E}[\mathrm{dsw}(\mathrm{ewmr},\pmb {\sigma})]$  . Therefore:
$$
\begin{array}{l} \Pr (\mathrm {d s w} (j, \sigma) \geq (1 + \varepsilon) \mathbb {E} [ \mathrm {d s w} (\mathrm {e w m r}, \sigma) ]) \\ \leq \Pr (\mathrm {d s w} (j, \sigma) \geq \mathbb {E} [ \mathrm {d s w} (j, \sigma) ] + \varepsilon \mathbb {E} [ \mathrm {d s w} (\mathrm {e w m r}, \sigma) ]) \\ = \Pr (\mathrm {d s w} (j, \boldsymbol {\sigma}) - \mathbb {E} [ \mathrm {d s w} (j, \boldsymbol {\sigma}) ] \geq \varepsilon \mathbb {E} [ \mathrm {d s w} (\mathrm {e w m r}, \boldsymbol {\sigma}) ]) \\ \leq \Pr (\mathrm {d s w} (j, \sigma) - \mathbb {E} [ \mathrm {d s w} (j, \sigma) ] \geq \varepsilon n \mu) \\ \leq e ^ {- 2 (\varepsilon n \mu) ^ {2} / n} \\ \leq e ^ {- 2 \varepsilon^ {2} n \mu^ {2}}. \\ \end{array}
$$
Let  $\mathcal{E}$  be the event that the welfare of the  $\mathrm{ewmr}(\sigma)$  alternative is not too small, and that no other alternative has unusually large welfare;  $\mathrm{dsw}(\mathrm{ewmr},\sigma)\geq (1 - \varepsilon)\mathbb{E}[\mathrm{dsw}(\mathrm{ewmr},\sigma)]$  and that for all  $j\neq \mathrm{ewmr}(\sigma)$ ,  $\mathrm{dsw}(j,\sigma)\leq (1 + \varepsilon)\mathbb{E}[\mathrm{dsw}(\mathrm{ewmr},\sigma)]$ . Mathematically,
$$
\mathcal {E} = \left\{\mathrm {d s w} (\mathrm {e w m r}, \boldsymbol {\sigma}) \geq (1 - \varepsilon) \mathbb {E} [ \mathrm {d s w} (\mathrm {e w m r}, \boldsymbol {\sigma}) ] \right\} \cap \left\{\forall j: \mathrm {d s w} (j, \boldsymbol {\sigma}) \leq (1 + \varepsilon) \mathbb {E} [ \mathrm {d s w} (\mathrm {e w m r}, \boldsymbol {\sigma}) ] \right\}.
$$
Using the two results above, the probability of the event  $\mathcal{E}$  can be lower bounded by the following:
$$
\operatorname * {P r} (\mathcal {E}) \geq 1 - e ^ {- 2 \varepsilon^ {2} n \mu^ {2}} - e ^ {- 2 \varepsilon^ {2} n \mu^ {2}} = 1 - 2 e ^ {- 2 \varepsilon^ {2} n \mu^ {2}}
$$
Fix any  $j$ . Under event  $\mathcal{E}$ , the distributional distortion of this alternative satisfies the following inequality:
$$
\begin{array}{l} \operatorname {d d i s t} (j, \boldsymbol {\sigma}) = \frac {\operatorname {d s w} (j , \boldsymbol {\sigma})}{\max  _ {j} \operatorname {d s w} (j , \boldsymbol {\sigma})} \\ \leq \frac {(1 + \varepsilon) \mathbb {E} [ \mathrm {d s w} (\mathrm {e w m r} , \boldsymbol {\sigma}) ]}{\max  _ {j} \mathrm {d s w} (j , \boldsymbol {\sigma})} \\ \leq \frac {1 + \varepsilon}{1 - \varepsilon} \cdot \frac {\operatorname {d s w} (\operatorname {e w m r} , \boldsymbol {\sigma})}{\max  _ {j} \operatorname {d s w} (j , \boldsymbol {\sigma})} \\ = \frac {1 + \varepsilon}{1 - \varepsilon} \cdot \operatorname {d d i s t} (\operatorname {e w m r}, \sigma) \\ \end{array}
$$
Under event  $\neg \mathcal{E}$ , the distributional distortion of  $j$  is at most 1 by definition. Therefore, we can upper bound the expected distributional distortion of  $j$  relative to the expected distributional distortion of  $\mathrm{ewmr}(\sigma)$  as follows:
$$
\begin{array}{l} \mathbb {E} [ \mathrm {d d i s t} (j, \sigma) ] = \Pr (\mathcal {E}) \cdot \mathbb {E} [ \mathrm {d d i s t} (j, \sigma) | \mathcal {E} ] + \Pr (\neg \mathcal {E}) \cdot \mathbb {E} [ \mathrm {d d i s t} (j, \sigma) | \neg \mathcal {E} ] \\ \leq \frac {1 + \varepsilon}{1 - \varepsilon} \cdot \Pr (\mathcal {E}) \cdot \mathbb {E} [ \mathrm {d d i s t} (\operatorname {e w m r}, \sigma) | \mathcal {E} ] + 2 e ^ {- 2 \varepsilon^ {2} n \mu^ {2}} \\ \leq \frac {1 + \varepsilon}{1 - \varepsilon} \cdot \left(\Pr (\mathcal {E}) \cdot \mathbb {E} [ \mathrm {d d i s t} (\mathrm {e w m r}, \boldsymbol {\sigma}) | \mathcal {E} ] + \Pr (\neg \mathcal {E}) \cdot \mathbb {E} [ \mathrm {d d i s t} (\mathrm {e w m r}, \boldsymbol {\sigma}) | \neg \mathcal {E} ]\right) + 2 e ^ {- 2 \varepsilon^ {2} n \mu^ {2}} \\ \leq \frac {1 + \varepsilon}{1 - \varepsilon} \cdot \mathbb {E} [ \mathrm {d d i s t} (\mathrm {e w m r}, \boldsymbol {\sigma}) ] + 2 e ^ {- 2 \varepsilon^ {2} n \mu^ {2}} \cdot \frac {1}{\mu} \mathbb {E} [ \mathrm {d d i s t} (\mathrm {e w m r}, \boldsymbol {\sigma}) ] \\ \leq \left(\frac {1 + \varepsilon}{1 - \varepsilon} + \frac {2}{\mu} \cdot e ^ {- 2 \varepsilon^ {2} n \mu}\right) \mathbb {E} [ \mathrm {d d i s t} (\mathrm {e w m r}, \boldsymbol {\sigma}) ]. \\ \end{array}
$$
Take  $\varepsilon = \frac{\varepsilon_0}{8} \leq \frac{1}{2}$ , which is equal to  $\frac{\log(n)}{2\sqrt{\mu n}}$  by our choice of  $n$ . Then:
$$
\begin{array}{l} \frac {1 + \varepsilon}{1 - \varepsilon} + \frac {2}{\mu} e ^ {- 2 \varepsilon^ {2} n \mu} \leq 1 + 4 \varepsilon + \frac {2}{\mu} e ^ {- 2 \varepsilon^ {2} n \mu} \\ \leq 1 + \frac {\varepsilon_ {0}}{2} + \frac {2}{\mu} \sqrt {n} ^ {- \log (n)} \\ \leq 1 + \varepsilon_ {0}. \\ \end{array}
$$
We have shown that for all  $j$
$$
\mathbb {E} [ \mathrm {d d i s t} (j, \sigma) ] \leq \left(1 + \varepsilon_ {0}\right) \mathbb {E} [ \mathrm {d d i s t} (\mathrm {e w m r}, \sigma) ],
$$
$$
\mathbb {E} [ \mathrm {d d i s t} (\mathrm {e d m r}, \boldsymbol {\sigma}) ] = \max  _ {j} \mathbb {E} [ \mathrm {d d i s t} (j, \boldsymbol {\sigma}) ] \leq (1 + \varepsilon_ {0}) \mathbb {E} [ \mathrm {d d i s t} (\mathrm {e w m r}, \boldsymbol {\sigma}) ].
$$
We have shown that the expected social welfare maximizing rule is an  $\varepsilon$ -approximation for the expected distributional distortion maximizing rule for sufficiently large  $n$ . We now show that the same holds for sufficiently large  $m$  when  $\mathcal{D}$  is continuous. The proof is very similar to the above, except that we use Chebyshev's inequality instead of Hoeffding's inequality.
Note that  $\mathrm{dsw}(j,\pmb {\sigma})$  is the sum of  $n$  independent random variables, and let  $s_{\varepsilon}^{2}$  be an upper bound on the variance of any one of these random variables. We can use Lemma F.2 to find a  $m_0$  that is suitably large. Specifically, we choose  $m_0$  such that  $m\geq m_0$  implies  $s_\varepsilon^2\leq \frac{\varepsilon_0^3\mu^3n}{256}$ .
Because the variance of a sum of independent random variables is the sum of the variances, we can bound  $\operatorname{Var}(\mathrm{dsw}(j,\sigma))\leq ns_{\varepsilon}^{2}$ . Applying Chebyshev's inequality then yields:
$$
\begin{array}{l} \Pr (\mathrm {d s w} (\operatorname {e w m r}, \sigma) \leq (1 - \varepsilon) \mathbb {E} [ \mathrm {d s w} (\operatorname {e w m r}, \sigma) ]) \\ = \Pr (\mathrm {d s w} (\text {e w m r}, \boldsymbol {\sigma}) - \mathbb {E} [ \mathrm {d s w} (\text {e w m r}, \boldsymbol {\sigma}) ] \leq - \varepsilon \mathbb {E} [ \mathrm {d s w} (\text {e w m r}, \boldsymbol {\sigma}) ]) \\ \leq \Pr (\mathrm {d s w} (\mathrm {e w m r}, \sigma) - \mathbb {E} [ \mathrm {d s w} (\mathrm {e w m r}, \sigma) ] \leq - \varepsilon n \mu) \\ \leq \frac {n s _ {\varepsilon} ^ {2}}{(\varepsilon n \mu) ^ {2}} \\ = \frac {s _ {\varepsilon} ^ {2}}{\varepsilon^ {2} \mu^ {2} n}. \\ \end{array}
$$
For any  $j \neq \mathrm{ewmr}(\sigma)$ , we must have  $\mathbb{E}[\mathrm{dsw}(j, \sigma)] \leq \mathbb{E}[\mathrm{dsw}(\mathrm{ewmr}, \sigma)]$ , and therefore:
$$
\begin{array}{l} \Pr (\mathrm {d s w} (j, \sigma) \geq (1 + \varepsilon) \mathbb {E} [ \mathrm {d s w} (\mathrm {e w m r}, \sigma) ]) \\ \leq \Pr (\mathrm {d s w} (j, \boldsymbol {\sigma}) \geq \mathbb {E} [ \mathrm {d s w} (j, \boldsymbol {\sigma}) ] + \varepsilon \mathbb {E} [ \mathrm {d s w} (\mathrm {e w m r}, \boldsymbol {\sigma}) ]) \\ = \Pr (\mathrm {d s w} (j, \boldsymbol {\sigma}) - E [ \mathrm {d s w} (j, \boldsymbol {\sigma}) ] \geq \varepsilon \mathbb {E} [ \mathrm {d s w} (\mathrm {e w m r}, \boldsymbol {\sigma}) ]) \\ \leq \Pr (\mathrm {d s w} (j, \sigma) - \mathbb {E} [ \mathrm {d s w} (j, \sigma) ] \geq \varepsilon n \mu) \\ \leq \frac {n s _ {\varepsilon} ^ {2}}{(\varepsilon n \mu) ^ {2}} \\ = \frac {s _ {\varepsilon} ^ {2}}{\varepsilon^ {2} \mu^ {2} n}. \\ \end{array}
$$
Using the same logic as in the first part of the proof, this means that we have
$$
\mathbb {E} [ \mathrm {d d i s t} (j, \boldsymbol {\sigma}) ] \leq \left(\frac {1 + \varepsilon}{1 - \varepsilon} + \frac {2 s _ {\varepsilon} ^ {2}}{\varepsilon^ {2} \mu^ {3} n}\right) \mathbb {E} [ \mathrm {d d i s t} (\mathrm {e w m r}, \boldsymbol {\sigma}) ]
$$
We can upper bound this by taking  $\varepsilon = \frac{\varepsilon_0}{8}$  to get:
$$
\begin{array}{l} \frac {1 + \varepsilon}{1 - \varepsilon} + \frac {2 s _ {\varepsilon} ^ {2}}{\varepsilon^ {2} \mu^ {3} n} \leq 1 + 4 \varepsilon + \frac {2 s _ {\varepsilon} ^ {2}}{\varepsilon^ {2} \mu^ {3} n} \\ \leq 1 + \frac {\varepsilon_ {0}}{2} + \frac {1 2 8 s _ {\varepsilon} ^ {2}}{\varepsilon_ {0} ^ {2} \mu^ {3} n} \\ \leq 1 + \varepsilon_ {0}. \\ \end{array}
$$
As in the first section of the proof, this implies that:
G Proof of Theorem 5.2
Lemma G.1 Let  $\mathcal{D} \sim \operatorname{Bern}(p)$  for  $0 < \delta < p < 1$ . For  $m$  i.i.d. draws from  $\mathcal{D}$ , the expected welfare of order statistics satisfy the following inequality:
- For  $k \leq pm - \delta m$ ,
$$
\mathbb {E} [ X _ {(k)} ] \leq e ^ {- 2 \delta^ {2} m}.
$$
- For  $k \geq pm + \delta m$ ,
$$
\mathbb {E} [ X _ {(k)} ] \geq 1 - e ^ {- 2 \delta^ {2} m}.
$$
Proof. Fix  $k \leq pm - \delta m$ . By Hoeffding's inequality, the probability that  $X_{(k)}$  is 1 is bounded by:
$$
\begin{array}{l} \Pr (X _ {(k)} = 1) = \Pr (\operatorname {B i n} (m, p) \geq p m + \delta m) \\ = \Pr (\operatorname {B i n} (m, p) - p m \geq \delta m) \\ \leq e ^ {- 2 \delta^ {2} m}. \\ \end{array}
$$
The only two values that  $X_{(k)}$  can take are 0 and 1, so this implies that:
The other direction for  $k \geq pm + \delta m$  follows from a symmetric argument.
Proof of Theorem 5.2. Fix  $0 < \alpha \leq 0.5$ . Let  $\mu = \frac{\alpha}{4}$  and  $\delta = \frac{\alpha^2}{4}$ . Further let  $\mathcal{D}_1 = \operatorname{Bern}(\mu - \delta)$  and  $\mathcal{D}_2 = \operatorname{Bern}(\mu)$ . Choose  $m$  such that
$$
e ^ {- 2 m \delta^ {2}} \leq \delta ,
$$
and let  $n = (m - 1)$ . Define the preference profile  $\sigma$  as follows. Let alternative 1 have rank  $(\mu - \delta)m$  for all  $n$  voters. Let all of the other rankings be evenly divided among the other  $m - 1$  voters, so that for every  $j \in [2:m]$ , alternative  $j$  is ranked  $k$  by 1 voters for every  $k \in [1:m]$  except for  $k = (\mu - \delta)m$ .
By Lemma G.1, under distribution  $\mathcal{D}_1$ , alternative 1 will have expected welfare of:
$$
\begin{array}{l} \mathbb {E} [ \mathrm {d s w} (1, \sigma) ] \leq n e ^ {- 2 m \delta^ {2}} \tag {6} \\ \leq n \delta , \\ \end{array}
$$
by the choice of  $m$ . By Lemma F.1 there exists at least one alternative  $j$  with expected social welfare of at least:
$$
\mathbb {E} [ \mathrm {d s w} (j, \sigma) ] \geq n \mu . \tag {7}
$$
Similarly for  $\mathcal{D}_2$ , by Lemma G.1 and the choice of  $m$ , the social welfare for alternative 1 is
$$
\begin{array}{l} \mathbb {E} [ \mathrm {d s w} (1, \sigma) ] \geq (1 - e ^ {- 2 m \delta^ {2}}) n (8) \\ \geq (1 - \delta) n. (9) \\ \end{array}
$$
For  $j \neq 1$ , let  $N_{j}$  be the set of voters giving  $j$  rank  $(\mu + \delta)m$  or lower (better). Note that there are a total of  $(\mu + \delta)nm$  such rankings across all voters, and  $n$  of those positions are occupied by alternative 1. Furthermore, note that the social welfare of any alternative  $j$  can be represented as  $\mathrm{dsw}(j, \sigma) = \sum_{i=1}^{n} u_{ij}$  where  $u_{ij}$  are  $n$  i.i.d. random variables that are supported on  $[0,1]$ . Therefore, for any  $j \neq 1$ ,
$$
| N _ {j} | \leq \frac {(\mu + \delta) n m - n}{m - 1} \leq n (\mu + \delta).
$$
Using this, we can then conclude that
$$
\begin{array}{l} \mathbb {E} [ \mathrm {d s w} (j, \sigma) ] = \sum_ {i = 1} ^ {n} E [ u _ {i j} ] \\ = \sum_ {i \in N _ {j}} \mathbb {E} \left[ u _ {i j} \right] + \sum_ {i \notin N _ {j}} \mathbb {E} \left[ u _ {i j} \right] \\ \leq 1 \cdot | N _ {j} | + e ^ {- 2 m \delta^ {2}} \cdot (n - | N _ {j} |) \\ \leq \left| N _ {j} \right| + n e ^ {- 2 m \delta^ {2}} \\ \leq n (\mu + \delta) + n \delta \\ = (\mu + 2 \delta) n. \\ \end{array}
$$
Under distribution  $\mathcal{D}_1$ , by (6) and (7) we have that alternative 1 has expected welfare at most  $\frac{\delta}{\mu} = \alpha$  fraction of the expected welfare of the expected welfare maximizing alternative. Under distribution  $\mathcal{D}_2$ , any alternative  $j \neq 1$  has expected welfare at most (using that  $\alpha, \delta \leq 1/2$ )
$$
\frac {\mu + 2 \delta}{1 - \delta} \leq 2 \mu + 4 \delta \leq \alpha
$$
fraction of the expected welfare of the expected welfare maximizing alternative. Therefore, we can conclude that under this preference profile, no alternative can be chosen that is an  $\alpha$ -SWMR for both  $\mathcal{D}_1$  and  $\mathcal{D}_2$  simultaneously.
Corollary G.2 If the distribution can depend on  $n$ ,  $m$ , then for every  $m$  there exists an  $n$ , a preference profile, and two distributions  $D_1^m$  and  $D_2^m$  such that no rule is  $\frac{\sqrt{\log(m)}}{m^{1/4}}$ -SWMR for both distributions.
Proof. Take  $\alpha = \frac{2\sqrt{\log(m)}}{m^{1/4}}$ . This is valid as  $\delta$  satisfies the necessary  $m$  inequality for this choice of  $\alpha$ .
H Proof of Theorem 5.4
Denote the mean and median of  $\mathcal{D}$  as  $\mu$ . Note that for any  $k \geq \lceil \frac{m}{2} \rceil$ , the expectation of the  $k$ th order statistic from  $m$  i.i.d. draws from a symmetric distribution  $\mathcal{D}$  satisfies the lower bound
$$
\mathbb {E} \left[ X _ {(k)} \right] \geq \mu . \tag {10}
$$
Similarly, for any  $k < \lceil \frac{m}{2} \rceil$ ,
$$
\mathbb {E} \left[ X _ {(k)} \right] \leq \mu . \tag {11}
$$
Among all voters and alternatives, there are at least  $\left\lceil \frac{mn}{2} \right\rceil$  pairs of voters and alternatives where the voter ranked the alternative in the top  $\left\lceil \frac{m}{2} \right\rceil$  of alternatives, regardless of the preference profile  $\sigma$ . Therefore, by the pigeonhole principle, there must exist at least one alternative that is ranked in the top  $\left\lceil \frac{m}{2} \right\rceil$  by at least  $\frac{1}{m} \left\lceil \frac{mn}{2} \right\rceil$  voters.
Define  $\operatorname{Top-Half}(\sigma)$  as the alternative chosen by the Top-Half approval scoring rule and let  $z$  be the number of voters that rank  $\operatorname{Top-Half}(\sigma)$  in the top  $\lceil \frac{m}{2} \rceil$ . By definition  $\operatorname{Top-Half}(\sigma)$  is the alternative that is ranked in the top  $\lceil \frac{m}{2} \rceil$  by the most voters, and so  $z \geq \frac{1}{m} \lceil \frac{mn}{2} \rceil$ . Using (10), the expected welfare of  $\operatorname{Top-Half}(\sigma)$  can therefore be lower bounded by
$$
\mathbb {E} [ \mathrm {d s w} (\operatorname {T o p - H a l f}, \sigma) ] \geq z \mu .
$$
Note that because  $\mathcal{D}$  is non-negative and symmetric with median  $\mu$ , its support must lie within the range  $[0, 2\mu]$ . By construction, for all  $j \neq \operatorname{Top-Half}(\sigma)$ , alternative  $j$  is ranked in the top  $\lceil \frac{m}{2} \rceil$  by at most  $z$  voters. Furthermore,  $\mathbb{E}[X_{(k)}] \leq \mu$  for all  $k < \lceil \frac{m}{2} \rceil$ . Then for any  $j \neq \operatorname{Top-Half}(\sigma)$ ,
$$
\mathbb {E} [ \mathrm {d s w} (j, \sigma) ] \leq z \cdot 2 \mu + (n - z) \cdot \mu = n \mu + z \mu
$$
Combining these two equations with the fact that  $z \geq \frac{1}{m} \left\lceil \frac{mn}{2} \right\rceil \geq \frac{n}{2}$ , we get the following desired result:
$$
\frac {\mathbb {E} [ \mathrm {d s w} (\mathrm {T o p - H a l f} , \boldsymbol {\sigma}) ]}{\mathbb {E} [ \mathrm {d s w} (\mathrm {e w m r} , \boldsymbol {\sigma}) ]} \geq \frac {z \mu}{n \mu + z \mu} \geq \frac {1}{3}.
$$
I Proof of Theorem 5.5
Proof of Theorem 5.5. Let  $P_{1}$  be the Unif(0,1) distribution and let  $P_{2}$  be the Bernoulli distribution with probability 0.5. Fix  $1 > \alpha_{0} > 0$ . Define  $\alpha = \frac{\alpha_{0}}{4}$ . Finally, take  $m, n$  such that  $\alpha \geq \max \left(\frac{2\log(m)\sqrt{m} + 2}{m}, \frac{\log(m)}{\sqrt{m}}, e^{-2\log(m)^{2}}\right)$ . Note that this is simply a technical choice for this proof.
Define the preference profile  $\sigma$  as follows, for constants (to be determined later)  $p,q$  satisfying  $\frac{1}{2}\leq q < p\leq 1$ . Suppose alternative 1 is ranked  $\frac{m}{2} -\log (m)\sqrt{m}$  by  $pn$  voters and ranked last  $(m^{th})$  by all other voters. Suppose alternative 2 is ranked first by  $qn$  voters and ranked  $\frac{m}{2} +\log (m)\sqrt{m}$  by the remaining  $(1 - q)n$  voters. For simplicity, we will assume that  $pn,qn,\log (m),\sqrt{m},\frac{m}{2}$  are integers, and further that  $n,pn,qn$  are all divisible by  $m - 2$ . Suppose all other rankings are distributed as evenly as possible among the remaining voters.
Suppose that the true underlying distribution is  $P_{1}$ . The expectation of the  $k$ th order statistic of a uniform distribution is  $\frac{k}{m + 1}$ . We then get the following lower bounds on the expected social welfare for alternatives 1 and 2:
$$
\begin{array}{l} \mathbb {E} [ \mathrm {d s w} (1, \boldsymbol {\sigma}) ] = p n \cdot \frac {m / 2 + \log (m) \sqrt {m}}{m + 1} + (1 - p) n \cdot \frac {1}{m + 1} \\ = \frac {p n m / 2 + p n \log (m) \sqrt {m} + (1 - p) n}{m + 1} \\ \leq \frac {(p + \alpha) n m}{2 (m + 1)}, \tag {12} \\ \end{array}
$$
$$
\begin{array}{l} \mathbb {E} [ \mathrm {d s w} (2, \sigma) ] = q n \cdot \frac {m}{m + 1} + (1 - q) n \cdot \frac {\frac {m}{2} - \log (m) \sqrt {m}}{m + 1} \\ = \frac {q n m + (1 - q) n m / 2 - (1 - q) n \log (m) \sqrt {m}}{m + 1} \\ \geq \frac {(1 + q - \alpha) n m}{2 (m + 1)}. \tag {13} \\ \end{array}
$$
We will now consider what happens when the true underlying distribution is  $P_{2}$ . Suppose that this is the case. Then by applying Lemma G.1 with  $\delta = \log (m) / \sqrt{m}$  we can bound the expected social welfare for alternative 1 by:
$$
\begin{array}{l} (p - \alpha) n \leq p n (1 - e ^ {- 2 \log^ {2} (m)}) \\ \leq \mathbb {E} [ \mathrm {d s w} (1, \boldsymbol {\sigma}) ] \\ \leq p n + (1 - p) n e ^ {- 2 \log^ {2} (m)} \\ \leq (p + \alpha) n \tag {14} \\ \end{array}
$$
and similarly for alternative 2 by:
$$
\begin{array}{l} (q - \alpha) n \leq q n (1 - e ^ {- 2 \log^ {2} (m)}) \\ \leq \mathbb {E} [ \mathrm {d s w} (2, \sigma) ] \\ \leq q n + (1 - q) n e ^ {- 2 \log^ {2} (m)} \\ \leq (q + \alpha) n. \tag {15} \\ \end{array}
$$
For the remaining  $m - 2$  alternatives  $j \notin \{1,2\}$ , the remaining expected welfare is divided evenly among them. Note that by the divisibility assumptions stated at the beginning, we can exactly evenly divide up the remaining rankings (and therefore expected welfare) among these  $m - 2$  alternatives. This allows us to upper bound the expected social welfare for the any alternative  $j$  for  $j \notin \{1,2\}$  by:
$$
\begin{array}{l} \mathbb {E} [ \mathrm {d s w} (j, \boldsymbol {\sigma}) ] = \frac {1}{m - 2} \left(\frac {n m}{2} - E [ \mathrm {d s w} (1, \boldsymbol {\sigma}) ] - E [ \mathrm {d s w} (2, \boldsymbol {\sigma}) ]\right) \\ \leq \frac {1}{m - 2} \left(\frac {n m}{2} - (p + q) n + 2 \alpha n\right) \\ \leq \frac {1}{m - 2} \left(\frac {n m}{2} - n\right) + \frac {2 \alpha n}{m - 2} \\ \leq \frac {n}{2} + \alpha n \\ \leq (q + \alpha) n, \tag {16} \\ \end{array}
$$
since  $q \geq 1/2$ . Note a symmetric inequality shows that  $E[\mathrm{dsw}(j, \sigma)] \geq (q - \alpha)n$
If we select alternative 1, then comparing it to alternative 2 under  $P_{1}$  by combining (12) and (13), shows that the expected welfare is at most a  $\frac{p + \alpha}{1 + q - \alpha}$  fraction of the expected welfare maximizing alternative.
If we select an alternative that is not alternative 1, then by combining (14), (15), and (16), we see that the expected welfare is at most a  $\frac{q + \alpha}{p - \alpha}$  fraction of the expected welfare of the expected-welfare-maximizing alternative.
Therefore, we have a construction such that for every alternative, under some distribution the alternative will be at most a  $\frac{q + \alpha}{p - \alpha}$ -EWMR or at most a  $\frac{p + \alpha}{1 + q - \alpha}$ -EWMR. From this, we can solve for  $p, q$  in the following minimization to make the social welfare approximation for either of these options as low as possible.