SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Proof. Let  $\lambda \in (1 / b, 1)$ . It is shown in [47] that Algorithm 2 is  $\left(\frac{\lambda}{1 - e^{-\lambda}}\right)$ -consistent and  $\left(\frac{1}{1 - e^{-(\lambda - 1 / b)}}\right)$ -robust. Even though the upper bound shown in [47] suggests that the consistency bound is achieved only when given an exactly accurate prediction  $y$  of the number of snow days, i.e.  $y = x$ , it suffices to have  $\mathcal{Q} = \mathbb{1}(x \geq b)$  for the algorithm to behave just the same. Therefore, with probability  $p$  we have that the cost of the algorithm is at most  $\left(\frac{\lambda}{1 - e^{-\lambda}}\right) \mathrm{OPT}(x, b)$ , and with probability  $1 - p$  it is at most  $\left(\frac{1}{1 - e^{-(\lambda - 1 / b)}}\right) \mathrm{OPT}(x, b)$ . With a simple function analysis, we have that  $\frac{1}{1 - e^{-u}} \leq \frac{1}{e - 1} + \frac{1}{u}$  for any  $u \in (0, 1)$ . Therefore
$$
\begin{array}{l} \mathbf {E} [ \mathsf {A L G} (x, b, \mathcal {Q}) ] \leq \frac {p \lambda}{1 - e ^ {- \lambda}} + \frac {1 - p}{1 - e ^ {- (\lambda - 1 / b)}} \\ \leq p \lambda \left(\frac {1}{e - 1} + \frac {1}{\lambda}\right) + (1 - p) \left(\frac {1}{e - 1} + \frac {1}{\lambda - 1 / b}\right) \\ = \frac {p \lambda}{e - 1} + \frac {1 - p}{\lambda - 1 / b} + \frac {1}{e - 1} + \left(1 - \frac {1}{e - 1}\right) p \\ = \frac {p (\lambda - 1 / b)}{e - 1} + \frac {1 - p}{(\lambda - 1 / b)} + \frac {1}{e - 1} + \left(1 - \frac {1 - 1 / b}{e - 1}\right) p, \\ \end{array}
$$
The previous expression is minimized when  $\lambda$  satisfies  $\frac{p(\lambda - 1 / b)}{e - 1} = \frac{1 - p}{(\lambda - 1 / b)}$ , i.e. for  $\lambda^{\star} = \frac{1}{b} + \sqrt{(e - 1)\frac{1 - p}{p}}$ . However, we are restrained to choose  $\lambda \in (1 / b, 1)$ , thus the optimal choice of  $\lambda$  is  $\min \{1, \lambda^{\star}\}$ . We have that
$$
\begin{array}{l} \lambda^ {\star} \leq 1 \iff \frac {p (\lambda - 1 / b)}{e - 1} = \frac {1 - p}{(\lambda - 1 / b)} \leq 1 \\ \Longleftrightarrow (e - 1) \frac {1 - p}{p} \leq \left(1 - \frac {1}{b}\right) ^ {2} \\ \Longleftrightarrow \frac {1}{p} \leq 1 + \frac {(1 - 1 / b) ^ {2}}{e - 1} \\ \Longleftrightarrow p \geq \left(1 + \frac {(1 - 1 / b) ^ {2}}{e - 1}\right) ^ {- 1}. \\ \end{array}
$$
Therefore, if  $p \geq \left(1 + \frac{(1 - 1 / b)^2}{e - 1}\right)^{-1}$ , then with  $\lambda = \lambda^{\star}$  we have
$$
\begin{array}{l} \mathbf {E} [ \mathsf {A L G} (x, b, \mathcal {Q}) ] \leq \sqrt {\frac {p (1 - p)}{e - 1}} + \sqrt {\frac {p (1 - p)}{e - 1}} + + \frac {1}{e - 1} + \left(1 - \frac {1 - 1 / b}{e - 1}\right) p \\ = \frac {1}{e - 1} + 2 \sqrt {\frac {p (1 - p)}{e - 1}} + \left(1 - \frac {1 - 1 / b}{e - 1}\right) p, \\ \end{array}
$$
and if  $p < \left(1 + \frac{(1 - 1 / b)^2}{e - 1}\right)^{-1}$ , then with  $\lambda = 1$  we obtain
$$
\begin{array}{l} \mathbf {E} [ \mathsf {A L G} (x, b, \mathcal {Q}) ] \leq \frac {p}{e - 1} + \frac {1 - p}{1 - 1 / b} + \frac {1}{e - 1} + \left(1 - \frac {1}{e - 1}\right) p \\ = \frac {1}{e - 1} + \frac {1}{1 - 1 / b} + \left(1 - \frac {1}{1 - 1 / b}\right) p \\ = \frac {1}{e - 1} + \frac {b}{b - 1} - \frac {1}{b - 1} p \\ = \frac {1}{e - 1} + \frac {b - p}{b - 1}. \\ \end{array}
$$
B Secretary problem with  $B$  predictions
B.1 Proof of Theorem 3.1
Proof. Since the secretary problem is purely ordinal, and since the applicants arrive in a uniformly random order, the success probability of ADATHRESH does not depend on the values  $x_{1}, \ldots, x_{N}$ , but only on  $B$  and  $N$ . Therefore, we denote  $q_{N,B}$  the success probability of ADATHRESH with a budget  $B$  and an input sequence of size  $N$ ,  $t^{\star}$  the position of the best applicant,  $x_{s:t} = (x_{s}, \ldots, x_{t})$  for any  $s \leq t$ ,  $S(x_{1:N}, B)$  the event that ADATHRESH with a budget  $B$  succeeds on the input  $x_{1:N}$ , and  $R_{B} = r_{B}N$ . We define the random variable
$$
\tau = \min  \left\{t \geq \lceil r _ {B} N \rceil + 1: x _ {t} > \max  \left\{x _ {1}, \dots , x _ {t - 1} \right\} \right\}.
$$
We will prove the claim of the theorem by induction on  $B$ .
For  $B = 0$ , it is established that for  $r_0 = 1 / \mathrm{e}$ , we have a success probability of at least  $q_{0} = 1 / \mathrm{e}$ . Let  $B \geq 1$ , and assume that the result is true for  $B - 1$ , we have
$$
q _ {N, B} = \mathbf {P} \left(S \left(x _ {1: N}, B\right)\right) = \sum_ {t = R _ {B}} ^ {N} \mathbf {P} \left(S \left(x _ {1: N}, B\right) \text {a n d} \tau = t\right). \tag {3}
$$
For any  $t$ , since the oracle  $\mathcal{Q}$  is error-free, then under  $\tau = t$ , we have that  $\mathcal{Q}(x_t) = 0$  is equivalent to  $t^\star = t$ , and  $\mathcal{Q}(x_t) = 0$  is equivalent to  $t^\star > t$ , thus we have
$$
\begin{array}{l} \mathbf {P} \left(S \left(x _ {1: N}, B\right) \text {a n d} \tau = t \text {a n d} Q \left(x _ {t}\right) = 0\right) = \mathbf {P} (\tau = t \text {a n d} t ^ {\star} = t) = \frac {1}{N} \mathbf {P} (\tau = t \mid t ^ {\star} = t) (4) \\ = \frac {1}{N} \mathbf {P} \left(\max  x _ {1: R _ {B}} > \max  x _ {R _ {B} + 1: t - 1}\right) (5) \\ = \frac {1}{N} \times \frac {R _ {B}}{t - 1} = \frac {R _ {B} / N}{t - 1}, (6) \\ \end{array}
$$
where 4 is true because the event  $S(x_{1:N}, B)$  is implied by  $\tau = t^{\star}$ , the second equality because  $t^{\star}$  is a uniform random variable in  $\{1, \dots, N\}$ , and 5 because, conditionally to  $t^{\star} = t$ , the event  $\tau = t$  is equivalent to not selecting any candidate in  $x_{R_B + 1:t - 1}$ , i.e. all those candidates have smaller values than  $\max x_{1:R_B}$ , which is equivalent to saying that the maximum element in  $x_{1:t - 1}$  is among the first  $R_B$  ones, and this is independent of the event  $t = t^{\star}$ . On the other hand, we have that
$$
\begin{array}{l} \mathbf {P} (S (x _ {1: N}, B) \text {a n d} \tau = t \text {a n d} \mathcal {Q} (x _ {t}) = 1) = \mathbf {P} (S (x _ {1: N}, B) \mid \tau = t, \mathcal {Q} (x _ {t}) = 1) \\ \times \mathbf {P} (\tau = t \text {a n d} t ^ {\star} > t). \\ \end{array}
$$
Conditionally to  $\{\tau = t, \mathcal{Q}(x_t) = 1\}$ ,  $x_t$  is the best candidate up to step  $t$ , and ADATHRESH used a prediction to observe that  $t^\star > t$ . The algorithm is therefore restarted with input  $x_{t+1,N}$  and budget  $B$ , and  $S(x_{1:N}, B) = S(x_{t+1:N}, B - 1)$ , which is independent of the history up to step  $t$ , hence
$$
\begin{array}{l} \mathbf {P} \left(S \left(x _ {1: N}, B\right) \mid \tau = t, Q \left(x _ {t}\right) = 1\right) = \mathbf {P} \left(S \left(x _ {t + 1: N}, B - 1\right)\right) \\ = q _ {N - t, B - 1} \geq q _ {B - 1} \\ \end{array}
$$
by the induction hypothesis. Secondly, we have that the two following events are equivalent
$$
\left\{\tau = t \text {a n d} t ^ {\star} > t \right\} = \left\{\max  x _ {t + 1: N} > x _ {t} > \max  x _ {1: R _ {B}} > \max  x _ {R _ {B} + 1: t - 1} \right\},
$$
and since the applicants are shuffled uniformly at random, the relative ranks inside any subset  $Y$  of  $\{x_1, \ldots, x_N\}$  are independent of how  $\max Y$  compares to applicants outside  $Y$ , therefore
$$
\begin{array}{l} \mathbf {P} (\tau = t \text {a n d} t ^ {\star} > t) = \mathbf {P} \left(\max  x _ {t + 1: N} > x _ {t} > \max  x _ {1: R _ {B}} > \max  x _ {R _ {B} + 1: t - 1}\right) \\ = \mathbf {P} (\max  x _ {1: N} \in x _ {t + 1: N}) \mathbf {P} (\max  x _ {1: t} = x _ {t}) \mathbf {P} (\max  x _ {1: t - 1} \in x _ {1: R _ {B}}) \\ = \frac {N - t}{N} \times \frac {1}{t} \times \frac {R _ {B}}{t - 1}, \\ \end{array}
$$
which gives
$$
\mathbf {P} \left(S \left(x _ {1: N}, B\right) \text {a n d} \tau = t \text {a n d} \mathcal {Q} \left(x _ {t}\right) = 1\right) \geq \left(1 - \frac {t}{N}\right) \frac {R _ {B}}{t (t - 1)} q _ {B - 1}. \tag {7}
$$
Using 4 and 7 we deduce that
$$
\begin{array}{l} \mathbf {P} \left(S \left(x _ {1: N}, B\right) \text {a n d} \tau = t\right) \geq \frac {R _ {B} / N}{t - 1} + \left(1 - \frac {t}{N}\right) \frac {R _ {B}}{t (t - 1)} q _ {B - 1} \\ = (1 - q _ {B - 1}) \frac {R _ {B} / N}{t - 1} + \frac {q _ {B - 1} R _ {B}}{t (t - 1)} \\ \end{array}
$$
and substituting into 3 gives that
$$
\begin{array}{l} q _ {N, B} \geq (1 - q _ {B - 1}) \frac {R _ {B}}{N} \sum_ {t = R _ {B}} ^ {N} \frac {1}{t - 1} + q _ {B - 1} R _ {B} \sum_ {t = R _ {B}} ^ {N} \frac {1}{t (t - 1)} \\ \geq \left(1 - q _ {B - 1}\right) \frac {R _ {B}}{N} \int_ {R _ {B} - 1} ^ {N} \frac {d u}{u} + q _ {B - 1} R _ {B} \sum_ {t = R _ {B}} ^ {N} \left(\frac {1}{t - 1} - \frac {1}{t}\right) \\ = \left(1 - q _ {B - 1}\right) \frac {R _ {B}}{N} \log \left(\frac {N}{R _ {B} - 1}\right) + q _ {B - 1} \left(\frac {R _ {B}}{R _ {B} - 1} - \frac {R _ {B}}{N}\right), \\ \end{array}
$$
since  $R_B < r_B N + 1$ , we have that  $\log (N / (R_B - 1)) \geq \log (1 / r_B)$ , and
$$
\frac {R _ {B}}{R _ {B} - 1} - \frac {R _ {B}}{N} \geq \left(1 + \frac {1}{R _ {B} - 1}\right) - \left(r _ {B} + \frac {1}{N}\right) \geq 1 - r _ {B}.
$$
It follows that
$$
q _ {N, B} \geq (1 - q _ {B - 1}) r _ {B} \log \left(1 / r _ {B}\right) + q _ {B - 1} (1 - r _ {B}) \tag {8}
$$
The left term of this inequality is maximal for  $r_B = \exp \left(-\frac{1}{1 - q_{B-1}}\right)$ , and we obtain that
$$
\begin{array}{l} q _ {N, B} \geq (1 - q _ {B - 1}) r _ {B} \times \frac {1}{1 - q _ {B - 1}} + q _ {B - 1} (1 - r _ {B}) \\ = r _ {B} + q _ {B - 1} \left(1 - r _ {B}\right) \\ = q _ {B - 1} + \left(1 - q _ {B - 1}\right) r _ {B} = q _ {B}, \\ \end{array}
$$
B.2 Proof of Theorem 3.2
The algorithm Consider a simple adaptation of ADATHRESH, where the predictions are not always blindly trusted. If the conditions  $t \geq r_B N$  and  $x_t > \max\{x_1, \ldots, x_{t-1}\}$  are met, then with probability  $1 - \lambda$  the algorithm chooses to query a prediction and trust it, and with probability  $\lambda$  it chooses to select the candidate  $x_t$  without querying a prediction.
We will prove Theorem 3.2 using two lemmas. In the first one, we show the lower bound on the success probability of the algorithm we presented, and in the second one, we compute the limit of this lower bound when  $B \to \infty$ . We define the function  $f:[0,1] \to (0,\infty)$  by  $f(1) = 1$  and for  $x \in (0,1)$
$$
f (x) = x + (1 - x) \exp \left(- \frac {1}{1 - x}\right).
$$
Lemma B.1. The adapted ADATHRESH with budget  $B$  has a success probability at least  $q_{B}(p)$  where  $q_{0}(p) = 1 / e$  and for any  $B \geq 1$
$$
q _ {B} (p) = \max  \{1 / e, p f \left(q _ {B - 1} (p)\right) \}.
$$
Moreover, if  $p \leq (ef(1/e))^{-1}$  then for any  $B$  we have  $q_B(p) = 1/e$ , and otherwise  $q_B(p) = pf(q_{B-1}(p))$  for any  $B \geq 1$ .
Proof. The success probability when  $B = 0$  is lower bounded by  $1 / e$ . Let  $B \geq 1$  and assume that the success probability of this algorithm given a budget  $B - 1$  is lower bounded by some constant  $q_{B - 1}(p)$  that is independent of  $N$ , if the algorithm chooses to trust the prediction, then if it inaccurate the algorithm fails with probability 1, and if it is accurate, which happens with a probability  $p$ , then similarly to (8) in the proof of Theorem 3.1, the success probability is lower bounded by
$$
(1 - q _ {B - 1} (p)) r _ {B} \log (1 / r _ {B}) + q _ {B - 1} (p) (1 - r _ {B}).
$$
On the other hand, if the algorithm chooses not to trust the prediction, then its success probability is lower bounded by  $1 / e$ . Therefore, if we denote  $q_{N,B}(p,\lambda)$  the success probability of the algorithm on an instance of size  $N$ , given a budget  $B$  of predictions that are each accurate with an independent probability  $p$ , then we have
$$
q _ {N, B} (p, \lambda) \geq \frac {\lambda}{e} + p (1 - \lambda) \left((1 - q _ {B - 1} (p)) r _ {B} \log (1 / r _ {B}) + q _ {B - 1} (p) (1 - r _ {B})\right)
$$
The left term is maximized for  $r_B = \exp \left(-\frac{1}{1 - q_{B-1}(p)}\right)$ , and it gives
$$
q _ {N, B} (p, \lambda) \geq \frac {\lambda}{e} + (1 - \lambda) p \left(q _ {B - 1} (p) + (1 - q _ {B - 1} (p)) \exp \left(- \frac {1}{1 - q _ {B - 1} (p)}\right)\right),
$$
this is a linear function of  $\lambda$ , and it is maximized for
$$
\lambda = \mathbb {1} \left(1 / e > p q _ {B - 1} (p) + p \left(1 - q _ {B - 1} (p)\right) \exp \left(- \frac {1}{1 - q _ {B - 1} (p)}\right)\right).
$$
With this choice of  $\lambda$ , we have that the success probability  $q_{N,B}(p,\lambda)$  is lower bounded by the constant  $q_B(p)$  defined by
$$
\begin{array}{l} q _ {B} (p) = \max  \left\{\frac {1}{e}, p q _ {B - 1} (p) + p (1 - q _ {B - 1} (p)) \exp \left(- \frac {1}{1 - q _ {B - 1} (p)}\right) \right\} \\ = \max  \left\{\frac {1}{e}, p f (q _ {B - 1} (p)) \right\}. \\ \end{array}
$$
Before showing the second claim of the lemma, let us first prove that  $f$  is increasing. We do this simply by computing its derivative
$$
f ^ {\prime} (x) = 1 - \left(\frac {1}{1 - x} + 1\right) \exp \left(- \frac {1}{1 - x}\right)
$$
which is positive because  $\exp (u) > u + 1$  for any  $u > 1$ , in particular for  $u = 1 / (1 - x)$ .
Now, if  $pf(1/e) \leq 1/e$ , i.e.  $p \leq (ef(1/e))^{-1} = \left(1 + (e - 1)\exp\left(-\frac{e}{e - 1}\right)\right)^{-1}$ , since  $f$  is increasing, we have by induction for any  $B \geq 0$  that  $q_B(p) = 1/e$ . In fact, it is true for  $B = 0$ , and if  $q_{B-1}(p) = 1/e$  then  $pf(q_{B-1}(p)) = pf(1/e) \leq 1/e$ , thus  $q_B(p) = 1/e$ .
On the other hand, if  $pf(1/e) > 1/e$ , then for any  $B \geq 1$  we have that  $pf(q_{B-1}(p)) > 1/e$  and thus  $q_B(p) = pf(q_{B-1}(p))$ . In fact, the property is true for  $B = 1$ , and if it is true for some  $B \geq 1$  then  $q_B(p) = pf(q_{B-1}(p)) > 1/e$ , thus  $pf(q_B(p)) \geq pf(1/e) > 1/e$ , the property remains true for  $B + 1$ , which concludes the induction.
Lemma B.2. If  $p > (ef(1/e))^{-1}$ , then the sequence  $q_B(p)$  is increasing and
$$
\lim_{B\to \infty}q_{B}(p) = 1 - \frac{1}{1 + W\left(\frac{p}{e(1 - p)}\right)},
$$
where  $W$  is the inverse of  $u \to ue^{u}$  on  $[0, \infty)$ , called the W-Lambert function.
Proof. Assume that  $p > (ef(1/e))^{-1}$ , let us prove by induction over  $B \geq 1$  that  $q_B(p) > q_{B-1}(p)$ . For  $B = 1$  we have that  $q_1(p) = pf(1/e) > 1/e = q_0(p)$  by the assumption on  $p$ . Now, let  $B \geq 1$  and assume that  $q_B(p) > q_{B-1}(p)$ . Since  $f$  is an increasing function on  $[0,1]$ , we have that
$$
q _ {B + 1} (p) = p f \left(q _ {B} (p)\right) > p f \left(q _ {B - 1} (p)\right) = q _ {B} (p).
$$
The sequence  $(q_B(p))_{B\geq 0}$  is therefore increasing and it is upper bounded by 1 since it is a lower bound on a probability. We deduce that it converges to some limit  $\ell$  that depends on  $p$ . Since  $f$  is continuous we have that  $\ell = pf(\ell)$ , solving this equation gives
$$
\begin{array}{l} \ell = p f (\ell) \iff \ell = p \ell + p (1 - \ell) \exp \left(- \frac {1}{1 - \ell}\right) \\ \Longleftrightarrow (1 - p) \frac {\ell}{1 - \ell} = p \exp \left(- \frac {1}{1 - \ell}\right) \\ \Longleftrightarrow (1 - p) \left(\frac {1}{1 - \ell} - 1\right) = \frac {p}{e} \exp \left(1 - \frac {1}{1 - \ell}\right) \\ \iff \left(\frac {1}{1 - \ell} - 1\right) \exp \left(\frac {1}{1 - \ell} - 1\right) = \frac {p}{e (1 - p)} \\ \Longleftrightarrow \frac {1}{1 - \ell} - 1 = W \left(\frac {p}{e (1 - p)}\right) \\ \iff \ell = 1 - \frac {1}{1 + W \left(\frac {p}{e (1 - p)}\right)}. \\ \end{array}
$$
Therefore  $\lim_{B\to \infty}q_B(p) = 1 - \frac{1}{1 + W\left(\frac{p}{e(1 - p)}\right)}.$
Observe that the algorithm we presented can also be adapted to the case where the accuracy  $p$  of the oracle is not constant and might depend on the time or even on the input sequence. In fact, as long as the accuracy of the oracle is known to the algorithm, it can trust the oracle if it is correct with a probability larger than  $(ef(1/e)^{-1}$ , and ignore if otherwise. This gives a success probability always lower bounded by  $1/e$ , with better guarantees depending on the accuracy of the oracle at times when it is queried.
B.3 ADATHRESH with memory
We present now an improved version of ADATHRESH that keeps in memory the best observed value so far, and that rejects all applicants below this value.
Algorithm 5: ADATHRESH with memory
Input: Budget  $B$ , sequence of applicants  $(x_{1},\ldots ,x_{N})$ , largest previously observed value  $M$   
1 Reject the first  $\lceil r_B N\rceil$  applicants;  
2 for  $t = \lceil r_B N\rceil +1,\dots ,N$  do  
3 if  $x_{t} > M$  and  $x_{t} > \max \{x_{1},\ldots ,x_{t - 1}\}$  then  
4 if  $B > 0$  then  
5 query a prediction  $\mathcal{Q}(x_t)$ ;  
6 if  $\mathcal{Q}(x_t) = 0$  then Return:  $t$ ;  
7 else Return ADATHRESH  $(B - 1,(x_{t + 1},\dots,x_N),x_t)$ ;  
8 else Return:  $t$ ;
Initially, the algorithm is launched with  $M = -\infty$ . While we do not give theoretical guarantees on the success probability of Algorithm 5, we show experimentally how it compares to Algorithm 3 in Section 5.
C Preemptive  $B$ -clairvoyant job scheduling
C.1 Proof of Theorem 4.1
Proof. Let ALG be any non-clairvoyant algorithm that can query up to  $B$  job sizes for instances of  $N$  jobs. It is shown in Theorem 2.8 in [46] that by choosing an input instance  $\mathcal{E}_n$  of  $n$  job sizes sampled independently from the exponential distribution, any algorithm  $\mathsf{A}$  verifies
$$
\mathbf {E} [ \mathbf {A} (\mathcal {E} _ {n}) ] \geq (2 - 4 / n) \mathbf {E} [ \mathbf {O P T} (\mathcal {E} _ {n}) ],
$$
with  $\mathbf{E}[\mathsf{OPT}(\mathcal{E}_n)] = n + n(n + 1) / 4$ . Let us consider in particular the algorithm  $\mathsf{A}_N$  which, when given an instance  $\mathcal{I}$  of  $n \leq N$  jobs, samples  $N - n$  positive numbers  $z_1, \ldots, z_{N - n}$  from the exponential distribution, then runs ALG with the input  $\mathcal{I} \cup \{z_i\}_{i=1}^{N - n}$ . Let us denote  $\mathsf{A}_N(\mathcal{I}|\{z_i\}_{i=1}^{N - n})$  the output of  $\mathsf{A}_N$  conditionally to  $\{z_i\}_{i=1}^{N - n}$ . Now, let  $\mathcal{E}$  be an instance of  $N$  job sizes chosen from the exponential distribution to be given as input for ALG. During its execution, ALG will query a number  $b \leq B$  of sizes for jobs in some subset  $I$  of  $\{1, \ldots, N\}$ . Denoting  $J = \{1, \ldots, N\} \setminus I$ , we have by Equation 1 that
$$
\begin{array}{l} \mathsf {A L G} (\mathcal {E}) \geq \sum_ {i \in J} x _ {i} + \sum_ {i <   j \in J} \left(D ^ {\mathsf {A L G}} (i, j) + D ^ {\mathsf {A L G}} (j, i)\right) \\ = \mathsf {A} _ {N} \big (\{x _ {j} \} _ {j \in J} | \{x _ {i} \} _ {i \in I} \big). \\ \end{array}
$$
In expectation, using that  $|J| \geq N - B = N - o(N)$  with probability 1, we obtain
$$
\begin{array}{l} \mathbf {E} [ \mathsf {A L G} (\mathcal {E}) ] \geq \mathbf {E} \left[ \mathbf {E} [ \mathsf {A} _ {N} \left(\left\{x _ {j} \right\} _ {j \in J} | \left\{x _ {i} \right\} _ {i \in I}\right) \mid n = | J | ] \right] \\ \geq \mathbf {E} \left[ | J | ^ {2} / 2 + 3 | J | / 2 - 5 \right] \\ = \frac {N ^ {2}}{2} + O (N). \\ \end{array}
$$
Using again that  $\mathbf{E}[\mathsf{OPT}(\mathcal{E})] = N + N(N - 1) / 4 = N^2 /4 + O(N)$ , we deduce that  $\frac{\mathbf{E}[\mathsf{ALG}(\mathcal{E})]}{\mathbf{E}[\mathsf{OPT}(\mathcal{E})]}\geq$ $2 + o(1)$ , which shows that the competitive ratio of ALG is at least 2.
C.2 Intermediate Lemma and notation
Notations In all the following, we fix an instance  $\mathcal{J}$  of job sizes  $x_{1} \leq \ldots \leq x_{N}$ , and we denote  $\sigma$  a uniformly random permutation of  $\{1, \ldots, N\}$ ,  $I = \{\sigma(i)\}_{i=1}^{B}$  and  $J = \{\sigma(j)\}_{j=B+1}^{N}$ . This is equivalent to choosing a set  $I$  of  $B$  jobs uniformly at random without replacement. We also denote  $I_{t}$  and  $J_{t}$  respectively the sets of unfinished jobs in  $I$  and  $J$  at time  $t$  of a run of PAR.
The next lemma gives a simple computational result that we will use in further proofs.
Lemma C.1. For any  $i \neq j$  we have  $\mathbf{E}[\min\{x_{\sigma(i)}, x_{\sigma(j)}\}] = \frac{2}{N(N-1)} \sum_{k=1}^{N}(N-k)x_k$ .
Proof. Since  $\sigma$  is a random permutation, we only need to demonstrate the result for  $i,j = 1,2$ . Since we assume that  $x_{1}\leq \ldots ,x_{N}$ , then for any  $k\leq N$  we have
$$
\begin{array}{l} \mathbf {P} (\min  \{x _ {\sigma (1)}, x _ {\sigma (2)} \} = x _ {k}) = \mathbf {P} (\min  \{\sigma (1), \sigma (2) \} = k) \\ = 2 \mathbf {P} (\sigma (1) = k \text {a n d} \sigma (2) > k) \\ = 2 \mathbf {P} (\sigma (1) = k) \mathbf {P} (\sigma (2) > k \mid \sigma (1) = k) \\ = \frac {2 (N - k)}{N (N - 1)}. \\ \end{array}
$$
$$
\begin{array}{l} \mathbf {E} [ \min  \{x _ {\sigma (1)}, x _ {\sigma (2)} \} ] = \sum_ {k = 1} ^ {N} \mathbf {P} (\min  \{x _ {\sigma (1)}, x _ {\sigma (2)} \} = x _ {k}) x _ {k} \\ = \frac {2}{N (N - 1)} \sum_ {k = 1} ^ {N} (N - k) x _ {k}. \\ \end{array}
$$
C.3 Proof of Lemma 4.2
Proof. Let us fix some update rule for  $\alpha$ . In the execution of  $\mathsf{PAR}$ ,  $\mathsf{OPT}$  is run on  $I$  and  $\mathsf{RR}$  on  $J$ , and both sets are disjoint, therefore we have  $D^{\mathsf{PAR}}(i,j) = D^{\mathsf{OPT}}(i,j)$  for any  $i \neq j \in I$ , and  $D^{\mathsf{PAR}}(i,j) = D^{\mathsf{RR}}(i,j)$  for any  $i \neq j \in J$ , we can write
$$
\begin{array}{l} \mathsf {P A R} (\mathcal {J}) = \sum_ {i = 1} ^ {N} x _ {i} + D ^ {\mathsf {P A R}} (I, I) + D ^ {\mathsf {P A R}} (J, J) + D ^ {\mathsf {P A R}} (I, J) + D ^ {\mathsf {P A R}} (J, I) \\ = \sum_ {i = 1} ^ {N} x _ {i} + D ^ {\mathrm {O P T}} (I, I) + D ^ {\mathrm {R R}} (J, J) + D ^ {\mathrm {P A R}} (I, J) + D ^ {\mathrm {P A R}} (J, I). \tag {9} \\ \end{array}
$$
Using Equation 2 and Lemma C.1, we have that
$$
\begin{array}{l} \mathbf {E} \left[ D ^ {\mathrm {O P T}} (I, I) \right] = \sum_ {1 \leq i <   j \leq B} \mathbf {E} \left[ \min  \left\{x _ {\sigma (i)}, x _ {\sigma (j)} \right\} \right] \\ = \sum_ {1 \leq i <   j \leq B} \frac {2}{N (N - 1)} \sum_ {k = 1} ^ {N} (N - k) x _ {k} \\ = \frac {B (B - 1)}{N (N - 1)} \sum_ {k = 1} ^ {N} (N - k) x _ {k}, \\ \end{array}
$$
and in the same way
$$
\begin{array}{l} \mathbf {E} \left[ D ^ {\mathbb {R R}} (J, J) \right] = \sum_ {B <   i <   j \leq N} 2 \mathbf {E} \left[ \min  \left\{x _ {\sigma (i)}, x _ {\sigma (j)} \right\} \right] \\ = \frac {2 (N - B) (N - B - 1)}{N (N - 1)} \sum_ {k = 1} ^ {N} (N - k) x _ {k} \\ = \left(2 - \frac {2 B (2 N - B - 1)}{N (N - 1)}\right) \sum_ {k = 1} ^ {N} (N - k) x _ {k}. \\ \end{array}
$$
Summing these two equations gives
$$
\begin{array}{l} \mathbf {E} [ D ^ {\mathrm {O P T}} (I, I) ] + \mathbf {E} [ D ^ {\mathbb {R R}} (J, J) ] = \left(2 + \frac {B (B - 1) - 2 B (2 N - B - 1)}{N (N - 1)}\right) \sum_ {k = 1} ^ {N} (N - k) x _ {k} \\ = \left(2 - \frac {B}{N} (4 - 3 \frac {B - 1}{N - 1})\right) \sum_ {k = 1} ^ {N} (N - k) x _ {k}, \\ \end{array}
$$
and taking the expectation in Equation 9 concludes the proof.
C.4 Proof of Theorem 4.3
We use here the same notations as in Section 4.3. Before demonstrating the theorem, let us justify why  $(\alpha_{t}^{\mathsf{RR}})_{t}$  can be computed online, i.e for any  $t > 0$ ,  $\alpha_{t}^{\mathsf{RR}}$  can be computed knowing only the events that occurred up to time  $t$ .
For any subset  $K$  of  $\{1,\dots ,N\}$ , we denote  $\mathbb{R}\mathbb{R}(K)$  the algorithm that runs  $\mathbb{R}\mathbb{R}$  on  $\{x_{i}\}_{i\in K}$  and ignores the other jobs. Running  $\mathbb{R}\mathbb{R}$  on all the input instance is equivalent to running concurrently  $\mathbb{R}\mathbb{R}(I)$  and  $\mathbb{R}\mathbb{R}(J)$  with respective rates  $(\alpha_{t}^{\mathbb{R}\mathbb{R}})_{t}$  and  $(1 - \alpha_{t}^{\mathbb{R}\mathbb{R}})_{t}$ . In fact, for any  $t > 0$ , the processing rate of any unfinished job in  $I$  is  $\frac{\alpha_t^{\mathbb{R}\mathbb{R}}}{|I_t^{\mathbb{R}\mathbb{R}}|} = \frac{1}{|I_t^{\mathbb{R}\mathbb{R}}| + |J_t^{\mathbb{R}\mathbb{R}}|}$ , and we have the same for jobs in  $J$ .
Consider running  $\mathsf{RR}$  on  $\{x_1,\ldots ,x_N\}$ . We denote  $t_i^{\mathsf{RR}}$  the completion time of job  $1\leq i\leq N$ . For any  $t > 0$ , the amount of  $\mathsf{RR}(I)$  executed until time  $t$  is  $\int_0^t\alpha_u^{\mathsf{RR}}du$ . Therefore if we denote  $T_{i}^{\mathsf{RR}}(I)$  the completion time of job  $i\in I$  when running  $\mathsf{RR}(I)$ , then we have
$$
T _ {i} ^ {\mathsf {R R}} (I) = \int_ {0} ^ {t _ {i} ^ {\mathsf {R R}} (I)} \alpha_ {u} ^ {\mathsf {R R}} d u.
$$