SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Consider sets  $S_{1}, \ldots, S_{k}$  from Theorem 2.1. Denote  $K = \{S_{1}, \ldots, S_{k}\}$ . The definition of the competition among sets  $S_{1}, \ldots, S_{k}$  (given in the beginning of Section 2) is exactly the same as the definition of competition in  $K$ . Our goal is to show that  $\mathbf{E}[\mu(\text{winner}(K))] \leq 2(\ln k + 1) \cdot \min_{S_{i} \in K} \mu(S_{i})$ . In the proof of Theorem 2.1, we will consider competitions in different set systems  $I \subseteq K$ . We show the following key lemma. We defer the proof of Lemma 2.5 to Appendix A.
Lemma 2.5. Consider a partitioning of the set system  $K = \{S_{1},\ldots ,S_{k}\}$  into  $m$  sets  $I_{1},\dots,I_{m}$ . Then, winner  $(K)\in \{\mathrm{winner}(I_1),\dots,\mathrm{winner}(I_m)\}$ .
2.3 Set Elimination with Exponential Clock
Consider a set elimination game on sets  $S_{1}, \ldots, S_{k}$ . It is determined by the sequence of random i.i.d. draws  $\omega_{1}, \omega_{2}, \ldots$ . Random variable  $\omega_{n}$  is chosen in round  $n$ . We assign every round a random time  $\tau_{n}$ . Let the time between two consecutive rounds be an exponential random variable with parameter  $\mu(\Omega)$ . Specifically, let  $\Delta \tau_{1}, \Delta \tau_{2}, \ldots$  be a sequence of i.i.d. exponential random variables with parameter  $\mu(\Omega)$  and each  $\tau_{n} = \tau_{n-1} + \Delta \tau_{n} = \Delta \tau_{1} + \dots + \Delta \tau_{n}$ . Note that all  $\Delta \tau_{n}$  are positive and  $\tau_{1}, \tau_{2}, \ldots$  is an increasing sequence with probability 1. The number of draws that occur by time  $t$  (i.e.,  $N_{t}(\Omega) = |\{n : \tau_{n} \leq t\}|$ ) is a Poisson process with parameter  $\mu(\Omega)$ . We now can think of the set elimination game as follows: The host of the game observes a Poisson process with parameter  $\mu(\Omega)$ . Whenever the process jumps (at time  $\tau_{n}$ ), the host picks an element  $\omega_{n}$  in  $\Omega$  with probability  $\operatorname{Pr}(\omega_{n} = \omega) = \mu(\omega) / \mu(\Omega)$  and eliminates some sets according to the rules of the game discussed above. Note that by assigning every round some time  $\tau_{n}$ , we do not change the game, the winner, and the cost of the game (because the sequence of random draws  $\omega_{1}, \omega_{2}, \ldots$  remains the same as before). This interpretation of the game allows us to introduce a hitting time  $h(S)$  of every subset  $S \subset \Omega$  with the following properties: (a) each  $h(S)$  is an exponential random variable with rate  $\mu(S)$ ; (b) hitting times of disjoint sets are mutually independent random variables.
Definition 2.6. For every subset  $X \subset \Omega$ , the hitting time  $h(X)$  is the time  $\tau_n$  when the first  $\omega_n$  is drawn from  $X$ :  $h(X) = \min \{\tau_n : \omega_n \in X\}$ . When the set contains one element  $\omega$ , we will write  $h(\omega)$  instead of  $h(\{\omega\})$ .
We also define the elimination time of each set  $S_{i}$ .
Definition 2.7. Consider any set elimination game with the measure space  $(\Omega, \mu)$  and  $k$  sets  $S_{1}, S_{2}, \ldots, S_{k}$  in  $\Omega$ . The elimination time  $e(S_{i})$  of set  $S_{i}$  is the time when set  $S_{i}$  is eliminated from the game, i.e.,  $e(S_{i}) = \min \{\tau_{n} : S_{i} \notin \mathcal{R}_{n}(K)\}$ . If  $S_{i}$  is the winner, then we let  $e(S_{i}) = \infty$  (because the winner is never eliminated).
Note that  $e(S_i) \geq h(S_i)$ . Sometimes,  $e(S_i)$  may be equal to  $h(S_i)$ , but  $e(S_i)$  and  $h(S_i)$  are not always the same. We now prove that hitting times for disjoint sets are independent. To this end, we split the Poisson process  $N_t(\Omega) = |\{n : \tau_n \leq t\}|$ . Let  $N_t(\omega) = |\{n : \tau_n \leq t \text{ and } \omega_n = \omega\}|$ . It is easy to see that  $N_t(\Omega) = \sum_{\omega \in \Omega} N_t(\omega)$  for every  $t$ . It is also true that each  $N_t(\omega)$  is a Poisson process with parameter  $\mu(\omega)$  and all  $N_t(\omega)$  (for  $\omega \in \Omega$ ) are mutually independent. This fact follows from the Coloring Theorem (see e.g., Kingman [1992], Coloring Theorem, page 53).
Theorem 2.8 (Coloring Theorem). Let  $\Pi_t$  be a Poisson process on the real line with rate  $\lambda$ . We color each event of the Poisson process randomly with one of  $M$  colors: The probability that a point receives the  $i$ -th color is  $p_i$ . The colors of different points are independent. Let  $\Pi_t(i)$  be the number of events of color  $i$  in the interval  $(0, t]$ . Then,  $\Pi_t(1), \ldots, \Pi_t(M)$  are independent Poisson processes. The rate of process  $\Pi_t(i)$  is  $\lambda p_i$ .
Lemma 2.9. For every  $\omega \in \Omega$ ,  $h(\omega)$  is an exponential random variable with parameter  $\mu(\omega)$ , and all random variables  $h(\omega)$  (for  $\omega \in \Omega$ ) are mutually independent.
Proof. Observe that  $h(\omega) = \min \{t : N_t(\omega) \geq 1\}$ . Thus,  $h(\omega)$  is an exponential random variable (the time of the first jump of a Poisson process) with rate  $\mu(\omega)$ . Also, since all  $N_t(\omega)$  (for  $\omega \in \Omega$ ) are mutually independent, all  $h(\omega)$  are also mutually independent.
Note that the set elimination game depends only on the hitting times for elements  $\omega$  in  $\Omega$ . This is the case because it matters only when every  $\omega$  is drawn the first time. At that time - the hitting time of  $\omega$  - all sets that contain  $\omega$  are eliminated unless all remaining sets contain this  $\omega$ . When the same  $\omega$  is drawn again, it does not eliminate any new sets. Also, note that for any set  $S \subset \Omega$ , the hitting time  $h(S) = \min_{\omega \in S} h(\omega)$ . Thus,  $h(S)$  is an exponential random variable with parameter  $\mu(S) = \sum_{\omega \in S} \mu(\omega)$ .
3 Proof of Main Result
We now present the proof of our main result, Theorem 2.1. We assume without loss of generality that  $S_{1}$  is the smallest set i.e.,  $\mu(S_{1}) \leq \mu(S_{i})$  for all  $i$ . Then, the expected cost of the game is at most:
$$
\mu \left(S _ {1}\right) + \sum_ {i = 2} ^ {k} \Pr \left(S _ {i} = \operatorname {w i n n e r} (K)\right) \mu \left(S _ {i}\right). \tag {3}
$$
We first provide some intuition for the proof by considering the case when  $S_{1}$  does not intersect with sets  $S_{2},\ldots ,S_{k}$ , i.e. sets  $S_{1}$  and  $S_{i}$  are disjoint for all  $i = 2,3,\dots ,k$ . We split all sets into two groups  $S_{1}$  and the rest of the sets  $S_{2},\ldots ,S_{k}$ . We know from Lemma 2.5 that the winner among all sets  $S_{1},\ldots ,S_{k}$  is either  $S_{1}$  or winner  $\left(\{S_2,\ldots ,S_k\}\right)$ . Denote  $I^{-} = \{S_{2},\ldots ,S_{k}\}$ . Each set  $S_{i}$  is eliminated at time  $e(S_i)$ . The set  $S_{1}$  is eliminated at its hitting time  $h(S_{1})$  unless it is the only remaining set at time  $h(S_{1})$  (because we are considering the case when  $S_{1}$  does not overlap with other sets). Thus,
$$
\operatorname {w i n n e r} (K) = \left\{ \begin{array}{l l} S _ {1}, & \text {i f} h \left(S _ {1}\right) > e \left(\operatorname {w i n n e r} \left(I ^ {-}\right)\right); \\ \operatorname {w i n n e r} \left(I ^ {-}\right), & \text {i f} e \left(\operatorname {w i n n e r} \left(I ^ {-}\right)\right) > h \left(S _ {1}\right). \end{array} \right. \tag {4}
$$
When the winner among  $S_{1}, \ldots, S_{k}$  is not  $S_{1}$ , we consider two cases of the winner  $S_{i}$ : (1)  $S_{i}$  is a surprise set; (2)  $S_{i}$  is a non-surprise set.
Definition 3.1. We say that  $S_{i}$  is a surprise set if  $e(S_{i}) \geq h(S_{1}) \geq L / \mu (S_{i})$ , where  $L = \ln k$ .
We call  $S_{i}$  a surprise set because the probability of the event  $e(S_{i}) \geq h(S_{1}) \geq L / \mu (S_{i})$  is small. We give a bound on the probability of  $e(S_{i}) \geq h(S_{1}) \geq L / \mu (S_{i})$  in Lemma 3.3. Here, we provide some intuition. By Lemma 2.9, the hitting time  $h(S_{i})$  is an exponential random variable with parameter  $\mu (S_{i})$ . Thus, the expected hitting time for  $S_{i}$  is  $1 / \mu (S_{i})$ . Consider a set  $S_{i}$  with a small measure  $(\mu (S_{i})$  is close to  $\mu (S_{1}))$ . If the hitting time  $h(S_1) \geq L / \mu (S_i)$ , then  $h(S_1)$  is much larger than its expected value  $1 / \mu (S_1)$ , which happens with a small probability. Consider a set  $S_{i}$  with a large measure  $\mu (S_{i}) \gg \mu (S_{1})$ . Then, the expected hitting time for  $S_{i}$  is  $1 / \mu (S_{i})$ , which is much smaller than the expected hitting time of  $S_{1}$ . Thus, the event  $e(S_{i}) \geq h(S_{1})$  occurs with a small probability.
Let us examine bound (3). Let Surprise be the set of all surprise sets. Note that Surprise is a random set. Then,
$$
\begin{array}{l} \sum_ {i = 2} ^ {k} \Pr \left(S _ {i} = \operatorname {w i n n e r} (K)\right) \mu \left(S _ {i}\right) \leq \sum_ {i = 2} ^ {k} \Pr \left(S _ {i} = \operatorname {w i n n e r} (K), S _ {i} \notin \text {S u r p r i s e}\right) \cdot \mu \left(S _ {i}\right) \tag {5} \\ + \sum_ {i = 2} ^ {k} \Pr \left(S _ {i} \in \text {S u r p r i s e}\right) \cdot \mu \left(S _ {i}\right). \\ \end{array}
$$
We show in the next section (Lemma 3.3) that the second sum is upper bounded by  $\mu(S_1)$ . We now bound the first sum. For every winner  $S_i$  which is not a surprise set, we have  $e(S_i) \geq h(S_1)$  (because  $S_i$  is the winner) and  $h(S_1) \leq L / \mu(S_i)$  (because  $S_i$  is not a surprise set). We also have  $S_i = \mathrm{winner}(I^-)$ , thus
$$
\Pr \left(S _ {i} = \operatorname {w i n n e r} (K), S _ {i} \notin \text {S u r p r i s e}\right) \leq \Pr \left(h \left(S _ {1}\right) \leq L / \mu \left(S _ {i}\right) \text {a n d} S _ {i} = \operatorname {w i n n e r} \left(I ^ {-}\right)\right).
$$
By Lemma 2.9, all hitting times  $h(S_i) = \min_{\omega \in S_i} h(\omega)$  for  $i \geq 2$  are independent from  $h(S_1)$ . Thus, winner  $(I^{-})$  is also independent of  $h(S_1)$  (winner  $(I^{-})$  depends only on the hitting times for sets  $S_i \in I^-$ ). Therefore,
$$
\begin{array}{l} \operatorname * {P r} \left(S _ {i} = \operatorname {w i n n e r} (K), S _ {i} \notin \text {S u r p r i s e}\right) \leq \operatorname * {P r} \left(h \left(S _ {1}\right) \leq L / \mu \left(S _ {i}\right)\right) \cdot \operatorname * {P r} \left(S _ {i} = \operatorname {w i n n e r} \left(I ^ {-}\right)\right) \\ = \underbrace {\left(1 - e ^ {- L \mu (S _ {1}) / \mu (S _ {i})}\right)} _ {\leq L \mu (S _ {1}) / \mu (S _ {i})} \cdot \Pr \left(S _ {i} = \mathrm {w i n n e r} (I ^ {-})\right) \\ \leq \Pr \left(S _ {i} = \operatorname {w i n n e r} \left(I ^ {-}\right)\right) \cdot L \cdot \mu \left(S _ {1}\right) / \mu \left(S _ {i}\right). \\ \end{array}
$$
We combine all bounds on terms of (5) and get the following bound on the expected cost of the game:
$$
\mu \left(S _ {1}\right) + \sum_ {i = 2} ^ {k} \Pr \left(S _ {i} = \operatorname {w i n n e r} \left(I ^ {-}\right)\right) \cdot L \cdot \mu \left(S _ {1}\right) + \mu \left(S _ {1}\right) = (L + 2) \cdot \mu \left(S _ {1}\right) = (\ln k + 2) \cdot \mu \left(S _ {1}\right).
$$
This concludes the proof of the theorem for the case when  $S_{1}$  does not overlap with  $S_{2},\ldots ,S_{k}$ . We now analyze surprise sets.
3.1 Surprise Sets
In this section, we prove a bound on the probability that a set  $S_{i}$  is a surprise set. We no longer assume that  $S_{1}$  does not intersect with other sets  $S_{i}$ . We first show a lemma about exponential random variables.
Lemma 3.2. Let  $X$  and  $Y$  be two independent exponential random variables with positive parameters  $\lambda_{X}$  and  $\lambda_{Y}$ , respectively. Then, for every  $T \geq 0$ , we have
$$
\Pr (Y \geq X \geq T) = \frac {\lambda_ {X}}{\lambda_ {X} + \lambda_ {Y}} \cdot e ^ {- (\lambda_ {X} + \lambda_ {Y}) T}. \tag {6}
$$
Proof. The desired probability can be easily found by computing  $\int_{T}^{\infty}(F_X(t) - F_X(T))f_Y(t)dt$ , where  $F_{X}(t) = 1 - e^{-\lambda_{X}t}$  is the cumulative distribution function of  $X$ , and  $f_{Y}(t) = \lambda_{Y}\cdot e^{-\lambda_{Y}t}$  is the probability density function of  $Y$ . Here, we give an alternative proof. Write,
$$
\begin{array}{l} \Pr \left(Y \geq X \geq T\right) = \Pr \left(Y \geq X \& \min  (X, Y) \geq T\right) \\ = \Pr \left(X \leq Y \mid \min  (X, Y) \geq T\right) \cdot \Pr \left(\min  (X, Y) \geq T\right). \\ \end{array}
$$
We have  $\operatorname*{Pr}\left(\min (X,Y)\geq T\right) = e^{-(\lambda_X + \lambda_Y)T}$ , because the minimum of two independent exponential random variables with parameters  $\lambda_{X}$  and  $\lambda_{Y}$  is an exponential random variable with parameter  $\lambda_{X} + \lambda_{Y}$ . Then,  $\operatorname*{Pr}\left(X\leq Y\mid \min (X,Y)\geq T\right) = \operatorname*{Pr}\left(X\leq Y\right)$  because the exponential distribution is memoryless; and  $\operatorname*{Pr}\left(X\leq Y\right) = \lambda_X / (\lambda_X + \lambda_Y)$ .
Lemma 3.3. For every set  $S_{i}$ , we have
$$
\Pr (S _ {i} \text {i s s u r p r i s e s e t}) \leq \frac {1}{k} \cdot \frac {\mu \left(S _ {1}\right)}{\mu \left(S _ {i}\right)}.
$$
Proof. First, we show that  $\min(e(S_i), h(S_1)) \leq h(S_i \setminus S_1)$ .
Claim 3.4. We always have  $\min(e(S_i), h(S_1)) \leq h(S_i \setminus S_1)$ .
Proof. Consider an arbitrary realization of the game and the time  $t = h(S_i \setminus S_1)$  when  $S_i \setminus S_1$  is hit. If by this time,  $S_1$  has already been hit then  $h(S_1) < t$ . Similarly, if by this time,  $S_i$  has already been eliminated then  $e(S_i) < t$ . Otherwise, both  $S_1$  and  $S_i$  are still remaining in the game at time  $t$ . Therefore, when we pick  $\omega \in S_i \setminus S_1$  at time  $t$ , set  $S_i$  gets eliminated (since  $\omega \in S_i$ ;  $\omega \notin S_1$ ; both  $S_1$  and  $S_i$  are remaining in the game). Thus, in this case,  $e(S_i) = t$ . This concludes the proof.
If  $S_{i}$  is a surprise set, then  $\min(e(S_{i}), h(S_{1})) = h(S_{1}) \geq L / \mu(S_{i})$ . By Claim 3.4, we have
$$
h \left(S _ {i} \backslash S _ {1}\right) \geq \min  \left(e \left(S _ {i}\right), h \left(S _ {1}\right)\right) = h \left(S _ {1}\right) \geq L / \mu \left(S _ {i}\right).
$$
Thus,  $\operatorname*{Pr}(S_i$  is surprise set)  $\leq \operatorname*{Pr}\left(h(S_i\setminus S_1)\geq h(S_1)\geq L / \mu (S_i)\right)$ . By Lemma 3.2 applied to the independent exponential random variables  $h(S_{1}),h(S_{i}\setminus S_{1})$ , and time  $T = L / \mu (S_i)$ , we have
$$
\Pr (S _ {i} \text {i s s u r p r i s e s e t}) \leq \frac {\mu (S _ {1})}{\mu (S _ {i} \setminus S _ {1}) + \mu (S _ {1})} \cdot e ^ {- \frac {L (\mu (S _ {i} \setminus S _ {1}) + \mu (S _ {1}))}{\mu (S _ {i})}} \leq \frac {1}{k} \cdot \frac {\mu (S _ {1})}{\mu (S _ {i})}.
$$
3.2 General Case
Proof of Theorem 2.1. We upper bound the expected cost of the game for arbitrary sets  $S_{1}, \ldots, S_{k}$ . As before, we assume that  $S_{1}$  is the smallest set. We remind the reader that each hitting time  $h(S_{i})$  is an exponential random variable with parameter  $\mu(S_{i})$ . In the proof, we will use the definitions of surprise sets (see Definitions 3.1). We also set  $L = \ln k$ . We define all sets  $S_{i}$  for  $i \neq 1$  that are not a surprise set to be non-surprise sets.
We separately upper bound the cost of the winner depending on whether the winner is (a) set  $S_{1}$ , (b) surprise set, (c) non-surprise set. Write
$$
\begin{array}{l} \mathbf {E} \left[ \mu (\operatorname {w i n n e r} (K)) \right] = \mathbf {E} \left[ \mu (\operatorname {w i n n e r} (K)) \cdot \mathbf {1} \{\operatorname {w i n n e r} (K) = S _ {1} \} \right] (a) \\ + \mathbf {E} [ \mu (\operatorname {w i n n e r} (K)) \cdot \mathbf {1} \{\text {w i n n e r i s s u r p r i s e s e t} \} ] (b) \\ + \mathbf {E} [ \mu (\text {w i n n e r} (K)) \cdot \mathbf {1} \{\text {w i n n e r i s n o n - s u r p r i s e s e t} \} ]. (c) \\ \end{array}
$$
Term (a) is upper bounded by  $\mu(S_1)$ . We bound term (b) using Lemma 3.3: The probability that a set is a surprise set is at most  $1/k \cdot \mu(S_1)/\mu(S_i)$ . Thus, the expected total measure of all sets (not only the surprise winner) is upper bounded by  $\frac{1}{k} \sum_{i=2}^{k} \frac{\mu(S_1)}{\mu(S_i)} \mu(S_i) < \mu(S_1)$ .
We now bound term (c). Define a new random variable: Let  $\mathrm{cost}(\omega)$  be the cost of the winner (i.e.,  $\mu(S_i)$ , where  $S_i$  is the winner) if (1) the winner is a non-surprise set, and (2)  $\omega$  is the first element that was chosen in  $S_1$ . We let  $\mathrm{cost}(\omega) = 0$ , otherwise. If  $\omega$  is the first element that was chosen in  $S_1$ , then  $h(S_1) = h(\omega)$ . So, the definition of  $\mathrm{cost}(\omega)$  can be written as follows:
$$
\operatorname {c o s t} (\omega) = \mu \left(\operatorname {w i n n e r} (K)\right) \cdot \mathbf {1} \{h \left(S _ {1}\right) = h (\omega) \} \cdot \mathbf {1} \{\operatorname {w i n n e r} (K) \notin S u r p r i s e \}.
$$
Since the hitting time  $h(S_1)$  is finite with probability 1, the term (c) equals
$$
(c) = \sum_ {\omega \in S _ {1}} \mathbf {E} [ \mathrm {c o s t} (\omega) ].
$$
Lemma 3.5, which we prove below, gives a bound of  $2L\mu(S_1)$  on the expression above. Combining upper bounds on terms (a), (b), and (c), we get
$$
\mathbf {E} \left[ \mu (\operatorname {w i n n e r} (K)) \right] \leq (1 + 2 L + 1) \mu (S _ {1}) = (2 \ln k + 2) \cdot \mu (S _ {1}).
$$
Lemma 3.5. For every  $\omega \in S_1$ , we have  $\mathbf{E}[\mathrm{cost}(\omega)] \leq 2L\mu(\omega)$ .
$$
\mathbf {E} [ \operatorname {c o s t} (\omega) ] = \mathbf {E} \left[ \mu (\operatorname {w i n n e r} (K)) \cdot \mathbf {1} \{h \left(S _ {1}\right) = h (\omega) \} \cdot \mathbf {1} \{\operatorname {w i n n e r} (K) \notin S u r p r i s e \} \right]. \tag {7}
$$
If  $S_{i}$  is a non-surprise set, then  $h(S_1) < L / \mu (S_i)$  or  $e(S_i) < h(S_1)$ . If  $S_{i}$  is the winner, then  $e(S_i)\geq h(S_1)$ . Thus, if  $S_{i}$  is a non-surprise winner, then  $h(S_1) < L / \mu (S_i)$ . This observations gives us the following upper bound on (7):
$$
\mathbf {E} \left[ \operatorname {c o s t} (\omega) \right] \leq \sum_ {i = 2} ^ {k} \mu \left(S _ {i}\right) \cdot \Pr \left(S _ {i} = \operatorname {w i n n e r} (K) \text {a n d} h (\omega) = h \left(S _ {1}\right) <   L / \mu \left(S _ {i}\right)\right). \tag {8}
$$
Define two set systems  $I_{\omega}^{-}$  and  $I_{\omega}^{+}$  of sets  $S_{i}$  containing and not containing  $\omega$ :
$$
I _ {\omega} ^ {-} = \left\{S _ {i}: \omega \notin S _ {i} \text {a n d} i \geq 2 \right\};
$$
$$
I _ {\omega} ^ {+} = \left\{S _ {i}: \omega \in S _ {i} \text {a n d} i \geq 2 \right\}.
$$
Note that  $K\equiv \{S_1,\ldots ,S_k\} = \{S_1\} \cup I_\omega^- \cup I_\omega^+$ . By Lemma 2.5,
$$
\operatorname {w i n n e r} (K) \in \left\{S _ {1}, \operatorname {w i n n e r} \left(I _ {\omega} ^ {-}\right), \operatorname {w i n n e r} \left(I _ {\omega} ^ {+}\right) \right\}.
$$
Observe that if  $S_{i}$  with  $i \geq 2$  is the winner, then  $S_{i} = \mathrm{winner}(I_{\omega}^{-})$  or  $S_{i} = \mathrm{winner}(I_{\omega}^{+})$ . We replace the condition  $S_{i} = \mathrm{winner}(K)$  with  $S_{i} \in \{\mathrm{winner}(I_{\omega}^{-}), \mathrm{winner}(I_{\omega}^{+})\}$  in (8) and get bound:
$$
\mathbf {E} \big [ \operatorname {c o s t} (\omega) \big ] \leq \sum_ {i = 2} ^ {k} \mu (S _ {i}) \cdot \Pr \Big (S _ {i} \in \{\operatorname {w i n n e r} (I _ {\omega} ^ {-}), \operatorname {w i n n e r} (I _ {\omega} ^ {+}) \} \text {a n d} h (\omega) <   \frac {L}{\mu (S _ {i})} \Big).
$$
The key observation now is that sets winner  $(I_{\omega}^{-})$  and winner  $(I_{\omega}^{+})$  are independent of  $h(\omega)$ . This is the case, because sets remaining in the competitions  $\mathcal{R}_n(I_\omega^-)$  and  $\mathcal{R}_n(I_\omega^+)$  do not change when we select  $\omega$ . The set  $\mathcal{R}_n(I_\omega^-)$  does not change in the round  $n$  when  $\omega$  is chosen because all sets  $S_i$  in  $\mathcal{R}_n(I_\omega^-) \subset I_\omega^-$  do not contain  $\omega$ . The set  $\mathcal{R}_n(I_\omega^+)$  does not change in this round because all sets  $S_i$  in  $\mathcal{R}_n(I_\omega^+) \subset I_\omega^+$  contain  $\omega$  and consequently when  $\omega$  is chosen, none of these sets is removed from  $\mathcal{R}_n(I_\omega^+)$  (otherwise,  $\mathcal{R}_n(I_\omega^+)$  would become empty). Thus,
$$
\mathbf {E} \big [ \mathrm {c o s t} (\omega) \big ] \leq \sum_ {i = 2} ^ {k} \mu (S _ {i}) \cdot \operatorname * {P r} \left(S _ {i} \in \{\mathrm {w i n n e r} (I _ {\omega} ^ {-}), \mathrm {w i n n e r} (I _ {\omega} ^ {+}) \}\right) \cdot \operatorname * {P r} \left(h (\omega) <   \frac {L}{\mu (S _ {i})}\right).
$$
Using that  $h(\omega)$  is an exponential random variable with parameter  $\mu(\omega)$ , we get (for every  $i$ )
$$
\mu (S _ {i}) \cdot \operatorname * {P r} \left(h (\omega) \leq \frac {L}{\mu (S _ {i})}\right) = \mu (S _ {i}) \cdot \left(1 - e ^ {- L \frac {\mu (\omega)}{\mu (S _ {i})}}\right) \leq \mu (S _ {i}) \cdot L \frac {\mu (\omega)}{\mu (S _ {i})} = \mu (\omega) L.
$$
$$
\mathbf {E} \big [ \operatorname {c o s t} (\omega) \big ] \leq \mu (\omega) L \cdot \sum_ {i = 2} ^ {k} \Pr \left(S _ {i} \in \{\operatorname {w i n n e r} \left(I _ {\omega} ^ {-}\right), \operatorname {w i n n e r} \left(I _ {\omega} ^ {+}\right) \}\right).
$$
The sum on the right hand side is at most 2. Thus,  $\mathbf{E}[\mathrm{cost}(\omega)]\leq 2L\mu (\omega)$
The authors are supported by NSF Awards CCF-1955351, CCF-1934931, EECS-29 2216970.
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