SlowGuess/NeurIPS_2023_OCR huggingface dataset:
The proof of Lemma 4 follows easily by plugging in  $\eta = (\gamma \cdot \sqrt{nT})^{-1}$ ,  $\alpha = (8kD\gamma^2)^{-1}$  and  $\epsilon = \mathcal{O}(1 / \sqrt{T})$ .
D.5 Implementation of Projected Mirror Descent
We conclude this section by considering the Projected Mirror Descent update step, namely
$$
y ^ {\prime} = \underset {y ^ {*} \in \mathcal {F P} (\mathcal {T})} {\arg \min } [ \eta \cdot \langle c, y ^ {*} \rangle + \cdot D _ {R _ {\mathcal {T}}} (y ^ {*}, y) ]
$$
that takes as input a fractional facility placement  $y \in \mathcal{FP}(\mathcal{T})$  and returns some other  $y' \in \mathcal{FP}(\mathcal{T})$  that minimizes a linear cost under vector  $c$  plus the Bregman Divergence between the initial and the new point under regularizer  $R_{\mathcal{T}}$ . Here,  $\eta > 0$  is a tuning parameter that balances the dynamics between the linear cost and the Bregman Divergence.
By letting  $c$  be the sub-gradient of the fractional connection cost over the observed sequence of clients, we can use this update step in order to approximate the FTRL objective; this is, in fact, the implementation we did for our experimental evaluation of Algorithm 2. In this section we will argue that the special structure of  $R_{\mathcal{T}}$  allows us to compute the update step in linear (to the size of the HST) time.
By definition of the Bregman Divergence, we have
$$
D _ {R _ {\mathcal {T}}} (x, y) = R _ {\mathcal {T}} (x) - R _ {\mathcal {T}} (y) - \langle \nabla R _ {\mathcal {T}} (y), x - y \rangle
$$
Substituting everything, we get that the update step of Projected Mirror Descent can be written as
$$
y^{\prime} = \operatorname *{arg  min}_{y^{*}\in \mathcal{FP}(\mathcal{T})}F(y^{*})
$$
$$
\begin{array}{l} F (y ^ {*}) = \eta \cdot \sum_ {v} c _ {v} \cdot y _ {v} ^ {*} + \sum_ {v \neq r} w _ {v} \cdot \left(y _ {v} ^ {*} + \delta_ {v}\right) \cdot \ln \left(\frac {y _ {v} ^ {*} + \delta_ {v}}{y _ {p (v)} ^ {*} + \delta_ {p (v)}}\right) \\ - \sum_ {v \neq r} w _ {v} \cdot (y _ {v} + \delta_ {v}) \cdot \ln \left(\frac {y _ {v} + \delta_ {v}}{y _ {p (v)} + \delta_ {p (v)}}\right) \\ - \sum_ {v \neq r} \left(w _ {v} \cdot \ln \left(\frac {y _ {v} + \delta_ {v}}{y _ {p (v)} + \delta_ {p (v)}}\right) + \frac {w _ {v}}{2} + \frac {w _ {v}}{2} \cdot \mathbb {1} [ v \in L (\mathcal {T}) ]\right) (y _ {v} ^ {*} - y _ {v}) \\ - \frac {w _ {r}}{2} \left(y _ {r} ^ {*} - y _ {r}\right) \\ \end{array}
$$
It is always the case that we update  $y^\prime$  from some  $y\in \mathcal{FP}(\mathcal{T})$ , so we can simplify the above expression to get
$$
\begin{array}{l} F (y ^ {*}) = \eta \cdot \sum_ {v} c _ {v} \cdot y _ {v} ^ {*} + \sum_ {v \neq r} w _ {v} \cdot \left(y _ {v} ^ {*} + \delta_ {v}\right) \cdot \ln \left(\frac {\frac {y _ {v} ^ {*} + \delta_ {v}}{y _ {p (v)} ^ {*} + \delta_ {p (v)}}}{\frac {y _ {v} + \delta_ {v}}{y _ {p (v)} + \delta_ {p (v)}}}\right) \\ - \sum_ {v} \frac {w _ {v}}{2} \cdot (1 + \mathbb {1} [ v \in L (\mathcal {T}) ]) \cdot \left(y _ {v} ^ {*} - y _ {v}\right) \\ \end{array}
$$
Recall that by definition,  $\mathcal{FP}(T)$  is the polytope
$$
\mathcal {F P} (\mathcal {T}) = \left\{y \in \mathbb {R} ^ {| V (\mathcal {T}) |}: \begin{array}{l l} y _ {v} = \sum_ {u \in \operatorname {c l d} (v)} y _ {u} & v \notin L (\mathcal {T}) \\ y _ {v} \in [ 0, 1 ] & v \in L (\mathcal {T}) \\ y _ {r} = k \end{array} \right.
$$
Since our objective is to minimize function  $F(\cdot)$  over  $\mathcal{FP}(\mathcal{T})$ , we can write down the KKT optimality conditions to obtain the following conditions about the minimizer  $y^{*}$ :
$$
\frac {y _ {v} ^ {*} + \delta_ {v}}{y _ {p (v)} ^ {*} + \delta_ {p (v)}} = \frac {y _ {v} + \delta_ {v}}{y _ {p (v)} + \delta_ {p (v)}} \cdot \exp \left(\frac {1}{w _ {v}} (\mu_ {p (v)} - \mu_ {v} - \eta c _ {v})\right)
$$
where  $\mu_v$  is the Lagrange multiplier for constraint  $y_v = \sum_{u \in \operatorname{cld}(v)} y_u$  and  $\mu_v = 0$  for  $v \in L(\mathcal{T})$ . To complete our computation of  $y^*$ , it remains to compute the Lagrange multipliers  $\mu$ .
Since  $y^{*}\in \mathcal{FP}(\mathcal{T})$ , it is not hard to verify that for any  $v\notin L(\mathcal{T})$  it holds
$$
\sum_ {u \in \operatorname {c l d} (v)} \frac {y _ {u} ^ {*} + \delta_ {u}}{y _ {v} ^ {*} + \delta_ {v}} = 1
$$
and using the KKT optimality condition, this implies that for any  $v \notin L(\mathcal{T})$
$$
\sum_ {u \in \operatorname {c l d} (v)} \frac {y _ {u} + \delta_ {u}}{y _ {v} + \delta_ {v}} \cdot \exp \left(\frac {1}{w _ {u}} (\mu_ {v} - \mu_ {u} - \eta c _ {u})\right) = 1
$$
or equivalently, since  $w_{v} = 2w_{u}$  for all  $u\in \operatorname {cld}(v)$
$$
\mu_ {v} = - \frac {w _ {v}}{2} \cdot \ln \left(\sum_ {u \in \operatorname {c l d} (v)} \frac {y _ {u} + \delta_ {u}}{y _ {v} + \delta_ {v}} \cdot \exp \left(- \frac {\mu_ {u} + \eta c _ {u}}{w _ {u}}\right)\right)
$$
Thus, starting from  $\mu_v = 0$  on the leaves, this expression provides as a bottom-up algorithm to compute all the Lagrange multipliers  $\mu$ . Using these multipliers and the KKT optimality conditions, we can then easily compute the ratios
for all vertices  $v \neq r$ . Finally, we can start from the root vertex  $r$ , for which we know that  $y_r^* = k$  and cascade these ratios downwards until we reach the leaves and we have compute all entries of  $y^*$ . Clearly, this is all done in linear time to the number of vertices.
Intuitively, this update step can be interpreted as an application of the Multiplicative Weights Update algorithm on every parent vertex  $v$  that decides how its mass should be split to its children. We repeat this process in a bottom-up manner, and then we simply start with  $k$  facilities on the root and begin splitting them based on these ratios while moving downwards.
E Analysis of Cut&Round (Proofs of Section 5)
In this chapter of the appendix we present all the omitted proofs from Section 5 concerning our online rounding scheme Cut&Round. To avoid repetition, from now on we fix an arbitrary HST  $\mathcal{T}$  and use  $\mathcal{FP}(\mathcal{T})$  to denote the set of all fractional placements of  $k$  facilities on the leaves of  $\mathcal{T}$ .
Roadmap. In section E.1, we argue about the correctness of Cut&Round; namely, we show that no matter the input, Cut&Round always returns a set of  $k$ -leaves of  $\mathcal{T}$  where the facilities are placed. Then, in section E.2 we establish the main property of Cut&Round and prove Lemma 5. Finally, in section E.3 we analyze the expected connection cost of Cut&Round's output and prove Item 1 of Theorem 4 (Lemma 6) while in section E.4 we analyze the expected moving cost of Cut&Round's output and prove Item 2 of Theorem 4 (Lemma 7).
E.1 Correctness of Cut&Round
We begin by proving the correctness of Cut&Round. Fix any  $y \in \mathcal{FP}(\mathcal{T})$  and any set of thresholds  $\alpha \in [0,1]^{|V(\mathcal{T})|}$ . Let  $F = \mathrm{Cut\& Round}(\mathcal{T},y,\alpha)$ . In this section, we will prove that  $|F| = k$ , i.e., we will argue that Cut&Round always returns a set of  $k$  leaves at which facilities must be placed, as it is expected to. In order to show this, we will need to analyze the  $Y_{v}$  variables produced by Cut&Round.
Claim 14. For any leaf  $v \in L(\mathcal{T})$ , it holds that  $Y_{v} \in \{0,1\}$ .
Proof. Observe that for any  $v \in V(\mathcal{T})$ , sub-routine Alloc sets  $Y_v$  to either  $\lfloor y_v \rfloor$  or  $\lfloor y_v \rfloor + 1$ . By definition of  $\mathcal{FP}(\mathcal{T})$ , we have  $y_v \in [0,1]$  for each leaf  $v \in L(\mathcal{T})$ . We distinguish between two different cases. If  $y_v \in [0,1)$ , then clearly  $Y_v \in \{0,1\}$ . If  $y_v = 1$ , then  $\delta(y_v) = 0$  and thus Alloc will always set  $Y_v = \lfloor y_v \rfloor = 1$ . Thus, the claim holds for all leaves  $v \in L(\mathcal{T})$ .
Claim 15. Let  $v \notin L(\mathcal{T})$  be any non-leaf vertex. Then,  $Y_{v} = \sum_{u \in \operatorname{cld}(v)} Y_{u}$ .
Proof. Fix any non-leaf vertex  $v \notin L(\mathcal{T})$ . We will analyze the inner loop of Cut&Round that iterates over  $v$ 's children. Initially, Cut&Round sets  $Y_{rem} = Y_v$  and  $y_{rem} = y_v$ . Then, we proceed to iteratively call Alloc, once per child vertex of  $v$ . Each time Alloc assigns some value  $Y_u$  to a child vertex  $u \in \operatorname{cld}(v)$ , we update  $Y_{rem}$  to  $Y_{rem} - Y_u$ ; thus, to prove our claim it suffices to argue that after we update the last child vertex, we have  $Y_{rem} = 0$ .
Since by definition of sub-routine Alloc we know that  $Y_v \in \{\lfloor y_v \rfloor, \lfloor y_v \rfloor + 1\}$ , we know that initially (before any child vertex is assigned a value  $Y_u$ ) it holds that  $Y_{rem} \in \{\lfloor y_{rem} \rfloor, \lfloor y_{rem} \rfloor + 1\}$ . In fact, a simple case analysis over the decision tree of sub-routine Alloc suffices to see that this invariant holds not only at the beginning, but even after we begin assigning values to the child vertices and update  $Y_{rem}$  and  $y_{rem}$ .
Since  $y \in \mathcal{FP}(\mathcal{T})$ , we know that  $y_v = \sum_{u \in \operatorname{cld}(v)} y_u$  and thus  $y_{rem} = y_u$  at the time we iterate over the last child vertex  $u \in \operatorname{cld}(v)$ . Furthermore, from the above discussion we know that  $Y_{rem} \in \{\lfloor y_u \rfloor, \lfloor y_u \rfloor + 1\}$ . Since  $\delta(y_u) = \delta(y_{rem})$ , it is easy to verify that in any case Alloc sets  $Y_u = Y_{rem}$  and thus after the last update we have  $Y_{rem} = 0$ , as desired.
Proof of Correctness. Recall that by definition, the output of Cut&Round is  $F = \{v \in L(\mathcal{T}) : Y_v = 1\}$ . Since from Claim 14 we know that  $Y_v \in \{0,1\}$  for all  $v \in L(\mathcal{T})$ , this implies that  $|F| = \sum_{v \in L(\mathcal{T})} Y_v$ . We apply Claim 15 to the root vertex  $r$ , then again to each  $u \in \operatorname{cld}(r)$  and so on until we reach the leaves. This gives us that  $Y_r = \sum_{v \in L(\mathcal{T})} Y_v$  and thus  $|F| = Y_r$ . Since by definition Cut&Round sets  $Y_r = k$ , we have proven that  $|F| = k$  as desired.
E.2 Proof of Lemma 5 (Computing the Allocation Probabilities)
In this section, we formally prove the main property of algorithm Cut&Round, as stated in Lemma 5. Fix any fractional facility placement  $y \in \mathcal{FP}(\mathcal{T})$  and let  $\alpha_v \sim \mathrm{Unif}(0,1)$  be independent uniformly random thresholds for all  $v \in V(\mathcal{T})$ . Let  $F = \mathrm{Cut}\& \mathrm{Round}(\mathcal{T},y,\alpha)$  be the output of algorithm Cut&Round on this set of inputs. Recall that algorithm Cut&Round sets the variables  $Y_{v}$  during its execution, for all  $v \in V(\mathcal{T})$ . As we have already discussed,  $Y_{v}$  is the total number of facilities in  $F$  on the leaves of the sub-tree rooted at  $v$ , i.e.  $Y_{v} = |T(v) \cap F|$ . We will prove that for any  $v \in V(\mathcal{T})$ , we have
We begin by writing down the following property for sub-routine Alloc:
Claim 16. Fix any fractional facility placement  $y \in \mathcal{FP}(\mathcal{T})$  and let  $\alpha_v \sim \mathrm{Unif}(0,1)$  for all  $v \in V(\mathcal{T})$ . For any vertex  $u \in V(\mathcal{T})$  of  $\mathcal{T}$ , let  $Y_u = \mathrm{Alloc}(y_u, y_{rem}, Y_{rem}, \alpha_u)$  be the number of facilities assigned to the sub-tree of  $u$  by Line 8 of Algorithm Cut&Round (Algorithm 4). Then,
$$
\mathbb {P} _ {\alpha} \left[ Y _ {u} = \lfloor y _ {u} \rfloor \right] = \left\{ \begin{array}{l l} 1 & i f Y _ {r e m} = \lfloor y _ {r e m} \rfloor a n d \delta (y _ {u}) \leq \delta (y _ {r e m}) \\ \frac {1 - \delta (y _ {u})}{1 - \delta (y _ {r e m})} & i f Y _ {r e m} = \lfloor y _ {r e m} \rfloor a n d \delta (y _ {u}) > \delta (y _ {r e m}) \\ 0 & i f Y _ {r e m} \neq \lfloor y _ {r e m} \rfloor a n d \delta (y _ {u}) > \delta (y _ {r e m}) \\ \frac {\delta (y _ {r e m}) - \delta (y _ {u})}{\delta (y _ {r e m})} & i f Y _ {r e m} \neq \lfloor y _ {r e m} \rfloor a n d \delta (y _ {u}) \leq \delta (y _ {r e m}) \end{array} \right.
$$
Proof. This claim is a direct consequence of sub-routine Alloc's description (Algorithm 5) and the fact that  $\alpha_v \sim \mathrm{Unif}(0,1)$  for all  $v \in V(\mathcal{T})$ .
Using this claim, we are now ready to prove Lemma 5.
Proof of Lemma 5. We prove the lemma via a top-down induction on the vertices of  $\mathcal{T}$  (decreasing level order). For the root vertex, we know that since  $y\in \mathcal{FP}(\mathcal{T})$  we have  $y_{r} = k$  and also by definition of Cut&Round we have  $Y_{r} = k$ . Thus, we get that  $Y_{r} = y_{r} = \lfloor y_{r}\rfloor$  with probability  $1 - \delta (y_r) = 1$  and the claim holds. Now, fix any non-leaf vertex  $v\notin L(\mathcal{T})$  and assume that  $Y_{v} = \lfloor y_{v}\rfloor$  with probability  $1 - \delta (y_v)$  and  $Y_{v} = \lfloor y_{v}\rfloor +1$  with probability  $\delta (y_v)$ . To complete our induction, we will now proceed to prove the claim for all the children vertices of  $v$ .
We begin by proving the claim for the first child of vertex  $v$ , and then we will show how the same arguments extend for all its children. Let  $u \in \operatorname{cld}(v)$  be the first child vertex of  $v$  that Cut&Round iterates over. Then, by definition of Cut&Round we have that  $Y_{rem} = Y_v$  and  $y_{rem} = y_v$ . Using the inductive hypothesis on  $v$ , this implies that  $Y_{rem} = \lfloor y_{rem} \rfloor$  with probability  $1 - \delta(y_{rem})$  and  $Y_{rem} = \lfloor y_{rem} \rfloor + 1$  with probability  $\delta(y_{rem})$ . Conditioning on the value of  $Y_{rem}$ , we get
$$
\begin{array}{l} \mathbb {P} _ {\alpha} \left[ Y _ {u} = \left\lfloor y _ {u} \right\rfloor \right] = \mathbb {P} _ {\alpha} \left[ Y _ {u} = \left\lfloor y _ {u} \right\rfloor \mid Y _ {r e m} = \left\lfloor y _ {r e m} \right\rfloor \right] \cdot (1 - \delta \left(y _ {r e m}\right)) \\ + \mathbb {P} _ {\alpha} \left[ Y _ {u} = \left\lfloor y _ {u} \right\rfloor \mid Y _ {r e m} = \left\lfloor y _ {r e m} \right\rfloor + 1 \right] \cdot \delta (y _ {r e m}) \\ \end{array}
$$
We distinguish between two different cases based on whether  $\delta(y_u) \leq \delta(y_{rem})$  or  $\delta(y_u) > \delta(y_{rem})$ . In any case, we can use Claim 16 to substitute the conditional probabilities on the above expression and easily get that
$$
\mathbb {P} _ {\alpha} \left[ Y _ {u} = \left\lfloor y _ {u} \right\rfloor \right] = 1 - \delta \left(y _ {u}\right)
$$
Thus, we have already proven the claim for the first child of  $v$ . However, to complete our induction, we need to prove the claim for all children of  $v$  and not just the first one. The only property that we used and holds specifically for the first child was that  $Y_{rem} = \lfloor y_{rem} \rfloor$  with probability  $1 - \delta(y_{rem})$  and  $Y_{rem} = \lfloor y_{rem} \rfloor + 1$  with probability  $\delta(y_{rem})$ . Let  $Y_{rem}'$  and  $y_{rem}'$  be the updated remaining facilities after the value  $Y_u$  of the first child has been assigned. If we can prove that  $Y_{rem}' = \lfloor y_{rem}' \rfloor$  with probability  $1 - \delta(y_{rem}')$  and  $Y_{rem}' = \lfloor y_{rem}' \rfloor + 1$  with probability  $\delta(y_{rem}')$ , then we can keep applying the same argument and inductively prove the claim for all the children of  $v$ .
By definition, we have that  $Y_{rem}' = Y_{rem} - Y_u$  and  $y_{rem}' = y_{rem} - y_u$ . Once again, we distinguish between two different cases.
- Let  $\delta(y_u) \leq \delta(y_{rem})$ . In that case, we get that  $\lfloor y_{rem}' \rfloor = \lfloor y_{rem} \rfloor - \lfloor y_u \rfloor$  and also that  $\delta(y_{rem}') = \delta(y_{rem}) - \delta(y_u)$ . Since we know that  $Y_{rem} \in \{\lfloor y_{rem} \rfloor, \lfloor y_{rem} \rfloor + 1\}$  and  $Y_u \in \{\lfloor y_u \rfloor, \lfloor y_u \rfloor + 1\}$ , this implies that
$$
\begin{array}{l} \mathbb {P} _ {\alpha} \left[ Y _ {r e m} ^ {\prime} = \left\lfloor y _ {r e m} ^ {\prime} \right\rfloor \right] = \mathbb {P} _ {\alpha} \left[ Y _ {r e m} = \left\lfloor y _ {r e m} \right\rfloor \cap Y _ {u} = \left\lfloor y _ {u} \right\rfloor \right] \\ + \mathbb {P} _ {\alpha} \left[ Y _ {r e m} = \left\lfloor y _ {r e m} \right\rfloor + 1 \cap Y _ {u} = \left\lfloor y _ {u} \right\rfloor + 1 \right] \\ \end{array}
$$
Using conditional probabilities and the inductive hypothesis on the distribution of  $Y_{rem}$ , we obtain
$$
\begin{array}{l} \mathbb {P} _ {\alpha} \left[ Y _ {r e m} ^ {\prime} = \left\lfloor y _ {r e m} ^ {\prime} \right\rfloor \right] = \mathbb {P} _ {\alpha} \left[ Y _ {u} = \left\lfloor y _ {u} \right\rfloor \mid Y _ {r e m} = \left\lfloor y _ {r e m} \right\rfloor \right] \cdot (1 - \delta \left(y _ {r e m}\right)) \\ + \mathbb {P} _ {\alpha} \left[ Y _ {u} = \left\lfloor y _ {u} + 1 \right\rfloor \mid Y _ {r e m} = \left\lfloor y _ {r e m} \right\rfloor + 1 \right] \cdot \delta (y _ {r e m}) \\ \end{array}
$$
Using Claim 16 to substitute the conditional probabilities, we finally get
$$
\mathbb {P} _ {\alpha} \left[ Y _ {r e m} ^ {\prime} = \left\lfloor y _ {r e m} ^ {\prime} \right\rfloor \right] = 1 - \delta \left(y _ {r e m}\right) + \delta \left(y _ {u}\right) = 1 - \delta \left(y _ {r e m} ^ {\prime}\right)
$$
- Let  $\delta(y_u) > \delta(y_{rem})$ . In that case, we get that  $\lfloor y_{rem}' \rfloor = \lfloor y_{rem} \rfloor - \lfloor y_u \rfloor - 1$  and also that  $\delta(y_{rem}') = 1 + \delta(y_{rem}) - \delta(y_u)$ . Since we know that  $Y_{rem} \in \{\lfloor y_{rem} \rfloor, \lfloor y_{rem} \rfloor + 1\}$  and  $Y_u \in \{\lfloor y_u \rfloor, \lfloor y_u \rfloor + 1\}$ , this implies that
$$
\mathbb {P} _ {\alpha} \left[ Y _ {r e m} ^ {\prime} = \left\lfloor y _ {r e m} ^ {\prime} \right\rfloor \right] = \mathbb {P} _ {\alpha} \left[ Y _ {r e m} = \left\lfloor y _ {r e m} \right\rfloor \cap Y _ {u} = \left\lfloor y _ {u} \right\rfloor + 1 \right]
$$
$$
\mathbb {P} _ {\alpha} \left[ Y _ {r e m} ^ {\prime} = \left\lfloor y _ {r e m} ^ {\prime} \right\rfloor \right] = \mathbb {P} _ {\alpha} \left[ Y _ {u} = \left\lfloor y _ {u} \right\rfloor + 1 \mid Y _ {r e m} = \left\lfloor y _ {r e m} \right\rfloor \right] \cdot (1 - \delta \left(y _ {r e m}\right))
$$
$$
\mathbb {P} _ {\alpha} \left[ Y _ {r e m} ^ {\prime} = \left\lfloor y _ {r e m} ^ {\prime} \right\rfloor \right] = \delta \left(y _ {u}\right) - \delta \left(y _ {r e m}\right) = 1 - \delta \left(y _ {r e m} ^ {\prime}\right)
$$
Thus, we have concluded the proof of Lemma 5.
E.3 Proof of Item 1 in Theorem 4 (Bounding the Expected Connection Cost)
Lemma 6. Let  $F = \mathrm{Cut} \& \mathrm{Round}(y, \alpha)$  where for all  $v \in V(\mathcal{T})$ ,  $\alpha_v \sim \mathrm{Unif}(0,1)$  independently. Then,
$$
\mathbb {E} _ {\alpha} \left[ C _ {R} (F) \right] = f _ {R} (y) \text {f o r a n y} R \subseteq L (\mathcal {T})
$$
Proof. Fix any  $y \in \mathcal{FP}(\mathcal{T})$  and let  $\alpha \in [0,1]^{|V(\mathcal{T})|}$  be a set of thresholds such that for each  $v \in V(\mathcal{T})$ ,  $\alpha_v$  is drawn independently at random from the uniform distribution, i.e.  $\alpha_v \sim \mathrm{Unif}(0,1)$ . Let  $F = \mathrm{Cut} \& \mathrm{Round}(\mathcal{T}, y, \alpha)$ . We will prove that for any set of clients  $R \subseteq L(\mathcal{T})$ , it holds that  $\mathbb{E}_{\alpha}[C_R(F)] = f_R(y)$ .
Recall that the  $Y_{v}$  variables set by Cut&Round denote the total number of facilities in  $F$  that are placed on the sub-tree rooted at vertex  $v$ , i.e.  $Y_{v} = |F \cap T(v)|$ . As argued in section E.1, we know that  $Y \in \mathcal{FP}(\mathcal{T}) \cap \mathbb{N}$ , i.e.  $Y$  is a valid integral facility placement. Thus, from Claim 1 of section C.1, we know that  $C_{R}(F) = f_{R}(Y)$ . This implies that by definition of the fractional connection cost under client request  $R$ , we have that
$$
C _ {R} (F) = \sum_ {j \in R} \sum_ {v \in P (j, r)} 2 ^ {\operatorname {l e v} (v) + 1} \cdot \max  \left(0, 1 - Y _ {v}\right)
$$
Thus, we get
$$
\begin{array}{l} \mathbb {E} _ {\alpha} \left[ C _ {R} (F) \right] = \sum_ {j \in R} \sum_ {v \in P (j, r)} 2 ^ {\operatorname {l e v} (v) + 1} \cdot \mathbb {E} _ {\alpha} \left[ \max  \left(0, 1 - Y _ {v}\right) \right] \\ = \sum_ {j \in R} \sum_ {v \in P (j, r)} 2 ^ {\operatorname {l e v} (v) + 1} \cdot \mathbb {P} _ {\alpha} \left[ Y _ {v} = 0 \right] \\ \end{array}
$$
where the first equality holds by linearity of expectation, and the second equality holds by the fact that  $Y_v \in \mathbb{N}$  for all  $v \in V(\mathcal{T})$ . Since  $Y_v \in \{\lfloor y_v \rfloor, \lfloor y_v \rfloor + 1\}$ , we know that for any  $v \in V(\mathcal{T}), Y_v$  can be 0 only if  $y_v \in [0,1)$ . Furthermore, from Lemma 5, we know that in the case of uniformly random thresholds, this happens with probability precisely  $1 - y_v$ . Combining these facts, we get  $\mathbb{P}_{\alpha}[Y_v = 0] = \max(0, 1 - y_v)$  and thus
$$
\begin{array}{l} \mathbb {E} _ {\alpha} \left[ C _ {R} (F) \right] = \sum_ {j \in R} \sum_ {v \in P (j, r)} 2 ^ {\operatorname {l e v} (v) + 1} \cdot \max  (0, 1 - y _ {v}) \\ = f _ {R} (y) \\ \end{array}
$$
concluding the proof of Lemma 6.
E.4 Proof of Item 2 in Theorem 4 (Bounding the Expected Moving Cost)
Lemma 7. Let  $F = \mathrm{Round\& Cut}(y, \alpha)$  and also let  $F' = \mathrm{Round\& Cut}(\mathcal{T}, y', \alpha)$  where  $\alpha_v \sim \mathrm{Unif}(0,1)$  for all  $v \in V(\mathcal{T})$ . Then,
$$
\gamma \cdot \mathbb {E} _ {\alpha} \left[ M _ {\mathcal {T}} \left(F, F ^ {\prime}\right) \right] \leq 4 \cdot | | y - y ^ {\prime} | | _ {\mathcal {T}}
$$
Proof. Fix any pair of fractional facility placements  $y, y' \in \mathcal{FP}(\mathcal{T})$  and let corresponding outputs of Cut&Round be denoted as  $F = \mathrm{Cut\&Round}(\mathcal{T}, y, \alpha)$  and  $F' = \mathrm{Cut\&Round}(\mathcal{T}, y', \alpha)$ . Observe that the same set of (uniformly random) thresholds  $\alpha_v$  is used in both cases, as this will play a crucial part in our analysis. To prove Lemma 7, we need to prove that
where the expectation is taken over the value of the uniformly random thresholds  $\alpha_{v}$ .
The proof of Lemma 7 is technically involved, and thus we will break down our approach into smaller sections to ease the presentation. We begin by proving the Lemma in the special case where the transition from  $y$  to  $y'$  has a very simple structure, which we now proceed to define:
Definition 10. We say that two fractional facility placements  $y, y' \in \mathcal{FP}(\mathcal{T})$  are  $\epsilon$ -neighboring if there are two leaves  $s, t \in L(\mathcal{T})$  with least common ancestor  $p \in V(\mathcal{T})$  such that the following hold:
1.  $y_v' = y_v - \epsilon$  for all  $v \in P(s, p) \setminus \{p\}$ .
2.  $y_{v}^{\prime} = y_{v} + \epsilon$  for all  $v\in P(t,p)\setminus \{p\}$
3.  $y_{v}^{\prime} = y_{v}$  for all other  $v\in V(\mathcal{T})$
Furthermore, we say that  $y, y'$  are strictly  $\epsilon$ -neighboring if  $\epsilon$  is sufficiently small to satisfy
1.  $\epsilon \leq \delta(y_v)$  for all  $v \in P(s, p) \setminus \{p\}$  with  $\delta(y_v) > 0$ .
2.  $\epsilon \leq 1 - \delta (y_v)$  for all  $v\in P(t,p)\setminus \{p\}$  with  $\delta (y_{v}) > 0$
3.  $\epsilon < 1$
Basically, if  $y$  and  $y'$  are  $\epsilon$ -neighboring then  $y'$  is obtained by pushing  $\epsilon$ -mass on  $y$  from  $s$  to  $t$  along the unique path that connects these two leaves. Furthermore, if  $\epsilon$  is sufficiently small so that for any  $v \in V(\mathcal{T})$  either  $\lfloor y_v \rfloor = \lfloor y_v' \rfloor$  or  $|y_v - y_v'| \leq 1$  and at least one of the two is integral, then we say that the two fractional facility placements are strictly  $\epsilon$ -neighboring. As we will shortly argue, Lemma 7 holds in the special case where  $y, y'$  are strictly  $\epsilon$ -neighboring.
Claim 17. If  $y, y' \in \mathcal{FP}(\mathcal{T})$  are strictly  $\epsilon$ -neighboring for some  $\epsilon \geq 0$ , then
$$
\gamma \cdot \mathbb {E} _ {\alpha} \left[ M _ {\mathcal {T}} \left(F, F ^ {\prime}\right) \right] \leq 4 \cdot | | y - y ^ {\prime} | | _ {\mathcal {T}}.
$$