SlowGuess/NeurIPS_2023_OCR huggingface dataset:
where  $(b)$  used that  $\mathbb{P}(R_{t - 1} > B\mid \mathcal{F}_{t - 1}) = 0\Rightarrow \mathcal{T}_t = \emptyset$  and that  $r_t((Y_{i,t})_{i\in \mathcal{N}_t},\emptyset) = 0$  , which implies that almost surely
$$
\begin{array}{l} \mathbb {I} \left(\mathbb {P} \left(R _ {t - 1} > B \mid \mathcal {F} _ {t - 1}\right) = 0\right) \cdot \mathbb {P} \left(R _ {t} \leq B, R _ {t - 1} > B \mid \mathcal {F} _ {t - 1}\right) \\ = \mathbb {I} \left(\mathbb {P} \left(R _ {t - 1} > B \mid \mathcal {F} _ {t - 1}\right) = 0\right) \mathbb {P} \left(R _ {t - 1} + r _ {t} \left(\left(Y _ {i, t}\right) _ {i \in \mathcal {N}}, \emptyset\right) \leq B, R _ {t - 1} > B \mid \mathcal {F} _ {t - 1}\right) \\ = \mathbb {I} \left(\mathbb {P} \left(R _ {t - 1} > B \mid \mathcal {F} _ {t - 1}\right) = 0\right) \cdot \mathbb {P} \left(R _ {t - 1} \leq B, R _ {t - 1} > B \mid \mathcal {F} _ {t - 1}\right) \\ = 0 \\ \end{array}
$$
and  $(c)$  used that  $b_{t}\geq B$  , which implies that almost surely
$$
\mathbb {P} \left(R _ {t} \leq B \mid R _ {t - 1} > B, \mathcal {F} _ {t}\right) \leq \mathbb {P} \left(R _ {t} \leq b _ {t} \mid R _ {t - 1} > B, \mathcal {F} _ {t}\right) \leq \Delta_ {t}.
$$
Rearranging this, we obtain a recurrence relation: for any  $t = 2,\ldots ,T$
$$
\mathbb {P} \left(R _ {t} > B\right) \geq \left(1 - \Delta_ {t}\right) \cdot \mathbb {P} \left(R _ {t - 1} > B\right). \tag {14}
$$
Using the recurrence relation repeatedly for all  $t \in [T]$ , we obtain
$$
\begin{array}{l} \mathbb {P} \left(R _ {T} > B\right) \geq \prod_ {i = 2} ^ {T} \left(1 - \Delta_ {t}\right) \cdot \mathbb {P} \left(R _ {1} > b _ {1}\right) \geq \prod_ {t = 1} ^ {T} \left(1 - \Delta_ {t}\right) \\ \Rightarrow \mathbb {P} \left(R _ {T} \leq B\right) \leq 1 - \prod_ {t = 1} ^ {T} \left(1 - \Delta_ {t}\right) \leq \delta \\ \end{array}
$$
as required. To prove the second claim, observe that equality is attained in all of the above inequalities if equality is attained in (14),  $(i)$ ,  $(ii)$  and  $(iii)$ , and that equality is attained in (14) if equality is attained in  $(a)$  and  $(c)$ . Finally, note that equality in  $(a)$  is attained if  $r_t \leq 0$ ,  $\forall t \in [T]$  and equality in  $(c)$  is attained if equality is attained in  $(i)$  and  $(iv)$ .
B Stochastic domination
Lemma B.1 (Stochastic domination under truncation). For any two independent real random variable  $X, Z$  and real number  $a, t \in \mathbb{R}$  such that  $\mathbb{P}(X < a) > 0$ , we have that
$$
\mathbb {P} (X + Z \geq t \mid X <   a) \leq \mathbb {P} (X + Z \geq t).
$$
Proof of Lemma B.1. Assume that  $\mathbb{P}(X\geq a) > 0$  , or else the proof is trivial. We first claim that  $\mathbb{P}(X + Z\geq t\mid X <   a)\leq \mathbb{P}(X + Z\geq t\mid X\geq a)$  . Note that this holds if and only if
$$
\frac {\mathbb {P} (X \geq t - Z , X <   a)}{\mathbb {P} (X <   a)} \leq \frac {\mathbb {P} (X \geq t - Z , X \geq a)}{\mathbb {P} (X \geq a)}.
$$
The above holds since its lhs and rhs satisfies
$$
\frac {\mathbb {P} (X \geq t - Z , X <   a)}{\mathbb {P} (X <   a)} = \frac {\mathbb {P} (X \geq t - Z , X <   a , a \geq t - Z)}{\mathbb {P} (X <   a)} \leq \mathbb {P} (a \geq t - Z)
$$
$$
\frac {\mathbb {P} (X \geq t - Z , X \geq a)}{\mathbb {P} (X \geq a)} = \frac {\mathbb {P} (X \geq t - Z , X \geq a , a <   t - Z)}{\mathbb {P} (X \geq a)} + \mathbb {P} (a \geq t - Z)
$$
It then follows from law of total probability that
$$
\begin{array}{l} \mathbb {P} (X + Z \geq t) = \mathbb {P} (X + Z \geq t \mid X <   a) \mathbb {P} (X <   a) + \mathbb {P} (X + Z \geq t \mid X \geq a) \mathbb {P} (X \geq a) \\ \geq \mathbb {P} (X + Z \geq t \mid X <   a) \mathbb {P} (X <   a) + \mathbb {P} (X + Z \geq t \mid X <   a) \mathbb {P} (X \geq a) \\ = \mathbb {P} (X + Z \geq t \mid X <   a) \\ \end{array}
$$
Proof of Lemma 3.2. If  $M_{t-1}^{(1)} = 0$ , (8) holds with equality since  $S_{t-1}^{\mathcal{T}}(0) < S_{t-1}^{\mathcal{T}}(1) - B \iff B < 0$ . So, assume  $M_{t-1}^{(1)} > 0$  from now on. By (15f) and the conditional distributions of multivariate Gaussian, we have
$$
\left[ s _ {t} ^ {\mathcal {T}} (0) \middle | S _ {t - 1} ^ {\mathcal {T}} (0), \mathcal {F} _ {t - 1} \right] = \left[ \mu_ {2} + V _ {2 1} V _ {1 1} ^ {- 1} \left(S _ {t - 1} ^ {\mathcal {T}} (0) - \mu_ {1}\right) + \left(V _ {2 2} - V _ {2 1} V _ {1 1} ^ {- 1} V _ {1 2}\right) ^ {1 / 2} Z \middle | S _ {t - 1} ^ {\mathcal {T}} (0), \mathcal {F} _ {t - 1} \right]
$$
where  $Z \sim N(0,1)$  is independent of  $S_{t-1}^{\mathcal{T}}(0)$  conditioned on  $\mathcal{F}_{t-1}$  and  $\mu, V$  are defined in (15f). Here, we used that  $V_{11} > 0$  since  $\sigma_{p,t}(0)^2, \sigma(0)^2 > 0$  by Definition 3.1, and  $M_{t-1}^{(1)} \neq 0$ . Using the above and that  $S_t^\mathcal{T}(0) = s_t^\mathcal{T}(0) + S_{t-1}^\mathcal{T}(0)$ , we have
$$
\begin{array}{l} \left[ S _ {t} ^ {\mathcal {T}} (0) \middle | S _ {t - 1} ^ {\mathcal {T}} (0), \mathcal {F} _ {t - 1} \right] = \left[ (V _ {2 1} V _ {1 1} ^ {- 1} + 1) S _ {t - 1} ^ {\mathcal {T}} (0) + \mu_ {2} - V _ {2 1} V _ {1 1} ^ {- 1} \mu_ {1} \right. \\ + \left(V _ {2 2} - V _ {2 1} V _ {1 1} ^ {- 1} V _ {1 2}\right) ^ {1 / 2} Z \left| S _ {t - 1} ^ {\mathcal {T}} (0), \mathcal {F} _ {t - 1} \right]. \\ \end{array}
$$
Since  $V_{21}V_{11}^{-1} + 1 > 0$  in the above, using also that  $b_{t} - S_{t-1}^{\mathcal{T}}(1), S_{t-1}^{\mathcal{T}}(1) - B \in \mathcal{F}_{t-1}$  and that  $s_{t}^{\mathcal{T}}(1)$  is independent of  $S_{t-1}^{\mathcal{T}}(0), S_{t}^{\mathcal{T}}(0)$ , (8) follows from Lemma B.1.
C Derivation of the decision rule
Proof of these facts follows from standard Bayesian analysis (see e.g. [15])
Lemma C.1 (Posterior distributions). We have for  $w = 0, 1, t \in [T]$
$$
\mu_ {p, t} (1) := \mathbb {E} \left[ \mu_ {\text {t r u e}} (1) \mid \mathcal {F} _ {t - 1} \right] = \frac {1}{\frac {1}{\sigma_ {0} (1) ^ {2}} + \frac {M _ {t - 1} ^ {(1)}}{\sigma (0) ^ {2}}} \left(\frac {\mu_ {0} (1)}{\sigma_ {0} (1) ^ {2}} + \frac {S _ {t - 1} ^ {\mathcal {T}} (1)}{\sigma (1) ^ {2}}\right) \tag {15a}
$$
$$
\mu_ {p, t} (0) := \mathbb {E} \left[ \mu_ {\text {t r u e}} (0) \mid \mathcal {F} _ {t - 1} \right] = \frac {1}{\frac {1}{\sigma_ {0} (0) ^ {2}} + \frac {M _ {t - 1} ^ {(0)}}{\sigma (0) ^ {2}}} \left(\frac {\mu_ {0} (0)}{\sigma_ {0} (0) ^ {2}} + \frac {S _ {t - 1} ^ {\mathcal {C}} (0)}{\sigma (0) ^ {2}}\right) \tag {15b}
$$
$$
\sigma_ {p, t} (w) ^ {2} := \mathbb {V} \left[ \mu_ {\text {t r u e}} (w) \mid \mathcal {F} _ {t - 1} \right] = \left(\frac {1}{\sigma_ {0} (w) ^ {2}} + \frac {M _ {t - 1} (w)}{\sigma (w) ^ {2}}\right) ^ {- 1} \tag {15c}
$$
$$
\left[ \mu_ {\text {t r u e}} (w) \mid \mathcal {F} _ {t - 1} \right] \sim N \left(\mu_ {p, t} (w), \sigma_ {p, t} (w) ^ {2}\right) \tag {15d}
$$
$$
\left[ s _ {t} ^ {\mathcal {T}} (1) \mid \mathcal {F} _ {t - 1} \right] \sim N \left(\mu_ {p, t} (1) \cdot m _ {t}, m _ {t} ^ {2} \cdot \sigma_ {p, t} (1) ^ {2} + m _ {t} \cdot \sigma (0) ^ {2}\right) \tag {15e}
$$
$$
\left[ \binom {S _ {t - 1} ^ {\mathcal {T}} (0)} {s _ {t} ^ {\mathcal {T}} (0)} \right| \mathcal {F} _ {t - 1} \sim N (\mu , V) \tag {15f}
$$
$$
\mu := \left( \begin{array}{c} \mu_ {p, t} (0) \cdot M _ {t - 1} ^ {(1)} \\ \mu_ {p, t} (0) \cdot m _ {t} \end{array} \right),
$$
$$
V := \left( \begin{array}{c c} (M _ {t - 1} ^ {(1)}) ^ {2} \sigma_ {p, t} (0) ^ {2} + M _ {t - 1} ^ {(1)} \sigma (0) ^ {2} & M _ {t - 1} ^ {(1)} m _ {t} \sigma_ {p, t} (0) ^ {2} \\ M _ {t - 1} ^ {(1)} m _ {t} \sigma_ {p, t} (0) ^ {2} & m _ {t} ^ {2} \sigma_ {p, t} (0) ^ {2} + m _ {t} \sigma (0) ^ {2} \end{array} \right).
$$
D Robustness to non-identically distributed and non-Gaussian outcomes
Proof of Theorem 3.3. To show the experiment by Algorithm 1 is  $(\delta, B)$ -RRC under Definition 3.4, it suffices to show that (1), (2) hold for each  $t \geq 1$ . Since (1), (2) hold for each  $t \geq 1$  if  $m_t = 0$ , we only need to show that for each  $t \geq 1$ , if  $m_t \neq 0$ , almost surely
$$
\mathbb {P} \left(S _ {t} ^ {\mathcal {T}} (1) - S _ {t} ^ {\mathcal {T}} (0) > B \mid \mathcal {F} _ {t}\right) > 0 \tag {16a}
$$
$$
\mathbb {P} \left(S _ {t} ^ {\mathcal {T}} (1) - S _ {t} ^ {\mathcal {T}} (0) \leq b _ {t} \mid \mathcal {F} _ {t - 1}, S _ {t - 1} ^ {\mathcal {T}} (1) - S _ {t - 1} ^ {\mathcal {T}} (0) > B\right) \leq \Delta_ {t}. \tag {16b}
$$
Note that for each  $t \geq 1$ , if  $m_t \neq 0$
$$
\begin{array}{l} \mathbb {P} \left(S _ {t} ^ {\mathcal {T}} (1) - S _ {t} ^ {\mathcal {T}} (0) \leq b _ {t} \mid \mathcal {F} _ {t - 1}\right) = \mathbb {P} \left(\frac {s _ {t} ^ {\mathcal {T}} (1) - S _ {t} ^ {\mathcal {T}} (0) - \tilde {\mu} _ {t}}{\tilde {\sigma} _ {t}} \leq z _ {t} \mid \mathcal {F} _ {t - 1}\right) \\ \stackrel {(a)} {\leq} \mathbb {P} \left(\frac {s _ {t} ^ {\mathcal {T}} (1) - S _ {t} ^ {\mathcal {T}} (0) - \breve {\mu} _ {t}}{\breve {\sigma} _ {t}} \leq z _ {t} \mid \mathcal {F} _ {t - 1}\right) \\ \stackrel {(b)} {\leq} \Phi \left(z _ {t}\right) \stackrel {(c)} {\leq} \Delta_ {t} \\ \end{array}
$$
where we used first inequality in (13) in (a), second inequality in (13) in (b), and (11) in (c).
We now show (16a) by induction. For  $t = 1$ , if  $m_1 \neq 0$ , Algorithm 1 ensures that
$$
\mathbb {E} \left(\mathbb {P} \left(s _ {1} ^ {\mathcal {T}} (1) - s _ {1} ^ {\mathcal {T}} (0) \leq b _ {1} \mid \mathcal {F} _ {1}\right)\right) = \mathbb {P} \left(s _ {1} ^ {\mathcal {T}} (1) - s _ {1} ^ {\mathcal {T}} (0) \leq b _ {1}\right) \leq \Delta_ {1} <   1
$$
by construction, which implies that
$$
\mathbb {P} \left(S _ {1} ^ {\mathcal {T}} (1) - S _ {1} ^ {\mathcal {T}} (0) > B \mid \mathcal {F} _ {1}\right) \geq \mathbb {P} \left(s _ {1} ^ {\mathcal {T}} (1) - s _ {1} ^ {\mathcal {T}} (0) > b _ {1} \mid \mathcal {F} _ {1}\right) > 0
$$
almost surely. If  $m_1 = 0$ , then  $\mathbb{P}\left(S_1^{\mathcal{T}}(1) - S_1^{\mathcal{T}}(0) > B\mid \mathcal{F}_1\right) = 1$  since  $B < 0$ . This proves the base case. For the inductive case, if  $m_t\neq 0$ , Algorithm 1 ensures that
$$
\mathbb {E} \left(\mathbb {P} \left(S _ {t} ^ {\mathcal {T}} (1) - S _ {t} ^ {\mathcal {T}} (0) \leq b _ {t} \mid \mathcal {F} _ {t}\right) \mid \mathcal {F} _ {t - 1}\right) = \mathbb {P} \left(S _ {t} ^ {\mathcal {T}} (1) - S _ {t} ^ {\mathcal {T}} (0) \leq b _ {t} \mid \mathcal {F} _ {t - 1}\right) \leq \Delta_ {t} <   1
$$
$$
\mathbb {P} \left(S _ {t} ^ {\mathcal {T}} (1) - S _ {t} ^ {\mathcal {T}} (0) > B \mid \mathcal {F} _ {t}\right) \geq \mathbb {P} \left(S _ {t} ^ {\mathcal {T}} (1) - S _ {t} ^ {\mathcal {T}} (0) > b _ {t} \mid \mathcal {F} _ {t}\right) > 0
$$
almost surely. If  $m_t = 0$ , we have that
$$
\mathbb {P} \left(S _ {t} ^ {\mathcal {T}} (1) - S _ {t} ^ {\mathcal {T}} (0) > B \mid \mathcal {F} _ {t}\right) = \mathbb {P} \left(S _ {t - 1} ^ {\mathcal {T}} (1) - S _ {t - 1} ^ {\mathcal {T}} (0) > B \mid \mathcal {F} _ {t - 1}\right) > 0
$$
from inductive hypothesis. This shows (16a).
To show (16b), note that under Definition 3.4,
$$
\begin{array}{l} \left[ s _ {t} ^ {\mathcal {T}} (1) - S _ {t} ^ {\mathcal {T}} (0) \mid \mathcal {F} _ {t - 1}, S _ {t - 1} ^ {\mathcal {T}} (0) <   S _ {t - 1} ^ {\mathcal {T}} (1) - B \right] \\ \stackrel {d} {=} s _ {t} ^ {\mathcal {T}} (1) - s _ {t} ^ {\mathcal {T}} (0) - \left[ S _ {t - 1} ^ {\mathcal {T}} (0) \mid \mathcal {F} _ {t - 1}, S _ {t - 1} ^ {\mathcal {T}} (0) <   S _ {t - 1} ^ {\mathcal {T}} (1) - B \right] \\ \end{array}
$$
On the rhs,  $s_t^T(1) - s_t^T(0)$  is independent of
$$
\left[ S _ {t - 1} ^ {\mathcal {T}} (0) \mid \mathcal {F} _ {t - 1}, S _ {t - 1} ^ {\mathcal {T}} (0) <   S _ {t - 1} ^ {\mathcal {T}} (1) - B \right]
$$
and that  $S_{t - 1}^{\mathcal{T}}(1) - B\in \mathcal{F}_{t - 1}$ . It follows from these, (16a) and Lemma B.1 that
$$
\begin{array}{l} \mathbb {P} \left(s _ {t} ^ {\mathcal {T}} (1) - s _ {t} ^ {\mathcal {T}} (0) - S _ {t - 1} ^ {\mathcal {T}} (0) \leq b _ {t} \mid \mathcal {F} _ {t - 1}, S _ {t - 1} ^ {\mathcal {T}} (0) <   S _ {t - 1} ^ {\mathcal {T}} (1) - B\right) \\ \leq \mathbb {P} \left(s _ {t} ^ {\mathcal {T}} (1) - S _ {t} ^ {\mathcal {T}} (0) \leq b _ {t} \mid \mathcal {F} _ {t - 1}\right) \\ \end{array}
$$
Therefore, for each  $t \geq 1$ , if  $m_t \neq 0$
$$
\mathbb {P} \left(S _ {t} ^ {\mathcal {T}} (1) - S _ {t} ^ {\mathcal {T}} (0) \leq b _ {t} \mid \mathcal {F} _ {t - 1}, S _ {t - 1} ^ {\mathcal {T}} (1) - S _ {t - 1} ^ {\mathcal {T}} (0) > B\right) \leq \Delta_ {t}
$$
as required. This concludes the proof.
When are (13) satisfied Fix any  $t\geq 1$  where  $m_t\neq 0$  . Note that
$$
\left[ s _ {t} ^ {\mathcal {T}} (1) - S _ {t} ^ {\mathcal {T}} (0) \mid \mathcal {F} _ {t - 1} \right] = \sum_ {i \in \mathcal {T} _ {t}} \left(Y _ {i, t} (1) - Y _ {i, t} (0)\right) - \sum_ {r \in [ t - 1 ]} \sum_ {i \in \mathcal {T} _ {r}} \left[ Y _ {i, r} (0) \mid Y _ {i, r} (1) \right]
$$
The summands on the rhs are independent random variables under Definition 3.4. We thus expect that when  $m_{t}$  or  $M_{t - 1}$  are sufficiently large,
$$
\frac {\left[ s _ {t} ^ {\mathcal {T}} (1) - S _ {t} ^ {\mathcal {T}} (0) \mid \mathcal {F} _ {t - 1} \right] - \mathbb {E} \left[ s _ {t} ^ {\mathcal {T}} (1) - S _ {t} ^ {\mathcal {T}} (0) \mid \mathcal {F} _ {t - 1} \right]}{\sqrt {\mathbb {V} \left[ s _ {t} ^ {\mathcal {T}} (1) - S _ {t} ^ {\mathcal {T}} (0) \mid \mathcal {F} _ {t - 1} \right]}} \approx N (0, 1)
$$
by central limit theorem under mild moment-growth conditions (e.g. Lyapunov's conditions). We thus expect that first condition in (13) holds when  $m_t$  or  $M_{t-1}^{(1)}$  are sufficiently large for each  $t \geq 1$ .
We now focus on the second condition in (13). Suppose that  $\Delta_t \leq 0.5$ , which implies  $z_t \leq 0$  by (11). Note that we can write
$$
\check {\mu} _ {t} = \sum_ {i \in \mathcal {T} _ {t}} \mathbb {E} \left(Y _ {i, t} (1) - Y _ {i, t} (0)\right) - \sum_ {r \in [ t - 1 ]} \sum_ {i \in \mathcal {T} _ {r}} \mathbb {E} \left[ Y _ {i, t} (0) \mid Y _ {i, t} (1) \right]
$$
$$
\check {\sigma} _ {t} ^ {2} = \sum_ {r \in [ t - 1 ]} \mathbb {V} \left(Y _ {i, t} (1) - Y _ {i, t} (0)\right) + \sum_ {i \in \mathcal {T} _ {r}} \mathbb {V} \left[ Y _ {i, t} (0) \mid \sigma \left(Y _ {i, t} (1)\right) \right]
$$
$$
\tilde {\mu} _ {t} = m _ {t} \left(\mu_ {p, t} (1) - \mu_ {p, t} (0)\right) - \mu_ {p, t} (0) M _ {t - 1} ^ {(1)}
$$
$$
\tilde {\sigma} _ {t} ^ {2} = m _ {t} \cdot \left(\sigma (1) ^ {2} + \sigma (0) ^ {2}\right) + M _ {t - 1} ^ {(1)} \cdot \sigma (0) ^ {2} + m _ {t} ^ {2} \cdot \sigma_ {p, t} (1) ^ {2} + \left(m _ {t} + M _ {t - 1} ^ {(1)}\right) ^ {2} \cdot \sigma_ {p, t} (0) ^ {2}.
$$
For  $t = 1$
$$
\check {\mu} _ {t} = \sum_ {i \in \mathcal {T} _ {t}} \mathbb {E} \left(Y _ {i, t} (1) - Y _ {i, t} (0)\right), \quad \check {\sigma} _ {t} ^ {2} = \sum_ {i \in \mathcal {T} _ {t}} \mathbb {V} \left(Y _ {i, t} (1) - Y _ {i, t} (0)\right)
$$
$$
\tilde {\mu} _ {t} = m _ {t} \left(\mu_ {0} (1) - \mu_ {0} (0)\right),
$$
$$
\tilde {\sigma} _ {t} ^ {2} = m _ {t} \cdot \left(\sigma (0) ^ {2} + \sigma (1) ^ {2}\right) + m _ {t} ^ {2} \cdot \left(\sigma_ {0} (1) ^ {2} + \sigma_ {0} (0) ^ {2}\right).
$$
So, second condition in (13) holds for  $t = 1$  if we have chosen prior and model parameters such that
This corresponds to that we choose prior and model parameters conservatively in the sense that we do not overestimate treatment effect or underestimate its variability. Now fix any  $t \geq 2$ . From the law of large number, we expect that for  $M_{t-1}^{(1)}$  sufficiently large
$$
\mu_ {p, t} (0) \approx \frac {1}{M _ {t - 1} ^ {(1)}} \sum_ {r \in [ t - 1 ]} \sum_ {i \in \mathcal {T} _ {r}} \mathbb {E} \left[ Y _ {i, t} (0) \right]
$$
$$
\frac {1}{M _ {t - 1} ^ {(1)}} \sum_ {r \in [ t - 1 ]} \sum_ {i \in \mathcal {T} _ {r}} \mathbb {E} \left[ Y _ {i, t} (0) \mid Y _ {i, t} (1) \right] \approx \frac {1}{M _ {t - 1} ^ {(1)}} \sum_ {r \in [ t - 1 ]} \sum_ {i \in \mathcal {T} _ {r}} \mathbb {E} \left[ Y _ {i, t} (0) \right]
$$
$$
\begin{array}{l} \mu_ {p, t} (1) - \mu_ {p, t} (0) \approx \frac {1}{M _ {t - 1} ^ {(1)}} \sum_ {r \in [ t - 1 ]} \sum_ {i \in \mathcal {T} _ {r}} \mathbb {E} \left[ Y _ {i, t} (1) - Y _ {i, t} (0) \right] \\ \frac {1}{M _ {t - 1} ^ {(1)}} \sum_ {r \in [ t - 1 ]} \sum_ {i \in \mathcal {T} _ {r}} \mathbb {V} \left[ Y _ {i, t} (0) \mid Y _ {i, t} (1) \right] \approx \frac {1}{M _ {t - 1} ^ {(1)}} \sum_ {r \in [ t - 1 ]} \sum_ {i \in \mathcal {T} _ {r}} \mathbb {E} \mathbb {V} \left[ Y _ {i, t} (0) \mid Y _ {i, t} (1) \right] \\ \leq \frac {1}{M _ {t - 1} ^ {(1)}} \sum_ {r \in [ t - 1 ]} \sum_ {i \in \mathcal {T} _ {r}} \mathbb {V} \left[ Y _ {i, t} (0) \right] \\ \end{array}
$$
So if the treatment effects increase or stay roughly constant throughout the experiments
$$
\sigma (0) ^ {2} \geq \frac {1}{M _ {t - 1} ^ {(1)}} \sum_ {r \in [ t - 1 ]} \sum_ {i \in \mathcal {T} _ {r}} \mathbb {V} [ Y _ {i, t} (0) ], \sigma (0) ^ {2} + \sigma (1) ^ {2} \geq \frac {1}{m _ {t}} \sum_ {i \in \mathcal {T} _ {t}} \mathbb {V} (Y _ {i, t} (1) - Y _ {i, t} (0))
$$
the second condition in (13) holds for each  $t \geq 2$  and the experiment produced by Algorithm 1 is  $(\delta, B)$ -RRC.
E Algorithm for general Bayesian models and costs
The following outcome model is a generalization of Definition 3.1. Here, experiment outcomes are allowed to be multivariate with each coordinate corresponds a different business metric.
Definition E.1 (General Bayesian model). Fix  $p, q \geq 1$ . The model parameter  $\theta_{\mathrm{true}} \in \mathbb{R}^q$  is generated from certain prior  $\pi_0$ . The experiment outcome of unit  $i$  at stage  $t$  are distributed independently and identically as
$$
\left(Y _ {i, t} (0), Y _ {i, t} (0)\right) \stackrel {\mathrm {i i d}} {\sim} p (\theta_ {\mathrm {t r u e}})
$$
where  $Y_{i,t}(0),Y_{i,t}(0)\in \mathbb{R}^q$  and  $p(\theta_{\mathrm{true}})$  is a probability distribution on  $\mathbb{R}^{p\times p}$ .
The following is a generalization of Definition 2.1. It allows for general experiment cost beyond treatment effect. The cost of treating unit  $i$  is now  $h_{it} = h_t(Y_{i,t}(1),Y_{i,t}(0))$  for some function  $h_t:\mathbb{R}^{p\times p}\mapsto \mathbb{R}$  chosen by the user. For instance,  $h_t$  can be chosen to compute the worst treatment effect across multiple business metrics.
Definition E.2 (General experiment cost). For each  $t \geq 1$ , let the experiment cost from stage- $t$  and treated unit  $i$  be  $h_{it} = h_t(Y_{i,t}(1), Y_{i,t}(0))$  where  $h_t: \mathbb{R}^{p \times p} \mapsto \mathbb{R}$  is any user-chosen function. Then define  $r_t := \sum_{i \in \mathcal{T}_t} h_{i,t}$ . We let  $r_t = 0$  if  $\mathcal{T}_t = \emptyset$ . Define the cumulative experiment cost up to stage  $t$  as  $R_t := \sum_{k \in [t]} r_k$ .
We now move to derive an explicit algorithm Algorithm 1 from Theorem 3.1 that output  $(m_t)_{t\geq 1}$  such that the experiment is  $(\delta, B)$ -RRC. Compared to Algorithm 1, the algorithm developed in this section will require Monte-Carlo simulations and generally gives more conservative ramp schedule.
We first review the Cantelli's inequality, which is an improved version of the well-known Chebyshev's inequality for one-sided tail bounds.
Lemma E.3 (Cantelli's inequality). For any  $\lambda \geq 0$ , and real-valued random variable  $X$  with finite variance,
$$
\mathbb {P} (X - \mathbb {E} (X) \geq \lambda) \leq \frac {1}{1 + \lambda^ {2} / \mathbb {V} (X)}
$$
Given that (i)  $\mathbb{P}\left(R_{t - 1} \geq B \mid \mathcal{F}_{t - 1}\right) > 0$  and that (ii)  $\mathbb{E}\left[R_t \mid R_{t - 1} \geq B, \mathcal{F}_{t - 1}\right] \geq b_t$ , a direct application of Cantelli's inequality shows that
$$
\begin{array}{l} \mathbb {P} \left(R _ {t} \leq b _ {t} \mid R _ {t - 1} > B, \mathcal {F} _ {t - 1}\right) \\ = \mathbb {P} \left(\mathbb {E} \left[ R _ {t} \mid R _ {t - 1} > B, \mathcal {F} _ {t - 1} \right] - R _ {t} \geq \mathbb {E} \left[ R _ {t} \mid R _ {t - 1} > B, \mathcal {F} _ {t - 1} \right] - b _ {t} \middle | R _ {t - 1} > B, \mathcal {F} _ {t - 1}\right) \\ \leq \left(1 + \frac {\left(\mathbb {E} \left[ R _ {t} \mid R _ {t - 1} > B , \mathcal {F} _ {t - 1} \right] - b _ {t}\right) ^ {2}}{\mathbb {V} \left(R _ {t} \mid R _ {t - 1} > B , \mathcal {F} _ {t - 1}\right)}\right) ^ {- 1} \\ \end{array}
$$
where  $\mathcal{F}_0$  denotes trivial  $\sigma$ -algebra.
Our strategy to construct an algorithm that selects ramp size  $m_t$  such that (1), (2) hold is as follows: we first verify that condition (i) holds; if not, set  $m_t = 0$  and otherwise find  $m_t$  such that the following two inequalities hold
$$
\mathbb {E} \left[ R _ {t} \mid R _ {t - 1} \geq B, \mathcal {F} _ {t - 1} \right] \geq b _ {t} \tag {17a}
$$
$$
\frac {1}{1 + \frac {\left(\mathbb {E} \left[ R _ {t} \mid R _ {t - 1} > B , \mathcal {F} _ {t - 1} \right] - b _ {t}\right) ^ {2}}{\mathbb {V} \left(R _ {t} \mid R _ {t - 1} > B , \mathcal {F} _ {t - 1}\right)}} \leq \Delta_ {t} \tag {17b}
$$
To accomplish this, note that by exchangeability of the outcomes under Definition E.1,
$$
\begin{array}{l} \mathbb {E} \left[ R _ {t} \mid R _ {t - 1} \geq B, \mathcal {F} _ {t - 1} \right] = \mathbb {E} \left[ r _ {t} \mid R _ {t - 1} \geq B, \mathcal {F} _ {t - 1} \right] + \mathbb {E} \left[ R _ {t - 1} \mid R _ {t - 1} \geq B, \mathcal {F} _ {t - 1} \right] \\ = m _ {t} \mathbb {E} \left[ h _ {i = 1, t} \mid R _ {t - 1} \geq B, \mathcal {F} _ {t - 1} \right] + \mathbb {E} \left[ R _ {t - 1} \mid R _ {t - 1} \geq B, \mathcal {F} _ {t - 1} \right] \\ \end{array}
$$
$$
\begin{array}{l} \mathbb {V} \left(R _ {t} \mid R _ {t - 1} \geq B, \mathcal {F} _ {t - 1}\right) = \mathbb {V} \left(r _ {t} \mid R _ {t - 1} \geq B, \mathcal {F} _ {t - 1}\right) + \mathbb {V} \left(R _ {t - 1} \mid R _ {t - 1} \geq B, \mathcal {F} _ {t - 1}\right) \\ + \operatorname {C o v} \left(r _ {t}, R _ {t - 1} \mid R _ {t - 1} \geq B, \mathcal {F} _ {t - 1}\right) \\ = m _ {t} \mathbb {V} \left(h _ {i = 1, t} \mid R _ {t - 1} \geq B, \mathcal {F} _ {t - 1}\right) \\ + m _ {t} \left(m _ {t} - 1\right) \operatorname {C o v} \left(h _ {i = 1, t}, h _ {i = 2, t} \mid R _ {t - 1} \geq B, \mathcal {F} _ {t - 1}\right) \\ + \mathbb {V} \left(R _ {t - 1} \mid R _ {t - 1} \geq B, \mathcal {F} _ {t - 1}\right) + \operatorname {C o v} \left(r _ {t}, R _ {t - 1} \mid R _ {t - 1} \geq B, \mathcal {F} _ {t - 1}\right) \tag {19} \\ \end{array}
$$
We thus require a Monte-Carlo procedure to output estimates  $\hat{\varphi}_t(0),\dots ,\hat{\varphi}_t^{(6)}$  for the following posterior quantities on the rhs of (18), (19)
$$
\mathbb {P} \left(R _ {t - 1} \geq B \mid \mathcal {F} _ {t - 1}\right) \leftarrow \hat {\varphi} _ {t} (0)
$$
$$
\mathbb {E} \left(h _ {i = 1, t} \mid R _ {t - 1} \geq B, \mathcal {F} _ {t - 1}\right) \leftarrow \hat {\varphi} _ {t} (1)
$$
$$
\mathbb {E} \left(R _ {t - 1} \mid R _ {t - 1} \geq B, \mathcal {F} _ {t - 1}\right) \leftarrow \hat {\varphi} _ {t} ^ {(2)}
$$
$$
\mathbb {V} \left(h _ {i = 1, t} \mid R _ {t - 1} \geq B, \mathcal {F} _ {t - 1}\right) \leftarrow \hat {\varphi} _ {t} ^ {(3)}
$$
$$
\operatorname {C o v} \left(h _ {i = 1, t}, h _ {i = 2, t} \mid R _ {t - 1} \geq B, \mathcal {F} _ {t - 1}\right) \leftarrow \hat {\varphi} _ {t} ^ {(4)}
$$
$$
\mathbb {V} \left[ R _ {t - 1} \mid R _ {t - 1} \geq B, \mathcal {F} _ {t - 1} \right] \leftarrow \hat {\varphi} _ {t} ^ {(5)}
$$
$$
\operatorname {C o v} \left(h _ {i = 1, t}, R _ {t - 1} \mid R _ {t - 1} \geq B, \mathcal {F} _ {t - 1}\right) \leftarrow \hat {\varphi} _ {t} ^ {(6)}
$$
where  $h_{i = 1,t}, h_{i = 2,t}$  denote costs from treating two units  $i = 1, 2$  at stage  $t$ . Recall that under (...), the outcome of the units are exchangeable. So  $i = 1, 2$  simply refers to any two distinct units. These quantities will be used to construct estimates of  $\mathbb{E}\left[R_t \mid R_{t - 1} > B, \mathcal{F}_{t - 1}\right]$  and  $\mathbb{V}(R_t \mid R_{t - 1} > B, \mathcal{F}_{t - 1})$  as functions of  $m_t$  chosen.
We now outline a procedure to construct  $\hat{\varphi}_t(0),\dots ,\hat{\varphi}_t^{(6)}$  . Firstly, suppose that we can obtain  $K$  samples from the posterior distribution
$$
\left[ \left(Y _ {i, r} (0)\right) _ {i \in \mathcal {T} _ {r}, r \in [ t - 1 ]}, Y _ {i = 1, t} (0), Y _ {i = 1, t} (1), Y _ {i = 2, t} (0), Y _ {i = 2, t} (1) \mid \mathcal {F} _ {t - 1} \right], \tag {20}
$$