SlowGuess/NeurIPS_2023_OCR huggingface dataset:
We select NN as the prediction and preliminary calibration models in both PTC-B and PTC-E algorithms and compare them to benchmarks. Here, because  $c$  is only 1-dimensional, we replaced the DCC and IDCC algorithms, which are based on deep neural networks, with another clustering method, K-Means. Figure 5 shows the performance of different algorithms when  $d$  increases ( $d = 1, 4,$  and 16), and the plot is shown by fixing  $z_{2}, \ldots, z_{d}$  to 0 in the test phase. The Ellipsoid method performs the worst and gives an unchanged uncertainty set for all different  $z$  due to its ignorance of the contextual information. All algorithms degrade with an increase in irrelevant information, but localized methods such as kNN and K-Means exhibit more severe deterioration. Our proposed methods, however, maintain a certain degree of stability.
Figure 5: Comparison in performance deterioration with increasing dimensions
A.2 Shortest Path Problem
In this section, we consider a shortest path problem on a  $5 \times 5$  grid with 40 edges, and the cost of traveling through edge  $i$  is  $c_{i}$  for  $i = 1, \dots, 40$ . After observing the covariates  $z$  (e.g., weather conditions, day of the week), we aim to select a route from the left-top vertex to the right-bottom one to minimize the value-at-risk travel time, i.e.,  $\mathrm{VaR}_{\alpha, z}(c^{\top} x)$ . The data generation process is as follows. First, we generate a 0-1 matrix  $\Theta \in \mathbb{R}^{40 \times d}$  once with random seed 0 and fix it to encode the parameters of the true model, where each entry is generated from a Bernoulli distribution with probability 0.5. Then, the covariate vector  $\boldsymbol{z}^{t} \in \mathbb{R}^{d}$  is generated from  $N(0, I_{d})$  for  $t = 1, \dots, T$ .
Figure 6: Box plot of scaled VaR on shortest path problems under different risk level settings
Given  $z^t$ , the cost on edge  $i$  is  $c_i^t = \left[\left(\frac{1}{\sqrt{d}} (\Theta z^t) + 3\right)^5 + 1\right] \cdot \epsilon_i$ , where  $\epsilon_i \sim \mathrm{Unif}\left[\frac{3}{4}, \frac{5}{4}\right]$  independently for  $i = 1, \dots, 40$ . We implement  $d = 10$  here. To simulate the uneliminated irrelevant information, we set the last two dimensions of  $z$  to be independent of  $c$ , so their corresponding column vectors in  $\Theta$  are set to be zero. In the test phase, we evaluate the performance on 500  $z$ 's generated from the marginal distribution of  $z$ . We select the Kernel Ridge method with the RBF kernel as our prediction model and the NN model as our preliminary calibration model in both PTC-B and PTC-E.
In Figure 6, we show the box plot version of Table 1 with scaled VaR. Here, to alleviate the impact of randomness on  $z$ , we scale the VaR by the optimal value of the corresponding conditional stochastic program with expectation objective, i.e.,  $\mathrm{VaR}_{\alpha} / \mathrm{Obj}^{E}$ , where  $\mathrm{Obj}^{E} = \max_{Ax = b, x \geq 0} \mathbb{E}[c|z]^{\top} x$ . Figure 6 once again demonstrates the superior performance of our algorithms in minimizing VaR as that in Table 1.
A.3 Fractional Knapsack Problem
In this section, we consider a fractional knapsack problem and follow the setup in Ho-Nguyen and Kılınc-Karzan [2022]. This problem can be interpreted as a maximization problem, wherein a consumer aims to maximize their utility by selecting items under a budget constraint. Specifically, to fit our formulation, the problem can be written as
$$
\min  _ {x \in \mathcal {X}} \operatorname {V a R} _ {\alpha , z} (- c ^ {\top} x), \text {w h e r e} \mathcal {X} := \{x \in [ 0, 1 ] ^ {n}: p ^ {\top} x \leq B \}.
$$
Here,  $c$  is the utility vector and we use a minimization version to represent a maximization problem.  $p \in \mathbb{R}^n$  is some fixed positive vector indicating the price of items, and  $B > 0$  is the budget. The same as the shortest path problem, we randomly generate a 0-1 matrix  $\Theta \in \mathbb{B}^{20 \times d}$  once with the last 2 columns entirely filled as zeros to indicate the independence of the last dimensions of  $z$  to  $c$ . Then, we generate multivariate independent  $z$  from  $\mathrm{Unif}[0,4]^d$ , and  $c = (\Theta z)^2 \cdot \epsilon$ , where  $\epsilon$  is multivariate independent and following  $\mathrm{Unif}\left[\frac{4}{5},\frac{6}{5}\right]^n$ . We conduct experiments on data generated under  $T = 5000$ ,  $d = 10$ , and  $n = 20$ , and we choose the Kernel Ridge method with RBF kernel as our prediction model, and the NN as our preliminary calibration model in both the PTC-B and PTC-E.
Given a new covariate  $z$ , the conditional RO algorithm will output an uncertainty set. Under different constraints, the worst-case scenario corresponding to the uncertainty set may vary, so we compare the performance under randomly generated constraints. We generate the price vector  $p$  with each entry as an integer uniformly sampled from [1, 1000]. Then, we randomly generated  $u \sim \mathrm{Unif}[0,1]$ , and the
Figure 7: Box plot of scaled VaR on knapsack problems under different risk level settings
budget  $B$  is sampled from  $\mathrm{Unif}[\max_j p_j, \mathbf{1}^\top p - u \cdot \max_j p_j]$ . Here, we randomly generated 10 sets of constraints  $\{(p^i, B^i)\}_{i=1}^{10}$  (with random seed 0) and fix them. In the test phase, the average VaR and coverage are taken over  $100z$ 's and 10 constraints (1000 instances in total).
Table 2 associated with Figure 7 displays the comparison results under varying risk levels and demonstrates the advantages of our proposed algorithms again via the lower VaR and proper coverage that closely matches the risk level, where the scaled VaR in Figure 7 is the VaR divided by  $\max_{x\in \mathcal{X}}\mathbb{E}[c|z]^{\top}x$ . The contextual-ignorance method, Ellipsoid, performs the worst with the total utilities remaining 0, which shows its excessive conservatism by specifying the worst-case utility of each item to be non-positive and selecting nothing. The DCC and IDCC, as well as our PTC-B and PTC-E methods, exhibit better performance than the kNN method due to our utilization of expressive NN models, which are not limited to utilizing only local information. By virtue of the flexibility of our algorithms in the model selection, the chosen Kernel Ridge regression model is more suitable than the clustering-based DCC and IDCC methods for this problem with continuous covariates, so our algorithms achieve relatively better performance than theirs.
Table 2: Average VaR and coverage on fractional knapsack problems under different risk level settings. The result is reported based on 500 simulation trials.
(a) Average VaR
α	Ellips.	kNN	DCC	IDCC	PTC-B	PTC-E
0.6	0	-1175	-1256	-1131	-1310	-1310
0.7	0	-1148	-1241	-1111	-1298	-1297
0.8	0	-1137	-1230	-1090	-1283	-1282
0.85	0	-1128	-1223	-1086	-1273	-1273
0.9	0	-1114	-1213	-1087	-1261	-1261
0.95	0	-1102	-1180	-1092	-1242	-1243
(b) Average coverage
α	Ellips.	kNN	DCC	IDCC	PTC-B	PTC-E
0.6	0.66	0.96	0.75	0.65	0.57	0.65
0.7	0.74	0.96	0.79	0.74	0.73	0.71
0.8	0.82	0.96	0.86	0.83	0.79	0.82
0.85	0.86	0.96	0.89	0.89	0.88	0.86
0.9	0.91	0.96	0.92	0.93	0.91	0.91
0.95	0.96	0.96	0.97	0.95	0.92	0.97
B Supplementary for Section 3
This section provides discussions about the individual coverage and proofs for theoretical statements in Section 3. In Section B.1, we will provide a brief literature review about global coverage and individual coverage, and then, provide an algorithm to achieve the individual coverage with
corresponding theoretical results. In Sections B.2, B.3, and B.4, we will show Propositions 1, 2, 3, respectively. The proof will utilize some useful lemmas in Section D.
B.1 Individual Coverage
The uncertainty quantification literature arises from two communities. It is referred to as "conformal prediction" in the field of statistics and "model calibration" in the field of machine learning. In the following, we briefly review the results of global and individual coverage in uncertainty quantification in those two communities, respectively.
Conformal prediction: Milestone works on the topic, Lei et al. [2018], Barber et al. [2023], focus on global coverage. Angelopoulos and Bates [2021], a survey paper on conformal prediction, also focuses on the global guarantee. For conditional/individual coverage, Foygel Barber et al. [2021], Vovk [2012] state an "impossible triangle" in uncertainty quantification: (i) the coverage result is built for conditional coverage; (ii) it makes no assumptions on the underlying distribution; (iii) it has finite-sample guarantee rather than asymptotic consistency. Essentially, in our paper, Algorithms 1 and 2 basically fail to meet (i), but they barely rely on any assumptions on the underlying distribution and enjoy a finite-sample guarantee, that is, achieving (ii) and (iii). Later in this section, we will introduce a nonparametric algorithm that achieves conditional coverage, and the coverage can be finite-sample, i.e., achieving (i) and (iii), while this result requires additional assumptions on the underlying distribution, which is a violation of (ii).
Model calibration/recalibration: Lakshminarayanan et al. [2017], Cui et al. [2020], Chung et al. [2021], Kuleshov et al. [2018], Zhao et al. [2020] are a few state-of-the-art papers on calibrating the uncertainty of a regression model. We note that none of these papers claim to achieve a conditional coverage guarantee. Rather, they ensure a global guarantee and strive for better empirical conditional coverage. In particular, Zhao et al. [2020] gives a pessimistic result on conditional coverage that it is impossible to verify whether an algorithm achieves individual coverage. However, the proof is based on a very special case, and it is possible to rule it out with some mild conditions.
In the following, we present Algorithm 3 under our predict-then-calibrate framework, which can achieve individual coverage.
Algorithm 3 Uncertainty Quantification with Individual Coverage
1: Input: Validation data  $\mathcal{D}_{val}$ , ML model  $\hat{f}$ , kernel choice  $K_{h}(\cdot, \cdot)$ , bandwidth  $h$ , target covariate  $z_{0}$
2: For each  $t \in \mathcal{D}_{val}$ , let
3: Define the estimation of the conditional distribution  $r|z = z_0$ :
$$
\hat {P} _ {r | z _ {0}} := \sum_ {t \in \mathcal {D} _ {v a l}} \frac {K _ {h} (z _ {t} , z _ {0}) \delta_ {r _ {t}}}{\sum_ {t \in \mathcal {D} _ {v a l}} K _ {h} (z _ {t} , z _ {0})},
$$
where  $\delta_r$  denotes the delta distribution on  $r$
4: Choose a minimal  $\eta > 0$  such that
$$
\hat {P} _ {r \mid z _ {0}} (\{r: \| r \| _ {2} \leq \eta \}) \geq \alpha
$$
5: Output: Uncertainty set  $\mathcal{U}_{\alpha}(z_0) = \left\{c:\| c - \hat{f} (z_0)\| _2\leq \eta \right\}$
Here, this uncertainty quantification algorithm is mainly based on a nonparametric estimation in Hall et al. [1999], Liu et al. [2023], which is also similar to the nonparametric estimation in Algorithm 4. The estimator in Step 3 in Algorithm 3 covers a wide range of non-parametric estimators with different kernels, including  $k$ -nearest neighbors, and it can approximate the conditional distributions of the residuals  $r|z = z_0$  with a sublinear convergence rate under some Lipschitz conditions. As a result, by using this estimator, we can directly calculate the size of the uncertainty set so that the estimated conditional distribution attains individual coverage. That is Step 4 in Algorithm 3. In the
following, we will provide a statement without proof since the analysis is almost the same as the proof of Proposition 4. We also refer to Hall et al. [1999], Liu et al. [2023] for a detailed analysis.
Proposition 5. Assume there exists a constant  $L'$  such that the following Lipschitz condition holds for any  $z, z'$
$$
T V \left(P _ {r \mid z}, P _ {r \mid z ^ {\prime}}\right) \leq L \| z - z ^ {\prime} \| _ {2} ^ {s},
$$
where  $TV(\cdot, \cdot)$  denotes the total variation distance between two distributions, and  $P_{r|z}, P_{r|z'}$  denotes the conditional distribution of the residuals given covariates  $z$  or  $z'$ , respectively. Then under mild boundedness assumptions on the distribution, the uncertainty set  $\mathcal{U}_{\alpha}(z_0)$  output from Algorithm 3 satisfies
$$
T V (\hat {P} _ {r \mid z _ {0}}, P _ {r \mid z _ {0}}) \leq O \left(T ^ {- \frac {s}{2 (s + d)}} \log T\right).
$$
Moreover, it is worth noting that, through this approximation of the conditional distributions, we can extend our framework in tractable optimizing conditional value-at-risk besides VaR in (3). Specifically, one can generate a number of independent samples  $\{\tilde{c}_k\}_{k=1}^K$  from the approximation of the conditional distribution  $c|z = z_0$ . Then to optimize the following empirical CVaR objective (Rockafellar et al. [2000])
$$
\min  _ {x, \gamma} \gamma + \frac {1}{K (1 - \alpha)} \sum_ {k = 1} ^ {K} \left(\tilde {c} _ {t} ^ {\top} x - \gamma\right) ^ {+}
$$
This is a convex program that has an equivalent linear program and is thus computationally tractable.
To close this section, we want to emphasize that the Lipschitz assumption in Proposition 5 cannot be verified from the data prior. Also, we prefer not to claim we have solved the conditional coverage problem for two reasons: (i) We do not want to give an impression to the robust optimization community that conditional coverage is an easy task to achieve. One should proceed with caution when applying the nonparametric algorithm; (ii) We don't want to give an impression to the statistics and ML communities working on uncertainty quantification problems that we underestimate the difficulty of the conditional coverage and lack of understanding of the related literature.
B.2 Proof for Proposition 1 and Corollary 1
In this section, we first show Proposition 1, then show Corollary 1 based on the result of Proposition 1.
Proposition 1. For a new sample  $(c, A, b, z)$  from distribution  $\mathcal{P}$ , denote the uncertainty sets output from Algorithm 1 and Algorithm 2 by  $\mathcal{U}_{\alpha}^{(1)}(z)$  and  $\mathcal{U}_{\alpha}^{(2)}(z)$ , respectively. The following inequalities hold for  $k = 1, 2$
$$
\alpha \leq \mathbb {P} \left(c \in \mathcal {U} _ {\alpha} ^ {(k)} (z)\right) \leq \alpha + \frac {1}{| \mathcal {D} _ {2} | + 1}
$$
Proof. Denote  $n = |\mathcal{D}_2|$  as the number of samples in the  $\mathcal{D}_2$ . In the following, we show for  $k = 1, 2$ ,
$$
\mathbb {P} \left(c \in \mathcal {U} _ {\alpha} ^ {(k)} (z)\right) = \frac {\lceil \alpha (n + 1) \rceil}{n + 1}, \tag {12}
$$
where  $\lceil \cdot \rceil$  denotes the ceiling function. If it holds, we can prove this proposition by the following inequalities
$$
\frac {\left[ \alpha (n + 1) \right]}{n + 1} \geq \frac {\alpha (n + 1)}{n + 1} = \alpha ,
$$
$$
\frac {\left[ \alpha (n + 1) \right]}{n + 1} \leq \frac {\alpha n + 2}{n + 1} = \alpha + \frac {1}{n + 1}.
$$
To show (12), we will use the exchangeability of the dataset  $\mathcal{D}_2$  and new sample  $(c, A, b, z)$ . Specifically, denote  $\{c_t\}_{t=1}^n$  as the samples in  $\mathcal{D}_2$ , and denote  $c_{n+1}$  as the new sample. Then, we define  $\eta_t$
for  $t = 1, \dots, n + 1$  with respect to Algorithms 1 or 2 as follows
$$
\eta_ {t} = \left\{ \begin{array}{l l} \min  \left\{\eta \geq 0: \underline {{c}} _ {t} (\eta) \leq c _ {t} \leq \bar {c} _ {t} (\eta) \right\}, & \text {i f} k = 1, \\ \min  \left\{\eta \geq 0 \sqrt {(c _ {t} - \hat {f} (z _ {t})) ^ {\top} \hat {\Sigma} ^ {- 1} (c _ {t} - \hat {f} (z _ {t}))} \leq \eta \hat {g} (z _ {t}) \right\}, & \text {i f} k = 2. \end{array} \right.
$$
In the proof below, we only show (12) for Algorithm 1, and the proof for Algorithm 2 can be obtained similarly. Based on the choice of the uncertainty set and definition of  $\eta$  in Algorithm 1, we have that  $\eta$  is the  $\frac{[\alpha(n + 1)]}{n}$ -upper quantile in  $\{\eta_t\}_{t = 1}^n$ . This choice of  $\eta$  also implies  $c_{t + 1}\in \mathcal{U}_\alpha^{(1)}$  if  $\eta_{n + 1}\leq \eta$ , which means
$$
\mathbb {P} \left(c _ {n + 1} \in \mathcal {U} _ {\alpha} ^ {(1)}\right) = \mathbb {P} \left(\eta_ {n + 1} \text {i s a m o n g t h e} \lceil \alpha (n + 1) \rceil \text {- s m a l l e s t o f} \{\eta_ {t} \} _ {t = 1} ^ {n + 1}\right). \tag {13}
$$
Then, it is sufficient to show
$$
\mathbb {P} \left(\eta_ {n + 1} \text {i s a m o n g t h e} \lceil \alpha (n + 1) \rceil \text {- s m a l l e s t o f} \{\eta_ {t} \} _ {t = 1} ^ {n + 1}\right) = \frac {\lceil \alpha (n + 1) \rceil}{n + 1}.
$$
To this end, because  $\{c_t\}_{t = 1}^{n + 1}$  are i.i.d., we have
$$
\begin{array}{l} \mathbb {P} \left(\eta_ {n + 1} \text {i s a m o n g t h e} \lceil \alpha (n + 1) \rceil - \text {s m a l l e s t o f} \{\eta_ {t} \} _ {t = 1} ^ {n + 1}\right) \tag {14} \\ = \mathbb {P} \left(\eta_ {t} \text {i s m o n a n g l e} \left\lceil \alpha (n + 1) \right\rceil - \text {s m i l l e s t o f} \left\{\eta_ {t} \right\} _ {t = 1} ^ {n + 1}\right) \\ \end{array}
$$
for all  $t = 1, \dots, n$ , which is also referred to as the exchangability. Then, we can finish the proof by the following equalities:
$$
\begin{array}{l} \mathbb {P} \left(\eta_ {n + 1} \text {i s a m o n g t h e} \lceil \alpha (n + 1) \rceil - \text {s m a l l e s t o f} \{\eta_ {t} \} _ {t = 1} ^ {n + 1}\right) \\ = \frac {1}{n + 1} \sum_ {t = 1} ^ {n + 1} \mathbb {P} \left(\eta_ {t} \text {i s a m o n g t h e} \lceil \alpha (n + 1) \rceil - \text {s m a l l e s t o f} \{\eta_ {t} \} _ {t = 1} ^ {n + 1}\right) \\ = \frac {1}{n + 1} \mathbb {E} \left[ \sum_ {t = 1} ^ {n + 1} 1 \left\{\eta_ {t} \text {i s a m o n g t h e} \lceil \alpha (n + 1) \rceil - \text {s m a l l e s t o f} \left\{\eta_ {t} \right\} _ {t = 1} ^ {n + 1} \right\} \right] \\ = \frac {\lceil \alpha (n + 1) \rceil}{n + 1}, \\ \end{array}
$$
where the first equality is obtained by the exchangeability (14), the second equality is obtained by the definition of the expectation, and the last equality is obtained by the fact that
$$
\sum_ {t = 1} ^ {n + 1} 1 \left\{\eta_ {t} \text {i s a m o n g t h e} \lceil \alpha (n + 1) \rceil \text {- s m a l l e s t o f} \left\{\eta_ {t} \right\} _ {t = 1} ^ {n + 1} \right\} = \lceil \alpha (n + 1) \rceil .
$$
Next, we show Corollary 1:
where the probability is with respect to the new sample  $(c,A,b,z)$  and  $\mathcal{D}_2$ .
Proof. By definition of the robust optimization problem (5), we have
$$
c ^ {\top} x ^ {*} (z) \leq O P T (z)
$$
if  $c\in \mathcal{U}_\alpha^{(k)}(z)$  for  $k = 1,2$  . Then, Proposition 1 implies
$$
\mathbb {P} \left(c ^ {\top} x ^ {*} (z) \leq \operatorname {O P T} (z)\right) \geq \alpha .
$$
B.3 Proof for Proposition 2
Proposition 2. For any  $\alpha \in (0.5,1)$
$$
\frac {1}{\Gamma (k)} \left(\gamma \left(k, \max  \left\{0, - d \log \left((1 - \alpha) \left(1 + \frac {1}{d}\right) ^ {d - k}\right) \right\}\right) - \gamma \left(k, \max  \left\{0, - d \log \left(\alpha \left(1 + \frac {1}{d}\right) ^ {d - k}\right) \right\}\right)\right),
$$
and it decreases monotonously with respect to  $k$ .
$$
c = \sum_ {i = 1} ^ {d} z _ {i} - d \epsilon , \tag {15}
$$
and  $z_{i}$  and  $\epsilon$  are drawn from an Exponential distribution  $\mathrm{Exp}(1)$  for all  $i = 1, \dots, d$ . For any fixed  $k = 1, \dots, d$ , the corresponding prediction model is  $f_{k}(z_{1:k}) = \sum_{i=1}^{k} z_{i}$ .
We first characterize the distribution of the objective vector  $c$  given  $z_{1:k}$  for any  $k = 1, \ldots, d$ . By the additivity of Gamma distributions and the fact that the exponential distribution is a special case of the Gamma distribution, we have given  $z_{1:k}$
$$
\begin{array}{l} \mathbb {P} (c \geq 0) = \int_ {0} ^ {\infty} \dots \int_ {0} ^ {\infty} 1 \left\{f _ {k} \left(z _ {1: k}\right) + \sum_ {i = k + 1} ^ {d} z _ {i} \geq d \cdot \epsilon \right\} \cdot \prod_ {i = k + 1} ^ {d} e ^ {- z _ {i}} \cdot e ^ {- \epsilon} \mathrm {d} \epsilon \mathrm {d} z _ {k + 1} \dots \mathrm {d} z _ {d} \\ = \int_ {0} ^ {\infty} \int_ {0} ^ {\infty} 1 \left\{f _ {k} \left(z _ {1: k}\right) + \tilde {z} \geq d \cdot \epsilon \right\} \cdot \frac {\tilde {z} ^ {d - k + 1} e ^ {- \tilde {z}}}{\Gamma (d - k)} \cdot e ^ {- \epsilon} \mathrm {d} \epsilon \mathrm {d} \tilde {z} \tag {16} \\ = \int_ {0} ^ {\infty} \left(1 - e ^ {- (f _ {k} (z _ {1; k}) + \tilde {z}) / d}\right) \cdot \frac {\tilde {z} ^ {d - k + 1} e ^ {- \tilde {z}}}{\Gamma (d - k)} \mathrm {d} \tilde {z} \\ = 1 - e ^ {- f _ {k} \left(z _ {1: k}\right) / d} \cdot \left(1 + \frac {1}{d}\right) ^ {k - d}, \\ \end{array}
$$
where the first line comes from the definition of the objective vector (15), the second line comes from changing variables that  $\tilde{z} = \sum_{i=k+1}^{d} z_i$  and the additivity of Gamma distributions, and others come from the direct calculation. Then, by taking the complement, we have
$$
\mathbb {P} (c \leq 0) = e ^ {- f _ {k} \left(z _ {1: k}\right) / d} \cdot \left(1 + \frac {1}{d}\right) ^ {k - d}. \tag {17}
$$
Next, we compute the probability that  $x_{\alpha}^{*}(z_{1:k}) = 0$ . In the one-dimensional setting with the feasible set  $\{x: -1 \leq x \leq 1\}$ , the optimal solution of (5) can be determined by the sign of the objective vector. Thus, if we are confident that  $c$  is non-negative, the optimal solution should be  $x_{\alpha}^{*}(z_{1:k}) = -1$ ; if we are confident that  $c$  is non-positive, the optimal solution is  $x_{\alpha}^{*}(z_{1:k}) = 1$ ; if we are not confident about the sign of the objective vector, we will be conservative and choose  $x_{\alpha}^{*}(z_{1:k}) = 0$ . That is,
$$
x _ {\alpha} ^ {*} \left(z _ {1: k}\right) = \left\{ \begin{array}{l l} 1, & \text {i f} \mathbb {P} (c \leq 0 \mid z _ {1: k}) \geq \alpha , \\ 0, & \text {i f} \mathbb {P} (c \leq 0 \mid z _ {1: k}), \mathbb {P} (c \geq 0 \mid z _ {1: k}) \leq \alpha , \\ - 1, & \text {i f} \mathbb {P} (c \geq 0 \mid z _ {1: k}) \geq \alpha . \end{array} \right. \tag {18}
$$
Then, plugging (16) and (17) into (18), we have  $\mathbb{P}(c\leq 0|z_{1;k}),\mathbb{P}(c\geq 0|z_{1;k})\leq \alpha$  only when
$$
\left. \right. \max  \left\{0, - d \log \left(\alpha \left(1 + \frac {1}{d}\right) ^ {d - k}\right)\right\} \leq f _ {k} \left(z _ {1: k}\right) \leq \max  \left\{0, - d \log \left((1 - \alpha) \left(1 + \frac {1}{d}\right) ^ {d - k}\right)\right\}. \tag {19}
$$
Integrating the probability on the set (19) with respect to  $z_{1:k}$ , we have
$$
\mathbb {P} \left(x _ {\alpha} ^ {*} \left(z _ {1: k}\right) = 0\right) = \tag {20}
$$
$$
\left. \frac {1}{\Gamma (k)} \left(\gamma \left(k, \max  \left\{0, - d \log \left((1 - \alpha) \left(1 + \frac {1}{d}\right) ^ {d - k}\right) \right\}\right) - \gamma \left(k, \max  \left\{0, - d \log \left(\alpha \left(1 + \frac {1}{d}\right) ^ {d - k}\right) \right\}\right)\right). \right. \tag {21}
$$
To see it decrease, we can apply Jensen's inequality to see the result.
B.4 Proof for Proposition 3
Proposition 3. For  $\alpha \geq 1/2$ , let the uncertainty set be
$$
\mathcal {U} _ {\alpha} (z) = \left\{ \begin{array}{l l} \left[ \sqrt {z} - \frac {1 - \sqrt {2 - 2 \alpha}}{2}, \infty\right), & \text {i f} z \in [ 0, 1 ], \\ (- \infty , - \sqrt {- z} + \frac {1 - \sqrt {2 - 2 \alpha}}{2} ], & \text {i f} z \in [ - 1, 0). \end{array} \right.
$$
$$
x (z) = \left\{ \begin{array}{l l} - 1, & \text {i f} z \geq \frac {3 - 2 \alpha - 2 \sqrt {2 - 2 \alpha}}{4}, \\ 0, & \text {i f} \frac {- 3 + 2 \alpha + 2 \sqrt {2 - 2 \alpha}}{4} \leq z \leq \frac {3 - 2 \alpha - 2 \sqrt {2 - 2 \alpha}}{4}, \\ 1, & \text {i f} z \leq \frac {- 3 + 2 \alpha + 2 \sqrt {2 - 2 \alpha}}{4}. \end{array} \right.
$$
Proof. We first show that  $\mathbb{P}(c\in \mathcal{U}_{\alpha}(z)) = \alpha$ . Given the prediction function  $\hat{f} (z) = \mathrm{sign}(z)\cdot \sqrt{|z|}$ , the residual is  $r = c - \hat{f} (z) = \epsilon \sqrt{|z|}$ . By calculation, we have that the marginal distribution of the objective vector  $c$  satisfies
$$
\mathbb {P} (r \leq r _ {0}) = \left\{ \begin{array}{l l} - 2 r _ {0} ^ {2} + 2 r _ {0} + \frac {1}{2}, & \text {i f} 0 \leq r _ {0} \leq 1 / 2, \\ 2 r _ {0} ^ {2} + 2 r _ {0} + \frac {1}{2}, & \text {i f} - 1 / 2 \leq r _ {0} \leq 0, \end{array} \right. \tag {22}
$$
and this distribution is the same as the marginal distribution of the objective vector given  $z \geq 0$  or  $z \leq 0$  since the distribution of  $r$  only depends on the absolute value of  $z$ . Then, we have