SlowGuess/NeurIPS_2023_OCR huggingface dataset:
A.3.1 Proof for Example 3.3.1 17
A.3.2 Proof of Example 3.3.2 17
A.3.3 Proof of Lemma A.1 18
B Additional numerical experiments 18
B.1 Comparison between one-shot and exact weighted approaches 19
B.2 Local coverage of cmc 19
B.3 Additional results for homogeneous missingness 19
B.4 Estimation error for one-bit matrix estimation 19
B.5 Additional results for heterogeneous missingness 20
B.6 Misspecified sampling probability 20
B.7 Additional results for the sales dataset 22
B.7.1 Comparison under misspecified missingness 22
B.7.2 Results with convex optimization 22
C Additional details of algorithms and extensions 23
C.1 Extension to likelihood-based scores and categorical matrices 23
C.2 Full conformalized matrix completion 24
C.3 Exact split conformalized matrix completion 25
The section collects the proofs for the technical results in the main text. Section A.1 and A.2 are devoted to the proof of Lemma 3.1 and Theorem 3.2, respectively. In the end, Section A.3 provides proofs for the coverage gap bounds in the two examples in Section 3.3.
Fix a set of locations  $\mathcal{B} \coloneqq \{(i_1, j_1), \ldots, (i_{n_{\mathrm{cal}} + 1}, j_{n_{\mathrm{cal}} + 1})\}$ , and an index  $1 \leq m \leq n_{\mathrm{cal}} + 1$ . By the definition of conditional probability, we have
$$
\begin{array}{l} \mathbb {P} \left(\left(i _ {*}, j _ {*}\right) = \left(i _ {m}, j _ {m}\right) \mid \mathcal {S} _ {\mathrm {c a l}} \cup \left\{\left(i _ {*}, j _ {*}\right) \right\} = \mathcal {B}, \mathcal {S} _ {\mathrm {t r}} = \mathcal {S} _ {0}\right) \\ = \frac {\mathbb {P} \left(\left(i _ {*} , j _ {*}\right) = \left(i _ {m} , j _ {m}\right) , \mathcal {S} _ {\mathrm {c a l}} = \mathcal {B} \backslash \left\{\left(i _ {m} , j _ {m}\right) \right\} \mid \mathcal {S} _ {\mathrm {t r}} = \mathcal {S} _ {0} , | \mathcal {S} | = n\right)}{\mathbb {P} \left(\mathcal {S} _ {\mathrm {c a l}} \cup \left\{\left(i _ {*}, j _ {*}\right) \right\} = \mathcal {B} \mid \mathcal {S} _ {\mathrm {t r}} = \mathcal {S} _ {0} , | \mathcal {S} | = n\right)} \\ = \frac {\mathbb {P} \left((i _ {*} , j _ {*}) = (i _ {m} , j _ {m}) , \mathcal {S} _ {\mathrm {c a l}} = \mathcal {B} \backslash \{(i _ {m} , j _ {m}) \} \mid \mathcal {S} _ {\mathrm {t r}} = \mathcal {S} _ {0} , | \mathcal {S} | = n\right)}{\sum_ {l = 1} ^ {n _ {\mathrm {c a l}} + 1} \mathbb {P} \left((i _ {*}, j _ {*}) = (i _ {l} , j _ {l}) , \mathcal {S} _ {\mathrm {c a l}} = \mathcal {B} \backslash \{(i _ {l} , j _ {l}) \} \mid \mathcal {S} _ {\mathrm {t r}} = \mathcal {S} _ {0} , | \mathcal {S} | = n\right)}. \\ \end{array}
$$
It then boils down to computing  $\mathbb{P}\left(S_{\mathrm{cal}} = S_1, (i_*, j_*) = (i_l, j_l) \mid S_{\mathrm{tr}} = S_0, |S| = n\right)$  for any fixed set  $S_1$  and fixed location  $(i_l, j_l) \notin S_0 \cup S_1$ . To this end, we have the following claim (we defer its
proof to the end of this section)
$$
\begin{array}{l} \mathbb {P} \left(\mathcal {S} _ {\mathrm {c a l}} = \mathcal {S} _ {1}, \left(i _ {*}, j _ {*}\right) = \left(i _ {l}, j _ {l}\right) \mid \mathcal {S} _ {\mathrm {t r}} = \mathcal {S} _ {0}, | \mathcal {S} | = n\right) \\ = \frac {1}{d _ {1} d _ {2} - n} \cdot \frac {\prod_ {(i , j) \in \mathcal {S} _ {1}} \frac {p _ {i j}}{1 - p _ {i j}}}{\sum_ {\mathcal {A} \in \Omega_ {\mathcal {S} _ {\mathrm {t r}}, n}} \prod_ {(i ^ {\prime} , j ^ {\prime}) \in \mathcal {A}} \frac {p _ {i ^ {\prime} j ^ {\prime}}}{1 - p _ {i ^ {\prime} j ^ {\prime}}}}, \tag {8} \\ \end{array}
$$
where  $\Omega_{\mathcal{S}_{\mathrm{tr}},n} = \{\mathcal{A}\subseteq [d_1]\times [d_2]:|\mathcal{A}| = n - |\mathcal{S}_{\mathrm{tr}}|,\mathcal{A}\cap \mathcal{S}_{\mathrm{tr}} = \emptyset \}$ . As a result, we obtain
$$
\begin{array}{l} \mathbb {P} \left(\left(i _ {*}, j _ {*}\right) = \left(i _ {m}, j _ {m}\right) \mid \mathcal {S} _ {\mathrm {c a l}} \cup \left\{\left(i _ {*}, j _ {*}\right) \right\} = \mathcal {B}, \mathcal {S} _ {\mathrm {t r}} = \mathcal {S} _ {0}\right) \\ = \frac {\prod_ {(i , j) \in \mathcal {B} \backslash \{(i _ {m} , j _ {m}) \}} \frac {p _ {i j}}{1 - p _ {i j}}}{\sum_ {l = 1} ^ {n _ {\mathrm {c a l}} + 1} \prod_ {(i , j) \in \mathcal {B} \backslash \{(i _ {l} , j _ {l}) \}} \frac {p _ {i j}}{1 - p _ {i j}}} \\ = \frac {h _ {i _ {m} , j _ {m}}}{\sum_ {l = 1} ^ {n _ {\mathrm {c a l}} + 1} h _ {i _ {l} , j _ {l}}}, \tag {9} \\ \end{array}
$$
where  $h_{ij} = (1 - p_{ij}) / p_{ij}$ . This finishes the proof.
Proof of Equation (8). Denote  $\mathbf{W}$  the matrix consisting of entries  $W_{ij}$  defined in Algorithm 1. The matrix  $\mathbf{Z}$  consists of  $Z_{ij}$  which are the indicators of non-missingness, i.e.  $Z_{ij} = \mathbb{1}\{(i,j)$  is observed\}. Fix any two disjoint subsets  $\mathcal{S}_0,\mathcal{S}_1\subseteq [d_1]\times [d_2]$  with  $|\mathcal{S}_0| + |\mathcal{S}_1| = n$  . We have
$$
\begin{array}{l} \mathbb {P} \left(\mathcal {S} _ {\mathrm {c a l}} = \mathcal {S} _ {1}, \mathcal {S} _ {\mathrm {t r}} = \mathcal {S} _ {0} \mid | \mathcal {S} | = n\right) \\ = \mathbb {P} \left(\operatorname {s u p p} (\mathbf {Z}) = \mathcal {S} _ {0} \cup \mathcal {S} _ {1}, \mathcal {S} _ {0} \subseteq \operatorname {s u p p} (\mathbf {W}), \mathcal {S} _ {1} \subseteq \operatorname {s u p p} (\mathbf {W}) ^ {c} \mid | \mathcal {S} | = n\right). \\ \end{array}
$$
Since  $\mathbf{W}$  and  $\mathbf{Z}$  are independent, one further has
$$
\begin{array}{l} \mathbb {P} \left(\operatorname {s u p p} (\mathbf {Z}) = \mathcal {S} _ {0} \cup \mathcal {S} _ {1}, \mathcal {S} _ {0} \subseteq \operatorname {s u p p} (\mathbf {W}), \mathcal {S} _ {1} \subseteq \operatorname {s u p p} (\mathbf {W}) ^ {c} \mid | \mathcal {S} | = n\right) \\ = \mathbb {P} \left(\mathcal {S} _ {0} \subseteq \operatorname {s u p p} (\mathbf {W}), \mathcal {S} _ {1} \subseteq \operatorname {s u p p} (\mathbf {W}) ^ {c}\right) \cdot \mathbb {P} \left(\operatorname {s u p p} (\mathbf {Z}) = \mathcal {S} _ {0} \cup \mathcal {S} _ {1} \mid | \mathcal {S} | = n\right) \\ = q ^ {| \mathcal {S} _ {0} |} (1 - q) ^ {| \mathcal {S} _ {1} |} \cdot \mathbb {P} (\operatorname {s u p p} (\mathbf {Z}) = \mathcal {S} _ {0} \cup \mathcal {S} _ {1} \mid | \mathcal {S} | = n) \\ = \frac {q ^ {\left| \mathcal {S} _ {0} \right|} (1 - q) ^ {\left| \mathcal {S} _ {1} \right|}}{\mathbb {P} \left(\left| \mathcal {S} \right| = n\right)} \prod_ {\left(i ^ {\prime}, j ^ {\prime}\right) \in \left[ d _ {1} \right] \times \left[ d _ {2} \right]} \left(1 - p _ {i ^ {\prime} j ^ {\prime}}\right) \prod_ {\left(i, j\right) \in \mathcal {S} _ {0} \cup \mathcal {S} _ {1}} \frac {p _ {i j}}{1 - p _ {i j}}, \tag {10} \\ \end{array}
$$
where the last two identities are based on direct computations.
Based on (10), one can further compute
$$
\begin{array}{l} \mathbb {P} \left(\mathcal {S} _ {\mathrm {t r}} = \mathcal {S} _ {0} \mid | \mathcal {S} | = n\right) \\ = \sum_ {\mathcal {A} \in \Omega_ {\mathcal {S} _ {\mathrm {t r}}, n}} \mathbb {P} \left(\mathcal {S} _ {\mathrm {c a l}} = \mathcal {A}, \mathcal {S} _ {\mathrm {t r}} = \mathcal {S} _ {0} \mid | \mathcal {S} | = n\right) \\ = \frac {q ^ {\left| \mathcal {S} _ {0} \right|} (1 - q) ^ {n - \left| \mathcal {S} _ {0} \right|}}{\mathbb {P} (\left| \mathcal {S} \right| = n)} \sum_ {\mathcal {A} \in \Omega_ {\mathcal {S} _ {\mathrm {t r}}, n}} \prod_ {\left(i ^ {\prime}, j ^ {\prime}\right) \in [ d _ {1} ] \times [ d _ {2} ]} \left(1 - p _ {i ^ {\prime} j ^ {\prime}}\right) \prod_ {(i, j) \in \mathcal {S} _ {0} \cup \mathcal {A}} \frac {p _ {i j}}{1 - p _ {i j}}, \tag {11} \\ \end{array}
$$
where the last identity uses Equation (10).
Now we are ready to prove (8). Recall that the new data point  $(i_{*},j_{*})\mid S$  is drawn uniformly from  $\mathrm{Unif}(\mathcal{S}^c)$ . Therefore one has
$$
\begin{array}{l} \mathbb {P} \left(\mathcal {S} _ {\mathrm {c a l}} = \mathcal {S} _ {1}, \left(i _ {*}, j _ {*}\right) = \left(i _ {n + 1}, j _ {n + 1}\right) \mid \mathcal {S} _ {\mathrm {t r}} = \mathcal {S} _ {0}, | \mathcal {S} | = n\right) \\ = \frac {\mathbb {P} \left(\mathcal {S} _ {\mathrm {c a l}} = \mathcal {S} _ {1} , \mathcal {S} _ {\mathrm {t r}} = \mathcal {S} _ {0} , \left(i _ {* *} , j _ {*}\right) = \left(i _ {n + 1} , j _ {n + 1}\right) \mid | \mathcal {S} | = n\right)}{\mathbb {P} \left(\mathcal {S} _ {\mathrm {t r}} = \mathcal {S} _ {0} \mid | \mathcal {S} | = n\right)} \\ = \frac {\mathbb {P} \left(\left(i _ {*}, j _ {*}\right) = \left(i _ {n + 1} , j _ {n + 1}\right) \mid \mathcal {S} = \mathcal {S} _ {1} \cup \mathcal {S} _ {0}\right) \mathbb {P} \left(\mathcal {S} _ {\mathrm {c a l}} = \mathcal {S} _ {1} , \mathcal {S} _ {\mathrm {t r}} = \mathcal {S} _ {0} \mid | \mathcal {S} | = n\right)}{\mathbb {P} \left(\mathcal {S} _ {\mathrm {t r}} = \mathcal {S} _ {0} \mid | \mathcal {S} | = n\right)} \\ = \frac {1}{d _ {1} d _ {2} - n} \cdot \frac {\prod_ {(i , j) \in \mathcal {S} _ {1}} \frac {p _ {i j}}{1 - p _ {i j}}}{\sum_ {\mathcal {A} \in \Omega_ {\mathcal {S} _ {\mathrm {t r}}, n}} \prod_ {\left(i ^ {\prime} , j ^ {\prime}\right) \in \mathcal {A}} \frac {p _ {i ^ {\prime} j ^ {\prime}}}{1 - p _ {i ^ {\prime} j ^ {\prime}}}}, \tag {12} \\ \end{array}
$$
where the last line follows from (10) and (11).
A.2 Proof of Theorem 3.2
First, fix any  $a \in [0,1]$ . Lemma 3.1, together with the weighted conformal prediction framework of Tibshirani et al. (2019), implies that
$$
\mathbb {P} \left(M _ {i _ {*} j _ {*}} \in \widehat {M} _ {i _ {*} j _ {*}} \pm q _ {i _ {*} j _ {*}} ^ {*} (a) \cdot \widehat {s} _ {i _ {*} j _ {*}} \Big | \mathcal {S} _ {\mathrm {t r}}, \mathcal {S} _ {\mathrm {c a l}} \cup \{(i _ {*}, j _ {*}) \}\right) \geq 1 - a,
$$
$$
q _ {i _ {*} j _ {*}} ^ {*} (a) = \operatorname {Q u a n t i l e} _ {1 - a} \left(\sum_ {(i, j) \in \mathcal {S} _ {\mathrm {c a l}} \cup \{(i _ {*}, j _ {*}) \}} w _ {i j} \cdot \delta_ {R _ {i j}}\right),
$$
$$
w _ {i j} = \frac {h _ {i j}}{\sum_ {(i ^ {\prime} , j ^ {\prime}) \in \mathcal {S} _ {\mathrm {c a l}} \cup \{(i _ {* *} , j _ {*}) \}} h _ {i ^ {\prime} j ^ {\prime}}}.
$$
Indeed, here  $a$  can be any function of the random variables we are conditioning on—that is,  $a$  may depend on  $n$ , on  $\mathcal{S}_{\mathrm{tr}}$ , and on  $\mathcal{S}_{\mathrm{cal}} \cup \{(i_{*}, j_{*})\}$ .
Next define  $a = \alpha +\Delta$  where  $\Delta$  is defined as in (6). We observe that, since each  $\widehat{h}_{ij}$  is a function of  $S_{\mathrm{tr}}$ , then  $\Delta$  (and thus also  $a$ ) can therefore be expressed as a function of  $n$ ,  $S_{\mathrm{tr}}$ , and  $S_{\mathrm{cal}} \cup \{(i_*,j_*)\}$ . Therefore, applying the work above, we have
$$
\mathbb {P} \left(M _ {i _ {*} j _ {*}} \in \widehat {M} _ {i _ {*} j _ {*}} \pm q _ {i _ {*} j _ {*}} ^ {*} (\alpha + \Delta) \cdot \widehat {s} _ {i _ {*} j _ {*}} \Big | \mathcal {S} _ {\mathrm {t r}}, \mathcal {S} _ {\mathrm {c a l}} \cup \{(i _ {*}, j _ {*}) \}\right) \geq 1 - \alpha - \Delta
$$
and thus, after marginalizing,
$$
\mathbb {P} \left(M _ {i _ {*} j _ {*}} \in \widehat {M} _ {i _ {*} j _ {*}} \pm q _ {i _ {*} j _ {*}} ^ {*} (\alpha + \Delta) \cdot \widehat {s} _ {i _ {*} j _ {*}}\right) \geq 1 - \alpha - \mathbb {E} [ \Delta ].
$$
Next, we verify that
$$
\widehat {q} \geq q _ {i _ {*} j _ {*}} ^ {*} (\alpha + \Delta)
$$
holds almost surely—if this is indeed the case, then we have shown that
$$
\mathbb {P} \left(M _ {i _ {*} j _ {*}} \in \widehat {M} _ {i _ {*} j _ {*}} \pm \widehat {q} \cdot \widehat {s} _ {i _ {*} j _ {*}}\right) \geq 1 - \alpha - \mathbb {E} [ \Delta ],
$$
which establishes the desired result. Thus we only need to show that  $\widehat{q} \geq q_{i * j *}^* (\alpha + \Delta)$ , or equivalently,
$$
\operatorname {Q u a n t i l e} _ {1 - \alpha} \left(\sum_ {(i, j) \in \mathcal {S} _ {\mathrm {c a l}}} \widehat {w} _ {i j} \cdot \delta_ {R _ {i j}} + \widehat {w} _ {\text {t e s t}} \cdot \delta_ {+ \infty}\right) \geq \operatorname {Q u a n t i l e} _ {1 - \alpha - \Delta} \left(\sum_ {(i, j) \in \mathcal {S} _ {\mathrm {c a l}} \cup \{(i _ {*}, j _ {*}) \}} w _ {i j} \cdot \delta_ {R _ {i j}}\right).
$$
$$
w _ {i j} ^ {\prime} = \frac {\widehat {h} _ {i j}}{\sum_ {\left(i ^ {\prime} , j ^ {\prime}\right) \in S _ {\mathrm {c a l}} \cup \left\{\left(i _ {*}, j _ {*}\right) \right\}} \widehat {h} _ {i ^ {\prime} j ^ {\prime}}} \tag {13}
$$
for all  $(i,j)\in S\cup \{(i_{*},j_{*})\}$ . Then by definition of  $\widehat{\pmb{w}}$ , we see that  $w_{ij}^{\prime}\geq \widehat{w}_{ij}$  for  $(i,j)\in S$ , and therefore,
$$
\begin{array}{l} \text {Q u a n t i l e} _ {1 - \alpha} \left(\sum_ {(i, j) \in \mathcal {S} _ {\mathrm {c a l}}} \widehat {w} _ {i j} \cdot \delta_ {R _ {i j}} + \widehat {w} _ {\text {t e s t}} \cdot \delta_ {+ \infty}\right) \\ \geq \operatorname {Q u a n t i l e} _ {1 - \alpha} \left(\sum_ {(i, j) \in \mathcal {S} _ {\mathrm {c a l}}} w _ {i j} ^ {\prime} \cdot \delta_ {R _ {i j}} + w _ {(i _ {*}, j _ {*})} ^ {\prime} \cdot \delta_ {+ \infty}\right) \geq \operatorname {Q u a n t i l e} _ {1 - \alpha} \left(\sum_ {(i, j) \in \mathcal {S} _ {\mathrm {c a l}} \cup \{(i _ {*}, j _ {*}) \}} w _ {i j} ^ {\prime} \cdot \delta_ {R _ {i j}}\right) \\ \end{array}
$$
holds almost surely. Therefore it suffices to show that
$$
\operatorname {Q u a n t i l e} _ {1 - \alpha} \left(\sum_ {(i, j) \in \mathcal {S} _ {\mathrm {c a l}} \cup \{(i _ {*}, j _ {*}) \}} w _ {i j} ^ {\prime} \cdot \delta_ {R _ {i j}}\right) \geq \operatorname {Q u a n t i l e} _ {1 - \alpha - \Delta} \left(\sum_ {(i, j) \in \mathcal {S} _ {\mathrm {c a l}} \cup \{(i _ {*}, j _ {*}) \}} w _ {i j} \cdot \delta_ {R _ {i j}}\right)
$$
holds almost surely. Indeed, we have
$$
\mathsf{d}_{\mathsf{TV}}\left(\sum_{(i,j)\in \mathcal{S}_{\mathrm{cal}}\cup \{(i_{*},j_{*})\}}w^{\prime}_{ij}\cdot \delta_{R_{ij}},\sum_{(i,j)\in \mathcal{S}_{\mathrm{cal}}\cup \{(i_{*},j_{*})\}}w_{ij}\cdot \delta_{R_{ij}}\right)\\ \leq \frac{1}{2}\sum_{(i,j)\in \mathcal{S}_{\mathrm{cal}}\cup \{(i_{*},j_{*})\}}|w^{\prime}_{ij} - w_{ij}| = \Delta ,
$$
where  $\mathrm{d}_{\mathrm{TV}}$  denotes the total variation distance. This completes the proof.
A.3 Proofs for examples in Section 3.3
Recall the definition of  $\Delta$ :
$$
\Delta = \frac {1}{2} \sum_ {(i, j) \in \mathcal {S} _ {\mathrm {c a l}} \cup \{(i _ {*}, j _ {*}) \}} \left| \frac {\widehat {h} _ {i j}}{\sum_ {(i ^ {\prime} , j ^ {\prime}) \in \mathcal {S} _ {\mathrm {c a l}} \cup \{(i _ {*} , j _ {*}) \}} \widehat {h} _ {i ^ {\prime} j ^ {\prime}}} - \frac {h _ {i j}}{\sum_ {(i ^ {\prime} , j ^ {\prime}) \in \mathcal {S} _ {\mathrm {c a l}} \cup \{(i _ {*} , j _ {*}) \}} h _ {i ^ {\prime} j ^ {\prime}}} \right|.
$$
Note that  $\Delta \leq 1$  by definition. We start with stating a useful lemma to bound the coverage gap  $\Delta$  using the estimation error of  $\hat{h}_{ij}$ .
Lemma A.1. By the definition of the coverage gap  $\Delta$ , we have
$$
\Delta \leq \frac {\sum_ {(i , j) \in \mathcal {S} _ {\mathrm {c a l}} \cup \{(i _ {*} , j _ {*}) \}} \left| \widehat {h} _ {i j} - h _ {i j} \right|}{\sum_ {\left(i ^ {\prime} , j ^ {\prime}\right) \in \mathcal {S} _ {\mathrm {c a l}}} \widehat {h} _ {i ^ {\prime} j ^ {\prime}}}. \tag {14}
$$
A.3.1 Proof for Example 3.3.1
Under the logistic model, one has for any  $(i,j)$
$$
\mathbb {P} ((i, j) \in \mathcal {S} _ {\mathrm {t r}}) = q \cdot p _ {i j} = \frac {q \cdot \exp (u _ {i} + v _ {j})}{1 + \exp (u _ {i} + v _ {j})}.
$$
In this case, if  $\widehat{\pmb{u}}$  and  $\widehat{\pmb{v}}$  are the constrained maximum likelihood estimators in Example 3.3.1, Theorem 6 in Chen et al. (2023) implies that
$$
\| \widehat {\mathbf {u}} - \mathbf {u} \| _ {\infty} = O _ {\mathbb {P}} \left(\sqrt {\frac {\log d _ {1}}{d _ {2}}}\right), \quad \| \widehat {\mathbf {v}} - \mathbf {v} \| _ {\infty} = O _ {\mathbb {P}} \left(\sqrt {\frac {\log d _ {2}}{d _ {1}}}\right),
$$
with the proviso that  $\| \pmb{u}\|_{\infty} + \| \pmb{v}\|_{\infty} \leq \tau < \infty$ ,  $d_2 \gg \sqrt{d_1} \log d_1$  and  $d_1 \gg (\log d_2)^2$ . Then for  $h_{ij} = \exp(-u_i - v_j)$  and  $\widehat{h}_{ij} = \exp(-\widehat{u}_i - \widehat{v}_j)$ , we have
$$
\max _ {i, j} | \widehat {h} _ {i j} - h _ {i j} | = \max _ {i, j} e ^ {- u _ {i} - v _ {j}} \left(e ^ {- (\widehat {u} _ {i} - u _ {i}) - (\widehat {v} _ {j} - v _ {j})} - 1\right) = O _ {\mathbb {P}} \left(\sqrt {\frac {\log d _ {1}}{d _ {2}}} + \sqrt {\frac {\log d _ {2}}{d _ {1}}}\right).
$$
Further, as  $\min_{i,j} h_{i,j} = \min_{i,j} \exp(-u_i - v_j) \geq e^{-\tau}$ , then with probability approaching one, for every  $(i,j) \in [d_1] \times [d_2]$ , one has  $\widehat{h}_{i,j} \geq h_{i,j} - |h_{ij} - \widehat{h}_{ij}| \geq e^{-\tau}/2 =: h_0$ . By the upper bound (14), we have
$$
\Delta \lesssim \sqrt {\frac {\log d _ {1}}{d _ {2}}} + \sqrt {\frac {\log d _ {2}}{d _ {1}}} \asymp \sqrt {\frac {\log \max \{d _ {1} , d _ {2} \}}{\min \{d _ {1} , d _ {2} \}}}.
$$
Further, as  $\Delta \leq 1$ , we have  $\mathbb{E}[\Delta] \lesssim \sqrt{\frac{\log \max\{d_1, d_2\}}{\min\{d_1, d_2\}}}$ .
A.3.2 Proof of Example 3.3.2
Define the link function  $\psi(t) = q(1 + e^{-\phi(t)})$ , where  $\phi$  is monotonic. Applying Theorem 1 in the paper (Davenport et al., 2014), we obtain that with probability at least  $1 - C_1 / (d_1 + d_2)$
$$
\frac {1}{d _ {1} d _ {2}} \| \widehat {\mathbf {A}} - \mathbf {A} \| _ {\mathrm {F}} ^ {2} \leq \sqrt {2} \widetilde {C} _ {\tau} \sqrt {\frac {k \left(d _ {1} + d _ {2}\right)}{d _ {1} d _ {2}}}, \tag {15}
$$
with the proviso that  $d_{1}d_{2}\geq (d_{1} + d_{2})\log (d_{1}d_{2})$ . Here  $\widetilde{C}_{\tau} = 2^{39 / 4}e^{9 / 4}(1 + \sqrt{6})\tau L_{\tau}\beta_{\tau}$  with
$$
L _ {\tau} = \sup  _ {- \tau \leq t \leq \tau} \frac {| \psi^ {\prime} (t) |}{\psi (t) (1 - \psi (t))}, \quad \text {a n d} \quad \beta_ {\tau} = \sup  _ {- \tau \leq t \leq \tau} \frac {\psi (t) (1 - \psi (t))}{| \psi^ {\prime} (t) | ^ {2}}. \tag {16}
$$
Denote this high probability event to be  $\mathcal{E}_0$ . Since  $\| \mathbf{A}\| _1\leq \sqrt{d_1d_2}\| \mathbf{A}\|_{\mathrm{F}}$ , on this event  $\mathcal{E}_0$ , we further have
$$
\frac {1}{d _ {1} d _ {2}} \| \widehat {\mathbf {A}} - \mathbf {A} \| _ {1} \leq C _ {\tau} \left(\frac {k (d _ {1} + d _ {2})}{d _ {1} d _ {2}}\right) ^ {1 / 4},
$$
where  $C_{\tau} = \zeta \sqrt{L_{\tau}\beta_{\tau}}$  and  $\zeta$  is a universal constant.
Recall that  $h_{ij} = \exp(-\phi(A_{ij}))$  and  $\| \mathbf{A} \|_{\infty} \leq \tau$ . On the same event  $\mathcal{E}_0$ , we further have
$$
\frac {1}{d _ {1} d _ {2}} \| \widehat {\mathbf {H}} - \mathbf {H} \| _ {1} \leq C _ {\tau} ^ {\prime} \left(\frac {k \left(d _ {1} + d _ {2}\right)}{d _ {1} d _ {2}}\right) ^ {1 / 4}, \tag {17}
$$
where  $C_{\tau}^{\prime} = 2e^{\phi (\tau)}\sqrt{\widetilde{C}_{\tau}}$
By the feasibility of the minimizer  $\widehat{\mathbf{A}}$  and the fact that  $\widehat{h}_{ij} = \exp (-\phi (\widehat{A}_{ij}))$ , we have  $\widehat{h}_{ij}\geq h_0 = e^{-\phi (\tau)}$  for all  $(i,j)$ . This together with the upper bound (14) implies that
$$
\Delta \leq \frac{1}{h_{0}n_{\mathrm{cal}}}\sum_{(i,j)\in \mathcal{S}_{\mathrm{cal}}\cup \{(i_{*},j_{*})\}}|h_{ij} - \widehat{h}_{ij}|\leq \frac{1}{h_{0}n_{\mathrm{cal}}}\| \widehat{\mathbf{H}} -\mathbf{H}\|_{1}.
$$
Define a second high probability event  $\mathcal{E}_1 = \{n_{\mathrm{cal}} \geq (1 - c)(1 - q) \| \mathbf{P} \|_1\}$ . Using the Chernoff bound, we have  $\mathbb{P}(\mathcal{E}_1) \geq 1 - C_0(d_1 d_2)^{-5}$ . Therefore, on the event  $\mathcal{E}_0 \cap \mathcal{E}_1$ , we have
$$
\Delta \leq \frac {d _ {1} d _ {2}}{c h _ {0} (1 - q) \| \mathbf {P} \| _ {1}} C _ {\tau} ^ {\prime} \left(\frac {k \left(d _ {1} + d _ {2}\right)}{d _ {1} d _ {2}}\right) ^ {1 / 4}.
$$
Under the assumptions that  $p_{ij} = 1 / (1 + e^{-\phi (A_{ij})})$ , and that  $\| \mathbf{A}\|_{\infty}\leq \tau$ , we know that  $p_{ij}\geq C_2$  for some constant that only depends on  $\tau$ . As a result, we have  $\| \mathbf{P}\| _1\geq C_2d_1d_2$ , which further leads to the conclusion that
$$
\Delta \leq \frac {1}{c C _ {2} h _ {0} (1 - q)} C _ {\tau} ^ {\prime} \left(\frac {k (d _ {1} + d _ {2})}{d _ {1} d _ {2}}\right) ^ {1 / 4}.
$$
on the event  $\mathcal{E}_0\cap \mathcal{E}_1$ . In addition, on the small probability event  $(\mathcal{E}_0\cap \mathcal{E}_1)$ , one trivially has  $\Delta \leq 1$ . Therefore simple combinations of the cases yields the desired bound on  $\mathbb{E}[\Delta]$ .
A.3.3 Proof of Lemma A.1
Reusing the definition of  $w'$  (13), one has
B Additional numerical experiments
In this section, we provide additional simulation results. In Section B.1, we compare the proposed one-shot weighted approach with the exact weighted conformal prediction. In Section B.2, the
(a) Exact/original (og) cmc vs 1-shot cmc.
(b) Local performance of coverage given  $p_{ij} = p_0$ .
local coverage of cmc-als conditioning on  $p_{ij}$  is evaluated. We present the simulation results for the convex relaxation (cvx) and the conformalized convex method (cmc-cvx) in both settings with homogeneous and heterogeneous missingness in Section B.3 and B.5. We further evaluate the performance of cmc-als when the sampling probability is misspecified in Section B.6.
B.1 Comparison between one-shot and exact weighted approaches
We conduct comparison in the setting with heterogeneous missingness and Gaussian noise as shown in Section 4.1.2 with  $k^{*} = 12$ . From the result in Figure 5a, the performance of two approaches are essentially the same while the one-shot cmc-a1s is more computationally efficient.
B.2 Local coverage of cmc
In Figure 5b, we evaluate the local performance of cmc-als when conditioning on  $\{p_{ij} = p_0\}$ , i.e. a subpopulation determined by the value of sampling probability. We use the setting with heterogeneous missingness and Gaussian noise as shown in Section 4.1.2 with  $k^* = 12$  and  $p_0$  ranging from 0.1 to 0.9 with a step size of 0.1. The conditional coverage is approximated via kernel smoothing: with the indicators  $A_{ij}$  for coverage, we use the weight  $K_{ij} = \phi_{p_0,h}(p_{ij})$  and calculate the conditional coverage by  $(\sum A_{ij}K_{ij}) / (\sum K_{ij})$ . Here  $\phi_{\mu,\sigma}$  is the density function of  $\mathcal{N}(\mu, \sigma^2)$  and  $h = 0.05$ . When  $p_0$  increases from 0.2 to 0.8, the conditional coverage increases and stays around 0.9. At  $p_0 = 0.1$ , 0.9, the coverage varies much, which is due to the small effective sample size for the edge values of  $p_{ij}$ . Moreover, for uniform sampling, since the rows and columns are generated i.i.d., there are no meaningful subgroups of the data to condition on.
B.3 Additional results for homogeneous misingness
In this section, we present the results for synthetic simulation with the convex relaxation cvx and the conformalized convex matrix completion method cmc-cvx. Setting 1, 2, 3, 4 are the same as introduced in Section4.1. The true rank is  $r^* = 8$  and the hypothesized rank varies from 4 to 24 with the stepsize 4.
The conformalized methods, regardless of the based algorithm adopted, have nearly exact coverage around  $1 - \alpha$ . But we can observe different behaviors between a1s and cvx since the convex relaxation is free of the choice of  $r$  until the projection of  $\widehat{\mathbf{M}}_{\mathrm{cvx}}$  onto the rank- $r$  subspace (Chen et al., 2020). As a result, when  $r > r^{*}$ , cvx tends to overestimate the strength of the noise. In Figure 6a, 6b and 6d, when  $r > r^{*}$ , cvx has coverage rate higher than the target level and the confidence interval is more conservative than conformalized methods. Since the accuracy of cvx is based on the large sample size, in Figure 6b, when the effective sample size is insufficient with small  $p_{ij}$ , the residuals from cvx have a large deviation from the standard distribution and the intervals are much larger than the oracle ones. Besides, in Figure 6c, when the noise has heavy tails than Gaussian variables, cvx overestimates the noise strength similar to a1s and is conservative in coverage. When the incoherence condition is violated in Figure 6d, if  $r < r^{*}$ , both cvx and a1s fit the missed factor by overestimating the noise strength and produce extremely large intervals.
B.4 Estimation error for one-bit matrix estimation