SlowGuess/NAACL_2021_OCR huggingface dataset:
Lexicon	BERT	Random
oracle (ψ)	15.54	15.54	15.54
proxy-lex (ˆψnaive+C)	11.21	7.83	12.34
T-boost (C only, ˆψnaive+C)	13.30	10.92	12.70
T-boost (ˆψnaive+C)	14.68	12.01	13.59
W-Adjust (ψproxy)	15.01	13.88	13.30
Table 4: ATE estimates from a  $C$ -restricted classifier (row 2) are preferable to  $\hat{T}$  but conservative compared to less restrictive methods (T-boost and W-Adjust).
C Ablating T-boost
$T$ -boost only uses a classifier to change the treatment status of an example on  $\hat{T} = 0$  examples. Intuitively, this is because  $\hat{T}$  is assumed to be a reasonable estimate of  $\tilde{T}$  and therefore has a low false positive rate. We investigate this by abating this part of the algorithm and directly setting  $\hat{T}^*$  to the classifier's prediction. Our results on lexicon-based  $\hat{T}$ 's (Table 5, we observed similar outcomes with other  $\hat{T}$ 's) suggest this can reduce performance because a large number of correctly labeled  $\hat{T} = 1$  examples are flipped.
Estimator	Estimate
oracle (ψ)	15.54
proxy-lex (ˆψnaive+C)	11.21
T-boost (T=0 only, ˆψnaive+C)	14.60
T-boost (all examples, ˆψnaive+C)	10.00
Table 5: It is advantageous to only relabel  $\hat{T} = 0$  examples.
D Lemma 1 (used by Theorems 2 and 3)
$$
\begin{array}{l} \mathbb {E} [ Y \mid W, \hat {T} = 1 ] = \mathbb {E} [ Y \mid W, T = 1 ] \Pr (T = 1 \mid W, \hat {T} = 1) \\ + \mathbb {E} [ Y \mid W, T = 0 ] \Pr (T = 0 \mid W, \hat {T} = 1) \\ \end{array}
$$
$$
\begin{array}{l} \mathbb {E} [ Y \mid W, \hat {T} = 0 ] = \mathbb {E} [ Y \mid W, T = 0 ] \Pr (T = 0 \mid W, \hat {T} = 0) \\ + \mathbb {E} [ Y \mid W, T = 1 ] \Pr (T = 1 \mid W, \hat {T} = 0) \\ \end{array}
$$
Proof. Apply the law of total probability and definition of conditional independence to the causal graph given in Figure 3.
$$
\epsilon_ {0} = P (T = 0 \mid \hat {T} = 1, \tilde {Z})
$$
$$
\epsilon_ {1} = P (T = 1 \mid \hat {T} = 0, \tilde {Z}, C)
$$
$$
p _ {1} = P (T = 1 | \hat {T} = 1, \tilde {Z})
$$
$$
p _ {0} = P (T = 0 | \hat {T} = 0, \tilde {Z})
$$
$$
E _ {1} = \mathbb {E} [ Y \mid \tilde {Z}, T = 1 ]
$$
$$
E _ {0} = \mathbb {E} [ Y \mid \tilde {Z}, T = 0 ]
$$
Now recall
$$
\hat {\psi} ^ {\mathrm {p r o x y}} = \mathbb {E} _ {W} [ \mathbb {E} [ Y \mid \hat {T} = 1, \tilde {Z} ] - \mathbb {E} [ Y \mid \hat {T} = 0, \tilde {Z} ] ]
$$
Now we write the inner part using Lemma D, collect terms, and use the law of total probability to write everything in terms of misclassification probabilities:
$$
\begin{array}{l} = \left(E _ {1} p _ {1} + E _ {0} \epsilon_ {0}\right) - \left(E _ {0} p _ {0} - E _ {1} \epsilon_ {1}\right) \\ = E _ {1} \left(p _ {1} + \epsilon_ {1}\right) + E _ {0} \left(\epsilon_ {0} - p _ {0}\right) \\ = E _ {1} \left(\left(1 - \epsilon_ {0}\right) + \epsilon_ {1}\right) + E _ {0} \left(\epsilon_ {0} - \left(1 - \epsilon_ {1}\right)\right) \\ = \left(E _ {1} - E _ {0}\right) \left(1 - \left(\epsilon_ {0} + \epsilon_ {1}\right)\right) \\ \end{array}
$$
which completes the proof  $\square$
F Theorem about the bias due to noisy proxies
Here we show that the naive estimand which does not adjust for the text,
$$
\psi^ {\text {n a i v e}} = \mathbb {E} [ Y | \hat {T} = 1 ] - \mathbb {E} [ Y | \hat {T} = 0 ], \tag {13}
$$
can be arbitrarily biased away from the effect of interest,  $\psi^{\mathrm{rea}}$ .
$$
\psi^ {n a i v e} = \mathbb {E} _ {W} \left[ \mathbb {E} [ Y \mid \tilde {T} = 1, W ] \alpha (W) - \mathbb {E} [ Y \mid \tilde {T} = 0, W ] \beta (W) \right]
$$
$$
\alpha (W) = \frac {P (\tilde {T} = 1 , \hat {T} = 1 \mid W)}{P (\hat {T} = 1)} - \frac {P (\tilde {T} = 1 , \hat {T} = 0 \mid W)}{P (\hat {T} = 0)}
$$
$$
\beta (W) = \frac {P (\tilde {T} = 0 , \hat {T} = 0 | W)}{P (\hat {T} = 0)} - \frac {P (\tilde {T} = 0 , \hat {T} = 1 | W)}{P (\hat {T} = 1)}
$$
The  $\alpha$  and  $\beta$  terms are related to the error of the proxy label. This theorem says that correlations between the outcome and errors in the proxy can induce bias. Intuitively, this is similar to bias from confounding, though it is mathematically distinct. This means that even a highly accurate proxy label can result in highly misleading estimates. Proof:
$$
\begin{array}{l} \mathbb {E} [ Y \mid \hat {T} = 1 ] = \mathbb {E} \left[ \mathbb {E} [ Y \mid \hat {T} = 1, W ] \mid \hat {T} = 1 \right] \\ = \mathbb {E} \left[ \mathbb {E} [ Y \mid \hat {T} = 1, W ] \frac {\Pr (W \mid \hat {T} = 1)}{\Pr (W)} \right] \\ = \mathbb {E} \left[ \mathbb {E} [ Y \mid \hat {T} = 1, W ] \frac {\Pr (\hat {T} = 1 \mid W)}{\Pr (\hat {T} = 1)} \right] \\ \end{array}
$$
Where the first equality is by the tower property, the second by inverse probability weighting, and the third Bayes' rule. We continue by invoking Lemma 1:
$$
\begin{array}{l} \mathbb {E} [ Y \mid \hat {T} = 1, W ] \frac {\Pr (\hat {T} = 1 \mid W)}{\Pr (\hat {T} = 1)} = \mathbb {E} [ Y \mid W, T = 1 ] \Pr (T = 1 \mid W, \hat {T} = 1) \frac {\Pr (\hat {T} = 1 \mid W)}{\Pr (\hat {T} = 1)} \\ + \mathbb {E} [ Y \mid W, T = 0 ] \Pr (T = 0 \mid W, \hat {T} = 1) \frac {\Pr (\hat {T} = 1 \mid W)}{\Pr (\hat {T} = 1)} \\ = \mathbb {E} [ Y \mid W, T = 1 ] \frac {\Pr (T = 1 , \hat {T} = 1 \mid W)}{\Pr (\hat {T} = 1)} \\ + \mathbb {E} [ Y \mid W, T = 0 ] \frac {\Pr (T = 0 , \hat {T} = 1 \mid W)}{\Pr (\hat {T} = 1)}. \\ \end{array}
$$
The analogous expression for  $\mathbb{E}[Y\mid \hat{T} = 0]$  ..
$$
\begin{array}{l} \mathbb {E} [ Y \mid \hat {T} = 0 ] = \mathbb {E} \left[ \mathbb {E} [ Y \mid W, T = 0 ] \frac {\Pr (T = 0 , \hat {T} = 0 \mid W)}{\Pr (\hat {T} = 0)} \right. \\ \left. + \mathbb {E} [ Y \mid W, T = 1 ] \frac {\Pr (T = 1 , \hat {T} = 0 \mid W)}{\Pr (\hat {T} = 0)} \right] \\ \end{array}
$$
And now plugging into the ATE formula:
$$
\begin{array}{l} \hat {\psi} ^ {\text {r e a .}} = \mathbb {E} [ \mathbb {E} [ Y; d o (\hat {T} = 1) ] - \mathbb {E} [ Y; d o (\hat {T} = 0) ] ] \\ = \mathbb {E} [ \mathbb {E} [ Y | \hat {T} = 1 ] - \mathbb {E} [ Y | \hat {T} = 0 ] ] \\ = \mathbb {E} \left[ \right. \\ \end{array}
$$
$$
\begin{array}{l} \mathbb {E} [ Y \mid W, T = 1 ] \left(\frac {\Pr (T = 1 , \hat {T} = 1 \mid W)}{\Pr (\hat {T} = 1)} - \frac {\Pr (T = 1 , \hat {T} = 0 \mid W)}{\Pr (\hat {T} = 0)}\right) \\ - \mathbb {E} [ Y \mid W, T = 0 ] \left(\frac {\Pr (T = 0 , \hat {T} = 0 \mid W)}{\Pr (\hat {T} = 0)} - \frac {\Pr (T = 0 , \hat {T} = 1 \mid W)}{\Pr (\hat {T} = 1)}\right) \\ \end{array}
$$
The result follows immediately.  $\square$
G Deriving semi-oracle, a Causal Estimator for when  $P(T|\hat{T})$  is known
This is an ATE estimator which assumes access to  $P(\hat{T} | T)$  instead of  $T$  but is still unbiased. We derive this estimator using the "matrix adjustment" technique of Wood-Doughty et al. (2018); Pearl(2012). We start by decomposing the joint distribution
$$
\begin{array}{l} P (Y, T, \hat {T}, C) = P (\hat {T} | Y, C, T) P (Y, C, T) \\ P (Y, C, \hat {T}) = \sum_ {T} P (\hat {T} | Y, C, T) P (Y, C, T) \\ \end{array}
$$
We can write this as a product between a matrix  $\mathbf{M}_{c,y}(\hat{T},T) = P(\hat{T}|Y,C,T)$  and vector  $\mathbf{V}_{c,y}(T) = P(Y,C,T)$ :
$$
\begin{array}{l} \mathbf {V} _ {c, y} (\hat {T}) = \sum_ {T} \mathbf {M} _ {c, y} (\hat {T}, T) \mathbf {V} _ {c, y} (T) \\ = \mathbf {M} _ {c, y} \mathbf {V} _ {c, y} \\ \end{array}
$$
For our binary setting  $\mathbf{M}_{c,y}$  is:
$$
\begin{array}{l} \mathbf {M} _ {c, y} = \left[ \begin{array}{c c} 1 - \delta_ {c, y} & \epsilon_ {c, y} \\ \delta_ {c, y} & 1 - \epsilon_ {c, y} \end{array} \right] \\ \mathbf {M} _ {c, y} ^ {- 1} = \frac {1}{1 - \epsilon_ {c , y} - \delta_ {c , y}} \left[ \begin{array}{l l} 1 - \epsilon_ {c, y} & - \epsilon_ {c, y} \\ - \delta_ {c, y} & 1 - \delta_ {c, y} \end{array} \right] \\ \epsilon_ {c, y} = P (\hat {T} = 0 | T = 1, C, Y) \\ \delta_ {c, y} = P (\hat {T} = 1 | T = 0, C, Y) \\ \end{array}
$$
Under fairly broad conditions,  $\mathbf{M}$  has an inverse, which allows us to reconstruct the joint distribution:
$$
P (Y, T, C) = \sum_ {\hat {T}} \mathbf {M} _ {c, y} ^ {- 1} (T, \hat {T}) \mathbf {V} _ {c, y} (\hat {T})
$$
From which we can recover the ATE
$$
\psi^ {m a t r i x} = \sum_ {c} \left[ \frac {P (Y , T = 1 , C)}{\sum_ {Y} P (Y , T = 1 , C)} - \frac {P (Y , T = 0 , C)}{\sum_ {Y} P (Y , T = 0 , C)} \right] P (C)
$$
Note also that this expression is similar to  $\tau_{ME}$  in Wood-Doughty et al. (2018) except their error terms are of the form  $P(T|\hat{T})$ .