SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Biased Age Classifier (Note: Target value was normalized age, made binary after)
Layer (type)	Output Shape
Conv2d-1	[-1, 2, 32, 32]
BatchNorm2d-2	[-1, 2, 32, 32]
LeakyReLU-3	[-1, 2, 32, 32]
Dropout-4	[-1, 2, 32, 32]
Conv2d-5	[-1, 4, 16, 16]
BatchNorm2d-6	[-1, 4, 16, 16]
LeakyReLU-7	[-1, 4, 16, 16]
Dropout-8	[-1, 4, 16, 16]
Conv2d-9	[-1, 8, 8, 8]
BatchNorm2d-10	[-1, 8, 8, 8]
LeakyReLU-11	[-1, 8, 8, 8]
Dropout-12	[-1, 8, 8, 8]
Flatten-13	[-1, 512]
Linear-14	[-1, 64]
LeakyReLU-15	[-1, 64]
Linear-16	[-1, 1]
Sigmoid-17	[-1, 1]
Ground Truth Age Classifier (Note: Target value was normalized age; made binary after)
Layer (type)	Output Shape
Conv2d-1	[-1, 8, 32, 32]
BatchNorm2d-2	[-1, 8, 32, 32]
LeakyReLU-3	[-1, 8, 32, 32]
Dropout-4	[-1, 8, 32, 32]
Conv2d-5	[-1, 16, 16, 16]
BatchNorm2d-6	[-1, 16, 16, 16]
LeakyReLU-7	[-1, 16, 16, 16]
Dropout-8	[-1, 16, 16, 16]
Conv2d-9	[-1, 32, 8, 8]
BatchNorm2d-10	[-1, 32, 8, 8]
LeakyReLU-11	[-1, 32, 8, 8]
Dropout-12	[-1, 32, 8, 8]
Flatten-13	[-1, 2048]
Linear-14	[-1, 64]
LeakyReLU-15	[-1, 64]
Linear-16	[-1, 1]
Sigmoid-17	[-1, 1]
The distribution mapper used default architecture of SDV's CTGAN  $^{5}$  version 0.6.0, except for in the ProgressiveGAN experiment where embedding_dim =512, generator_dim = (512,512) were passed as arguments.
For the networks we trained, we utilized the Adam optimizer with learning rate between 0.002 and 0.0001.
The linear classifier utilized Scikit-Learn's LinearSVC (for latent editing) and RidgeClassifier for the biased Shapes classifier.
A.5 Proof of Lemma 1
Proof. First, note that if  $1^{|\mathcal{Y}|} \in \operatorname{cone}(E)$ , then likewise  $\frac{1}{\mathcal{Y}} 1^{|\mathcal{Y}|} \in \operatorname{cone}(E)$ .
Let  $\mathbf{z}' \sim \mathbb{P}_{z|C_{\phi} = i}$ ; i.e.,  $z$  is a draw from the distribution of noise such that the classifiers prediction of the generated sample corresponding to  $z'$  is group  $i$ .
Let  $(C' \circ G_{\theta})_* \mathbb{P}_{z|C_{\phi} = i}$  be the pushforward distribution of the perfect classifier  $C'$ 's output when conditioned on the generator's output of draws from  $\mathbb{P}_{z|C_{\phi} = i}$ . Then,
$$
\begin{array}{l} \left(C ^ {\prime} \circ G _ {\theta}\right) _ {*} \mathbb {P} _ {z | C _ {\phi} = i} = \left[ P r (\mathbf {y} = 1 | C _ {\theta} = i), P r (\mathbf {y} = 2 | C _ {\theta} = i), \dots , P r (\mathbf {y} = N | C _ {\theta} = i) \right] \\ = E _ {:, i} \\ \end{array}
$$
$^{5}$ https://sdv.dev/SDV/user_guide/single_table/ctgan.html#how-to-modify-the-ctgan-hyperparameters
$$
\operatorname {c o n e} \left(\left\{\left(C ^ {\prime} \circ G _ {\theta}\right) _ {*} \mathbb {P} _ {z | C _ {\phi} = i}, \dots , \left(C ^ {\prime} \circ G _ {\theta}\right) _ {*} \mathbb {P} _ {z | C _ {\phi} = | y |} \right\}\right) = \operatorname {c o n e} (E)
$$
Therefor, following from above,
$$
\frac {1}{\mathcal {Y}} 1 ^ {| \mathcal {Y} |} \in \operatorname {c o n e} \left(\left\{\left(C ^ {\prime} \circ G _ {\theta}\right) _ {*} \mathbb {P} _ {z | C _ {\phi} = i}, \dots , \left(C ^ {\prime} \circ G _ {\theta}\right) _ {*} \mathbb {P} _ {z | C _ {\phi} = | \mathcal {Y} |} \right\}\right)
$$
This means that  $\exists \lambda_1, \lambda_2, \dots, \lambda_{|\mathcal{V}}$  such that the following holds:
$$
\begin{array}{l} \lambda_ {1} \left(C ^ {\prime} \circ G _ {\theta}\right) _ {*} \mathbb {P} _ {z | C _ {\phi} = i} + \dots + \lambda_ {| \mathcal {Y} |} \left(C ^ {\prime} \circ G _ {\theta}\right) _ {*} \mathbb {P} _ {z | C _ {\phi} = | \mathcal {Y} |} = [ \frac {1}{| \mathcal {Y} |}, \dots , \frac {1}{| \mathcal {Y} |} ] \\ = \operatorname {U n i f} (\mathcal {Y}) \\ \end{array}
$$
This is equivalent to saying that:
$$
C ^ {\prime} (G _ {\theta} (\mathbf {z})) \sim \operatorname {U n i f} (\mathcal {Y})
$$
for  $\mathbf{z} \sim \sum_{i=1}^{|\mathcal{Y}|} \lambda_i \mathbb{P}_{z|C_\phi=i} = \mathbb{Q}^\lambda$ . Thus, by definition  $\mathbb{Q}^\lambda$  is a Fair Noise Distribution.
A.6 Proof of Lemma 2
Proof. Note that the sign of the coefficient of the cross product  $E_{\cdot,1} \times E_{\cdot,2}$  is  $P(\mathbf{y} = 1|\hat{\mathbf{y}} = 1)P(\mathbf{y} = 2|\hat{\mathbf{y}} = 2) - P(\mathbf{y} = 1|\hat{\mathbf{y}} = 2)P(\mathbf{y} = 2|\hat{\mathbf{y}} = 1)$ . Also note that  $E_{\cdot,1} \times [0.5,0.5]$  is  $0.5P(\mathbf{y} = 1|\hat{\mathbf{y}} = 1) - 0.5P(\mathbf{y} = 2|\hat{\mathbf{y}} = 1)$ .
$$
\begin{array}{l} P (\mathbf {y} = 1 | \hat {\mathbf {y}} = 1) P (\mathbf {y} = 2 | \hat {\mathbf {y}} = 2) > P (\mathbf {y} = 1 | \hat {\mathbf {y}} = 1) 0. 5 \\ > 0 \\ \end{array}
$$
$$
\begin{array}{l} 0 <   P (\mathbf {y} = 1 | \hat {\mathbf {y}} = 2) P (\mathbf {y} = 2 | \hat {\mathbf {y}} = 1) \\ <   0. 5 P (\mathbf {y} = 2 | \hat {\mathbf {y}} = 1) \\ \end{array}
$$
Thus, the coefficient of  $E_{:,1} \times E_{:,2}$  is greater than  $E_{:,1} \times [0.5, 0.5]$ , while there signs are equal. This implies that  $[0.5, 0.5]$  is in between  $E_{:,1}$  and  $E_{:,2}$ . Thus,  $[0.5, 0.5] \in \operatorname{cone}(E)$ . The rest of the proof follows directly from Lemma 1.
A.7 Proof of Proposition 1
Proof. Note that  $\mathbb{P}_E^\lambda$  has density  $\left[\sum_{i}\lambda_{i}Pr(\mathbf{y} = 1|\hat{\mathbf{y}} = i),\ldots ,\sum_{i}\lambda_{i}Pr(\mathbf{y} = N|\hat{\mathbf{y}} = i)\right]$ . For ease of notation let us refer to  $\sum_{i}\lambda_{i}Pr(\mathbf{y} = m|\hat{\mathbf{y}} = i)$  as  $r_m^\lambda$ .
$$
\begin{array}{l} \operatorname {K L} \left\{\mathbb {P} _ {E} ^ {\lambda} | | \operatorname {U n i f} (\mathcal {Y}) \right\} = \sum_ {m} r _ {m} ^ {\lambda} \log \left(\frac {r _ {m} ^ {\lambda}}{u}\right) \\ = \sum_ {m} \left(r _ {m} ^ {\lambda} \log \left(r _ {m} ^ {\lambda}\right) - r _ {m} ^ {\lambda} \log \left(\frac {1}{| \mathcal {Y} |}\right)\right) \\ = \sum_ {m} r _ {m} ^ {\lambda} \log (r _ {m} ^ {\lambda}) - \sum_ {m} r _ {m} ^ {\lambda} \log (\frac {1}{| \mathcal {Y} |}) \\ \end{array}
$$
Note that  $\log \left(\frac{1}{N}\right)$  is constant for each term in the second summation. Thus,
$$
\begin{array}{l} = \sum_ {m} r _ {m} ^ {\lambda} \log \left(r _ {m} ^ {\lambda}\right) - \log \left(\frac {1}{N}\right) \sum r _ {m} ^ {\lambda} \\ = \sum_ {m} r _ {m} ^ {\lambda} \log (r _ {m} ^ {\lambda}) - \log \left(\frac {1}{N}\right), \\ \end{array}
$$
As  $\log \left(\frac{1}{N}\right)$  does not depend on  $r_m^{\lambda}$ ,
$$
\begin{array}{l} \underset {\lambda} {\operatorname {a r g m i n}} \mathrm {K L} \left\{\mathbb {P} _ {E} ^ {\lambda} | | \text {U n i f} (\mathcal {Y}) \right\} = \underset {\lambda} {\operatorname {a r g m i n}} \sum_ {m} r _ {m} ^ {\lambda} \log \left(r _ {m} ^ {\lambda}\right) \\ = \operatorname * {a r g m i n} _ {\lambda} - H (\mathbb {P} _ {E} ^ {\lambda}) \\ = \operatorname * {a r g m a x} _ {\lambda} H (\mathbb {P} _ {E} ^ {\lambda}) \\ \end{array}
$$
A.8 Proof of Proposition 2
$$
\begin{array}{l} \left(C ^ {\prime} \circ G _ {\theta}\right) _ {*} \mathbb {P} _ {z | C _ {\phi} = i} = \left[ P r (\mathbf {y} = 1 | C _ {\theta} = i), P r (\mathbf {y} = 2 | C _ {\theta} = i), \dots , P r (\mathbf {y} = N | C _ {\theta} = i) \right] \\ \Rightarrow \sum_ {i} \lambda_ {i} \left(C ^ {\prime} \circ G _ {\theta}\right) _ {*} \mathbb {P} _ {z | C _ {\phi} = i} = \left[ \sum_ {i} \lambda P r (\mathbf {y} = 1 | C _ {\theta} = i), \sum_ {i} P r (\mathbf {y} = 2 | C _ {\theta} = i), \right. \\ \ldots , \sum_ {i} \lambda P r (\mathbf {y} = N | C _ {\theta} = i) ] \\ \Longrightarrow \mathbb {P} _ {E} = \mathbb {Q} ^ {\lambda} \\ \end{array}
$$
A.9 Proof of Theorem 1
The first statement follows directly from Proposition 1 and Proposition 2.
If  $C_{\phi} = C'$ , then  $\{ \frac{E_{:,1}}{|E_{:,1}|}, \dots, \frac{E_{:,|\mathcal{Y}|}}{|E_{:,|\mathcal{Y}|}|} \}$  forms a standard basis of  $\mathbb{R}^{|\mathcal{Y}|}$ , and therefor  $1^{|\mathcal{Y}|}$  is in  $\operatorname{cone}(E)$ . Thus,  $Q^{\lambda*}$  is a Fair Noise Distribution by Lemma 1.