Surrogate Membership for Inferred Metrics in Fairness Evaluation

Abstract

As artificial intelligence becomes more embedded into daily activities, it is imperative to ensure models perform well for all subgroups. This is particularly important when models include underprivileged populations. Binary fairness metrics, which compare model performance for protected groups to the rest of the model population, are an important way to guard against unwanted bias. However, a significant drawback of these binary fairness metrics is that they require protected group membership attributes. In many practical scenarios, protected status for individuals is sparse, unavailable, or even illegal to collect. This paper extends binary fairness metrics from deterministic membership attributes to their surrogate counterpart under the probabilistic setting. We show that it is possible to conduct binary fairness evaluation when exact protected attributes are not immediately available but their surrogate as likelihoods is accessible. Our inferred metrics calculated from surrogates are proved to be valid under standard statistical assumptions. Moreover, we do not require the surrogate variable to be strongly related to protected class membership; inferred metrics remain valid even when membership in the protected and unprotected groups is equally likely for many levels of the surrogate variable. Finally, we demonstrate the effectiveness of our approach using publicly available data from the Home Mortgage Disclosure Act and simulated benchmarks that mimic real-world conditions under different levels of model disparity.

Appendix - Comparison to Weighted Fairness Statistic

Our inferred metrics are similar in approach to an estimator described in [6]. In this section, we re-write our inferred metrics and the weighted estimator so they can be compared directly and present a mathematical argument for why the weighted estimator is biased toward 0 under the regularity conditions described.

First, our estimator $m(X^\top )-m(X^\bot )$ is derived from the WOLS estimator of the value $\beta _1$ from Eq. 5.

$$\begin{aligned} m_{wols}(X^\top )-m_{wols}(X^\bot )=\frac{\sum _z n_z (m_z-\bar{m})(P_z (x\in X^\top )-\bar{P}(x\in X^\top ))}{\sum _z n_z(P_z(x\in X^\top )-\bar{P}(x\in X^\top )\big )^2} \end{aligned}$$

(8)

where $\bar{m}$ is the overall mean for the model metric and $\bar{P}(x\in X^\top )$ is the overall mean for the probability of being in the protected group.

The weighted estimator described in [6] is:

$$\begin{aligned} m_w(X^\top )-m_w(X^\bot )=\frac{\sum _x P_z(x\in X^\top )m(x)}{\sum _x P_z(x\in X^\top )}-\frac{\sum _x P_z(x\in X^\bot )m(x)}{\sum _x P_z(x\in X^\bot )} \end{aligned}$$

(9)

where

• m(x) is the value of the model metric for each individual (e.g. if the m is statistical parity, $m(x)=I(ML(x)=1)$)

• $\sum _x$ indicates a sum over all N individuals for which we are calculating fairness metrics

• $P_z(x\in X^\top )$ is the probability that each individual is in the protected group given their surrogate class membership $z \in Z$.

In the proof below, we re-write these equations to show that they are the same except for one term in the denominator. Specifically, we re-write our inferred metric as:

$$\begin{aligned} m_{wols}(X^\top )-m_{wols}(X^\bot )=\frac{N \sum _x m(x) P_z(x\in X^\top ) - \sum _x m(x) \sum _x P_z(x\in X^\top )}{N\sum _x P_z(x\in X^\top )^2 - \Big (\sum _x P_z(x\in X^\top )\Big )^2} \end{aligned}$$

(10)

We re-write the weighted estimator from [6] as:

$$\begin{aligned} m_w(X^\top )-m_w(X^\bot )=\frac{N \sum _x m(x) P_z(x\in X^\top ) - \sum _x m(x) \sum _x P_z(x\in X^\top )}{N\sum _x P_z(x\in X^\top ) - \Big (\sum _x P_z(x\in X^\top )\Big )^2} \end{aligned}$$

(11)

where $N=\sum _z n_z$ (N is the total number of individuals for which we are calculating fairness metrics).

Equation 10 and Eq. 11 are the same except for the first term in the denominator. We argue here that this difference implies that the weighted estimator is biased toward 0 under the conditions described in (Sect. 3). This means that the weighted estimator will show smaller differences between groups than are actually present in the data.

First, note that $P_z(x\in X^\top )$ is a probability, and therefore bounded between (0,1)

$$ P_z(x\in X^\top )<1 \implies N>\sum _x P_z(x\in X^\top ) \implies N\sum _x P_z(x\in X^\top ) > \Big (\sum _x P_z(x\in X^\top )\Big )^2 $$

This means that the sign of the weighted estimator (whether it is negative or positive) is determined by the numerator of the equation.

Now, because $P_z(x\in X^\top )$ is a probability,

$$P_z(x\in X^\top )^2<P_z(x\in X^\top ) \forall x \implies \sum _x P_z(x\in X^\top ) < \sum _x P_z(x\in X^\top )^2$$

This shows that the first term in the denominator is smaller for our inferred estimator, and therefore:

$$\begin{aligned} &\Big | \frac{N \sum _x m(x) P_z(x\in X^\top ) - \sum _x m(x) \sum _x P_z(x\in X^\top )}{N\sum _x P_z(x\in X^\top )^2 - \big (\sum _x P_z(x\in X^\top )\big )^2}\Big | \\ &\quad > \Big |\frac{N \sum _x m(x) P_z(x\in X^\top ) - \sum _x m(x) \sum _x P_z(x\in X^\top )}{N\sum _x P_z(x\in X^\top ) - \big (\sum _x P_z(x\in X^\top )\big )^2}\Big | \end{aligned}$$

which means:

$$|m_{wols}(X^\top )-m_{wols}(X^\bot )| > |m_w(X^\top )-m_w(X^\bot )|$$

In (Sect. 3) we refer to a set of conditions where WOLS is unbiased that follows from [1] and the Gauss-Markov theorem. The weighted estimator is always smaller in absolute value and must therefore be biased toward 0 under the same conditions.

1.1 Re-Writing the Weighted Estimator

In order to compare the weighted estimator with our inferred estimator, we re-write the weighted estimator for the case where there are two groups, and one surrogate variable Z that acts as a predictor. Now, $P_z(x\in X^\bot )=1-P_z(x \in X^\top )$, so that:

$$m_w(X^\top )-m_w(X^\bot )=\frac{\sum _x m(x)P_z(x\in X^\top )}{\sum _x P_z(x\in X^\top )}-\frac{\sum _x m(x)(1-P_z(x\in X^\top ))}{\sum (1-P_z(x\in X^\top ))}$$

Multiply each of these fractions to get a common denominator.

$$m_w(X^\top )-m_w(X^\bot )=\frac{\sum m(x)P_z(x\in X^\top )\sum (1-P_z(x\in X^\top )) - \sum m(x)(1-P_z(x\in X^\top ))\sum P_z(x\in X^\top )}{\sum P_z(x\in X^\top )\sum (1-P_z(x\in X^\top ))}$$

Then, starting with the numerator, we expand the parentheses and distribute the sums, which gives the following:

$$\begin{aligned} {\begin{matrix} N\sum m(x) P_z(x\in X^\top ) &{}-\sum m(x)P_z(x\in X^\top )\sum P_z(x\in X^\top )\\ &{}-\sum m(x)\sum P_z(x\in X^\top ) +\sum m(x)P_z(x\in X^\top )\sum P_z(x\in X^\top )\\ \end{matrix}} \end{aligned}$$

(12)

The second and fourth terms cancel, so that:

$$m_w(X^\top )-m_w(X^\bot )=\frac{N\sum _x m(x) P_z(x\in X^\top )-\sum _x m(x)\sum _x P_z(x\in X^\top )}{\sum _x(1-P_z(x\in X^\top ))\sum _x P_z(x\in X^\top )}$$

Following the same process for the denominator gives us the following form for the weighted estimator:

$$m_w(X^\top )-m_w(X^\bot )=\frac{N \sum _x m(x) P_z(x\in X^\top )-\sum _x m(x)\sum _x P_z(x\in X^\top )}{N\sum _x P_z(x\in X^\top )-\big (\sum _x P_z(x\in X^\top )\big )^2}$$

1.2 Re-Writing the Inferred Estimator

We can follow the same process to re-write the estimator for the difference between $m_{wols}(X^\top )-m_{wols}(X^\bot )$, and express our inferred fairness metric in terms of the individual values m(x).

As before, start with the numerator, expand the terms in parentheses and distribute the sums, which gives us the following expression.

$$\begin{aligned} {\begin{matrix} \sum _z n_z(m_z-\bar{m})(P_z(x\in X^\top )-\bar{P}(x\in X^\top )) =&{}\sum _z n_z m_z\bar{P}(x\in X^\top )\\ &{}-\bar{m}\sum _zn_zP_z(x\in X^\top )\\ &{}-\bar{P}(x\in X^\top )\sum _z n_zm_z+ \bar{m}\bar{P}(x\in X^\top )\sum _z n_z \end{matrix}} \end{aligned}$$

(13)

Observe the following:

• We require m to be an arithmetic mean, therefore, $m_z=\frac{1}{n_z}\sum _z m(x)$, and $n_z m_z=\sum _{z}m_{x}$

• $\bar{m}=\frac{1}{N} \sum _x m(x)$

• $\bar{P}(x\in X^\top )=\frac{1}{N}\sum _x P(x\in X^\top )=\frac{1}{N}\sum _z n_z P_z(x\in X^\top )$

Taking each term in the numerator separately, we re-write them as:

• $\sum _z n_z m_z\bar{P}(x\in X^\top )=\sum _z n_z(m_z P_z(x\in X^\top ))= \sum _x m(x) P_z(x\in X^\top )$

• $\bar{m}\sum _z n_z P_z(x\in X^\top )=\frac{1}{N}\sum _x m(x) \sum _x P(x\in X^\top )= N\bar{m}\bar{P}(x\in X^\top )$

• $\bar{P}\sum _z n_z m_z=\frac{1}{N}\sum _x P(x\in X^\top ) \sum _x m(x) = N\bar{m}\bar{P}(x\in X^\top )$

• $\bar{m}\bar{P}(x\in X^\top )\sum _z n_z=N\bar{m}\bar{P}(x\in X^\top )$

This lets us collect three of the four terms in the numerator and leaves us with:

$$\begin{aligned} {\begin{matrix} m(X^\top )-m(X^\bot )=\frac{\sum _x m(x) P_z(x\in X^\top )-N\bar{m}\bar{P}(x\in X^\top )}{\sum _z n_z(P_z(x\in X^\top )-\bar{P}(x\in X^\top ))^2} \end{matrix}} \end{aligned}$$

(14)

For the denominator, we again expand the parentheses and collect the sums to give the following:

$$\begin{aligned} {\begin{matrix} \sum _z n_z(P_z(x\in X^\top )-\bar{P}(x\in X^\top ))^2 &{}= \sum _z n_z P_z(x\in X^\top )^2\\ &{}- 2\sum _z n_z P_z(x\in X^\top )\bar{P}(x\in X^\top ) + \sum _z n_z \bar{P}(x \in X^\top )^2 \end{matrix}} \end{aligned}$$

(15)

Again, taking each term separately, we simplify as follows:

• $\sum _z n_z P_z(x\in X^\top )^2 = \sum _z \sum _{x\in z} P_z(x\in X^\top )^2= \sum _x P_z(x\in X^\top )^2$

• $- 2\sum _z n_z P_z(x\in X^\top )\bar{P}(x\in X^\top )=-2N\bar{P}(x\in X^\top )^2$

• $\sum _z n_z \bar{P}(x\in X^\top )^2=\bar{P}_(x\in X^\top )^2\sum _z n_z = N\bar{P}(x\in X^\top )^2$

This gives us the expression:

$$m_{wols}(x\in X^\top )-m_{wols}(X^\bot )=\frac{\sum _xm(x) p (x\in X^\top )-N\bar{m}\bar{P}}{\sum _x P_z(x\in X^\top )^2- N\bar{P}(x\in X^\top )^2}$$

Multiplying the above fraction by $\frac{N}{N}$, gives us the form of the equation as written in (10).

$$m_{wols}(x\in X^\top )-m_{wols}(X^\bot )=\frac{N\sum _x m(x) p (x\in X^\top )-\sum _x m(x) \sum _x P_z(x\in X^\top )}{N\sum _x P_z(x\in X^\top )^2- \big (\sum _x P_z(x\in X^\top )\big )^2}$$