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Alternative participation approach: Assign randomly rather than rank #79

@MaxGhenis

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@MaxGhenis

The SNAP README states:

Households are then ranked by the probability of participation, and, for each state's subgroup, their weights are summed until they reach the administrative level.

This approach would work well if the regression models have good fit, i.e. probabilities are either close to 0 or close to 1. But if probabilities are fuzzier, this approach could overstate the correlation between predictors and participation.

As a simple example, suppose P(SNAP) = max(1 - (income / 50000), 0). A ranking approach might assign participation to everyone with income below $25k, but this would overestimate participation among those with income ~$24k and underestimate it among those around $26k, and in general would overstate the correlation between income and participation.

An alternative is to assign participation randomly depending on predicted probabilities. So if the predicted probabilities align with the administrative total (sum(prob) = participation), you assign if their probability exceeds a U(0,1) random number. This would preserve the fuzziness and avoid excessive correlation.

This gets a bit more complicated when the administrative total doesn't equal the sum of probabilities, which it probably won't, but numpy.random.choice can be used here. For example, suppose there are three filers with predicted SNAP participation probabilities of 10%, 50%, and 80%, and the administrative target is two participants. This function will select two filers using the appropriate probability (notebook):

from numpy.random import choice

def weighted_sample(ids, prob_participate, participation_target):
    return choice(ids, p=prob_participate / prob_participate.sum(),
                  size=participation_target,replace=False)

For example:

import numpy as np

ids = np.array([1, 2, 3])
prob_participate = np.array([0.1, 0.5, 0.8])
participation_target = 2

weighted_sample(ids, prob_participate, participation_target)
# array([2, 3])

[2, 3] is the most likely outcome, but 1 will also be selected some of the time since it has nonzero probability. Under the ranking system it might never get assigned participation.

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