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0 pointsyummyfajitas9y ago0 comments

Clearly statistics terminology is confusing you. The definition of bias is E[\hat{\theta} - \theta]. The definition of disparate impact is a predictor computing different means/quantiles for different protected classes.

https://en.wikipedia.org/wiki/Bias_of_an_estimator

https://en.wikipedia.org/wiki/Disparate_impact

To understand this intuitively, here's a simple thought experiment.

Consider Captain Hindsight, a predictor which returns the right answer 100% of the time. By definition, E[\hat{theta} - \theta] = 0, i.e. zero bias. (Also zero variance.)

Now suppose that blacks have a higher recidivism rate (hardly implausible, ProPublica's analysis suggests they do with p < 0.01).

Captain Hindsight - being 100% accurate and having no bias - must predict that blacks have a higher recidivism rate. Yet because Captain Hindsight predicts a higher recidivism rate for blacks, he now has disparate impact.

Seriously, you are calling standard mathematical terminology Orwellian? What's your angle here?

0 comments

zyxley9y ago

Your thought experiment here is incorrect, given that the analysis compares COMPAS results to actual recidivism rates and shows over- and under-prediction in comparison to them.

yummyfajitasOP9y ago

The thought experiment is a mathematical proof that the two concepts are causally unrelated, nothing more. I really suggest you brush up on your basic math - you seem to not be following along.

Your claims about the emirical means of recidivism rates do not prove what you think they prove. Different races might be misclassified at different rates for a variety of reasons - e.g., one race might be affected more by some high-variance predictor, or there could be composition effects (e.g. the pdf of blacks|high score might be different than whites|high score).

The way to factor out whether they scores are biased is to do the cox survival analysis with interaction terms. Which they did. You just don't like the result.

Could you clearly lay out the statistical argument that you believe implies that E[\hat{\theta} - \theta] > 0?

zyxley9y ago

> Could you clearly lay out the statistical argument that you believe implies that E[\hat{\theta} - \theta] > 0?

Again: I don't care about your domain-specific definition of "bias". I care about whether the end result of this secret algorithm is unequal and inaccurate treatment of different demographic groups.

1 more reply

jdp239y ago

Clearly the idea that words have meanings beyond statistics terminology is confusing you.

And no, I'm not calling standard mathematical terminology Orwellian. What I'm calling Orwellian is your describing a biased system as unbiased (by attempting to reframe the discussion around a specific statistical definition, chosen by you).

j / k navigate · click thread line to collapse

0 pointsyummyfajitas9y ago0 comments

https://en.wikipedia.org/wiki/Bias_of_an_estimator

https://en.wikipedia.org/wiki/Disparate_impact

To understand this intuitively, here's a simple thought experiment.

Consider Captain Hindsight, a predictor which returns the right answer 100% of the time. By definition, E[\hat{theta} - \theta] = 0, i.e. zero bias. (Also zero variance.)

Now suppose that blacks have a higher recidivism rate (hardly implausible, ProPublica's analysis suggests they do with p < 0.01).

Seriously, you are calling standard mathematical terminology Orwellian? What's your angle here?

0 comments

zyxley9y ago

Your thought experiment here is incorrect, given that the analysis compares COMPAS results to actual recidivism rates and shows over- and under-prediction in comparison to them.

yummyfajitasOP9y ago

The thought experiment is a mathematical proof that the two concepts are causally unrelated, nothing more. I really suggest you brush up on your basic math - you seem to not be following along.

The way to factor out whether they scores are biased is to do the cox survival analysis with interaction terms. Which they did. You just don't like the result.

Could you clearly lay out the statistical argument that you believe implies that E[\hat{\theta} - \theta] > 0?

zyxley9y ago

> Could you clearly lay out the statistical argument that you believe implies that E[\hat{\theta} - \theta] > 0?

1 more reply

jdp239y ago

Clearly the idea that words have meanings beyond statistics terminology is confusing you.

j / k navigate · click thread line to collapse