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> Why not? If you define poverty as being below the Xth percentile of wealth, then it is tautological that we cannot reach 100% non-poverty. This is indeed how

> Why not? If you define poverty as being below the Xth percentile of wealth, then it is tautological that we cannot reach 100% non-poverty. This is indeed how it is defined in many wellfare states eg in Europe. "People are considered at risk of monetary poverty when their equivalised disposable income (after social transfers) is below the at-risk-of-poverty threshold. This is set at 60 % of the national median equivalised disposable income after social transfers." http://ec.europa.eu/eurostat/statistics-explained/index.php/Europe_2020_indicators_-_poverty_and_social_exclusion http://ec.europa.eu/eurostat/statistics-explained/index.php/... There will always be poor people with this definition.

Al-Khwarizmi 9y ago
No. "60% of the national median" is not the same thing as fixing a percentile, and it is perfectly possible to not have poverty under that definition. For example, in a country with ten people whose incomes are 7, 7, 8, 8, 10, 10, 12, 50, 47212 and 4000000000000000, there is no poverty after that definition (the median would be 10). As you can see, it doesn't even need to be a very egalitarian society!
zaarn 9y ago
"60% of the Median" can easily be a zero number of elements from the set. In this case, median means "sorted by income, at which point have we divided the entire dataset in half". For a dataset with more than 2 elements (so 3 elements minimum), you can construct a dataset for which any below-100% of the median of the dataset is not represented in the dataset. The easiest example would be using "99% of the median" and [9.999, 10, 11]. Median is 10, 99% of the median is 9.9, the smallest sample is 9.999. 60% of the median income basically means "the lowest 50% income bracket should not have more than 40% deviation". Or rather; the lowest possible income is bigger than 60% of the lowest earner of the top 50% of society.

Al-Khwarizmi 9y ago

No. "60% of the national median" is not the same thing as fixing a percentile, and it is perfectly possible to not have poverty under that definition. For example, in a country with ten people whose incomes are 7, 7, 8, 8, 10, 10, 12, 50, 47212 and 4000000000000000, there is no poverty after that definition (the median would be 10). As you can see, it doesn't even need to be a very egalitarian society!

zaarn 9y ago

"60% of the Median" can easily be a zero number of elements from the set. In this case, median means "sorted by income, at which point have we divided the entire dataset in half". For a dataset with more than 2 elements (so 3 elements minimum), you can construct a dataset for which any below-100% of the median of the dataset is not represented in the dataset. The easiest example would be using "99% of the median" and [9.999, 10, 11]. Median is 10, 99% of the median is 9.9, the smallest sample is 9.999. 60% of the median income basically means "the lowest 50% income bracket should not have more than 40% deviation". Or rather; the lowest possible income is bigger than 60% of the lowest earner of the top 50% of society.