3 ms·

I did not compare raw counts, I compared per capita counts.

by flanked-evergl 2y ago

sdenton4 2y ago
That's not what I said? Reread the comment, you might learn something.
- flanked-evergl 2y ago
  I did re-read. Quoting you: "There are far more cities with Dem mayors than Republican mayors, so comparing raw counts is pretty meaningless. (Similar to counting comparing total homicide instead of per capita.)" Quoting me: "I did not compare raw counts, I compared per capita counts." I did not learn anything I did not know before re-reading.
  - sdenton4 2y ago
    You're using raw counts /of cities/ in a top ten list, when the actual leadership of cities is heavily skewed towards Dem mayors. I point out that this is similar to comparing raw counts instead of per-capita.
    - flanked-evergl 2y ago
      Thanks for clarifying your objection. Here is a list of the most violent US cities, reckoned per capita, without first selecting for being the most high crime cities in absolute terms: https://worldpopulationreview.com/us-city-rankings/most-violent-cities-in-america https://worldpopulationreview.com/us-city-rankings/most-viol... Here is the political affiliation of the city governments: 1. St. Louis, MO: Democrat 2. Detroit, MI: Democrat 3. Baltimore, MD: Democrat 4. Memphis, TN: Democrat 5. Little Rock, AR: Democrat 6. Milwaukee, WI: Democrat 7. Rockford, IL: Democrat 8. Cleveland, OH: Democrat 9. Stockton, CA: Democrat 10. Albuquerque, NM: Democrat 11. Springfield, MO: Independent 12. Indianapolis, IN: Democrat 13. Oakland, CA: Democrat 14. San Bernardino, CA: Democrat 15. Anchorage, AK: Independent 16. Nashville, TN: Democrat 17. Lansing, MI: Democrat 18. New Orleans, LA: Democrat 19. Minneapolis, MN: Democrat 20. Chicago, IL: Democrat
      - sdenton4 2y ago
        Let v = high violence, d = dem mayor, r = republican mayor. You seem to want to compare P(v|d) to P(v|r). We have little to no data for comparison because there are so few examples of cities with republican mayors. Those that exist are amongst the smaller American cities. This means you need to disentangle the effect of city size on violence rates from the effect of the mayor's party. I repeat: There are 10 Republican mayors amongst the 50 largest cities, and furthermore, these are heavily concentrated at the low end. https://en.m.wikipedia.org/wiki/List_of_mayors_of_the_50_lar https://en.m.wikipedia.org/wiki/List_of_mayors_of_the_50_lar... It turns out the relationship between crime and population is non linear, and that nonlinearity is also true outside of the US and it's political context. https://crimesciencejournal.biomedcentral.com/articles/10.11 https://crimesciencejournal.biomedcentral.com/articles/10.11... So, what you're seeing is bigger cities with higher per capita rates due to the underlying relationship between population density and crime. It also happens to be the case that in the US, urban areas strongly prefer Democratic mayors.
        flanked-evergl 2y ago
        You could be right, but I have my doubts that you are.

I did not compare raw counts, I compared per capita counts.

sdenton4 2y ago

That's not what I said? Reread the comment, you might learn something.

flanked-evergl 2y ago
I did re-read. Quoting you: "There are far more cities with Dem mayors than Republican mayors, so comparing raw counts is pretty meaningless. (Similar to counting comparing total homicide instead of per capita.)" Quoting me: "I did not compare raw counts, I compared per capita counts." I did not learn anything I did not know before re-reading.
- sdenton4 2y ago
  You're using raw counts /of cities/ in a top ten list, when the actual leadership of cities is heavily skewed towards Dem mayors. I point out that this is similar to comparing raw counts instead of per-capita.
  - flanked-evergl 2y ago
    Thanks for clarifying your objection. Here is a list of the most violent US cities, reckoned per capita, without first selecting for being the most high crime cities in absolute terms: https://worldpopulationreview.com/us-city-rankings/most-violent-cities-in-america https://worldpopulationreview.com/us-city-rankings/most-viol... Here is the political affiliation of the city governments: 1. St. Louis, MO: Democrat 2. Detroit, MI: Democrat 3. Baltimore, MD: Democrat 4. Memphis, TN: Democrat 5. Little Rock, AR: Democrat 6. Milwaukee, WI: Democrat 7. Rockford, IL: Democrat 8. Cleveland, OH: Democrat 9. Stockton, CA: Democrat 10. Albuquerque, NM: Democrat 11. Springfield, MO: Independent 12. Indianapolis, IN: Democrat 13. Oakland, CA: Democrat 14. San Bernardino, CA: Democrat 15. Anchorage, AK: Independent 16. Nashville, TN: Democrat 17. Lansing, MI: Democrat 18. New Orleans, LA: Democrat 19. Minneapolis, MN: Democrat 20. Chicago, IL: Democrat
    - sdenton4 2y ago
      Let v = high violence, d = dem mayor, r = republican mayor. You seem to want to compare P(v|d) to P(v|r). We have little to no data for comparison because there are so few examples of cities with republican mayors. Those that exist are amongst the smaller American cities. This means you need to disentangle the effect of city size on violence rates from the effect of the mayor's party. I repeat: There are 10 Republican mayors amongst the 50 largest cities, and furthermore, these are heavily concentrated at the low end. https://en.m.wikipedia.org/wiki/List_of_mayors_of_the_50_lar https://en.m.wikipedia.org/wiki/List_of_mayors_of_the_50_lar... It turns out the relationship between crime and population is non linear, and that nonlinearity is also true outside of the US and it's political context. https://crimesciencejournal.biomedcentral.com/articles/10.11 https://crimesciencejournal.biomedcentral.com/articles/10.11... So, what you're seeing is bigger cities with higher per capita rates due to the underlying relationship between population density and crime. It also happens to be the case that in the US, urban areas strongly prefer Democratic mayors.
      - flanked-evergl 2y ago
        You could be right, but I have my doubts that you are.