Thursday, January 27, 2011

Earmarks revisited

Harry Reid is trying the "earmarks don't increase spending" line. I commented on that idea here.

Monday, January 24, 2011

Note toward a future blog post

Bradford Plumer, like Matt Yglesias, is a fan of the Chinese idea (I'm taking it for granted that they're right about this, I don't know it myself) of having good mayors promoted up through the mayorly ranks.
If the mayor of a small city—say, out in China's provinces—does a good job, then when his term is up he may get appointed as mayor of an even bigger city. Being able to govern a city is a useful skill, after all, so why not people who have shown some talent at it move up and try their hand at governing bigger cities, rather than restricting the job to whatever local dogcatcher can prove residency? Maybe voters don't want meddling outsiders. But they can make that choice themselves, no?

Sounds good. But I'll go with "no." In the American party system I'm thinking that turns mayorships into almost purely patronage positions. Every medium-sized city mayor in America decides to try to succeed Daley. Who will win? Well, there will be a Democratic primary to decide it. But with dozens of potential candidates who are poorly known to the local voters, someone is going to do some serious vetting: the incumbent's machine, or the governor if of the same party, or US Senators if of the same party... or the President if of the same party.

The U.S. has some 50,000 municipalities. The coordination problem of hundreds or thousands of mayors trying to move around the country and up the totem pole, when the voters "hiring" them in this odd labor market can't possibly have information about most of them, are huge. So, formally or informally, we'd end up with the Chinese model: the party hierarchy would provide the coordination.

American urban government is often nothing to write home about. But I'm unpersuaded that it would be improved by concentrating selection in the national political parties; and I think that's what this would necessarily amount to.