I’m not sure if using the Lindy effect for forecasting x-risks makes sense. The Lindy effect states that with 50% probability, things will last as long as they already have. Here is an example for AI timelines.
The Lindy rule works great on average, when you are making one-time forecasts of many different processes. The intuition for this is that if you encounter a process with lifetime T at time t<T, and t is uniformly random in [0,T], then on average T = 2*t.
However, if you then keep forecasting the same process over time, then once you surpass T/2 your forecast becomes worse and worse as time goes by. Just when t is very close to T is when you are most confident that T is a long time away. If forecasting this particular process is very important (eg: because it’s an x-risk), then you might be in trouble.
Suppose that some x-risk will materialize at time T, and the only way to avoid it is doing a costly action in the 10 years before T. This action can only be taken once, because it drains your resources, so if you take it more than 10 years before T, the world is doomed.
This means that you should act iff you forecast that T is less than 10 years away. Let’s compare the Lindy strategy with a strategy that always forecasts that T is <10 years away.
If we simulate this process with uniformly random T, for values of T up to 100 years, the constant strategy saves the world more than twice as often as the Lindy strategy. For values of T up to a million years, the constant strategy is 26 times as good as the Lindy strategy.
Don’t both strategies start out with forecasts < 10 years and act immediately? Actually, calculating the utility-maximizing decision depends on the distribution forecast, not just the best-guess point estimate.
I’m not sure if using the Lindy effect for forecasting x-risks makes sense. The Lindy effect states that with 50% probability, things will last as long as they already have. Here is an example for AI timelines.
The Lindy rule works great on average, when you are making one-time forecasts of many different processes. The intuition for this is that if you encounter a process with lifetime T at time t<T, and t is uniformly random in [0,T], then on average T = 2*t.
However, if you then keep forecasting the same process over time, then once you surpass T/2 your forecast becomes worse and worse as time goes by. Just when t is very close to T is when you are most confident that T is a long time away. If forecasting this particular process is very important (eg: because it’s an x-risk), then you might be in trouble.
Suppose that some x-risk will materialize at time T, and the only way to avoid it is doing a costly action in the 10 years before T. This action can only be taken once, because it drains your resources, so if you take it more than 10 years before T, the world is doomed.
This means that you should act iff you forecast that T is less than 10 years away. Let’s compare the Lindy strategy with a strategy that always forecasts that T is <10 years away.
If we simulate this process with uniformly random T, for values of T up to 100 years, the constant strategy saves the world more than twice as often as the Lindy strategy. For values of T up to a million years, the constant strategy is 26 times as good as the Lindy strategy.
Don’t both strategies start out with forecasts < 10 years and act immediately? Actually, calculating the utility-maximizing decision depends on the distribution forecast, not just the best-guess point estimate.