As a worked example, if I start off assuming that chance of there being n busses is 1/2^n (nice and simple, adds up to 1), then the posterior is 1/n(ln(2))(2^n) - multiply the two distributions, then divide by the integral (ln(2)) so that it adds up to 1.
No, that’s not the posterior distribution—clearly, the number of buses cannot be lower than 1546, but that distribution has material probability mass on low integers. I’m not quite sure how you got that equation.
But regardless, I think this shows where we disagree. That prior has mean 2… that’s a pretty strong assumption about the distribution of n. If you want to avoid that kind of assumption, you can get posterior distributions but not a posterior expectation.
As a worked example, if I start off assuming that chance of there being n busses is 1/2^n (nice and simple, adds up to 1), then the posterior is 1/n(ln(2))(2^n) - multiply the two distributions, then divide by the integral (ln(2)) so that it adds up to 1.
No, that’s not the posterior distribution—clearly, the number of buses cannot be lower than 1546, but that distribution has material probability mass on low integers. I’m not quite sure how you got that equation.
But regardless, I think this shows where we disagree. That prior has mean 2… that’s a pretty strong assumption about the distribution of n. If you want to avoid that kind of assumption, you can get posterior distributions but not a posterior expectation.
Sorry, I meant to add in an example where for simplicity you saw the bus numbered 1.
Agreed it’s a terrible prior, it’s just an easy one for a worked example.