Musings

You can think of it a Bayesian prior that gives the probability that “we” are any particular one of the copies. (For various reasons, Jim and I prefer not to call the xerographic distribution a prior, but that’s just semantics.)

The whole point of the Bayesian theory is to provide a degree of robustness against badly-chosen priors. You’re not being Bayesian if you don’t avail yourself of Bayes’ Theorem!

My own personal philosophical belief (and Jim Hartle’s) is that we should not insist upon a uniform xerographic distribution.

You and Jim argue that the xerographic prior is a poor one, and we can quantify that.

A “good” prior is one that allows you to converge as quickly as possible to the “correct” probability distribution, through the application of Bayes’ Theorem. The xerographic distribution yields the prior discussed above, where $$P_0 = 1-\delta$$ with $\delta$ very tiny. Applying Bayes’ Theorem, we see that $P_n$ stays near 1 until $$n\sim \frac{\log\delta}{\log\epsilon}$$ microseconds, at which point it abruptly drops to near zero.

That’s very fast, but a prior that converges even faster is to start with $P_0$ near 0. That converges for $$n\sim O(1)$$ (with disappointing results for the BB observer, but he’s not around to care).

So I agree with your conclusion. I just disagree that it is merely a “philosophical belief.”

Thanks for taking the time to reply, and sorry if my math was unreadable - writing latex on my tablet was too tricky - let me write it again so we definitely understand each other.

I wrote that the relevant probability was,$$p(\text{model} | \text{data}) \propto p(\text{data} | \text{model}) \times p(\text{model})$$I then wrote that, $$p(\text{data} | \text{model}) = p(\text{data} | \text{model}, \text{we are BB}) p(\text{we are BB} | \text{model}) + p(\text{data} | \text{model}, \text{we are NOT BB}) p(\text{we are NOT BB} | \text{model})$$The second probabilities, $p(\text{we are NOT BB} | \text{model}) = 1 - p(\text{we are BB} | \text{model})$, are the Xerographic distributions.

We know that $p(\text{data} | \text{model}, \text{we are BB}) \simeq 0$ and that $p(\text{data} | \text{model}, \text{we are NOT BB}) \simeq 1$.

Sean Caroll’s Xerographic distribution is that $p(\text{we are NOT BB} | \text{model})\simeq 1$. So he should conclude that either his Xerographic distribution was wrong or he should reject his model.

Right now, I find Sean Caroll’s argument convincing/cogent. You say that the Xerographic distribution is wrong and that $p(\text{we are NOT BB} | \text{model})\simeq 0$. The BB arguments therefore have no power to help us pick models.

You say that the Xerographic is wrong because we should pick a prior, $p(\text{we are NOT BB} | \text{model})$ such that our posterior $p(\text{we are NOT BB} | \text{model}, \text{data})$ converges as quickly as possible? I don’t follow that. Shouldn’t we pick priors that best represent our state of knowledge prior to seeing the data?

Maybe I’ve misunderstood. Let me ask it another way. The Xerox distribution appears on its own in my equations, i.e. it is not conditioned on the data. Why should it be?

Best,

Andrew.

PS Prof. Caroll, Srednicki, sorry if I am misrepresenting your opinions/arguments.