This is another write-up of a fact that is generally known, but that I haven't seen proven explicitly: the fact that SIA does not depend upon the reference class.

Specifically:

Assume there are a finite number of possible universes $U_{i}$ . Let $R$ be a reference class of finitely many agents in those universes, and assume you are in $R$ . Let $R_{s}$ be the reference class of agents subjectively indistinguishable from you. Then SIA using $R$ is independent of $R$ as long as $R_{s} \subset R$ .

Proof:

Let ${U_{i}}_{i \in I}$ be a set of universes for some indexing set $I$ , and $P$ a probability distribution over them. For a universe $U_{i}$ , let $R (U_{i})$ be the number of agents in the reference class $R$ in $U_{i}$ .

Then if $P_{R}$ is the probability distribution from SIA using $R$ :

$P_{R} (U_{i}) = \frac{R (U_{i}) P (U_{i})}{\sum_{j \in I} R (U_{j}) P (U_{j})}$ .

We now wish to update on our own subjective experience $s u b$ . Since there are $R (U_{i})$ agents in our reference class, and $R_{s} (U_{i})$ have subjectively indistinguishable experiences, this updates the probabilities by weights $P_{R} (U_{i}) \to P_{R} (U_{i}) \times \frac{R_{s} (U_{i})}{R (U_{i})}$ , which is just $R_{s} (U_{i}) P (U_{i}) / (\sum_{j \in I} R (U_{j}) P (U_{j}))$ . After normalising, this is:

$\begin{matrix} P_{R} (U_{i} | s u b) & = \frac{P (U_{i}) R_{s} (U_{i})}{\sum_{k \in I} P (U_{k}) R_{s} (U_{k})} \times \frac{(\sum_{j \in I} R (U_{j}) P (U_{j}))}{(\sum_{j \in I} R (U_{j}) P (U_{j}))} = \frac{P (U_{i}) R_{s} (U_{i})}{\sum_{k \in I} P (U_{k}) R_{s} (U_{k})} = P_{R_{s}} (U_{i}) . \end{matrix}$

Thus this expression is independent of $R$ .

Given some measure theory (and measure theoretic restrictions on $R$ to make sure expressions like $\int_{I} R (U_{i}) P (U_{i})$ converge), the result extends to infinite classes of universes, with $\int$ in the proof instead of $\sum$ .