sj9999

Karma: 9

sj9999 12 May 2022 20:02 UTC
1 point
AF
on: Bayesianism versus conservatism versus Goodhart
These might be some typos:
From the following, the second $V_{3}$ should be $V_{2}$ :
What can we say about $U$ ? Well, if we fix a $v_{3} \in V_{3}$ , then it will appear in $(\frac{N_{1}}{n_{1}}) (\frac{N_{2}}{n_{2}}) (\frac{N_{3} - 1}{n_{3} - 1})$ of the $U_{σ}$ terms (i.e. all choices of the $V_{1}$ and $V_{3}$ variables, and all possible choices of the other $n_{3} - 1$ variables in the other $N_{3} - 1$ variables in $V_{3}$ ).
as in
(i.e. all choices of the $V_{1}$ and $V_{2}$ variables, and all possible choices of the other $n_{3} - 1$ variables in the other $N_{3} - 1$ variables in $V_{3}$ ).
In the following, the second term $(\prod_{v_{1} \in σ_{3}} H (v_{2}))$ should instead be multiplied over $v_{2} \in σ_{2}$ .
Define $σ = (σ_{1}, σ_{2}, σ_{3})$ as selecting $n_{i}$ variables from $V_{i}$ , and
$U_{σ} = (\sum v_{1} \in σ_{1} v_{1}) \cdot (\prod v_{1} \in σ_{3} H (v_{2})) + \sum v_{3} \in σ_{3} - e^{- v_{3}} .$

I believe the following should be “from positive to negative” rather than “from negative to positive”.
How about a fixed $v_{2} \in V_{2}$ ? Well, if $v_{2}$ goes below $0$ , that will kill off $(\frac{N_{1}}{n_{1}}) (\frac{N_{2} - 1}{n_{2} - 1}) (\frac{N_{3}}{n_{3}})$ of the $U_{σ}$ . So if all other $v_{2}^{'} \in V_{2}$ are positive, sending $v_{2}$ from negative to positive will multiply the expected value by $1 - n_{2} / N_{2}$ .

And here, it seems “(ignoring the $V_{3}$ )” might be “(ignoring the $v$ )”.
In some ways we cannot afford to be sloppy: assume that $v$ should be in $V_{2}$ but isn’t; so the true utility is $U = U_{V_{1}, V_{2} \cup {v}, V_{3}}$ , but a $U^{'} = U_{V_{1}, V_{2}, V_{3}}$ -maximiser might sacrifice $v$ to increase $U^{'}$ ; thus (ignoring the $V_{3}$ ) maximising $U^{'}$ may set $U$ to $0$ .
Thanks to Rupert McCallum for help in identifying some of these typos.

sj9999 23 Apr 2022 23:57 UTC
LW: 10 AF: 6
AF
on: Generalised models as a category
I think these might be some typos you could correct:
$Q_{1} (E_{1}) \leq Q_{0} (r^{- 1} (E_{0}))$ , or both measures are undefined.
The $E_{0}$ should be $E_{1}$ .
For such an $r$ -stable set, $Q_{0} (E_{0}) \leq Q_{1} (r (E_{0}))$ and $Q_{1} (r (E_{1}) \leq Q_{0} (r^{- 1} r (E_{0})) = Q_{0} (E_{0})$ , thus $Q_{0} (E_{0}) = Q_{1} (r (E_{0}))$ .
There is a missing parenthesis and the $E_{1}$ should be $E_{0}$ : $Q_{1} (r (E_{0})) \leq Q_{0} (r^{- 1} r (E_{0})) = Q_{0} (E_{0})$