Comment by sj9999 (Sai Joseph) on Bayesianism versus conservatism versus Goodhart · 2022-05-12T20:02:50.951Z · LW · GW

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 $U_{\sigma }$ 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 $\left({\prod }_{v_1\in {\sigma }_3}H\right(v_2\left)\right)$ should instead be multiplied over $v_2\in {\sigma }_2$.  

Define $\sigma =({\sigma }_1,{\sigma }_2,{\sigma }_3)$ as selecting $n_i$ variables from $V_i$, and

 

$$U_{\sigma }=\left(\sum _{v_1\in {\sigma }_1}{v}_1\right)\cdot \left(\prod _{v_1\in {\sigma }_3}H\right(v_2\left)\right)+\sum _{v_3\in {\sigma }_3}-e^{-v_3}.$$

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 $U_{\sigma }$. So if all other $v_2^{\prime }\in V_2$ 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 $U=U_{V_1,V_2\cup \left\{v\right\},V_3}$, but a $U^{\prime }=U_{V_1,V_2,V_3}$-maximiser might sacrifice $v$ to increase $U^{\prime }$; thus (ignoring the $V_3$) maximising $U^{\prime }$ may set $U$ to $0$.

 

Thanks to Rupert McCallum for help in identifying some of these typos. 

Comment by sj9999 (Sai Joseph) on Generalised models as a category · 2022-04-23T23:57:14.678Z · LW · GW

I think these might be some typos you could correct: 

$Q_1\left(E_1\right)\le Q_0\left(r^{-1}\right(E_0\left)\right)$, or both measures are undefined.

The $E_0$ should be  $E_1$.

For such an $r$-stable set, $Q_0\left(E_0\right)\le Q_1\left(r\right(E_0\left)\right)$ and $Q_1\left(r\right(E_1)\le Q_0(r^{-1}r\left(E_0\right))=Q_0(E_0)$, thus $Q_0\left(E_0\right)=Q_1\left(r\right(E_0\left)\right)$.

There is a missing parenthesis and the $E_1$ should be $E_0$:    $Q_1\left(r\right(E_0\left)\right)\le Q_0\left(r^{-1}r\right(E_0\left)\right)=Q_0\left(E_0\right)$