sj9999 comments on Bayesianism versus conservatism versus Goodhart

[sj9999]

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.