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sj9999

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sj9999

sj9999 — LessWrong

sj9999

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sj9999

Replying toBayesianism versus conservatism versus Goodhart

sj99994y

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... (read more)

Replying toGeneralised models as a category

sj99994y

Generalised models as a category

I think these might be some typos you could correct:

, 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 $E_{0}$ : $Q_{1} (r (E_{0})) \leq Q_{0} (r^{- 1} r (E_{0})) = Q_{0} (E_{0})$