Improved formalism for corruption in DIRL
discussion post by Vadim Kosoy 98 days ago | discuss

We give a treatment of advisor corruption in DIRL, more elegant and general than our previous formalism.

The following definition replaces the original Definition 5.

# Definition

Consider a meta-universe $$\upsilon=(\mu,r)$$ and $$\beta:(0,\infty)\rightarrow(0,\infty)$$. A metapolicy $$\alpha$$ is called $$\beta$$-rational for $$\upsilon$$ (as opposed to before, we assume $$\alpha$$ is an $${\mathcal{I}}$$-metapolicy rather than an $${{\bar{{\mathcal{I}}}}}$$-metapolicy; this is purely for notational convenience and it is straightforward to generalize the definition) when there exists $$\{L^\alpha_t: \operatorname{hdom}{\mu} \times {\mathcal{A}}\rightarrow [0,\infty]\}_{t \in (0,\infty)}$$ s.t.

1. For any $$h \in \operatorname{hdom}{\mu}$$ and $$t \in (0,\infty)$$, there is $$a \in {\mathcal{A}}$$ s.t. $$L^\alpha_t(ha)=0$$.

2. $$\alpha_t(h)(a)=\exp(-\beta(t)L^\alpha_t(ha)) \max_{a^* \in {\mathcal{A}}} \alpha_t(h)(a^*)$$

3. For any $$\pi \in \Pi$$ and $$t \in (0,\infty)$$

$\lim_{t \rightarrow \infty}\min({\underset{x\sim\mu_t\bowtie\pi}{\operatorname{E}}}[\sum_{n=0}^\infty e^{-n/t} L^\alpha_t(x_{:n+1/2})]-{\underset{x\sim\mu_t\bowtie\pi}{\operatorname{E}}}[\sum_{n=0}^\infty e^{-n/t}({\operatorname{V}}^\upsilon_t(x_{:n})-{\operatorname{Q}}^\upsilon_t(x_{:n+1/2}))],0)=0$

In condition ii, $$\exp(-\infty)$$ is understood to mean 0. Conditions i+ii can be seen as the definition of $$L^\alpha_t$$ given $$\alpha_t$$. A notable special case of condition iii is when for any $$x \in {({\mathcal{A}}\times {\mathcal{O}})^\omega}$$

$\sum_{n=0}^\infty e^{-n/t} L^\alpha_t(x_{:n+1/2}) \geq \sum_{n=0}^\infty e^{-n/t}({\operatorname{V}}^\upsilon_t(x_{:n})-{\operatorname{Q}}^\upsilon_t(x_{:n+1/2}))$

As a simple example, we can have a set of corrupt states $$\{{\mathcal{C}}_t \subseteq {({\mathcal{A}}\times {\mathcal{O}})^*}\}_{t\in(0,\infty)}$$ in which the behavior of the advisor becomes arbitrary, but for each $$h \in {\mathcal{C}}_t$$ there is $$g \in {({\mathcal{A}}\times {\mathcal{O}})^*}\times {\mathcal{A}}$$ s.t. $$g \sqsubset h$$ and $$L^\alpha_t(g)=\infty$$ (i.e., to corrupt the advisor one has to take an action that the advisor would never take). As opposed to before, this formalism can also account for partial corruption, e.g. if for each $$h \not\in {\mathcal{C}}_t$$ and $$a \in {\mathcal{A}}$$, we have $$L^\alpha_t(ha) \geq {\operatorname{V}}^\upsilon_t(h) - {\operatorname{Q}}^\upsilon_t(ha)$$ (like in strict $$\beta$$-rationality) whereas for $$h \in {\mathcal{C}}_t$$, we only have $$L^\alpha_t(ha) \geq {\operatorname{V}}^\upsilon_t(h) - {\operatorname{Q}}^\upsilon_t(ha) - \delta$$ for some constant $$\delta > 0$$, then to ensure $$\beta$$-rationality, it is sufficient that for each $$h = a_0o_0a_1o_1 \ldots \in {\mathcal{C}}_t$$:

$\sum_{n=0}^{\max\{m \mid h_{:m} \not\in{\mathcal{C}}_t\}} e^{-n/t}(L^\alpha_t(h_{:n}a_n) - {\operatorname{V}}^\upsilon_t(h_{:n}) - {\operatorname{Q}}^\upsilon_t(h_{:n}a_n)) \geq \frac{\delta e^{-(\max\{m \mid h_{:m} \not\in{\mathcal{C}}_t\}+1)/t}}{1-e^{-1/t}}$

# Theorem

Consider $${\mathcal{H}}= \{\upsilon^k\}_{k \in {\mathbb{N}}}$$ a countable family of $${\mathcal{I}}$$-meta-universes and $$\beta: (0,\infty) \rightarrow (0,\infty)$$ s.t. $$\beta(t) = \omega(t^{2/3})$$. Let $$\{\alpha^k\}_{k \in {\mathbb{N}}}$$ be a family of $${\mathcal{I}}$$-metapolicies s.t. for every $$k \in {\mathbb{N}}$$, $$\alpha^k$$ is $$\beta$$-rational for $$\upsilon^k$$. Define $$\bar{{\mathcal{H}}}:=\{\bar{\upsilon}^k[\alpha^k]\}_{k \in {\mathbb{N}}}$$. Then, $$\bar{{\mathcal{H}}}$$ is learnable.

# Proof of Theorem

We don’t spell out the proof in detail, but only the modifications with respect to the original.

As in the proof of the original theorem, we can assume without loss of generality that $${\mathcal{H}}$$ is finite. Define $$\pi^*$$ the same way as in Lemma A, but with $$L_t$$ redefined as

$L_t(ha):={\underset{k\sim\zeta_t(h)}{\operatorname{E}}}[L^{\alpha^k}_t(ha)]$

Similarly, define $$\pi^!$$ the same way as in the proof of Lemma A, but with $$L_t$$ redefined as

$L_t(ha):={\underset{k\sim\zeta^{!k}_t(h)}{\operatorname{E}}}[L^{\alpha^k}_t(ha)]$

As in the proof of Lemma A, we have

$\frac{1}{N}\sum_{k < N}({\operatorname{EU}}_{\bar{\upsilon}^k[\alpha^k]}^{*}(t) - {\operatorname{EU}}_{\bar{\upsilon}^k[\alpha^k]}^{\pi^{!k}}(t))=\sum_{n=0}^\infty e^{-n/t} {\underset{(k,x)\sim\rho^!_t}{\operatorname{E}}}[{\operatorname{V}}^{\upsilon^k[\alpha^k]}_t(x_{:n})-{\operatorname{Q}}^{\upsilon^k[\alpha^k]}_t(x_{:n}\pi^{!k}(x_{:n}))]$

Using condition iii in the Definition, we conclude that for some function $$\delta:(0,\infty)\rightarrow[0,\infty)$$ with $$\lim_{t\rightarrow\infty}\delta(t)=0$$

$\frac{1}{N}\sum_{k < N}({\operatorname{EU}}_{\bar{\upsilon}^k[\alpha^k]}^{*}(t) - {\operatorname{EU}}_{\bar{\upsilon}^k[\alpha^k]}^{\pi^{!k}}(t)) \leq \sum_{n=0}^\infty e^{-n/t} {\underset{(k,x)\sim\rho^!_t}{\operatorname{E}}}\left[[[\pi^{!k}(x_{:n})\ne\bot]]L^{\alpha^k}_t(\underline{x_{:n}}\pi^{!k}(x_{:n}))+[[\pi^{!k}(x_{:n})=\bot]]{\underset{a\sim\alpha^k(\underline{x_{:n}})}{\operatorname{E}}}[L^{\alpha^k}_t(\underline{x_{:n}}a)]\right]+\delta(t)$

We can now repeat the same arguments as in the proof of Lemma A to get

$\frac{1}{N}\sum_{k < N}({\operatorname{EU}}_{\bar{\upsilon}^k[\alpha^k]}^{*}(t) - {\operatorname{EU}}_{\bar{\upsilon}^k[\alpha^k]}^{\pi^{*}}(t)) \leq \left(\frac{1}{t}+1+\frac{8 {\lvert {\mathcal{A}}\rvert}^3 \ln{N}}{e(1-e^{-1})^2}\right)\frac{t^{2/3}}{\beta(t)}+\frac{N-1}{t^{1/3}}+\delta(t)$

The desired result follows.

### NEW DISCUSSION POSTS

What does the Law of Logical
 by Alex Appel on Smoking Lesion Steelman III: Revenge of the Tickle... | 0 likes

To quote the straw vulcan:
 by Stuart Armstrong on Hyperreal Brouwer | 0 likes

I intend to cross-post often.
 by Scott Garrabrant on Should I post technical ideas here or on LessWrong... | 1 like

I think technical research
 by Vadim Kosoy on Should I post technical ideas here or on LessWrong... | 2 likes

I am much more likely to miss
 by Abram Demski on Should I post technical ideas here or on LessWrong... | 1 like

Note that the problem with
 by Vadim Kosoy on Open Problems Regarding Counterfactuals: An Introd... | 0 likes

Typos on page 5: *
 by Vadim Kosoy on Open Problems Regarding Counterfactuals: An Introd... | 0 likes

Ah, you're right. So gain
 by Abram Demski on Smoking Lesion Steelman | 0 likes

> Do you have ideas for how
 by Jessica Taylor on Autopoietic systems and difficulty of AGI alignmen... | 0 likes

I think I understand what
 by Wei Dai on Autopoietic systems and difficulty of AGI alignmen... | 0 likes

>You don’t have to solve
 by Wei Dai on Autopoietic systems and difficulty of AGI alignmen... | 0 likes

 by Vadim Kosoy on Delegative Inverse Reinforcement Learning | 0 likes

My confusion is the
 by Tom Everitt on Delegative Inverse Reinforcement Learning | 0 likes

> First of all, it seems to
 by Abram Demski on Smoking Lesion Steelman | 0 likes

> figure out what my values
 by Vladimir Slepnev on Autopoietic systems and difficulty of AGI alignmen... | 0 likes