Visual walkthrough — Outer vs inner alignment

This page rebuilds the central result of Outer vs Inner Alignment — the alignment gap decomposition — using nothing but pictures and plain words. We will not use a single symbol before it earns a place on a diagram. By the end you will see why one honest-looking equation splits every AI failure into exactly two families.

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Step 1 — What is a "policy", and what is a "score"?

WHAT. Before any math, two words. A policy is just a way of behaving — a complete rule that says, in every situation, what the AI does. Think of it as one dot on a big flat map of "all possible behaviours." Call the map the behaviour space. Every dot is one policy.

A score is a number we attach to a policy telling us how good that behaviour is. Different observers give different scores to the same dot.

WHY these two ideas first. The whole result is a comparison of dots using scores. If you don't have "a point on a map" and "a height above each point," none of the later arrows mean anything.

PICTURE. The map is the flat grey sheet. Each labelled dot is one policy (a behaviour). We will pile heights on top of these dots in the next steps.

Here $\pi$ (Greek letter "pi") is just a name for "some behaviour." It carries no meaning beyond "a dot."

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Step 2 — Two different heights: what we want vs what we asked for

WHAT. We now put two different height-fields over the very same map.

• $U^{\ast }$ — read "you-star" — is the true utility: the height we actually care about. Tall = genuinely good for us. We usually cannot write it down.
• $\mathcal{L}$ — read "the training objective" — is the score we actually wrote into the computer, drawn so that tall = good (a reward, i.e. minus the usual loss). Tall = scores well on our test.

WHY two fields. This is the heart of the parent note: the thing we want and the thing we measure are two different landscapes over the same behaviours. Drawing them as two surfaces makes their disagreement literally visible as a mismatch in shape.

PICTURE. Same flat map, two coloured surfaces floating above it. Where the blue ($U^{\ast }$) surface peaks and where the orange ($\mathcal{L}$) surface peaks are not the same spot — that gap is the entire story.

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Step 3 — Three special dots we will keep pointing at

WHAT. From the two landscapes we pick out exactly three policies.

• ${\pi }^{\ast }$ = the dot where the blue surface is highest → the best behaviour for what we want.
• ${\pi }_{\mathcal{L}}^{\ast }$ = the dot where the orange surface is highest → the best behaviour for what we asked for.
• ${\pi }_{\text{learned}}$ = the dot the training process actually lands on.

The symbol $\arg \max$ (read "arg-max") just means "the input where the height is biggest." Not the height itself — the location. That is exactly why we need it: we care which behaviour, not how tall.

${\pi }^{\ast }=\underset{\text{the dot that is highest}}{\underbrace{\arg \underset{\pi }{\max }}}\underset{\text{using the blue (true) surface}}{\underbrace{U^{\ast }(\pi )}}$

${\pi }_{\mathcal{L}}^{\ast }=\underset{\text{the dot that is highest}}{\underbrace{\arg \underset{\pi }{\max }}}\underset{\text{using the orange (asked-for reward) surface}}{\underbrace{\mathcal{L}(\pi )}}$

WHY these three and no others. They are the only anchors we need: the ideal, the best-we-could-hope-if-training-were-perfect, and reality. Every failure is a distance between two of these.

PICTURE. The two surfaces again, with three flags planted: blue flag at the blue peak (${\pi }^{\ast }$), orange flag at the orange peak (${\pi }_{\mathcal{L}}^{\ast }$), red flag wherever training stopped (${\pi }_{\text{learned}}$).

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Step 4 — Measuring the miss: the alignment gap

WHAT. We now measure everything with one honest ruler: the blue surface $U^{\ast }$, because that is the only thing we truly care about. Read the true height at the ideal dot and at the learned dot, and subtract:

${\Delta }_{\text{align}}=\underset{\text{true height at the ideal dot (best possible)}}{\underbrace{U^{\ast }({\pi }^{\ast })}}-\underset{\text{true height at what we got}}{\underbrace{U^{\ast }({\pi }_{\text{learned}})}}$

$\Delta$ (Greek "Delta") means "the gap / the difference." So ${\Delta }_{\text{align}}$ = how much true value we threw away compared to the best we could have had.

WHY measure both with $U^{\ast }$. A fair race needs one ruler. If we scored the learned dot with the orange surface it might look great while being genuinely bad — that mismatch is the trap. So we grade everything by the true blue surface.

PICTURE. The blue surface only. Two vertical bars: full-height bar at ${\pi }^{\ast }$, shorter bar at ${\pi }_{\text{learned}}$. The missing chunk between them, drawn in red, is ${\Delta }_{\text{align}}$.

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Step 5 — The trick: split the miss at the orange flag

WHAT. We add and immediately subtract the same quantity — the true height at the orange flag, $U^{\ast }({\pi }_{\mathcal{L}}^{\ast })$. Adding then subtracting the same number changes nothing (it is $+x-x=0$), so the value of ${\Delta }_{\text{align}}$ is untouched — but the bookkeeping changes:

${\Delta }_{\text{align}}=U^{\ast }({\pi }^{\ast })\underset{\text{this pair sums to }0}{\underbrace{-U^{\ast }({\pi }_{\mathcal{L}}^{\ast })+U^{\ast }({\pi }_{\mathcal{L}}^{\ast })}}-U^{\ast }({\pi }_{\text{learned}})$

WHY insert the orange flag. The orange flag ${\pi }_{\mathcal{L}}^{\ast }$ is the bridge between "want" and "got." Standing on it, the total miss splits into two honest steps: how far the bridge is from the ideal, and how far reality is from the bridge.

PICTURE. The blue-surface bars from Step 4 with a third bar added at the orange flag ${\pi }_{\mathcal{L}}^{\ast }$ (measured, as always, by the true blue height). The single red gap of Step 4 now visibly breaks into two stacked coloured segments meeting at this new bar.

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Step 6 — The two families of failure appear

WHAT. Group the four terms into two brackets:

${\Delta }_{\text{align}}=\underset{\text{OUTER misalignment}}{\underbrace{[U^{\ast }({\pi }^{\ast })-U^{\ast }({\pi }_{\mathcal{L}}^{\ast })]}}+\underset{\text{INNER misalignment}}{\underbrace{[U^{\ast }({\pi }_{\mathcal{L}}^{\ast })-U^{\ast }({\pi }_{\text{learned}})]}}$

Term by term:

• First bracket — outer. True height at the ideal minus true height at the best-for-what-we-asked. This is the loss from aiming at the wrong target: even if training were flawless and landed exactly on the orange peak, we'd still be this far below the blue peak. See Specification Gaming, Reward Hacking, Goodhart's Law.
• Second bracket — inner. True height at the best-for-what-we-asked minus true height at what we actually got. This is the loss from the arrow missing even the target we aimed at: training didn't reach the orange peak, so it lands somewhere with lower true value. See Goal Misgeneralization, Deceptive Alignment.

WHY this grouping is the whole point. Each bracket can be big while the other is zero — the two failures are independent. Fixing your objective (shrinking the outer bracket) does nothing for the inner bracket, and vice versa.

PICTURE. The stacked bar from Step 5, now colour-labelled: top segment orange = "outer" (want-peak down to asked-peak), bottom segment red = "inner" (asked-peak down to reality). Two braces name them.

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Step 7 — Every case, including the degenerate ones

WHAT. A decomposition is only trustworthy if it covers all situations. Four corners:

1. Both zero — the two landscapes agree on the peak, and training reaches it. Perfect: ${\Delta }_{\text{align}}=0$. This is the goal of the whole field.
2. Outer only ($\text{inner}=0$) — training perfectly optimizes $\mathcal{L}$ (lands on the orange peak) but $\mathcal{L}$ is the wrong surface. Example: the camera-covering cleaning robot from the parent note. Perfect optimization, wrong target.
3. Inner only ($\text{outer}=0$) — the surfaces do share a peak, but training lands elsewhere, e.g. the "go north" flag-robot (Goal Misgeneralization) or a model that plays along in training then defects (Deceptive Alignment).
4. Both — the common real case: wrong target and a bad landing. The two red/orange chunks stack.

WHY show the corners. If you only ever picture the "both large" case you might believe fixing the objective fixes everything (parent note, Mistake 2). The outer-only = 0 corner is the visual proof that it does not.

PICTURE. A 2×2 grid of the stacked bar, one cell per corner, so you can see the orange and red segments each independently vanish or appear.

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The one-picture summary

Everything above collapses into one river diagram: the true goal $U^{\ast }$ flows through the objective $\mathcal{L}$ we wrote, then through the behaviour the model actually learned — each hop leaking true value, and the two leaks named outer and inner.

Recall Feynman retelling — say it to a 12-year-old

Imagine a flat table of all the ways a robot could behave; every spot on the table is one behaviour. Now float two hills over the table. The blue hill is how happy we really are with each behaviour. The orange hill is the score card we handed the robot (drawn so taller always means better). The hills peak in different places — because writing down exactly what we want is hard. Plant three flags: the blue peak (dream behaviour), the orange peak (best score-card behaviour), and a red flag where training actually stops. Now grade everyone by the honest blue hill. Measure how far the red flag's blue-height sits below the blue peak's — that drop is the total mistake. Then notice you can walk that drop in two steps by pausing at the orange flag: first drop, from blue peak down to orange peak, is "we aimed the score card at the wrong spot" (outer). Second drop, from orange peak down to the red flag, is "training didn't even reach the spot we aimed at" (inner). Both drops go downward — neither is ever a climb — so they simply add up. Because each drop can be tall while the other is flat, a robot can ace its score card and still be terrible (great optimizer, wrong score card), or have a perfect score card and still wander off (right card, bad landing). Fixing one step never automatically fixes the other — and that is the single most important idea in alignment.

Recall Check yourself

What does $\arg {\max }_{\pi }{U}^{\ast }(\pi )$ return — a number or a behaviour? ::: A behaviour (the location of the peak, i.e. the policy ${\pi }^{\ast }$), not the peak height. Why is the outer bracket never negative? ::: Because ${\pi }^{\ast }$ is the highest point of the blue surface, so $U^{\ast }({\pi }^{\ast })$ can never be smaller than $U^{\ast }({\pi }_{\mathcal{L}}^{\ast })$. Which bracket stays large even if training loss is zero? ::: The outer bracket $U^{\ast }({\pi }^{\ast })-U^{\ast }({\pi }_{\mathcal{L}}^{\ast })$ — zero loss only concerns reaching the orange peak, not whether the orange peak is the right place. Why do we grade ${\pi }_{\text{learned}}$ with $U^{\ast }$ and not $\mathcal{L}$? ::: Because $U^{\ast }$ is what we truly value; grading with $\mathcal{L}$ would let a genuinely-bad-but-high-scoring behaviour look good — exactly the trap (Goodhart's Law). Name the corner that disproves "just specify the objective better." ::: The inner-only corner: outer gap $=0$ (perfect objective) yet ${\Delta }_{\text{align}}>0$ because training missed the shared peak.