In-context learning mechanisms: Worked examples

This page is the practice arena for the parent note. There we built the Bayesian task-inference view (demos shift a posterior $p (c ∣ D_{k})$ over hidden tasks) and the attention-as-one-gradient-step view. Here we run those two engines on every case class ICL can throw at you — and we compute real numbers so you can feel the machinery move.

Before we start, a one-line refresher of the two symbols we will reuse constantly:

Recall What do

$c$ , $D_{k}$ , and $p (c ∣ D_{k})$ mean? $c$ is a hidden "task" the model could be doing (e.g. "classify sentiment"). $D_{k}$ is the ==bag of $k$ demonstrations== you put in the prompt. $p (c ∣ D_{k})$ is the model's ==belief that task $c$ is the one you want==, after reading those demos. $k$ -shot ::: a prompt carrying $k$ demonstrations before the query ( $k = 0$ means zero examples).

We also need one measurement that keeps returning throughout this page: how far apart two tasks are.

The scenario matrix

Every ICL question you will ever see lands in one of these cells. Each worked example below is tagged with the cell(s) it covers.

Cell	Case class	What breaks / what to watch	Covered by
C1	$k = 0$ (zero-shot), degenerate	no demos → posterior = prior only	Ex 1
C2	$k$ growing, clean demos	posterior collapses exponentially in $k$	Ex 2
C3	Saturation / limiting $k \to \infty$	diminishing returns, formal plateau	Ex 2, Ex 3, Ex 9
C4	Two tasks nearly identical (small KL)	slow collapse — needs many shots	Ex 3
C5	Wrong / flipped labels	format survives, mapping degrades	Ex 4
C6	Ordering & recency bias	reordering flips the answer	Ex 5
C7	Attention = one GD step (numeric)	dot-product prediction sign & size	Ex 6
C8	Real-world word problem	choosing shot count under a budget	Ex 7
C9	Exam twist / limiting degenerate	two indistinguishable tasks, KL $= 0$	Ex 8
C10	Unequal priors ( $p (c^{*}) \neq = 0.5$ )	prior helps or fights the evidence	Ex 10

We will use one running toy world so numbers are comparable across examples.

Intuition A note on "why only two tasks?"

To keep every number by hand we compare exactly two candidate tasks. With $m > 2$ classes the same logic runs through a softmax over $m$ competitors. A softmax is just the normalisation "each candidate's score divided by the sum of all candidates' scores", so the numbers form a valid probability distribution over the $m$ tasks: $p (c^{*} ∣ D_{k}) = \frac{π _{c^{*}} \prod _{i} p ( d _{i} ∣ c ^{*} )}{\sum _{c} π _{c} \prod _{i} p ( d _{i} ∣ c )}, π_{c} \equiv p (c) (the pretraining prior on task c) .$ Here $π_{c}$ (the Greek letter "pi", used as a name for the ==prior weight of task $c$ ==) is exactly the prior probability the model assigns to task $c$ before seeing demos. The collapse now needs $c^{*}$ to beat every rival, so the effective rate is set by the closest competitor (smallest pairwise KL). Every two-task conclusion below is the "hardest pair" slice of that bigger softmax — the general multi-class case is never easier than the toughest pair we solve here.

Example 1 — Zero-shot: the degenerate case (C1)

Forecast: guess before reading — with zero evidence, is the model sure, unsure, or biased?

Write the posterior with $k = 0$ . The product $\prod_{i = 1}^{0}$ over zero demos is empty, so it equals $1$ . Why this step? An empty product is $1$ by definition — "multiply nothing" changes nothing. This forces the formula to fall back on the prior only.
Plug in. $p (c^{*} ∣ D_{0}) = \frac{0.5 \cdot 1}{0.5 \cdot 1 + 0.5 \cdot 1} = \frac{0.5}{1.0} = 0.5.$ Why this step? With no evidence, Bayes returns the prior untouched.

Answer: $p (c^{*} ∣ D_{0}) = 0.5$ — a coin-flip. The model is maximally unsure between the two tasks and relies entirely on whatever the pretraining prior favours.

Verify: Symmetric priors, no data ⇒ posterior must equal prior $= 0.5$ . Units are a probability in $[0, 1]$ . ✓

Example 2 — Clean demos, growing $k$ : watch the collapse (C2, C3)

Forecast: does the belief rise linearly, jump then plateau, or fall?

Per-demo likelihood ratio. $r = \frac{p ( d ∣ c ^{*} )}{p ( d ∣ c ^{'} )} = \frac{0.8}{0.4} = 2.$ Why this step? Because priors are equal, the posterior depends only on the accumulated ratio $r^{k}$ — this is the compact form of Step 1.
Posterior formula in terms of $r$ . With equal priors, $p (c^{*} ∣ D_{k}) = \frac{r ^{k}}{r ^{k} + 1} .$ Why this step? Divide top and bottom of the two-task posterior by $p (c^{'}) \prod p (d ∣ c^{'})$ ; every factor collapses into $r^{k}$ .
Evaluate.
- $k = 1 : \frac{2}{3} = 0.6667$
- $k = 2 : \frac{4}{5} = 0.8000$
- $k = 3 : \frac{8}{9} = 0.8889$
- $k = 5 : \frac{32}{33} = 0.9697$ Why this step? Just powers of $2$ over "power $+ 1$ ".

Answer: belief climbs $0.5 \to 0.67 \to 0.80 \to 0.89 \to 0.97$ — big early jumps, then saturation toward $1$ .

Verify: The parent claims $lo g \frac{p ( c ^{*} )}{p ( c ^{'} )}$ grows $\approx k \cdot KL$ . Here $lo g r^{k} = k lo g 2$ , a straight line in $k$ (yellow line in the figure) — exactly linear log-odds, exponential collapse of the complement. ✓

Example 3 — Nearly identical tasks: slow collapse (C3, C4)

Forecast: more shots or fewer shots than Example 2?

Ratio. $r = \frac{0.55}{0.45} = \frac{11}{9} \approx 1.2222.$ Why this step? Small gap ⇒ $r$ close to $1$ ⇒ weak evidence per demo.
Solve $\frac{r ^{k}}{r ^{k} + 1} \geq 0.9$ . This means $r^{k} \geq 9$ , i.e. $k \geq \frac{ln 9}{ln r} = \frac{2.1972}{0.2007} \approx 10.95.$ Why this step? $\frac{x}{x + 1} \geq 0.9 \Leftrightarrow x \geq 9$ ; take logs to isolate $k$ .
Round up. $k = 11$ shots. Why this step? $k$ must be a whole number of demos, and we need to exceed the threshold, so ceil.

Answer: 11 shots — versus only 4 shots for the same $0.9$ target in Example 2 ( $2^{k} \geq 9 \Rightarrow k \geq 3.17$ ; check: $k = 4$ gives $16/17 = 0.941 \geq 0.9$ , $k = 3$ gives $8/9 = 0.889 < 0.9$ , so the minimum is $4$ ).

Verify: $r^{11} = (11/9)^{11} \approx 9.35 \geq 9$ ✓, and $r^{10} \approx 7.65 < 9$ ✓ so 11 is the true minimum. The small-KL cell C4 genuinely needs many more demos — this is why "hard/novel" tasks demand more shots.

Example 4 — Flipped labels: format survives, mapping degrades (C5)

Forecast: does flipping labels destroy task location, mapping, both, or neither?

Task-identity posterior. Uses $r_{format} = 2$ : $p (task ∣ D_{4}) = \frac{2 ^{4}}{2 ^{4} + 1} = \frac{16}{17} = 0.9412.$ Why this step? Format/label-space/input-distribution survive label flips — flipping a label doesn't change that it looks like the task.
Mapping posterior. Uses $r_{map} = 1$ : $p (correct map ∣ D_{4}) = \frac{1 ^{4}}{1 ^{4} + 1} = \frac{1}{2} = 0.5.$ Why this step? With equal-and-opposite label evidence, the ratio product is $1$ ; the mapping belief stays at the prior.

Answer: task identity $= 0.941$ (strong), correct mapping $= 0.5$ (coin-flip). The model still knows what game it's playing but must fall back on its pretraining prior for the actual answer.

Verify: matches the parent's Example 2 claim exactly — "labels don't matter" for task location, but mapping collapses to prior, so on novel tasks (weak prior) accuracy tanks. Probabilities in $[0, 1]$ ✓. See Fine-tuning vs Prompting for why weight updates would fix the mapping.

Example 5 — Ordering & recency bias (C6)

Forecast: does the model answer 2 (the true match for B) or something recency-driven?

Set the content-match scores from the rule above. Query key is B. The demo B→2 is an exact key match → content $= 1$ . Suppose C is a lexical near-miss to B (adjacent letter / similar embedding) → partial match → content $= 0.5$ . A is neither → content $\approx 0$ . Why this step? We must state where every number comes from: content $\in {0, 0.5, 1}$ is our declared match rule (exact / partial / none), not a magic constant. Real heads learn a graded similarity; we discretise it to three levels so the arithmetic is transparent.
Distances from the query (query is position 4). C→3 is distance 1, B→2 distance 2, A→1 distance 3, so $w_{C} = ρ^{1}, w_{B} = ρ^{2}, w_{A} = ρ^{3}$ . Why this step? Recency bias means the nearest demo tokens dominate the residual stream.
Score = content × recency. $score (B \to 2) = 1 \times ρ^{2} = 0.36$ ; $score (C \to 3) = 0.5 \times ρ^{1} = 0.30$ ; $score (A \to 1) = 0 \times ρ^{3} = 0$ . Why this step? Combine the two declared factors; the thin gap $0.36$ vs $0.30$ is exactly what makes ICL fragile.

Answer: predicts 2 (score $0.36$ ), but the margin over the recency lure C→3 is only $0.36 - 0.30 = 0.06$ .

Reorder test: put C→3 last (distance 1) and B→2 far back (distance 3): now $score (B \to 2) = 1 \times ρ^{3} = 0.216$ and $score (C \to 3) = 0.5 \times ρ^{1} = 0.30$ . Now 3 wins — the answer flipped purely from ordering.

Verify: $0. 6^{2} = 0.36$ , $0. 6^{3} = 0.216$ , $0.5 \times 0.6 = 0.30$ ; before-reorder max is B ( $0.36$ ), after-reorder max is C ( $0.30$ ). Different winners ✓ — a clean demonstration of cell C6.

Example 6 — Attention is one gradient step: a full numeric run (C7)

Recall The parent result we are using (linear-attention ↔ one GD step)

Fit a linear model $\overset{y}{^} = W x$ to demos ${(x_{i}, y_{i})}$ with squared-error loss, starting from $W_{0} = 0$ . One gradient-descent step with step size $η$ gives $W_{1} = η \sum_{i} y_{i} x_{i}^{⊤}$ , so its prediction on a query is $\overset{y}{^} = W_{1} x_{k + 1} = η \sum_{i} y_{i} (x_{i}^{⊤} x_{k + 1})$ . A single linear attention head that uses $value_{i} = y_{i}$ , $key_{i} = x_{i}$ , $query = x_{k + 1}$ and dot-product similarity outputs $\sum_{i} y_{i} (x_{i}^{⊤} x_{k + 1})$ — the identical formula. So the forward pass "trains" a virtual linear model in its activations.

Forecast: will $\overset{y}{^}$ be positive, negative, or zero?

Dot products. $x_{1}^{⊤} x_{3} = 1 \cdot 2 + 0 \cdot 3 = 2$ . $x_{2}^{⊤} x_{3} = 0 \cdot 2 + 1 \cdot 3 = 3.$ Why this step? The dot product is the "similarity" the attention head uses to weight each value.
Weighted sum of values. $\overset{y}{^} = 1 \cdot (2) + (- 1) \cdot (3) = 2 - 3 = - 1.$ Why this step? $value_{i} = y_{i}$ , so we add each label times its similarity — exactly $η \sum_{i} y_{i} (x_{i}^{⊤} x_{3})$ .
Cross-check via the GD weight matrix — mind the shapes. Here each $x_{i}$ is a $2 \times 1$ column and each label $y_{i}$ is a scalar ( $1 \times 1$ ), so the outer product $y_{i} x_{i}^{⊤}$ is (scalar) $\times$ ( $1 \times 2$ row) $=$ a $1 \times 2$ row. Summing two $1 \times 2$ rows keeps a $1 \times 2$ row: $W_{1} = η \sum_{i} y_{i} x_{i}^{⊤} = (+ 1) [1, 0] + (- 1) [0, 1] = [1, - 1] .$ Then $W_{1} x_{3} = [1, - 1] [2, 3]^{⊤} = 2 - 3 = - 1$ — a $1 \times 2$ times a $2 \times 1$ gives the $1 \times 1$ scalar prediction, dimensions matching a single-output linear model $\overset{y}{^} = W x$ . This is the same $1 \times 2$ operator the attention read-out applies (values $y_{i}$ against keys $x_{i}$ ), which is why the two routes must agree. Why this step? Checking shapes proves $y_{i} x_{i}^{⊤}$ is the correct orientation: it maps an input column to a scalar output, exactly like the attention head that reads keys $x_{i}$ and emits value-weighted scalars.

Answer: $\overset{y}{^} = - 1$ . The forward pass literally computed one Gradient Descent step's prediction — no weights of the real network moved.

Verify: both routes yield $- 1$ ✓; sign is negative because the query leans toward $x_{2}$ (dot $3 > 2$ ), whose label is $- 1$ .

Example 7 — Real-world word problem: budget your shots (C8)

Forecast: will the final demo add a lot or a little?

Max demos in budget. $⌊ 300/60 ⌋ = 5$ demos. Why this step? Integer division — you cannot use a partial demo.
Accuracy at $k = 5$ . $\frac{2 ^{5}}{2 ^{5} + 1} = \frac{32}{33} = 0.9697.$ Why this step? Plug into the collapse law $acc (k) = \frac{2 ^{k}}{2 ^{k} + 1}$ .
Marginal gain of demo 5. $acc (5) - acc (4) = \frac{32}{33} - \frac{16}{17} = 0.9697 - 0.9412 = 0.0285.$ Why this step? Cost–benefit: 60 tokens bought only $\approx 2.9$ percentage points — the saturation warned about in cell C3.

Answer: fit 5 demos, reach 96.97%, but the 5th demo adds only ~2.85 points for 60 tokens.

Verify: $32/33 = 0.96970$ , $16/17 = 0.94118$ , difference $0.02852$ ✓. Consistent with the parent mistake "more shots always help" being false. For a permanent fix you'd choose fine-tuning or add retrieval instead of stuffing demos.

Example 8 — Exam twist: two indistinguishable tasks, $KL = 0$ (C9)

Forecast: will more demos ever help here?

Per-demo ratio. $r = \frac{0.5}{0.5} = 1$ for all demos. Why this step? Identical likelihoods ⇒ no evidence to prefer either task.
Posterior for any $k$ . $\frac{1 ^{k}}{1 ^{k} + 1} = \frac{1}{2}$ — frozen at the prior forever. Why this step? $r^{k} = 1$ regardless of $k$ ; the limit as $k \to \infty$ is still $0.5$ .
KL per demo. $KL (p_{c^{*}} ∥ p_{c^{'}}) = \sum p lo g \frac{p}{p} = \sum p \cdot 0 = 0.$ Why this step? When two distributions are equal, KL is exactly zero — the parent's "collapse rate $\approx k \cdot KL$ " gives zero collapse.

Answer: $p (c^{*} ∣ D_{k}) = 0.5$ for all $k$ ; the limit is $0.5$ ; $KL = 0$ . No number of demos can disambiguate tasks that pretraining made identical. This is the degenerate limit of cell C9 — and it explains why some tasks are simply unreachable by ICL and need different scaffolding or new weights.

Verify: $r = 1 \Rightarrow r^{k} = 1 \Rightarrow$ posterior $= 0.5$ for $k = 1, 10, 1000$ ✓; KL of a distribution with itself is $0$ ✓ (link to Bayesian Inference).

Example 9 — The formal limit $k \to \infty$ : proving the plateau (C3)

Forecast: does the belief reach exactly $1$ , approach $1$ forever, or level off below $1$ ?

Rewrite the posterior by dividing by $r^{k}$ . $\frac{r ^{k}}{r ^{k} + 1} = \frac{1}{1 + r ^{- k}} .$ Why this step? Isolating $r^{- k}$ exposes the only term that depends on $k$ ; everything else is constant.
Take the limit. As $k \to \infty$ with $r > 1$ , $r^{- k} \to 0$ , so $\frac{1}{1 + r ^{- k}} \to \frac{1}{1 + 0} = 1.$ Why this step? $r > 1$ means $r^{- k}$ is a shrinking geometric sequence; its limit is $0$ . So the belief approaches $1$ but never equals it for finite $k$ — an asymptote.
Quantify the error. $err (k) = 1 - \frac{1}{1 + r ^{- k}} = \frac{r ^{- k}}{1 + r ^{- k}} \leq r^{- k} .$ For $r = 2$ this is $\leq 2^{- k}$ : the error at least halves every demo. Why this step? "Geometric decay of the error" is the precise meaning of exponential collapse — and dividing by ever-larger $r^{k}$ is why gains shrink as $k$ grows (the plateau).

Answer: $lim_{k \to \infty} p (c^{*} ∣ D_{k}) = 1$ , approached with error $\leq r^{- k}$ (for $r = 2$ : err $\leq 2^{- k}$ , e.g. $2^{- 5} = 0.03125 \geq$ the true $1/33 \approx 0.0303$ ). The curve saturates at $1$ from below — never touching it, gains shrinking geometrically. This is the formal content of cell C3.

Verify: at $k = 5$ , true error $1 - 32/33 = 1/33 = 0.03030$ , and the bound $2^{- 5} = 0.03125 \geq 0.03030$ ✓; limit is $1$ ✓.

Example 10 — Unequal priors: when pretraining helps or fights (C10)

Forecast: will the demos still win, and will it take more or fewer shots than the equal-prior case?

General posterior with priors. Divide top and bottom by $p (c^{'}) \prod p (d ∣ c^{'})$ : $p (c^{*} ∣ D_{k}) = \frac{p ( c ^{*} ) r ^{k}}{p ( c ^{*} ) r ^{k} + p ( c ^{'} )} = \frac{π r ^{k}}{π r ^{k} + ( 1 - π )}, π \equiv p (c^{*}) .$ Why this step? The prior-odds $\frac{π}{1 - π}$ multiplies the evidence, so a bad prior is a handicap the demos must overcome — this generalises the equal-prior formula (recover it by setting $π = 0.5$ ).
(a) Zero-shot ( $k = 0$ ). $\frac{0.2 \cdot 1}{0.2 \cdot 1 + 0.8} = \frac{0.2}{1.0} = 0.2.$ Why this step? With no demos the belief is the prior $0.2$ — the model starts out believing the wrong task.
(b) Overturn the prior. Need $\frac{0.2 \cdot 2 ^{k}}{0.2 \cdot 2 ^{k} + 0.8} \geq 0.5$ , i.e. $0.2 \cdot 2^{k} \geq 0.8$ , i.e. $2^{k} \geq 4$ , so $k \geq 2$ . Why this step? $\frac{a}{a + b} \geq \frac{1}{2} \Leftrightarrow a \geq b$ ; the demos must supply likelihood-odds $r^{k}$ big enough to cancel the prior-odds $\frac{0.8}{0.2} = 4$ .
Sanity at $k = 2$ . $\frac{0.2 \cdot 4}{0.2 \cdot 4 + 0.8} = \frac{0.8}{1.6} = 0.5$ exactly; at $k = 3$ , $\frac{0.2 \cdot 8}{0.2 \cdot 8 + 0.8} = \frac{1.6}{2.4} = 0.6667$ . Why this step? Confirms $k = 2$ is the crossover and the belief keeps climbing after.

Answer: (a) zero-shot $= 0.2$ ; (b) need $k = 2$ demos just to reach the coin-flip $0.5$ , and more to be confident; (c) posterior $= \frac{π r ^{k}}{π r ^{k} + ( 1 - π )}$ . A hostile prior costs extra shots; a favourable prior (large $π$ ) would let fewer demos — or even zero — suffice. This is cell C10, and it explains why the same demos work instantly on tasks the model already expects, but sluggishly on tasks its pretraining resists.

Verify: $0.2/ (0.2 + 0.8) = 0.2$ ✓; $2^{2} = 4 \geq 4$ so $k = 2$ crossover, with belief exactly $0.5$ ✓ and $0.6667$ at $k = 3$ ✓.

Recall Quick self-test

With equal priors and per-demo ratio $r = 2$ , what is $p (c^{*} ∣ D_{3})$ ? ::: $\frac{2 ^{3}}{2 ^{3} + 1} = \frac{8}{9} \approx 0.889$ . Write the KL divergence between the true and rival task. ::: $KL = \sum_{d} p (d ∣ c^{*}) lo g \frac{p ( d ∣ c ^{*} )}{p ( d ∣ c ^{'} )}$ , always $\geq 0$ , zero iff the two distributions are identical. Why does the KL $= 0$ case never improve? ::: Every per-demo likelihood ratio is $1$ , so the posterior stays pinned at the prior no matter how many demos you add. With prior $π = p (c^{*})$ , what is the posterior after $k$ demos (ratio $r$ )? ::: $\frac{π r ^{k}}{π r ^{k} + ( 1 - π )}$ . In Ex 6, why is the ICL prediction negative? ::: The query is more similar (larger dot product, $3 > 2$ ) to the demo whose label is $- 1$ . How would the two-task analysis change for $m > 2$ classes? ::: It becomes a softmax over all $m$ tasks; collapse speed is set by the closest competitor (smallest pairwise KL), so it is never easier than the hardest pair.

Practice done. Return to the parent and re-read the two mechanistic stories — every cell above is one of those two engines run to its edge.