Worked examples — Concept-based explanations

This page drills TCAV until no case surprises you. Before we compute, one promise: every symbol used here is defined the moment it appears. If you have not yet met the dot product or the gradient, read the two boxes below — they are the whole toolkit.

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The scenario matrix

Every TCAV situation you will ever meet falls into one of these cells. The worked examples below are tagged with the cell(s) they cover.

#	Case class	What is unusual	Example
A	Sign positive — concept agrees with class	$\nabla S\cdot {\mathbf{v}}_c>0$ on most inputs	Ex 1
B	Sign negative — concept opposes class	$\nabla S\cdot {\mathbf{v}}_c<0$	Ex 2
C	Zero / perpendicular — concept irrelevant	dot product $\approx 0$	Ex 3
D	Degenerate CAV — the direction is meaningless	${\mathbf{v}}_c$ near zero / unnormalized	Ex 4
E	Random-baseline boundary — signal vs noise	TCAV just above 0.5	Ex 5
F	Statistical-significance / p-value	many random trials	Ex 6
G	Real-world word problem — spurious concept	grass vs stripes	Ex 7
H	Limiting values — TCAV = 0 and = 1 exactly	all-agree / all-oppose	Ex 8
I	Exam twist — sign flips under CAV negation	${\mathbf{v}}_c\to -{\mathbf{v}}_c$	Ex 9

We will do them in order. Cells A, B, C, H, I are pure geometry — they get figures.

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Example 1 — Cell A: the concept points uphill

1. Compute the dot product $\nabla S\cdot {\mathbf{v}}_c=(3)(1)+(4)(0)=3$. Why this step? The sign of this single number is the whole test — positive means "uphill along the concept."
2. Read the sign. $3>0$, so the indicator $1\{\cdot >0\}=1$: this image is counted. Why this step? TCAV over many images is just the average of these 0/1 votes.

Verify: Geometrically (figure), the gradient arrow $(3,4)$ makes an angle with ${\mathbf{v}}_c=(1,0)$ whose cosine is $\frac{3}{\sqrt{3^2+4^2}}=\frac{3}{5}=0.6>0$. A positive cosine ⇔ positive dot product. Consistent. ✓

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Example 2 — Cell B: the concept opposes the class

1. Dot product: $(-2)(1)+(5)(0)=-2$. Why this step? Only the component along ${\mathbf{v}}_c$ survives; the $5$ is perpendicular to ${\mathbf{v}}_c$ and contributes nothing.
2. Sign: $-2<0\Rightarrow 1=0$. This image is not counted; the concept pushes the score down here. Why this step? A concept can actively work against a class — TCAV near 0 is a real, informative outcome, not "no data."

Verify: $\cos \theta =\frac{-2}{\sqrt{(-2{)}^2+5^2}}=\frac{-2}{\sqrt{29}}\approx -0.371<0$. Negative cosine ⇒ obtuse angle ⇒ the arrows genuinely disagree. ✓

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Example 3 — Cell C: perpendicular = concept irrelevant

1. Dot product: $(1)(0)+(0)(7)=0$. Why this step? This is the degenerate boundary between counted and not-counted.
2. Apply the indicator. The condition is strict: $1\{\cdot >0\}$. Since $0$ is not $>0$, this image is not counted. Why this step? The convention matters at the exact boundary — many students miscount here.
3. Interpret at the class level. If every gradient were perpendicular to ${\mathbf{v}}_c$, TCAV $=0$, meaning "moving toward the concept doesn't change the score at all." The concept is orthogonal to what the model cares about.

Verify: $\cos \theta =0/(\Vert \nabla S\Vert \Vert {\mathbf{v}}_c\Vert )=0\Rightarrow \theta =90^{\circ }$. A right angle — no agreement, no disagreement. ✓

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Example 4 — Cell D: a degenerate CAV

1. Normalize the CAV. $\Vert \mathbf{w}\Vert =\sqrt{0.0001^2+0.0001^2}=0.0001\sqrt{2}$. Unit CAV: ${\mathbf{v}}_c=\frac{\mathbf{w}}{\Vert \mathbf{w}\Vert }=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$. Why this step? TCAV only reads the sign of the dot product; scaling $\mathbf{w}$ scales the number but never flips its sign. A tiny magnitude is not "weak usage."
2. Recompute with the unit CAV. $\nabla S\cdot {\mathbf{v}}_c=(3)(\frac{1}{\sqrt{2}})+(4)(\frac{1}{\sqrt{2}})=\frac{7}{\sqrt{2}}\approx 4.95>0$. Why this step? Same sign as the raw $0.0007$ — proving the panic was baseless.

Verify: raw dot product $=0.0001(3)+0.0001(4)=0.0007>0$; normalized $=7/\sqrt{2}\approx 4.9497>0$. Both positive → same vote. ✓

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Example 5 — Cell E: is 0.55 really above the noise floor?

1. Compute the z-score (how many std-devs above the random mean): $z=\frac{0.55-0.50}{0.06}=\frac{0.05}{0.06}\approx \mathrm{0.833.}$ Why this step? The std-dev of the random baseline is the ruler for "how big is big." Comparing to 0.5 alone ignores the noise scale.
2. Judge. $z\approx 0.83$ is well under the usual $z\approx 1.96$ ($p<0.05$) bar. So $0.55$ is not significant — likely noise. Why this step? This is exactly Mistake 2 from the parent note, made quantitative.

Verify: $(0.55-0.50)/0.06=0.8333\dots$; since $0.833<1.96$, fail to reject the null. ✓

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Example 6 — Cell F: the exact p-value

1. Plug into the formula $p=\frac{1+{\sum }_{i=1}^{n}1\{{\text{TCAV}}_{\text{rand},i}\ge {\text{TCAV}}_c\}}{n+1}=\frac{1+0}{20+1}=\frac{1}{21}\approx \mathrm{0.0476.}$ Why this step? The "$1+$" and "$n+1$" (Laplace/add-one) stop $p$ from ever being exactly $0$ — you can never prove zero chance from finite trials.
2. Compare. $0.0476<0.05\Rightarrow$ significant. The stripe concept is genuinely used. Why this step? This is the pass/fail gate that turns a raw number into a claim.

Verify: $1/21=0.047619\dots <0.05$. Significant. ✓ Note: with only $n=19$ trials, $p=1/20=0.05$ would not be $<0.05$ — the trial count matters.

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Example 7 — Cell G: the real-world spurious concept

1. Z-score each concept against the random baseline (ruler $=0.05$):
   • stripes: $(0.30-0.50)/0.05=-4.0$
   • grass: $(0.90-0.50)/0.05=+8.0$
   • four-legs: $(0.55-0.50)/0.05=+1.0$ Why this step? Raw scores mislead; z-scores tell us which are beyond noise.
2. Interpret.
   • grass ($z=+8$): strongly, significantly used — a spurious concept.
   • stripes ($z=-4$): significantly anti-correlated — the model even mildly avoids stripes!
   • four-legs ($z=+1$): within noise, inconclusive. Why this step? Presence $\ne$ usage. Zebras have stripes, but this model classifies via background — exactly the parent note's warning.
3. Danger. Deploy this on a zebra photographed on sand and it fails: the grass cue vanishes.

Verify: $(0.30-0.50)/0.05=-4$, $(0.90-0.50)/0.05=8$, $(0.55-0.50)/0.05=1$. Only grass and stripes exceed $|1.96|$. ✓ Links to Adversarial Examples (spurious cues) and Model Editing (ablate the grass CAV).

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Example 8 — Cell H: the two limiting values

1. All positive. Every vote is $1$, so $\text{TCAV}=100/100=1$. Geometry: every gradient arrow lies in the same half-plane as ${\mathbf{v}}_c$ (angle $<90^{\circ }$). Why this step? $1$ is the hard ceiling — "the concept helps this class every single time."
2. All negative. Every vote is $0$, so $\text{TCAV}=0/100=0$. Geometry: every gradient arrow is in the opposite half-plane (angle $>90^{\circ }$). Why this step? $0$ is the floor — "the concept opposes this class every time." Not the same as "irrelevant" (Ex 3, where it hovers at $0.5$).

Verify: $100/100=1$ and $0/100=0$; both lie in $[0,1]$. The midpoint $0.5$ is the random-half-plane split. ✓

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Example 9 — Cell I: the exam twist (negate the CAV)

1. Effect on one dot product. $\nabla S\cdot (-{\mathbf{v}}_c)=-(\nabla S\cdot {\mathbf{v}}_c)$. Every positive becomes negative and vice-versa. Why this step? Negating one vector negates the dot product's sign — this is the crux.
2. Effect on the count. Every image that voted $1$ now votes $0$ (strictly-positive flips to negative), so the new score is $1-0.87=0.13$ only if no dot products were exactly zero. Assuming none are (generic case), ${\text{TCAV}}_{-{\mathbf{v}}_c}=0.13$. Why this step? The complement $1-0.87$ works because zero-dot-products (which flip $1\to$ still 0, since $0>0$ is false both ways) are measure-zero.
3. Are they wrong? No — they measured the anti-concept "not-striped." Direction is a convention; report it consistently.

Verify: $1-0.87=0.13$. Negation maps score $s\mapsto 1-s$ (ignoring exact zeros). ✓

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Key Takeaways

1. TCAV reads only the sign of $\nabla S\cdot {\mathbf{v}}_c$ — magnitude and CAV length never change a vote (Ex 4).
2. The three sign cases — positive (helps, Ex 1), negative (opposes, Ex 2), zero (irrelevant, Ex 3) — are geometrically an acute, obtuse, and right angle.
3. Limits: TCAV $=1$ (always helps) and $=0$ (always opposes); random $\approx 0.5$ (Ex 8).
4. A score is only meaningful relative to the random baseline — always z-score it (Ex 5, 7) and compute the p-value (Ex 6).
5. Presence $\ne$ usage: a zebra model may live on grass, not stripes (Ex 7).
6. Negating the CAV maps $s\mapsto 1-s$ — it measures the anti-concept (Ex 9).

See also Probing Classifiers (does the concept exist?) versus TCAV (is it used?), and Feature Attribution Methods / Layer-wise Relevance Propagation for the pixel-level view these concept scores sit above.