Lesson 38: Mock Exam — Phase 1 Theory

Discussion prompt

Before we open Lesson 38: Mock Exam — Phase 1 Theory: without looking back, what was the main idea of Phase 1 Review & Mock Prep, and what could you do by the end of it that you could not do before?

Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.

Comparison matrix

From How this mock is built: refill the what you'll do column from what you know. The rest of the table is as it appeared.

Discussion prompt

Our running matrix for this question is A = [[2, 1], [1, 2]] — symmetric, so the theorem applies. Let's find its eigenpairs by hand.

That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.

Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.

Discussion prompt

Explain The eigenvalue equation by analogy to something with no Machine Learning in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.

Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.

Put in order

Put the moves of Solve the characteristic equation into the order they have to happen.

1. Write A − λI for our matrix
2. Take the 2×2 determinant ad − bc
3. Expand (2−λ)² and set to zero
4. Factor the quadratic

Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Subtract λ from each diagonal entry of [[2,1],[1,2]].

Annotate

From Solve the characteristic equation — read this one piece at a time. What is each part doing?

On: \( A - \lambda I = \begin{bmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{bmatrix} \)

• Subtract λ from each diagonal entry of [[2,1],[1,2]].
• det = (2−λ)(2−λ) − (1)(1). The off-diagonal product is 1·1 = 1.
• (2−λ)² = 4 − 4λ + λ², so the equation is λ² − 4λ + 4 − 1 = 0.

Predict first

Two free checks pin eigenvalues before you find eigenvectors: they must sum to the trace and multiply to the determinant.

Commit before you compute: what does Sanity-check with trace and determinant come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.

Correct: Determinant check: product of eigenvalues = det

Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. det(A) = 2·2 − 1·1 = 3, and 1·3 = 3.

Discussion prompt

Work backwards. The example finished here:

Determinant check: product of eigenvalues = det

What was it asked to do, and what must it have been given? Reconstruct the problem from its answer.

Hint: Every quantity in the result had to enter somewhere. Account for each one.

Discussion prompt

Find the eigenvectors, entry by entry — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.

Hint: It starts with: For λ = 3: solve (A − 3I)v = 0

Draw it

Draw what Find the eigenvectors, entry by entry just did — the shape of it, not the line-by-line working. One picture, labels only where you need them. Then check it against the steps: anything you could not draw is a step you followed rather than understood.

Discussion prompt

Use eigh (the symmetric solver — it returns sorted real eigenvalues and orthonormal eigenvectors). Runnable as written:

What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.

Hint: Anything you would have to invent to get started is a thing the problem must supply.

Comparison matrix

From Confirm in numpy: every row here is a choice with a cost. Fill the value (verified) column, then say which row you would actually pick and what you give up for it.

Predict first

A student writes this, and it looks reasonable:

A = [[2, 1], [1, 2]] — the diagonal is 2, 2, so the eigenvalues are 2 and 2.

It is wrong. Say what breaks — and say it before you turn the page.

Correct: The diagonal equals the eigenvalues ONLY for a triangular (or diagonal) matrix.

Solve det(A − λI) = 0 — the off-diagonals matter.

Why: The diagonal equals the eigenvalues ONLY for a triangular (or diagonal) matrix. Here the off-diagonal 1s couple the axes, so the true spectrum is {1, 3}, not {2, 2}. It even fails the det check: 2·2 = 4 ≠ det = 3.

Put in order

These are the steps of The eigenpair recipe (any 2×2), scrambled. Put them back in order before the next slide shows you.

1. Characteristic equation: det(A − λI) = 0 → λ² − (trace)λ + (det) = 0
2. Solve for λ (factor or quadratic formula); confirm Σλ = trace, Πλ = det
3. Each λ: solve (A − λI)v = 0 for the direction v, then normalize
4. Symmetric A: eigenvalues are real and eigenvectors come out orthogonal — check v_i · v_j = 0
5. In code: np.linalg.eigh for symmetric, np.linalg.eig for general matrices

Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.

Discussion prompt

The eigenpair recipe (any 2×2) works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".

Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.

Eliminate the wrong options

Which property is guaranteed for a real symmetric matrix but NOT for a general square matrix?

3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.

• A. Real eigenvalues and orthogonal eigenvectors
• B. All eigenvalues strictly positive
• C. Eigenvalues equal to the diagonal entries
• D. Exactly one eigenvector

Survives elimination: A

Why: The spectral theorem: a real symmetric matrix has real eigenvalues and an orthonormal eigenbasis. We saw it directly — [[2,1],[1,2]] gave real λ = {1,3} with perpendicular eigenvectors. A general matrix can have complex eigenvalues and non-orthogonal eigenvectors.

Check your understanding

Which property is guaranteed for a real symmetric matrix but NOT for a general square matrix?

• A. Real eigenvalues and orthogonal eigenvectors (correct)
• B. All eigenvalues strictly positive
• C. Eigenvalues equal to the diagonal entries
• D. Exactly one eigenvector

Answer: A

Why: The spectral theorem: a real symmetric matrix has real eigenvalues and an orthonormal eigenbasis. We saw it directly — [[2,1],[1,2]] gave real λ = {1,3} with perpendicular eigenvectors. A general matrix can have complex eigenvalues and non-orthogonal eigenvectors.

Why B tempts people

Strictly positive eigenvalues is positive-DEFINITENESS — a stronger, separate condition. A symmetric matrix can have negative eigenvalues (e.g. [[0,1],[1,0]] has λ = {−1, 1}).

Why C tempts people

Diagonal entries equal the eigenvalues only for triangular matrices. Our symmetric [[2,1],[1,2]] has diagonal {2,2} but eigenvalues {1,3} — the exact trap from the previous slide.

Why D tempts people

A symmetric n×n matrix has a FULL set of n orthogonal eigenvectors, not one. Even repeated eigenvalues still yield an orthogonal eigenbasis.

Discussion prompt

Explain What maximum likelihood asks to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.

Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.

Discussion prompt

Maximize over μ — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.

Hint: It starts with: Differentiate ℓ with respect to μ

Predict first

Maximize over σ² is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.

Correct: Differentiate ℓ with respect to σ²

Why: Treat σ² as one variable s. The −(n/2)log s term gives −n/(2s); the last term gives +Σ(dᵢ−μ)²/(2s²).

A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.

Predict first

The table runs: Σ dᵢ / n = μ̂ | 40 / 8 = 5 | 5.0 · Σ(dᵢ−μ̂)² | 32 | 32.0 · σ̂² = Σ(dᵢ−μ̂)²/n | 32 / 8 = 4 | 4.0

In Compute σ̂² by hand, then verify, given the rows so far: what is the next one — the row where quantity is σ̂ = √σ̂²?

Correct: σ̂ = √σ̂² | 2 | 2.0

Why: The relationship between the columns, not the individual numbers, is what generates the next row. d.mean() is ÷n and (( )**2).mean() is also ÷n — exactly the MLE, matching the hand result 40/8 and 32/8.

Draw it

Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.

Predict first

A student writes this, and it looks reasonable:

Variance means the sample variance, so divide the sum of squares by n − 1 = 7.

It is wrong. Say what breaks — and say it before you turn the page.

Correct: That's the unbiased sample variance (Bessel's correction), a different estimator.

The MLE of the variance divides by n = 8.

Why: That's the unbiased sample variance (Bessel's correction), a different estimator. The question asked for the MLE, whose derivative gives exactly ÷n. Mixing them up is the most common MLE slip.

Predict first

For d = [2,4,4,4,5,5,7,9], the maximum-likelihood Gaussian estimates (μ̂, σ̂²) are:

Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.

Correct: (5.0, 4.0)

Why: μ̂ = Σd/n = 40/8 = 5. σ̂² = Σ(d−μ̂)²/n = 32/8 = 4. The MLE variance divides by n. Verified: numpy returns (5.0, 4.0).

Check your understanding

For d = [2,4,4,4,5,5,7,9], the maximum-likelihood Gaussian estimates (μ̂, σ̂²) are:

• A. (5.0, 4.0) (correct)
• B. (5.0, 4.571)
• C. (5.0, 2.0)
• D. (40, 32)

Answer: A

Why: μ̂ = Σd/n = 40/8 = 5. σ̂² = Σ(d−μ̂)²/n = 32/8 = 4. The MLE variance divides by n. Verified: numpy returns (5.0, 4.0).

Why B tempts people

32/7 ≈ 4.571 is the ÷(n−1) unbiased sample variance, not the MLE. The MLE derivative gives ÷n — the exact trap from the previous slide.

Why C tempts people

2.0 is σ̂ (the standard deviation), not σ̂² (the variance). The question asked for the variance: σ̂² = σ̂² = 4, and √4 = 2.

Why D tempts people

40 = Σd (unnormalized) and 32 = Σ(d−μ̂)² (unnormalized). Both still need dividing by n = 8 to become the estimates 5 and 4.

Put in order

Put the moves of Why mixed signs mean a saddle into the order they have to happen.

1. Walk along the first eigenaxis z = [1, 0]
2. Walk along the second eigenaxis z = [0, 1]
3. Up one axis, down another ⇒ saddle

Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. f ≈ ½·5·1² = +2.5 > 0 — moving this way INCREASES f.

Sort into buckets

These are the pieces of Lesson 38: Mock Exam — Phase 1 Theory, out of order. Put each one back under the part of the lesson it belongs to.

Q1 — Linear Algebra

Why symmetric matrices are special; The eigenvalue equation; Solve the characteristic equation

Q2 — MLE (Probability)

What maximum likelihood asks; Likelihood → log-likelihood; Maximize over μ

Q3 — Optimization

Zero gradient isn't enough; The second-derivative test, by eigenvalue sign; Why mixed signs mean a saddle

s1

Q1 — Linear Algebra is where Lesson 38: Mock Exam — Phase 1 Theory puts Why symmetric matrices are special, The eigenvalue equation, Solve the characteristic equation. Knowing which part of the lesson a problem belongs to is most of knowing which method to reach for.

s2

Q2 — MLE (Probability) is where Lesson 38: Mock Exam — Phase 1 Theory puts What maximum likelihood asks, Likelihood → log-likelihood, Maximize over μ. Knowing which part of the lesson a problem belongs to is most of knowing which method to reach for.

s3

Q3 — Optimization is where Lesson 38: Mock Exam — Phase 1 Theory puts Zero gradient isn't enough, The second-derivative test, by eigenvalue sign, Why mixed signs mean a saddle. Knowing which part of the lesson a problem belongs to is most of knowing which method to reach for.

Fill in the blanks

From Classify from eigenvalues in code — one line has had its right-hand side removed. Put it back.

import numpy as np
H = np.array([[5., 0.], [0., -2.]]) # a stationary point's Hessian
vals = np.linalg.eigvalsh(H)
print("Hessian eigenvalues:", vals) # [-2. 5.]
sign = ("min" if (vals > 0).all() else
"max" if (vals < 0).all() else "saddle")
print("classification:", sign)

Why: vals is what everything below it consumes, so the wrong expression here fails later and somewhere else. eigvalsh sorts ascending. (vals>0).all() is False (−2 fails) and (vals<0).all() is False (5 fails), so the classifier falls through to 'saddle'.

Sort into buckets

Sort these by value (verified), from memory, without looking back at Classify from eigenvalues in code. Telling them apart on the spot is the skill; the table is only where the answer happens to be written down.

[-2., 5.]

eigvalsh(H)

False

(vals > 0).all(); (vals < 0).all()

saddle

classification

g1

value (verified) is "[-2., 5.]" for eigvalsh(H) — that is what the table on "Classify from eigenvalues in code" records, and it is the single property separating this group from the rest.

g2

value (verified) is "False" for (vals > 0).all(), (vals < 0).all() — that is what the table on "Classify from eigenvalues in code" records, and it is the single property separating this group from the rest.

g3

value (verified) is "saddle" for classification — that is what the table on "Classify from eigenvalues in code" records, and it is the single property separating this group from the rest.

Check your understanding

At a point where ∇f = 0, the Hessian has eigenvalues {+5, −2}. The point is:

• A. A saddle point (indefinite Hessian) (correct)
• B. A local minimum
• C. A local maximum
• D. A global minimum

Answer: A

Why: Mixed-sign eigenvalues make the Hessian indefinite: the surface curves up along the +5 axis and down along the −2 axis. That's a saddle — verified, our classifier returned 'saddle'.

Why B tempts people

A local minimum needs ALL eigenvalues positive (positive-definite Hessian). The −2 is a genuine downhill direction, so it can't be a min.

Why C tempts people

A local maximum needs all eigenvalues negative. The +5 is an uphill direction, so it can't be a max either.

Why D tempts people

It isn't even a local minimum, let alone global — the negative eigenvalue gives an immediate descent direction away from the point.

Discussion prompt

Compute both directions by hand — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.

Hint: It starts with: D(P‖Q) with P = [.5,.5], Q = [.25,.75]

Fill in the blanks

From Verify KL in numpy — one line has had its right-hand side removed. Put it back.

import numpy as np
P = np.array([0.5, 0.5])
Q = np.array([0.25, 0.75])
kl = lambda p, q: float(np.sum(p * np.log2(p / q)))
print("D(P||Q) =", round(kl(P, Q), 6))
print("D(Q||P) =", round(kl(Q, P), 6))
print("symmetric?", kl(P, Q) == kl(Q, P))

Why: Q is what everything below it consumes, so the wrong expression here fails later and somewhere else. The numbers match the hand computation to the digit, and the equality test returns False — confirming asymmetry.

Check your understanding

Which statement about the KL divergence D(P‖Q) is correct?

• A. It is ≥ 0 and generally asymmetric (correct)
• B. It is a symmetric distance metric
• C. It can be negative
• D. It equals H(P) − H(Q)

Answer: A

Why: KL ≥ 0 by Gibbs' inequality (zero only when P = Q) and is generally asymmetric — we computed D(P‖Q) = 0.2075 ≠ 0.1887 = D(Q‖P) on the same pair.

Why B tempts people

KL is not symmetric (our two directions differed) and violates the triangle inequality, so it is a divergence, not a metric.

Why C tempts people

Individual terms Pᵢlog(Pᵢ/Qᵢ) can be negative (our second term was −0.2925), but the full P-weighted SUM is always ≥ 0.

Why D tempts people

That's backwards and mismatched: cross-entropy H(P,Q) = H(P) + KL, so KL = H(P,Q) − H(P), not H(P) − H(Q).

Step through it

Step through Count concordant pairs by hand one row at a time. What is driving the change, and what would the row after the last one be?

1. Step 1: example is a
2. Step 2: example is b
3. Step 3: example is c
4. Step 4: example is d
5. Step 5: example is e
6. Step 6: example is f

Predict first

A student writes this, and it looks reasonable:

AUC-ROC = 0.9 means the model gets 90% of predictions correct.

It is wrong. Say what breaks — and say it before you turn the page.

Correct: Accuracy needs a threshold and counts labels; AUC is threshold-free and measures RANKING.

AUC-ROC = 0.9 means a random positive outranks a random negative 90% of the time.

Why: Accuracy needs a threshold and counts labels; AUC is threshold-free and measures RANKING. A model can have AUC 0.9 yet 50% accuracy at a badly chosen threshold, or on a 99%-negative set.

Discussion prompt

The rule this trap just fixed:

AUC-ROC = 0.9 means a random positive outranks a random negative 90% of the time.

Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?

Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.

Check your understanding

A model has AUC-ROC = 0.9. This means:

• A. A random positive ranks above a random negative 90% of the time (correct)
• B. The model is 90% accurate
• C. Precision is 0.9
• D. 90% of predictions exceed probability 0.9

Answer: A

Why: AUC-ROC equals P(random positive scored above random negative) — the Mann–Whitney identity we counted directly (8/9 concordant pairs = 0.889). It's a threshold-free ranking measure.

Why B tempts people

Accuracy depends on a chosen threshold and counts correct labels; AUC is threshold-free and measures ordering. The two can differ wildly, especially under class imbalance.

Why C tempts people

Precision is a single-threshold quantity (TP / predicted-positive). AUC aggregates ranking across ALL thresholds, so it isn't any one precision value.

Why D tempts people

AUC ignores absolute probability values entirely — only the relative ordering of positive vs negative scores matters, so it says nothing about how many exceed 0.9.

Predict first

Take three points and the linear kernel k(x,z) = xᵀz, whose Gram matrix is XXᵀ. Then contrast with a symmetric matrix that is NOT a valid Gram matrix.

Commit before you compute: what does The linear kernel passes; a fake one fails come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.

Correct: The 'bad' symmetric matrix has eigenvalues [−1, −1, 2] — a negative → NOT a kernel

Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Symmetry is not enough. Two negative eigenvalues mean no feature map φ can reproduce it, so this symmetric function fails Mercer.

Annotate

Walk the callouts on The linear kernel passes; a fake one fails. Each one is a place this is easy to get subtly wrong.

• Every eigenvalue is non-negative, so the Gram matrix is PSD. The linear kernel is genuinely an inner product (it literally IS xᵀz).
• Symmetry is not enough. Two negative eigenvalues mean no feature map φ can reproduce it, so this symmetric function fails Mercer.

Predict first

A student writes this, and it looks reasonable:

k(x, z) = k(z, x), so it's symmetric — therefore it's a valid kernel.

It is wrong. Say what breaks — and say it before you turn the page.

Correct: Symmetry is necessary but NOT sufficient.

Check that every Gram matrix is positive semidefinite, not just symmetric.

Why: Symmetry is necessary but NOT sufficient. Our 'bad' matrix is perfectly symmetric yet has eigenvalues {−1, −1, 2} — a negative one — so no φ produces it and it fails Mercer.

Predict first

A function k(x, z) is a valid kernel if and only if:

Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.

Correct: Its Gram matrix is positive semidefinite for every dataset

Why: Mercer's theorem: k corresponds to an inner product φ(x)·φ(z) exactly when every Gram matrix is PSD (all eigenvalues ≥ 0). We saw the linear kernel pass ([0,1,3]) and a symmetric non-kernel fail ([−1,−1,2]).

The rating matters as much as the answer: confident-and-wrong is the combination that survives revision, because nothing about it feels like it needs revisiting.

Check your understanding

A function k(x, z) is a valid kernel if and only if:

• A. Its Gram matrix is positive semidefinite for every dataset (correct)
• B. It is symmetric in x and z
• C. It is always non-negative
• D. It equals a Euclidean distance

Answer: A

Why: Mercer's theorem: k corresponds to an inner product φ(x)·φ(z) exactly when every Gram matrix is PSD (all eigenvalues ≥ 0). We saw the linear kernel pass ([0,1,3]) and a symmetric non-kernel fail ([−1,−1,2]).

Why B tempts people

Symmetry is necessary but not sufficient — our 'bad' matrix was symmetric yet indefinite, so it's not a kernel. PSD is the missing requirement.

Why C tempts people

Kernel values can be negative — the linear kernel xᵀz is negative for opposing vectors. Non-negativity is neither required nor sufficient.

Why D tempts people

A distance is not a kernel; the RBF kernel puts a squared distance INSIDE an exponential (e^{−γ‖x−z‖²}) precisely to manufacture a PSD Gram matrix.

Predict first

The table runs: 1 | 0.2891 | 0.4525 | underfit (high bias) · 3 | 0.0200 | 0.0629 | well-tuned

In Watch overfitting happen, given the rows so far: what is the next one — the row where degree is 9?

Correct: 9 | 0.0000 | 1.6056 | overfit (high variance)

Why: The relationship between the columns, not the individual numbers, is what generates the next row. A line can't bend to a sine, so it's wrong everywhere.

Comparison matrix

From Watch overfitting happen: refill the train MSE column from what you know. The rest of the table is as it appeared.

Check your understanding

A model has very low training error but high test error. This indicates:

• A. High variance (overfitting) (correct)
• B. High bias (underfitting)
• C. High irreducible noise
• D. The model is well-tuned

Answer: A

Why: Low train / high test error is the classic high-variance signature — the model fits training noise and fails to generalize. Our degree-9 polynomial did exactly this: train 0.0000, test 1.6056.

Why B tempts people

High bias (underfitting) shows up as high error on BOTH train and test — like our degree-1 fit (0.2891 and 0.4525), not a low-train/high-test split.

Why C tempts people

Irreducible noise raises both train and test error together; it does not by itself create a train–test GAP. The gap is what signals variance.

Why D tempts people

A large train–test gap is the opposite of well-tuned. The tuned model here was degree 3, with both errors low and close (0.0200 vs 0.0629).

Discussion prompt

Explain Why the gradient is just p − y to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.

Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.

Discussion prompt

Softmax the logits — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.

Hint: It starts with: Softmax: exponentiate then normalize

Draw it

Draw what Softmax the logits just did — the shape of it, not the line-by-line working. One picture, labels only where you need them. Then check it against the steps: anything you could not draw is a step you followed rather than understood.

Put in order

Put the moves of Subtract the one-hot truth into the order they have to happen.

1. One-hot the true class 2
2. Gradient = p − y, componentwise
3. Result: [0.0900, 0.2447, −0.3348]

Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Class 2 (last position) is the truth, so y = [0, 0, 1].

Annotate

From Subtract the one-hot truth — read this one piece at a time. What is each part doing?

On: \( \nabla_z L = [\,0.0900,\; 0.2447,\; -0.3348\,] \)

• Class 2 (last position) is the truth, so y = [0, 0, 1].
• The full derivation (softmax Jacobian times the CE derivative) telescopes to exactly p − y. Subtract y from p entry by entry.
• Only the true-class entry is negative (push its logit UP); the two wrong classes are positive (push DOWN). The three components sum to 0, as they must.

Predict first

For softmax outputs p and one-hot labels y, the gradient of cross-entropy loss with respect to the logits z is:

Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.

Correct: p − y

Why: The softmax Jacobian times the CE derivative telescopes to ∇_z L = p − y. Verified: z=[1,2,3], class 2 gave [0.0900, 0.2447, −0.3348] from both hand and PyTorch autograd.

Check your understanding

For softmax outputs p and one-hot labels y, the gradient of cross-entropy loss with respect to the logits z is:

• A. p − y (correct)
• B. y − p
• C. p(1 − p)
• D. −log p

Answer: A

Why: The softmax Jacobian times the CE derivative telescopes to ∇_z L = p − y. Verified: z=[1,2,3], class 2 gave [0.0900, 0.2447, −0.3348] from both hand and PyTorch autograd.

Why B tempts people

y − p is the negation — the direction of ASCENT. Gradient DESCENT uses p − y (or equivalently steps along −(p−y) = y−p); the gradient itself is p − y.

Why C tempts people

p(1−p) is the derivative of a single sigmoid's output, not the logit-gradient of softmax cross-entropy. It confuses the activation's slope with the loss gradient.

Why D tempts people

−log p is (up to indexing) the cross-entropy LOSS value itself, not its gradient. The question asked for the derivative, which is p − y.

Discussion prompt

Suppose a two-sided z-test gives statistic z = 2.05. The p-value is the probability, under H₀, of a statistic at least that extreme:

What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.

Hint: Anything you would have to invent to get started is a thing the problem must supply.

Discussion prompt

Work backwards. The example finished here:

p = 0.0404 = P(|Z| ≥ 2.05 | H₀)

What was it asked to do, and what must it have been given? Reconstruct the problem from its answer.

Hint: Every quantity in the result had to enter somewhere. Account for each one.

Predict first

A student writes this, and it looks reasonable:

A study reports p = 0.04. Conclusion: there's a 4% probability the null hypothesis is true.

It is wrong. Say what breaks — and say it before you turn the page.

Correct: This reverses the conditioning.

p = P(data this extreme | H₀ true) — a statement about the data, not the hypothesis.

Why: This reverses the conditioning. The p-value is P(data this extreme | H₀), a statement about the data given the null — not P(H₀ | data). Our example computed p = 0.0404 as a tail area, not a hypothesis probability.

Sort into buckets

Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.

Holds up

Rule of the room: no notes, no calculator, attempt cold. The struggle is the encoding — revealing early throws the value away.; Our running matrix for this question is A = [[2, 1], [1, 2]] — symmetric, so the theorem applies. Let's find its eigenpairs by hand.; An eigenvector v is a direction A only scales: Av = λv. The scale factor λ is the eigenvalue. Rearrange to (A − λI)v = 0.

Breaks

A = [[2, 1], [1, 2]] — the diagonal is 2, 2, so the eigenvalues are 2 and 2.; Variance means the sample variance, so divide the sum of squares by n − 1 = 7.

sound

These are stated as this lesson states them — each one survives the edge cases Lesson 38: Mock Exam — Phase 1 Theory puts it through.

flawed

Each of these is lifted from a trap in this deck: reasonable-sounding, and wrong in a way that only shows up once you rely on it.

Eliminate the wrong options

A study reports p = 0.04. Which interpretation is correct?

3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.

• A. If H₀ were true, data this extreme would occur only 4% of the time
• B. There is a 4% probability the null hypothesis is true
• C. There is a 96% probability the alternative hypothesis is true
• D. The effect size is 0.04

Survives elimination: A

Why: The p-value is P(data this extreme | H₀ true) — a tail probability under the null. Our z = 2.05 example gave p = 0.0404 as exactly that tail area, a statement about the data, not the hypothesis.

Check your understanding

A study reports p = 0.04. Which interpretation is correct?

• A. If H₀ were true, data this extreme would occur only 4% of the time (correct)
• B. There is a 4% probability the null hypothesis is true
• C. There is a 96% probability the alternative hypothesis is true
• D. The effect size is 0.04

Answer: A

Why: The p-value is P(data this extreme | H₀ true) — a tail probability under the null. Our z = 2.05 example gave p = 0.0404 as exactly that tail area, a statement about the data, not the hypothesis.

Why B tempts people

This reverses the conditional. p is P(data | H₀), not P(H₀ | data); getting the latter requires a prior and Bayes' rule, which the p-value never uses — the exact trap.

Why C tempts people

Same reversal as B plus a complement error: 1 − p is not P(H₁ true). The p-value says nothing about the posterior probability of either hypothesis.

Why D tempts people

The p-value is not an effect size. A tiny effect can give a small p with enough data, and a large effect can give a large p with too little — they are different quantities.

Discussion prompt

Explain Project: precision, recall, F1 from scratch by analogy to something with no Machine Learning in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.

Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.

Step through it

Step through Milestone 3 — verify against sklearn one row at a time. What is driving the change, and what would the row after the last one be?

1. Step 1: metric is precision
2. Step 2: metric is recall
3. Step 3: metric is F1

Predict first

All three milestones in one program — build the counts, derive the metrics, and check sklearn in the same run:

Commit before you compute: what does The full evaluator, assembled come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.

Correct: Both lines print 0.6 0.75 0.6667

Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Your hand-built confusion-matrix metrics and sklearn agree to the last digit — you built classification evaluation from the counts up.

Comparison matrix

From The full evaluator, assembled: every row here is a choice with a cost. Fill the F1 column, then say which row you would actually pick and what you give up for it.

Discussion prompt

The exam's value is in the debrief. For each wrong answer, write three sentences: the correct reasoning, the exact step where your path went wrong, and the one-line result to memorize.

That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.

Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.

Comparison matrix

From Rank your gaps for Phase 2: refill the the fastest fix column from what you know. The rest of the table is as it appeared.

Draw it

One page, no notation unless you need it: draw how these connect — Q1 — Linear Algebra · Q2 — MLE (Probability) · Q3 — Optimization · Q4 — Information Theory · Q5 — Evaluation Metrics · Q6 — SVM & Kernels. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.