UP.T1 — The Lemma of the Course

Flow-Based Generative Models · UFRJ · 2026.2

01Why this session exists

Where this course is going

• Normalizing Flows (U1). Build the transport out of invertible layers with a tractable inverse and Jacobian. Exact likelihood; the price is architectural restriction.
• Continuous Normalizing Flows (U2). Let the transport be the flow map of a Neural ODE. Invertibility from well-posedness, volume change tracked continuously; the price is numerical integration.
• Flow Matching (U3). Learn the velocity field by regression against targets you can sample. Likelihood is not required for training.

Today is none of those. Today is the foundation all three stand on.

Eight minutes for the whole of block A. Do not explain any of the three phrases — they are place-markers, not content. If a hand goes up asking what a velocity field is, the answer is “U3.T1” and move on. The point of the slide is that the audience can locate today on a map.

The promise made today

We prove one result. Then we announce, on one slide, the three later sessions that use it.

When each of those three moments arrives, I will point back at this board.

Everything else in this session exists so that the promise can be stated precisely.

Say the joint-audience sentence out loud here: PGMAT has had the bootcamp, PESC has the self-study packet, and this session is where the two populations merge. It depends on UP.A and on nothing else — say that explicitly, because half the room is wondering whether they are behind.

Then state the three house rules: the Lemma is cited by name for the rest of the course; every expectation in every loss carries its sampling subscripts; and conditional expectations are identified up to almost-sure equality, regression functions up to \(p_Y\)-almost-everywhere equality — a pointwise \(m(y)\) means a version was chosen. Point at the primer §5 drill as the contract for the second.

02Conditional expectation, done properly

The value, the function, and the random variable

\(m(y)\) is a value · \(m\) is a function · \(m(Y) = \mathbb{E}[X \mid Y]\) is a random variable. Ask it of every conditional expectation you meet: which panel is this?

Board work, 8 to 25 minutes. Draw the three-object diagram before showing this slide, on the running example \(Y \sim \mathrm{Unif}\{1,2,3\}\), \(X \mid Y = k \sim \mathcal{N}(k^2, 1)\) — it is the primer’s own example, so it costs nothing to set up.

Then, on the board and in full:

1. the tower rule, \(\mathbb{E}_{y \sim p_Y}[m(y)] = \mathbb{E}_{x \sim p_X}[x]\), proved with densities in four lines; name Fubini at the line that moves the \(y\)-integration inside, and say once that the density identity only holds almost everywhere. One honest sentence on what the density calculations suppress (versions, Fubini, density-free definition), and then never again.
2. the pull-out property, one line: once you condition on Y, a function of Y behaves like a constant — given measurability and integrability.

Leave both on a SIDE board. They are the two ingredients of the orthogonality calculation in the next block, and the room should be able to look up at them while that proof runs.

The three objects, separated in time

The previous slide shows the three objects side by side. Side by side is the wrong relation: they are not three alternatives, they are one object being BUILT, and each step changes what is fixed and what is free. This scene puts that order back.

Four moments to name as they pass:

1. y is fixed. The dot is a NUMBER.
2. y ranges. The dot traces a curve, and the curve is a FUNCTION — still nothing random anywhere on the screen.
3. Only now does Y become random, and it is the INPUT that becomes random.
4. The die settles six times, and the image lands on 1, 4 or 9. That is the RANDOM VARIABLE.

The closing frame carries \(\mathbb{E}_{y \sim \mathrm{Unif}\{1,2,3\}}[m(y)] = 14/3\), which is the tower rule on this example — the side board is where they should look while it appears. (The scene writes \(m\) since the 2026-09-17 re-labelling; the published video shows the old \(f^{*}\) until the animation batch re-renders it.)

03The projection Lemma

The statement

Let \(m(y) = \mathbb{E}[X \mid Y = y]\), and write \(f^{*} := m\). For every square-integrable \(f(Y)\): \[ \begin{aligned} &\mathbb{E}_{(x, y) \sim p_{XY}}\big[\, \Vert f(y) - x \Vert^{2} \,\big] \\[2pt] &\qquad\;=\; \mathbb{E}_{(x, y) \sim p_{XY}}\big[\, \Vert m(y) - x \Vert^{2} \,\big] \;+\; \mathbb{E}_{y \sim p_Y}\big[\, \Vert f(y) - m(y) \Vert^{2} \,\big] . \end{aligned} \] So \(m\) minimizes the population risk: \(m(Y)\) is the unique minimizer, \(\mathbb{P}\)-a.s.; \(m(y)\) is unique \(p_Y\)-a.e.

Square-integrable, throughout: \(\mathbb{E}_{x \sim p_X}[\Vert x\Vert^{2}] < \infty\) and \(\mathbb{E}\Vert f(Y)\Vert^{2} < \infty\).

The proof is on the board: one expansion, and one cross term that dies.

25 to 50 minutes, all board, no gaps. This is the one proof of the day that gets none.

Order: add and subtract m(y) inside the norm; expand into three terms; kill the cross term by writing the joint expectation as \(\mathbb{E}_{y \sim p_Y}[\mathbb{E}_{x \sim p(\cdot \mid y)}[\cdot]]\) (tower rule), pulling \(a(y)\) and \(m(y)\) out of the inner expectation, and using \(m(y) = \mathbb{E}_{x \sim p(\cdot \mid y)}[x]\); read off the minimization at once. Step 0 (the terms are finite) is conditional Jensen plus the tower rule — say it in one breath.

Draw the right triangle. X above the plane, f*(Y) its foot, f(Y) a third vertex in the plane. Say the words “orthogonal projection” and leave the picture up.

Then the corollary, three lines, deliberately anticlimactic: (a) the two POPULATION objectives differ by a constant free of \(\theta\), so the same minimizer set and identical gradients wherever they exist; (b) any surrogate Z with E[Z|Y] = E[X|Y] = m(Y) gives the same minimizer set and the same gradients wherever they exist, the offset being \(C_Z - C_X\). Say the phrase “same conditional mean” and make them repeat it. Then the finite-sample caution, out loud: this is an identity between population risks; two empirical losses on one dataset do NOT differ by a constant, and their minibatch gradients need not coincide. PS1.2(c) is built to show exactly that.

Do NOT say what X, Y, Z will be. The abstraction is the product.

What the picture says

The vertical edge is the irreducible \(L^{2}\) error among predictors that use only \(Y\). The horizontal edge measures how far the chosen predictor is from the optimal one.

Use this slide only if the board triangle came out badly, or as the thing left on screen while the corollary is proved. Here \(\Vert Z\Vert_{L^2} = (\mathbb{E}\Vert Z\Vert^2)^{1/2}\), so the edges are distances and the squares add.

The angle that never opens

The previous slide is one competitor. The Lemma is about all of them, and its visual content is an invariance: f(Y) goes anywhere it likes in the plane, and the right angle does not open. Point at the blue bar and say “watch that one” — it does not move once.

Two moments worth stopping on:

• The pause where f(Y) = f*(Y). The orange bar reads 0.0000 and the hypotenuse reads 0.5731, equal to the blue bar. That is the minimizer, and it is the one place the error is only the floor.
• The closing frame. The bars are MEASURED, not drawn: L2 norms by Monte Carlo on the PS1.2 pair, and the maximum standardized cross-term deviation is 0.71 standard errors across all 241 positions. Under a deliberately wrong projection the same diagnostic reaches about 65 standard errors.

Say that last part out loud. It is the difference between a picture that illustrates a theorem and a picture that was checked against one.

04The announcement

This Lemma will be used exactly three times

1. U3.T1 — after the continuity-equation argument identifies the marginal velocity field as a conditional expectation of conditional velocities, to recognize that field as the \(L^{2}\)-optimal regression target given the current state.
2. U3.T2 — to prove the CFM theorem: swapping the unknown marginal target for a conditional surrogate changes the population objective only by a \(\theta\)-independent constant and therefore leaves the \(\theta\)-gradients unchanged wherever those gradients exist (corollary (b)).
3. U3.T2, detour D2 — to derive denoising score matching: same projection argument, different random variables.

PROTECTED SLIDE. It is a named, versioned artifact, and it is shown again — grayed out, with the current use highlighted — in U3.T1 and in U3.T2. Do not edit the wording without editing those two sessions. Wording as of L3-UP-T1 v0.5 (item 1 matches L1 v1.2 §2: the continuity equation identifies the field, the Lemma characterizes it; item 2 says “wherever those gradients exist”; the old closing line “Tattoo it somewhere” is gone — L2-UP v0.7 sync note 9 tells U3 to take this text).

Item 1 is now a literal application of the Lemma, not an analogy: the field is a conditional expectation (that is U3.T1’s theorem), and the Lemma says a conditional expectation is the L²-optimal regression target.

Eight minutes, and they are not filler. Read the three items SLOWLY.

Then say the thing that makes the device work: “you are not supposed to understand items 1 to 3 yet. You are supposed to recognize them when they arrive.” Say it in those words. A promise that is not announced as a promise is just a confusing slide.

If the session is running late, this is the block that does NOT get compressed. Compress block E instead.

05Pushforwards, and what a model is

Pushforward

\(\psi_{\#}\, p\) is the law of \(\psi(x)\) when \(x \sim p\): \[ \mathbb{E}_{y \sim \psi_{\#} p}\big[ g(y) \big] = \mathbb{E}_{x \sim p}\big[ g(\psi(x)) \big] \quad \text{for every bounded measurable } g . \]

It also gives the sampling rule: draw \(x \sim p\) and return \(\psi(x)\); the result has law \(\psi_{\#}\, p\). As a set formula, \((\psi_{\#} p)(A) = p(\psi^{-1}(A))\) — a probability law always, a density not necessarily.

You built one in the primer: \(x = \mu + \sigma\varepsilon\) is \(\mathcal{N}(\mu, \sigma^2) = \psi_{\#}\,\mathcal{N}(0,1)\).

The symbol \(\psi_{\#}\) is the course standard, after (Lipman et al. 2024) §3.3 — which states it for diffeomorphisms; any measurable \(\psi\) will do here.

The reparameterization callback is worth ten seconds of pause. They wrote the first pushforward of the course before the word existed, and they wrote it as a trick for getting gradients through a sampler.

A transport-based generative model is a pushforward of noise

\[ p_\theta = \big(\psi^{\theta}\big)_{\#}\, p_0 , \qquad p_0 = \mathcal{N}(0, I_d) \]

One sentence per element, and no more. The source is fixed, not learned, and cheap to sample from. The transport \(\psi^\theta\) is parameterized by \(\theta\) — U1 builds it from invertible layers, later units obtain it from a learned field. The model law \(p_\theta\) is induced by the pushforward; it need not have a closed-form density, and whether it has one, and whether that density can be evaluated, depends on the transport. And \(q\) is the data distribution: in the usual learning setting its density is not available, what we have is samples \(x_1 \sim q\).

Then state the time convention of the course, out loud, for the first time in a live session: t runs over [0,1], t = 0 is noise, t = 1 is data. Say that the diffusion literature runs it the other way and that the translation lives in exactly one document. Then never revisit it.

Maximum likelihood is a KL

With \(\mathcal{J}(\theta) = \mathbb{E}_{x_1 \sim q}[\log p_\theta(x_1)]\) and \(C_q = \mathbb{E}_{x_1 \sim q}[\log q(x_1)]\), \[ \begin{gathered} \mathrm{KL}\big(q \,\Vert\, p_\theta\big) \;=\; C_q - \mathcal{J}(\theta) , \\[2pt] \text{so}\quad \operatorname*{arg\,max}_{\theta}\; \mathcal{J}(\theta) \;=\; \operatorname*{arg\,min}_{\theta}\; \mathrm{KL}\big(q \,\Vert\, p_\theta\big) . \end{gathered} \]

Under: \(q\) and \(p_\theta\) with densities; \(q \ll p_\theta\) (necessary for a finite KL, not sufficient); \(C_q\) and \(\mathcal{J}(\theta)\) finite. Two lines: expand the divergence; the term with no \(\theta\) is \(C_q\). The empirical MLE objective is the sample average of \(\mathcal{J}\).

58 to 75 minutes. Prove it live; it really is two lines. If the clock is against you, sketch it and say “the details are two lines, and they are in the notes” — this is the block the L3 designates as compressible, precisely because the notes carry it in full.

The problem this creates

The data sits in the first argument — mass-covering pressure, in the primer’s restricted-family heuristic.

But maximum likelihood needs \(\log p_\theta(x_1)\), and the pushforward definition alone gives no tractable way to evaluate it.

Three distinct things: the density exists (assumed above) · it has an expression · it can be evaluated tractably. The objective needs the third.

The bold sentence is the bridge out of the session. For the invertible transports of U1 and U2, evaluation means relating the data point to its source point and accounting for the volume change — inverse maps and Jacobians in U1, the density change tracked through the ODE in U2. Do not say this for an arbitrary measurable map.

06What comes next

Why likelihoods will be hard

Unit	Strategy	The price
U1 Normalizing Flows	invertible by construction: inverse and Jacobian tractable	architectural restriction
U2 Continuous NFs	flow map of a Neural ODE; invertibility from well-posedness; volume change tracked continuously	numerical integration, everywhere
U3 Flow Matching	change the objective: regress a velocity field on targets you can sample	no likelihood needed for training — the Lemma returns, three times

Sampling: draw from the source and transport the sample. Density evaluation: track how probability density changes under that transport.

Three phrases, no formulas, and this is the block that absorbs any overrun. If there is no time, say the U3 row alone and go to PS1.2.

Nothing beyond what is already on the board: no path, no KL between anything new.

PS1.2 — the Lemma, in closed form and on a machine

\(X \sim \mathcal{N}(\mu_0, \sigma_0^{2} I_d)\), and \(Y = X + \sigma\varepsilon\).

(a) Compute \(\mathbb{E}[X \mid Y = y]\) and \(\operatorname{Cov}(X \mid Y = y)\) in closed form, from the toolkit sheet.

(b) In \(d = 1\): train a small network on samples \((y_i, x_i)\) with the empirical square loss. Plot it against \(m(y) = 0.8\, y\) on the evaluation grid.

(c) Replace the target \(X\) by \(Z = X + \eta\), \(\eta \sim \mathcal{N}(0, \tau^{2})\), \(\eta \perp (X, Y)\). On paper: \(\mathcal{L}_Z = \mathcal{L}_X + \tau^{2}\), and the finite-sample difference is not a constant. Then several seeds, two sample sizes: same population target; summarize the finite-sample behaviour across seeds.

This exercise has a secret identity, to be revealed in U3.

Two minutes. Part (c) is the corollary AND its caution, made visible: the two population risks differ by exactly tau² = 1 (C_X = 0.2, C_Z = 1.2); the empirical losses do not differ by a constant — the realized cross term with the eta_i is what keeps them apart; the noisier Z target should generally show more finite-sample variability, and larger n should generally stabilize both fits — trends across seeds, not per-seed rules. The required sentence is in the notes; read it once.

Say clearly that nothing here needs the bootcamp: the assignment ships labs/ps1-2-starter.py, a self-contained loop with the architecture and optimizer fixed, designed for a laptop CPU; --sweep runs part (c)’s protocol. Within one seed the X-run and the Z-run share data and initialization. PESC students are not left guessing.

Do NOT say what the secret identity is. The reveal is scheduled in U3.T2, detour D2, and the solution key must not leak it either.

References

Lipman, Yaron, Marton Havasi, Peter Holderrieth, Neta Shaul, Matt Le, Brian Karrer, Ricky T. Q. Chen, David Lopez-Paz, Heli Ben-Hamu, and Itai Gat. 2024. “Flow Matching Guide and Code.” https://arxiv.org/abs/2412.06264.