Diffusion models forward - reverse process: Worked examples

Two quantities do all the work on this page, so pin them down before we start:

Definition The three quantities you must never confuse

$β_{t}$ — the noise-injection fraction of one step: the fraction of the variance that is thrown away and replaced by fresh Gaussian noise at step $t$ . Small $β_{t}$ (near $1 0^{- 4}$ early, up to $0.02$ late) means a gentle step.
$α_{t} = 1 - β_{t}$ — the survival factor of one step. How much of last step's signal survives this step (whatever is not injected as noise).
$\overset{α}{ˉ}_{t} = α_{1} α_{2} \dots α_{t}$ — the survival factor from the very start (the bar means "cumulative product from step 1 to $t$ "). How much of the original $x_{0}$ survives after $t$ steps. Convention: the empty product $\overset{α}{ˉ}_{0} = 1$ (before any step, 100% of the signal is intact).

Everything else is built from these via the direct-sampling law (parent note): $x_{t} = \overset{α}{ˉ}_{t} x_{0} + 1 - \overset{α}{ˉ}_{t} ϵ, ϵ \sim N (0, I)$ Read it as: shrink the picture by $\overset{α}{ˉ}_{t}$ , add noise scaled by $1 - \overset{α}{ˉ}_{t}$ . The two scalings are locked together — squared, they sum to 1.

\overset{α}{ˉ}_{t} + (1 - \overset{α}{ˉ}_{t}) = 1

the signal-energy fraction plus the noise-energy fraction always equal the whole. This is why the variance stays

\approx 1

\overset{α}{ˉ}_{0} = ?

1

— the empty product; before any step, all of

x_{0}

survives.

The scenario matrix

Every worked example below is a cell in this table. Together they touch every corner of the topic.

Cell	Scenario class	What breaks / what to watch	Example
A	Early forward step ( $t$ small, $\overset{α}{ˉ}_{t} \to 1$ )	Almost no noise; check "nearly identity"	Ex 1
B	Late forward step ( $t \to T$ , $\overset{α}{ˉ}_{t} \to 0$ )	Signal gone; $x_{T} \approx ϵ$	Ex 2
C	Building $\overset{α}{ˉ}_{t}$ from a $β$ -schedule	Cumulative product, not sum	Ex 3
D	One reverse denoising step, generic $t$	Mean formula + injected noise	Ex 4
E	Degenerate reverse step $t = 1$	No noise added ( $z = 0$ ); uses $\overset{α}{ˉ}_{0} = 1$ ; final output	Ex 5
F	Perfect model ( $ϵ_{θ} = ϵ$ )	Reverse exactly inverts forward	Ex 6
G	Broken / zero model ( $ϵ_{θ} = 0$ )	What the mean does with no denoising	Ex 7
H	Signal-to-noise reasoning (word problem)	"At which $t$ is the digit half-gone?"	Ex 8
I	Exam twist: solve for $β_{t}$ from data	Invert $\overset{α}{ˉ}$ relation	Ex 9

A — Early forward step: barely any noise

Worked example Ex 1 (Cell A)

MNIST pixel $x_{0} = 0.80$ . Schedule gives $\overset{α}{ˉ}_{1} = 0.9999$ . Draw noise $ϵ = 1.5$ . Compute $x_{1}$ and say how far it moved from $x_{0}$ .

Forecast: guess before computing — will $x_{1}$ be near $0.80$ or wildly off?

Signal term $\overset{α}{ˉ}_{1} x_{0} = 0.9999 \cdot 0.80 = 0.99995 \cdot 0.80 = 0.79996$ . Why this step? This is the shrunken original. With $\overset{α}{ˉ}_{1} \approx 1$ the shrink is invisible.
Noise term $1 - \overset{α}{ˉ}_{1} ϵ = 0.0001 \cdot 1.5 = 0.01 \cdot 1.5 = 0.015$ . Why this step? $1 - \overset{α}{ˉ}_{1} = 0.01$ is tiny, so even a fat noise draw barely nudges the pixel.
Add $x_{1} = 0.79996 + 0.015 = 0.81496$ . Why this step? Direct-sampling law is just signal + noise.

Verify: displacement $∣ x_{1} - x_{0} ∣ = 0.01496 < 0.02$ . A pixel changing by ~1.5% is "nearly identical" — matches the parent note's "noise barely visible" at $t = 1$ . ✓

This is the $\overset{α}{ˉ}_{t} \to 1$ limit: the forward map is almost the identity.

B — Late forward step: signal destroyed

Figure — Diffusion models forward - reverse process

Figure s01 — the two forward scales versus $t$ . The horizontal axis is the timestep $t$ from $0$ to $1000$ ; the vertical axis is a pure scale between $0$ and $1$ . The red curve is the signal scale $\overset{α}{ˉ}_{t}$ : it starts at $1$ (full signal) and sinks toward $0$ . The black curve is the noise scale $1 - \overset{α}{ˉ}_{t}$ : it starts at $0$ and climbs toward $1$ . They cross where signal and noise carry equal scale. Ex 1 lives at the far left edge (red near 1, black near 0 — barely noised); Ex 2 lives at the far right edge (red near 0, black near 1 — signal destroyed). The whole forward process is a journey from left to right along this picture.

C — Building $\overset{α}{ˉ}_{t}$ from a schedule (the cumulative product)

Worked example Ex 3 (Cell C)

Tiny toy schedule $β = (0.1, 0.2, 0.5)$ for $T = 3$ . Compute $α_{t}$ and $\overset{α}{ˉ}_{t}$ for all three steps.

Forecast: will $\overset{α}{ˉ}_{3}$ be closer to $0.5$ or to $0.1$ ?

Per-step survival $α_{t} = 1 - β_{t}$ : $α_{1} = 0.9, α_{2} = 0.8, α_{3} = 0.5$ . Why this step? Each $β_{t}$ is the fraction of variance injected; $1 - β_{t}$ is the fraction kept.
Cumulative $\overset{α}{ˉ}_{1} = 0.9$ . Why this step? After one step, cumulative = per-step (and $\overset{α}{ˉ}_{1} = \overset{α}{ˉ}_{0} \cdot α_{1} = 1 \cdot 0.9$ ).
$\overset{α}{ˉ}_{2} = 0.9 \times 0.8 = 0.72$ . Why this step? Multiply, because signal that survives step 1 must also survive step 2.
$\overset{α}{ˉ}_{3} = 0.72 \times 0.5 = 0.36$ . Why this step? One more surviving fraction stacked on.

Verify: $\overset{α}{ˉ}_{3} = 0.36 < α_{3} = 0.5$ — cumulative survival is always $\leq$ any single step's survival, as a product of numbers in $(0, 1)$ must be. Also $0.9 \geq 0.72 \geq 0.36$ is monotonically decreasing, as required. ✓

This schedule $β = (0.1, 0.2, 0.5)$ is our house schedule; Examples 3, 4, 5 and 7 all use it so the numbers stay comparable.

D — One generic reverse step

Recall the reverse mean (parent note): $μ_{θ} (x_{t}, t) = \frac{1}{α _{t}} (x_{t} - \frac{β _{t}}{1 - α ˉ _{t}} ϵ_{θ} (x_{t}, t)), \tilde{β}_{t} = \frac{1 - α ˉ _{t - 1}}{1 - α ˉ _{t}} β_{t} .$

Notice $\tilde{β}_{t}$ needs $\overset{α}{ˉ}_{t - 1}$ . At $t = 1$ this is $\overset{α}{ˉ}_{0} = 1$ (our convention above), which makes $\tilde{β}_{1} = 0$ — the reverse step at $t = 1$ is deterministic. We hit exactly that in Ex 5.

Worked example Ex 4 (Cell D)

At $t = 2$ using the house schedule: $α_{2} = 0.8, β_{2} = 0.2, \overset{α}{ˉ}_{2} = 0.72, \overset{α}{ˉ}_{1} = 0.9$ . Take scalar $x_{2} = 1.0$ , model output $ϵ_{θ} = 0.5$ , and reverse-noise draw $z = - 1.0$ . Compute $x_{1}$ .

Forecast: denoising removes estimated noise — will $x_{1}$ be pulled below or above $x_{2} = 1.0$ ?

Noise-removal coefficient $\frac{β _{2}}{1 - α ˉ _{2}} = \frac{0.2}{0.28} = \frac{0.2}{0.52915} = 0.37796$ . Why this step? This scales how much of the predicted noise we subtract; larger $β_{t}$ ⇒ more aggressive removal.
Inside the bracket $x_{2} - 0.37796 \cdot ϵ_{θ} = 1.0 - 0.37796 \cdot 0.5 = 1.0 - 0.18898 = 0.81102$ . Why this step? Strip the estimated noise off the noisy sample.
Rescale by $1/ α_{2}$ $μ_{θ} = \frac{0.81102}{0.8} = \frac{0.81102}{0.89443} = 0.90674$ . Why this step? The forward step had shrunk the signal by $α_{t}$ ; dividing undoes that shrink.
Reverse variance $\tilde{β}_{2} = \frac{1 - α ˉ _{1}}{1 - α ˉ _{2}} β_{2} = \frac{0.1}{0.28} \cdot 0.2 = 0.35714 \cdot 0.2 = 0.071429$ , so $\tilde{β}_{2} = 0.26726$ . Why this step? This sets how much fresh randomness we inject to keep the reverse chain stochastic.
Sample $x_{1} = μ_{θ} + \tilde{β}_{2} z = 0.90674 + 0.26726 \cdot (- 1.0) = 0.63948$ . Why this step? Draw an actual sample from the reverse Gaussian, not just its mean.

Verify: $μ_{θ} = 0.9067 < x_{2} = 1.0$ ✓ (denoising pulled it down, as forecast, because $ϵ_{θ} > 0$ ). And $μ_{θ}$ landed back near the un-shrunk scale — sensible since $x_{2}$ itself was near $x_{0}$ -scale. ✓

E — Degenerate case $t = 1$ : no noise injected

Worked example Ex 5 (Cell E)

Final step $t = 1$ , house schedule: $α_{1} = 0.9, β_{1} = 0.1, \overset{α}{ˉ}_{1} = 0.9$ , and $\overset{α}{ˉ}_{0} = 1$ . Take $x_{1} = 0.64$ , model output $ϵ_{θ} = 0.2$ . Compute the returned image $x_{0}$ .

Forecast: will there be any random wobble in this output?

Confirm no noise $\tilde{β}_{1} = \frac{1 - α ˉ _{0}}{1 - α ˉ _{1}} β_{1} = \frac{0}{0.1} \cdot 0.1 = 0$ . Why this step? $\overset{α}{ˉ}_{0} = 1$ forces the numerator to zero — this is the degenerate case.
Removal coefficient $\frac{β _{1}}{1 - α ˉ _{1}} = \frac{0.1}{0.1} = \frac{0.1}{0.31623} = 0.31623$ . Why this step? Same structure as Ex 4, evaluated at the schedule's first entry.
Bracket $x_{1} - 0.31623 \cdot 0.2 = 0.64 - 0.063246 = 0.576754$ . Why this step? Remove the last dab of predicted noise.
Rescale $μ_{θ} = \frac{0.576754}{0.9} = \frac{0.576754}{0.94868} = 0.60795$ . Why this step? Undo the first-step shrink.
Output $x_{0} = μ_{θ}$ (no $z$ term, since $\tilde{β}_{1} = 0$ ). Why this step? $t = 1$ ⇒ deterministic; this is the returned pixel.

Verify: output $x_{0} = 0.6080$ , a clean deterministic number with zero variance contribution. If we had wrongly added $\tilde{β}_{1} z$ we would inject noise into the finished image — but $\tilde{β}_{1} = 0$ makes that term vanish automatically, exactly as the $\overset{α}{ˉ}_{0} = 1$ convention guarantees. ✓

F — Perfect model: reverse exactly inverts forward

Worked example Ex 6 (Cell F)

Same house schedule, $t = 1$ ( $α_{1} = 0.9, β_{1} = 0.1, \overset{α}{ˉ}_{1} = 0.9$ ) — no new numbers, so keep the earlier examples in mind. Suppose the network is perfect: it returns the true noise, $ϵ_{θ} = ϵ$ , and we look at the mean (skip the stochastic term, DDIM-style). Start from $x_{0} = 0.80$ with true noise $ϵ = 1.5$ . First forward-noise to get $x_{1}$ , then reverse. Do we recover $x_{0}$ ?

Forecast: if the model knows the exact noise, should denoising land back on $0.80$ exactly?

Forward $x_{1} = 0.9 \cdot 0.80 + 0.1 \cdot 1.5 = 0.94868 \cdot 0.80 + 0.31623 \cdot 1.5 = 0.75895 + 0.47434 = 1.23329$ . Why this step? Apply the direct-sampling law to build a real noisy sample. Note $\overset{α}{ˉ}_{1} = 0.9 = 0.94868$ (not $0.8$ ).
Reverse mean with $ϵ_{θ} = ϵ = 1.5$ : removal coeff $\frac{0.1}{0.1} = 0.31623$ ; bracket $= 1.23329 - 0.31623 \cdot 1.5 = 1.23329 - 0.474342 = 0.758948$ . Why this step? Subtract exactly the noise that was added.
Rescale $μ_{θ} = \frac{0.758948}{0.9} = \frac{0.758948}{0.94868} = 0.80000$ . Why this step? Undo the shrink.

Verify: $μ_{θ} = 0.80000$ vs true $x_{0} = 0.80$ — exact recovery. Why exact here? At $t = 1$ the mean formula reduces to $x_{0} = (x_{1} - 1 - \overset{α}{ˉ}_{1} ϵ) / \overset{α}{ˉ}_{1}$ , which is the algebraic inverse of the forward law (because $\overset{α}{ˉ}_{1} = α_{1}$ and $β_{1} / 1 - \overset{α}{ˉ}_{1} = β_{1} = 1 - \overset{α}{ˉ}_{1}$ when $\overset{α}{ˉ}_{1} = α_{1}$ ). At $t = 1$ perfect noise prediction inverts forward exactly. ✓

G — Broken / zero model: what happens with no denoising?

H — Word problem: at which $t$ is the digit "half gone"?

Worked example Ex 8 (Cell H — real-world SNR reasoning)

"Half gone" means signal energy = noise energy, i.e. $\overset{α}{ˉ}_{t} = 1 - \overset{α}{ˉ}_{t}$ . Using a schedule where $\overset{α}{ˉ}_{t} = e^{- 0.002 t}$ (a common continuous approximation), find the timestep $t^{⋆}$ where the digit is half destroyed, and give the signal-to-noise ratio there.

Forecast: guess whether $t^{⋆}$ is early (near 0) or late (near 1000).

Set the condition $\overset{α}{ˉ}_{t^{⋆}} = 0.5$ . Why this step? Equal signal and noise energy ⇔ each is half of the total $1$ .
Solve $e^{- 0.002 t^{⋆}} = 0.5 \Rightarrow - 0.002 t^{⋆} = ln 0.5 \Rightarrow t^{⋆} = \frac{- ln 0.5}{0.002} = \frac{0.693147}{0.002} = 346.57$ . Why this step? Take logs — the natural log is the tool that undoes the exponential, letting us solve for $t$ in the exponent.
SNR $= \frac{α ˉ}{1 - α ˉ} = \frac{0.5}{0.5} = 1$ . Why this step? Signal-to-noise ratio compares the two scales; here they're equal, so SNR $= 1$ .

Verify: $t^{⋆} \approx 347$ — squarely in mid-schedule (not near 0 or 1000), matching the parent note's progression where $t = 250$ is still recognisable and $t = 1000$ is pure noise. SNR $= 1$ is the exact crossover: below $t^{⋆}$ signal dominates, above it noise does. So a digit is "half destroyed" roughly one-third of the way through the diffusion, and fully gone only near the end. ✓

Figure s02 — the half-gone crossover. Horizontal axis is timestep $t$ ; vertical axis is energy fraction (0 to 1). The red curve is signal energy $\overset{α}{ˉ}_{t}$ falling from 1; the black curve is noise energy $1 - \overset{α}{ˉ}_{t}$ rising from 0. The dashed vertical line marks $t^{⋆} \approx 347$ where the two curves cross at height $0.5$ — the "half-gone" point of Ex 8, where SNR $= 1$ . Left of the line the digit is signal-dominated (recognisable); right of the line it is noise-dominated.

I — Exam twist: recover a single $β_{t}$ from cumulative products

Recall

Recall Cover check

Signal energy fraction after $t$ steps ::: $\overset{α}{ˉ}_{t}$ What is $β_{t}$ ? ::: the noise-injection fraction — the share of variance thrown away and replaced by fresh Gaussian noise at step $t$ Value of $\overset{α}{ˉ}_{0}$ and why it matters ::: $1$ (empty product); it makes $\tilde{β}_{1} = 0$ so the last reverse step adds no noise Why is no noise added at reverse step $t = 1$ ? ::: it is the final output; $\tilde{β}_{1} = 0$ automatically because $\overset{α}{ˉ}_{0} = 1$ A zero-output network's reverse mean does what to the sample? ::: multiplies it by $1/ α_{t} > 1$ — amplifies, never denoises $\overset{α}{ˉ}_{t}$ from $\overset{α}{ˉ}_{t - 1}$ ? ::: multiply by one new $α_{t}$ : $\overset{α}{ˉ}_{t} = \overset{α}{ˉ}_{t - 1} α_{t}$ "Half-gone" condition ::: $\overset{α}{ˉ}_{t} = 1 - \overset{α}{ˉ}_{t}$ , i.e. $\overset{α}{ˉ}_{t} = 0.5$ , SNR $= 1$ Why does perfect noise prediction invert forward exactly at $t = 1$ ? ::: because $\overset{α}{ˉ}_{1} = α_{1}$ , so the mean formula is the algebraic inverse of the forward law

See also: Langevin dynamics (the reverse step as noisy gradient ascent), Variational autoencoders (the loss's variational roots), and Classifier-free guidance (steering the reverse mean). Parent: Diffusion models forward - reverse process.