Noise scheduling: Question bank — AI-ML deep dive

Before we start, the symbols this page leans on — all defined in the parent note, repeated and earned from scratch here so no symbol is used unexplained:

Definition The stage: what is being noised, and over how many steps

$x_{0}$ ::= the clean starting sample (e.g. a photograph flattened into a list of $d$ numbers — one per pixel/channel). The number $d$ is the dimension: how many values make up the data.
$x_{t}$ ::= the sample after $t$ noising steps — same shape as $x_{0}$ , just progressively blurrier.
$T$ ::= the total number of diffusion steps in the schedule (e.g. $T = 1000$ ). Time runs $t = 0$ (clean) up to $t = T$ (pure noise). Every " $/ T$ " you see below is just measuring progress as a fraction of the whole journey.
$ϵ \sim N (0, I)$ ::= a standard Gaussian noise vector, same dimension $d$ as the data. Here $N (μ, Σ)$ means "Gaussian with mean $μ$ and covariance $Σ$ ", and $I$ is the $d \times d$ identity matrix — its diagonal-of-ones shape says each coordinate gets independent, equal-variance noise (no coordinate is special). So $ϵ$ is $d$ independent draws from a bell curve of variance $1$ .

Definition The cosine kernel — every symbol first

The cosine schedule needs one shape function. Read the argument inside-out: $f (t) := cos (a number in [0, 1] (\frac{t / T + s}{1 + s}) \cdot \frac{π}{2})^{2}$

$t / T$ ::= progress fraction (0 at start, 1 at end) — this is why $T$ appears here.
$s$ ::= a small offset constant (typically $s = 0.008$ ). Its whole job: nudge the fraction away from exactly $0$ at $t = 0$ so the earliest $β_{t}$ doesn't collapse to zero (which, as the traps below show, is singular). Dividing by $(1 + s)$ re-normalizes so the fraction still lands in $[0, 1]$ .
the outer $(\cdot)^{2}$ ::= squares the cosine so $f (t) \geq 0$ always. Parenthesization to remove ambiguity: the whole bracketed number is multiplied by $\frac{π}{2}$ first, the cosine is taken of that, and then the result is squared — i.e. $f (t) = (cos θ)^{2}$ with $θ = \frac{t / T + s}{1 + s} \cdot \frac{π}{2}$ .
The cosine schedule then sets $\overset{α}{ˉ}_{t} = f (t) / f (0)$ , so $f$ is what shapes the smooth S-curve.

Picture the two schedules and their SNR side by side before diving into the traps:

True or false — justify

β_{t}

the total amount of noise present at step

t

False.

β_{t}

is only the fresh noise added at that single step; the accumulated noise is measured by

1 - \overset{α}{ˉ}_{t}

, which grows from products of all past steps.

β_{t}

increases every step, then

\overset{α}{ˉ}_{t}

must decrease every step.

True. Each factor

α_{t} = 1 - β_{t}

is below

1

, so multiplying by it always shrinks the running product

\overset{α}{ˉ}_{t}

, regardless of whether

β_{t}

itself rises or falls.

A larger

β_{T}

at the end guarantees

\overset{α}{ˉ}_{T} \approx 0

False.

\overset{α}{ˉ}_{T}

depends on the whole product, not the last term. A tiny final

β_{T}

can still give

\overset{α}{ˉ}_{T} \approx 0

if earlier steps already destroyed the signal.

The linear and cosine schedules can share the same

β_{1}

and

β_{T}

yet reach very different

\overset{α}{ˉ}_{t}

t = 500

True. Endpoints of

β

pin the boundaries, but the path between them (the whole product) is what sets mid-schedule

\overset{α}{ˉ}_{t}

; cosine keeps signal alive longer in the middle.

SNR (t)

is always a decreasing function of

t

for any valid schedule.

True. As

\overset{α}{ˉ}_{t}

falls monotonically, the numerator drops and denominator

1 - \overset{α}{ˉ}_{t}

rises, so the ratio strictly decreases — signal only ever gets harder to hear.

Doubling every

β_{t}

simply doubles the total noise at each step.

False. Noise accumulates multiplicatively through

\prod (1 - β_{s})

, not additively; doubling

β_{t}

shifts the whole

\overset{α}{ˉ}_{t}

curve nonlinearly and can collapse the signal far earlier.

The cosine schedule was designed mainly to make sampling faster.

False. Its purpose is a smoother

\overset{α}{ˉ}_{t}

decay (no sharp early drop), giving more stable gradients and better SNR distribution; the speed benefit is a downstream consequence, not the design goal.

\overset{α}{ˉ}_{t}

can equal exactly

1

t = 0

in the mathematics.

True but dangerous.

\overset{α}{ˉ}_{0} = 1

is the honest starting value (the empty product), yet it makes

SNR (0) = \frac{1}{0} = \infty

and

λ (0) = lo g SNR (0)

undefined, so SNR-parametrized schedules clip it to

1 - ε

Spot the error

"Set

β_{1} = 0

so we never damage the original image." — what breaks?

With

β_{1} = 0

we get

α_{1} = 1 - β_{1} = 1

; the reverse-step factor

\frac{1}{α _{1}} = \frac{1}{1}

is fine here, but

\overset{α}{ˉ}_{1} = α_{1} = 1

forces

SNR (1) = \frac{1}{1 - 1} = \infty

, so

λ (1) = lo g SNR (1) = + \infty

— a mathematical singularity in the log-SNR loss weighting. A tiny

β_{1} \approx 1 0^{- 4}

is used instead so

\overset{α}{ˉ}_{1} < 1

and

λ (1)

is finite.

"SNR schedule: interpolate

λ (t) = lo g SNR

linearly from

λ_{m a x} = + \infty

t = 0

down to

λ_{m i n}

." — find the mistake.

λ_{m a x} = + \infty

is not a finite number, so you cannot interpolate from it. You must regularize: cap

\overset{α}{ˉ}_{0}

1 - ε

, which makes

λ_{m a x} = lo g \frac{α ˉ _{0}}{1 - α ˉ _{0}} = lo g \frac{1 - ε}{ε}

finite.

"To recover

\overset{α}{ˉ}_{t}

from log-SNR

λ (t)

, use

\overset{α}{ˉ}_{t} = e^{λ (t)}

." — correct it.

Wrong: that ignores the denominator. Since

λ (t) = lo g \frac{α ˉ _{t}}{1 - α ˉ _{t}}

, exponentiating gives

\frac{α ˉ _{t}}{1 - α ˉ _{t}} = e^{λ}

; solving for

\overset{α}{ˉ}_{t}

gives the sigmoid,

\overset{α}{ˉ}_{t} = \frac{e ^{λ}}{1 + e ^{λ}} = σ (λ)

, which correctly stays in

[0, 1]

"The variance of the summed noise across two steps is

σ_{1}^{2} + σ_{2}^{2}

." — what's wrong?

Variances of independent Gaussians add directly:

σ_{1}^{2} + σ_{2}^{2}

. Taking square roots first is the error; only the final standard deviation is a square root of the summed variances.

"Cosine schedule keeps

\overset{α}{ˉ}_{t} = cos (\cdot)

." — fix the notation.

It uses the kernel

f (t) = cos (\cdot)^{2}

(cosine squared), then normalizes:

\overset{α}{ˉ}_{t} = f (t) / f (0)

. Squaring keeps

f (t) \geq 0

and produces the smooth S-curve; a bare cosine would go negative.

"Increasing

T

from 1000 to 4000 with the same

β

endpoints improves quality for free." — critique.

More steps stretch the same total destruction across finer increments, but if the schedule already wastes late steps on near-pure noise, extra steps just cost compute — the SNR distribution matters more than raw step count.

"Since noise variance is

(1 - \overset{α}{ˉ}_{t})

, the signal variance must be

\overset{α}{ˉ}_{t}

." — is the claim complete?

Incomplete: signal variance is

\overset{α}{ˉ}_{t} ∥ x_{0} ∥^{2}

and noise variance is

(1 - \overset{α}{ˉ}_{t}) d

across

d

dimensions. The clean

\overset{α}{ˉ}_{t}

1 - \overset{α}{ˉ}_{t}

split is the coefficient, not the full variance.

Why questions

Why do we track

\overset{α}{ˉ}_{t}

instead of just the list of

β_{t}

(or

α_{t}

Because

\overset{α}{ˉ}_{t}

collapses the entire history into one number giving direct sampling

q (x_{t} ∣ x_{0}) = N (\overset{α}{ˉ}_{t} x_{0}, (1 - \overset{α}{ˉ}_{t}) I)

— we jump to any step without simulating every intermediate one.

Why does a network learn best when

λ (t) = lo g SNR

decreases linearly?

It makes the denoising difficulty roughly constant per step, so no timestep is disproportionately easy or hard, which stabilizes gradients and balances the training signal across

t

Why is the sum-of-Gaussians fact (

σ_{1}^{2} + σ_{2}^{2}

) the linchpin of the whole derivation?

It lets us merge the many independent per-step noises into a single Gaussian, collapsing the recurrence

x_{t} = α_{t} x_{t - 1} + \dots

into the closed form

x_{t} = \overset{α}{ˉ}_{t} x_{0} + 1 - \overset{α}{ˉ}_{t} ϵ

Why does the linear schedule "waste" late steps?

Its signal collapses (

\overset{α}{ˉ}_{t} \to 0

) well before

t = T

, so many final steps operate on essentially pure noise where there is almost nothing left to learn — computation with little information gain.

Why must

\overset{α}{ˉ}_{T}

be near

0

but not necessarily exactly

0

We need

x_{T}

indistinguishable from the prior

N (0, I)

so sampling can start from pure noise;

\overset{α}{ˉ}_{T} < 1 0^{- 2}

is close enough, and exact

0

isn't required for the prior match.

Why does a learned variance schedule help over a fixed one?

Because the optimal noise level can be data-dependent — smooth regions and high-frequency regions want different destruction rates — so letting

β_{t} = σ_{θ} (x_{t}, t)

adapts per input rather than forcing one global recipe.

Why does cosine beat linear on FID even at the same step count

T

Its smoother SNR schedule makes each step equally informative, so the network isn't starved of signal early or wasted late — better use of the same

T

steps means fewer artifacts.

Edge cases

What is

SNR (0)

with the honest value

\overset{α}{ˉ}_{0} = 1

SNR (0) = \frac{1}{1 - 1} = \infty

and

λ (0) = lo g SNR (0) = + \infty

— an undefined boundary, which is exactly why SNR schedules clip

\overset{α}{ˉ}_{0}

1 - ε

What happens to

SNR (t)

\overset{α}{ˉ}_{t} \to 0

at the very end?

It tends to

\frac{0}{1} = 0

, so

λ (t) = lo g SNR (t) \to - \infty

— the picture is fully drowned in noise, matching the prior

N (0, I)

If every

β_{t}

were equal to some constant

c

, what is

\overset{α}{ˉ}_{t}

\overset{α}{ˉ}_{t} = (1 - c)^{t}

— a pure exponential decay; even a constant schedule collapses the signal geometrically, which is why designers shape

β_{t}

instead of leaving it flat.

Degenerate case: what if

β_{t} = 1

for some

t

Then

α_{t} = 1 - β_{t} = 0

, that factor zeroes the product, and

\overset{α}{ˉ}_{s} = 0

for all

s \geq t

— the signal is annihilated in one step, an over-aggressive extreme to avoid.

Boundary: with the sigmoid inversion, what is

\overset{α}{ˉ}_{0}

when

λ_{m a x} = lo g \frac{1 - ε}{ε}

\overset{α}{ˉ}_{0} = σ (λ_{m a x}) = 1 - ε

— just below one, giving the tiny-noise-at-start behavior with a finite, well-defined

λ (0)

Limiting behaviour: as

s \to 0

in the cosine offset, what fails?

With

s = 0

the kernel

f (0) = cos (0)^{2} = 1

is fine, but

β_{1} = 1 - \overset{α}{ˉ}_{1} / \overset{α}{ˉ}_{0}

can hit near-zero at

t = 0

; the offset

s = 0.008

exists precisely to keep the earliest

β_{t}

from vanishing.

Recall One-line summary of the traps

$β_{t}$ = fresh noise per step; $α_{t} = 1 - β_{t}$ = per-step survival; $\overset{α}{ˉ}_{t}$ = surviving signal (a product, empty at $t = 0$ so $\overset{α}{ˉ}_{0} = 1$ ); $SNR$ and $λ = lo g SNR$ tie them together and blow up at $t = 0$ unless regularized. Almost every trap here is "additive vs multiplicative" or "single step vs accumulated".

Connections

Noise scheduling — parent note this bank interrogates
4.5.11-Forward-and-reverse-diffusion-process — where the recurrence $x_{t} = α_{t} x_{t - 1} + \dots$ lives
4.5.13-Denoising-objective — $λ = lo g SNR$ sets the loss weighting these traps reference
4.5.15-Score-matching — SNR magnitude questions connect here
4.5.18-Fast-sampling — the "wasted late steps" trap motivates step-skipping