Linear Beta Schedule and Alpha-Bar

Problem Statement

Build the linear noise schedule used by the original DDPM paper and derive the two quantities every later formula depends on: the per-step signal retention alpha_t and the cumulative signal retention alpha_bar_t.

Background

The forward diffusion process corrupts data one small step at a time:

$q(x_t \mid x_{t-1}) = \mathcal{N}\!\left(x_t;\ \sqrt{1-\beta_t}\,x_{t-1},\ \beta_t I\right)$

The variances $\beta_1, \dots, \beta_T$ are the noise schedule. DDPM uses a linear schedule: $T$ values spaced evenly (endpoints included) from beta_start to beta_end, with the defaults $\beta_1 = 10^{-4}$ and $\beta_T = 0.02$ for $T = 1000$.

From the betas we define

$\alpha_t = 1 - \beta_t, \qquad \bar{\alpha}_t = \prod_{s=1}^{t} \alpha_s$

$\bar{\alpha}_t$ is the running product, not a running sum. It is what makes the forward process collapse into a single closed-form Gaussian, so getting it right is the foundation of everything else in diffusion.

Your Task

Implement:

def linear_schedule(T, beta_start=1e-4, beta_end=0.02):

Return a tuple (betas, alphas, alpha_bars) of three 1-D NumPy arrays, each of length T, using 0-based indexing (element i corresponds to timestep $t = i+1$ in the maths above).

Input Format

• T (int): number of diffusion steps, T >= 1.
• beta_start, beta_end (float): first and last beta. Both endpoints must appear in the array.

Output Format

A tuple of three NumPy arrays (betas, alphas, alpha_bars).

Sample

b, a, ab = linear_schedule(5, 0.1, 0.5)
print(np.round(b, 4).tolist())
print(np.round(a, 4).tolist())
print(np.round(ab, 4).tolist())

[0.1, 0.2, 0.3, 0.4, 0.5]
[0.9, 0.8, 0.7, 0.6, 0.5]
[0.9, 0.72, 0.504, 0.3024, 0.1512]

Betas are evenly spaced from 0.1 to 0.5, alphas are 1 - beta, and alpha_bars is their cumulative product, so it decays quickly toward zero.