DDPM Posterior Mean and Variance from x0

mediumPROParameterizations and the Reverse PosteriorDiffusion ModelsDiffusion Models 1

Problem Statement

The reverse step samples from the tractable posterior q(x_{t-1} | x_t, x_0). Compute its mean and variance from x_t, the (predicted) x0, and the schedule.

Background

Given beta_t, alpha_bar_t, and alpha_bar_{t-1} (with alpha_t = 1 - beta_t), the posterior is Gaussian with

$\tilde{\mu}_t = \frac{\sqrt{\bar{\alpha}_{t-1}}\,\beta_t}{1 - \bar{\alpha}_t}\, x_0 + \frac{\sqrt{\alpha_t}\,(1 - \bar{\alpha}_{t-1})}{1 - \bar{\alpha}_t}\, x_t$

$\tilde{\beta}_t = \frac{1 - \bar{\alpha}_{t-1}}{1 - \bar{\alpha}_t}\, \beta_t$

Your Task

Implement:

def posterior(x_t, x0, beta_t, alpha_bar_t, alpha_bar_prev):

Return a dict with "mean" (a list, rounded to 4 decimals) and "variance" (a float, rounded to 6 decimals).

Input Format

x_t, x0: lists of length D.
beta_t, alpha_bar_t, alpha_bar_prev (float).

Output Format

A dict with a list and a float.

Sample

print(posterior([1.0, 1.0], [0.5, 0.5], 0.2, 0.5, 0.64))

Output:

{'mean': [0.804, 0.804], 'variance': 0.144}

Example

Input

print(posterior([1.0, 1.0], [0.5, 0.5], 0.2, 0.5, 0.64))

Output

{'mean': [0.804, 0.804], 'variance': 0.144}

Reasoning

coef_x0 = sqrt(0.64)0.2/(1-0.5) = 0.80.2/0.5 = 0.32; coef_xt = sqrt(0.8)(1-0.64)/0.5 = 0.89440.72 = 0.644. mean = 0.320.5 + 0.6441 = 0.804. variance = (1-0.64)/(1-0.5)*0.2 = 0.144.

Constraints

len(x_t) == len(x0); schedule values in (0, 1).
alpha_t = 1 - beta_t; use the two coefficients above.
Mean rounded to 4 decimals; variance to 6 decimals.

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Concept primerPREVIEW

1. Background Knowledge

In Denoising Diffusion Probabilistic Models (DDPMs), the forward process gradually adds Gaussian noise to data, while the reverse process learns to denoise. A key insight is that the posterior distribution $q(x_{t-1} \mid x_t, x_0)$ is analytically tractable and Gaussian. This means we can compute its mean and variance in closed form without sampling.

The posterior mean $\tilde{\mu}_t$ is a weighted combination of the current noisy sample $x_t$ and the predicted clean sample $x_0$ . The weights depend on the noise schedule parameters: $\beta_t$ (the noise added at step $t$ ), $\bar{\alpha}_t$ (the cumulative product of $\alpha_i = 1 - \beta_i$ up to step $t$ ), and $\bar{\alpha}_{t-1}$ (the same cumulative product up to step $t-1$ ). Intuitively, as $t$ increases, $\bar{\alpha}_t$ decreases, meaning more noise has been added, so the weight on $x_0$ (the "target") increases relative to $x_t$ .

The posterior variance $\tilde{\beta}_t$ is a scalar (isotropic) that scales the identity covariance matrix. It is always smaller than $\beta_t$ because knowing $x_0$ reduces uncertainty about $x_{t-1}$ . This tractable posterior is what allows DDPMs to train efficiently: the model predicts $x_0$ (or equivalently $\epsilon$ ), and the posterior provides the target distribution for the reverse step.

2. Algorithm Approach

This is a direct formula evaluation problem. No iterative or optimization algorithm is needed. The approach is:

Parse the input parameters: $x_t$ , $x_0$ (vectors), and $\beta_t$ , $\bar{\alpha}_t$ , $\bar{\alpha}_{t-1}$ (scalars).
Compute the two scalar coefficients for the mean:

Coefficient for $x_0$ : $\frac{\sqrt{\bar{\alpha}_{t-1}}\,\beta_t}{1 - \bar{\alpha}_t}$
Coefficient for $x_t$ : $\frac{\sqrt{\alpha_t}\,(1 - \bar{\alpha}_{t-1})}{1 - \bar{\alpha}_t}$ where $\alpha_t = 1 - \beta_t$ .

Compute the mean vector as the element-wise weighted sum of $x_0$ and $x_t$ using these coefficients.
Compute the variance scalar: $\tilde{\beta}_t = \frac{1 - \bar{\alpha}_{t-1}}{1 - \bar{\alpha}_t}\,\beta_t$ .
Round the mean elements to 4 decimal places and the variance to 6 decimal places.
Return a dictionary with keys "mean" and "variance".

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solution.py

Python 3.13

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