求助：从Numpy到Scipy的Wald分布参数转换与拟合问题

Hey there! I totally get the frustration when your fitted parameters don't match what you simulated—let's break down why this happens with the Wald (inverse Gaussian) distribution and how to fix it.

The Root Cause: Different Parameterizations #

Unlike the normal distribution, where Scipy's parameterization (loc = mean, scale = standard deviation) aligns with common intuitive parameters, the Wald distribution has two common parameterization schemes:

• Scipy's scheme: Uses loc (location parameter) and scale (scale parameter).
• Standard textbook/simulation scheme: Uses μ (mean of the distribution) and λ (shape parameter, which controls spread—higher λ = narrower distribution).

These aren't interchangeable, so you need to convert between them to get back your original simulation parameters.

Conversion Formulas #

If you simulated your data using the μ (mean) and λ (shape) parameters, here's how to map them to Scipy's loc/scale, and vice versa:

• To get Scipy parameters from μ and λ:

  loc = μ - (1 / λ)
  scale = 1 / λ
  

• To convert Scipy's fitted loc/scale back to μ and λ:

  μ = loc + scale
  λ = 1 / scale
  

Example Code to Verify #

Let's walk through a concrete example matching your scenario (assuming you simulated Wald data with μ=81 and λ=7, like your normal distribution parameters):

import numpy as np
from scipy.stats import wald

# Your original simulation parameters
mu_true = 81
lambda_true = 7

# Convert to Scipy's Wald parameters
loc_true = mu_true - 1 / lambda_true
scale_true = 1 / lambda_true

# Generate simulated data
np.random.seed(42)
data = wald.rvs(loc=loc_true, scale=scale_true, size=10000)

# Fit the data with Scipy
loc_fit, scale_fit = wald.fit(data)
print(f"Fitted Scipy parameters: loc = {loc_fit:.2f}, scale = {scale_fit:.2f}")

# Convert back to original μ and λ
mu_fit = loc_fit + scale_fit
lambda_fit = 1 / scale_fit
print(f"Restored original parameters: μ = {mu_fit:.2f}, λ = {lambda_fit:.2f}")

Running this code will output values very close to your original (81,7)—fixing the mismatch you saw!

Quick Validation Tip #

If you're unsure about the parameterization, remember that the variance of a Wald distribution is μ³ / λ. You can calculate the variance of your simulated data and cross-check it against the converted parameters to confirm everything lines up.

内容的提问来源于stack exchange，提问作者BumbleTee