Distance of high-dimensional normal distribution has mean √D, variance 0.7

from In high-dimensional space, the normal distribution is almost uniformly distributed on the hypersphere Distance of high-dimensional normal distribution has mean √N, variance 0.7

• The Euclidean distance from the origin of a data point that follows a D-dimensional normal distribution $N_D(0,1)$ roughly follows a distribution with mean $\sqrt{D}$standard deviation $1/\sqrt{2}$.
  • Not exactly chi-square distribution.
  • Surprisingly small standard deviation
    • So [In high-dimensional space, the normal distribution is almost uniformly distributed on the hypersphere

:

mean= 9.972699059944992 sd= 0.70375223347664
mean= 19.98802691399663 sd= 0.7060739521568569
mean= 29.992667664100654 sd= 0.7078926693796469
mean= 39.99309275039724 sd= 0.7074454497781785

py

import numpy as np
import matplotlib.pyplot as plt

# of samples and number of dimensions
num_samples = 100000
dimensions = [100, 400, 900, 1600] # Example dimensions

# Plot the distribution of distances
plt.figure(figsize=(10, 6))
for d in dimensions:
    # Generate random data following a d-dimensional normal distribution
    data = np.random.normal(0, 1, (num_samples, d))
    
    # Calculate the distance of each data point from the origin
    distances = np.linalg.norm(data, axis=1)

    # Mean and variance
    print("mean=", distances.mean(), "sd=", distances.std())
    
    # Plot a histogram
    plt.hist(distances, bins=30, alpha=0.5, label=f'{d} dimensions', density=True)

# Graph setup
plt.xlabel('Distance from Origin')
plt.ylabel('Density')
plt.legend()
plt.title('Distribution of Distances from Origin in High-Dimensional Space')
plt.show()

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