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pythonsamplingnormal-distribution

Sequential Sampling

To sample from N(1,2) with sample size 100 and calculating the mean of this sample we can do this:

import numpy as np

s = np.random.normal(1, 2, 100)
mean = np.mean(s)

Now if we want to produce 10000 samples and save mean of each of them we can do:

sample_means = []
for x in range(10000):
    sample = np.random.normal(1, 2, 100)
    sample_means.append (sample.mean())

How can I do it when we want to sample sequentially from N(1,2) and estimate the distribution mean sequentially?

Solution

  • IIUC you meant accumulative

    sample = np.random.normal(1,2,(10000, 100))
sample_mean = []
for i,_ in enumerate(sample):
    sample_mean.append(sample[:i+1,:].ravel().mean())

    Then sample_mean contains the accumulative samples mean

    sample_mean[:10]
[1.1185342714036368,
 1.3270808654923423,
 1.3266440422140355,
 1.2542028664103761,
 1.179358517854582,
 1.1224645540064788,
 1.1416887857272255,
 1.1156887336750463,
 1.0894328800573165,
 1.0878896099712452]