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?
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]
Maybe list comprehension?
sample_means = [np.random.normal(1, 2, 100).mean() for i in range(10000)]
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