Z-Transformation of Binomial Sampling Distribution

 2026-05-25

#statistics #binomial #z-transform #normal-distribution

In [1]:

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns

Let's create a binomial distribution with $p=0.7$.

In [2]:

data = np.random.choice([1, 0], 30000, p=[0.7, 0.3])
data[:20]

Out[2]:

array([0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1])

In [3]:

data_mean = np.mean(data)
data_var = np.var(data)

In [4]:

data_pmf = pd.Series(data).value_counts(normalize=True).sort_index()
data_pmf

Out[4]:

0    0.297467
1    0.702533
Name: proportion, dtype: float64

In [5]:

fig, ax = plt.subplots()
ax.bar(data_pmf.index, data_pmf)
ax.set_xticks([0, 1])
plt.show()

No description has been provided for this image

Looks about right.

z-transformation using "normal" sampling¶

As you can see, CLT also works on binomial distribution.

In [6]:

means = [np.mean(np.random.choice(data, 100)) for _ in range(5000)]
bins = np.arange(np.min(means), np.max(means), 0.005)

In [7]:

fig, ax = plt.subplots(1, 2, figsize=(12, 4))
sns.kdeplot(means, ax=ax[0])
sns.rugplot(means, color="red", height=0.05, ax=ax[0])
ax[1].hist(means, bins=bins, rwidth=0.9)
plt.show()

But does bootstraping work on binomial distribution?

In [8]:

sample = np.random.choice(data, 100)
means_bootstrap = [
    np.mean(np.random.choice(sample, len(sample), replace=True)) for _ in range(5000)
]
bins_bootstrap = np.arange(np.min(means_bootstrap), np.max(means_bootstrap), 0.005)

In [9]:

fig, ax = plt.subplots(1, 2, figsize=(12, 4))
sns.kdeplot(means_bootstrap, ax=ax[0])
sns.rugplot(means_bootstrap, color="red", height=0.05, ax=ax[0])
ax[1].hist(means_bootstrap, bins=bins_bootstrap, rwidth=0.9)
plt.show()

Yes, it works. Because bootstrap will work where the CLT works.

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