## What is central limit theorem?

The distribution of the sum of independent samples consisting of `n`

points drawn from an arbitrary distribution approach a normal distribution as `n`

increases.

If the distribution of the values has a mean and standard deviation, the distribution of sum is approximately given by $ N(n\mu, n\sigma^2)$

Some points to keep in mind:

- The values are to be drawn independently
- The values have to come from same distribution
- The underlying distribution should have finite mean and variance
- The rate convergence to the normal distribution depends on the skewness of the parent distribution.

We start with some crazy distribution that has got nothing to do with a normal distribution. Sample points from that distribution with some arbitrary sample size, following which we plot the sample mean (or sample sum) on a frequency table -- repeat this lot of times (tending to infinity) we end up getting a normal distribution of sample means!

**The Central Limit Theorem explains the prevalence of normal distributions in the natural world.** This limit is central to the ideas of hypothesis testing and helpful for estimating confidence intervals.

Below a simple python experiment to show this in action.

```
import random as rand
import numpy as np
from scipy import stats
# High DPI rendering for mac
%config InlineBackend.figure_format = 'retina'
# Plot matplotlib plots with white background:
%config InlineBackend.print_figure_kwargs={'facecolor' : "w"}
```

```
import matplotlib.pyplot as plt
import seaborn as sns
plot_params = {
'font.size' : 10,
'axes.titlesize' : 10,
'axes.labelsize' : 10,
'axes.labelweight' : 'bold',
'xtick.labelsize' : 10,
'ytick.labelsize' : 10,
}
plt.rcParams.update(plot_params)
sns.color_palette('colorblind')
```

```
from numpy.random import default_rng
rng = default_rng(42)
```

```
dice = np.arange(1,7) # Dice numbers possible
probabilities = [0.2, 0.3, 0.0, 0.2, 0.0, 0.3] #Weighted probabilites for the numbers
```

Define a function to draw samples from the dice and calculate the mean.

```
def sample_draw_mean(_trials=1000, _sample_size=1):
sample_mean_trials = []
# Sample a number from the distribution equal to trials
for i in range(_trials):
sample = rng.choice(dice, size=_sample_size, p=probabilities, replace=True)
sample_mean_trials.append(np.mean(sample))
return sample_mean_trials
```

Drawing `sample_size`

=1 from the distribution multiple times, i.e. equal to `num_of_trials`

variable

```
num_of_trials = 1000
sample_size = 1
sns.histplot(sample_draw_mean(_trials=num_of_trials, _sample_size=sample_size), bins=len(dice), stat='density', kde=True);
plt.title('Visualize sample mean distribution for {} sample drawn {} times'.format(sample_size, num_of_trials), fontsize=15);
```

```
num_of_trials = 1000
sample_size = 4
sns.histplot(sample_draw_mean(_trials=num_of_trials, _sample_size=sample_size), bins=len(dice), stat='density', kde=True);
plt.title('Visualize sample mean distribution for {} sample drawn {} times'.format(sample_size, num_of_trials), fontsize=15);
```

```
num_of_trials = 1000
sample_size = 20
sns.histplot(sample_draw_mean(_trials=num_of_trials, _sample_size=sample_size), bins=len(dice), stat='density', kde=True);
plt.title('Visualize sample mean distribution for {} sample drawn {} times'.format(sample_size, num_of_trials), fontsize=15);
```

As we keep plotting the frequency distribution for the sample mean it starts to approach the normal distribution!

```
def normal_distribution(x, mean=0, sigma=1):
out = (1/np.sqrt(2 * np.pi * sigma ** 2)) * np.exp(-1/2 * ((x - mean)/sigma)**2)
return(out)
```

```
num_of_trials = 1000
sample_size = 20
fig, ax = plt.subplots(1,1, figsize=(5,5))
sample_means = np.sort(sample_draw_mean(_trials=num_of_trials, _sample_size=sample_size))
# Plot histogram density
sns.histplot(sample_means, bins=len(dice), stat='density', kde=False, ax=ax)
# Plot normal distribution
ax.plot(sample_means, normal_distribution(sample_means, np.mean(sample_means), np.std(sample_means)), color='black', linestyle='--', label='Normal Distribution')
# Plot sample mean
ax.axvline(np.mean(sample_means), color='red', linestyle='--', linewidth=2.0, label='Sample Mean')
ax.set_xlabel('Dice number')
plt.title('Visualize sample mean distribution for {} sample drawn {} times'.format(sample_size, num_of_trials), fontsize=15);
plt.tight_layout()
```

```
beta = 5.0
num_of_trials = 1000
sample_size_list = [1, 10, 100, 500]
```

```
def generate_mean_samples(_beta, _iter, _sample_size):
samples_mean = []
for i in range(_iter):
sample_numbers = np.random.exponential(_beta, _sample_size)
samples_mean.append(np.mean(sample_numbers))
return(samples_mean)
```

```
sample_plot_list = []
for n in sample_size_list:
sample_plot_list.append((n, generate_mean_samples(beta, num_of_trials, n)))
```

```
fig, ax = plt.subplots(2,2, figsize=(10,10))
ax = ax.flatten()
for i, entry in enumerate(sample_plot_list):
sns.histplot(entry[1], stat='density', alpha=0.6, kde=False, ax=ax[i])
ax[i].set_title('Sample size: {}'.format(entry[0]))
sample_mean = np.mean(entry[1])
sample_std = np.std(entry[1])
normal_x = np.sort(entry[1])
# Plot normal distribution
ax[i].plot(normal_x, normal_distribution(normal_x,sample_mean,sample_std), linewidth=4.0, color='black', linestyle='--', label='Normal Distribution')
# Sample mean
ax[i].axvline(sample_mean, color='red', linestyle='--', linewidth=4.0, label='Sample Mean')
ax[i].set_xlabel('Sample Mean')
plt.suptitle(r'Visualize sample mean distribution for Exponential distribution $\beta$={}, Sampled {} times'.format(beta, num_of_trials));
plt.legend(bbox_to_anchor=(1.04,1), loc="upper left")
#plt.tight_layout()
```

**Estimate coin toss probability**

A coin is flipped 30 times, you get 22 heads. Find if the coin is fair or not. That is, if the probability of getting heads-tails is 50%.

This can be solved by estimating the probability of getting heads / tails provided the above condition is met.

Since we can model the coin toss process (a priori model) using Bernoulli's distribution, we will estimate the probability of 22 heads considering a fair coin. This will be our Null Hypothesis.

**Null hypothesis:**
The null hypothesis is a model of the system based on the assumption that the apparent effect was actually due to chance.

Assuming a bernoulli distribution:

$$X_{i} \sim B(p)$$

$$ P(N_H=22) = \binom nx p^{22}(1-p)^{30-22} $$

By central limit theorem: $$ \sum_{i=1}^{30}{X_{i}} \sim N(30p, 30(1-p)) $$

From maximum likelihood estimate, more detailts on MLE can be found here.:

$$ \hat{p} = 0.73 $$

**Estimate 95% confidence interval:**

- Assuming a normal distribution: $$ \mu \pm 1.96 \sigma $$

$$ 30\hat{p} \pm 1.96 \sqrt{ 30 * (1-\hat{p}) } $$

$$ 22 \pm 1.96 \sqrt{( 30 * 0.26 )} = (16.4, 27.58) $$

```
rng = np.random.default_rng(42)
```

Define a `numpy.random.choice`

function to simulate coin tosses. This can repeated to 30 times.

```
sampling_30 = rng.choice([0,1], replace=True, size=30) # we can randint(2) here as well.
```

`np.where`

is used to find the entries with heads, that way for each 30 coin tosses we can estimate how many heads are there. In this case we are treating heads as 1 and tails as 0

```
len(np.where(sampling_30 == 1)[0]) # or just sum the list since all we have is 1 / 0
```

```
sum(sampling_30)
```

Setup the problem to perform multiple trails of 30 coin tosses, when done with the trials we will keep an account of how many of those trials had 22 heads.

```
heads_condition = 22
num_heads_list = []
constraint_satisy = 0
num_trials = 5000
```

```
for _ in range(num_trials):
sampling_30 = rng.choice([0,1], replace=True, size=30, p=[0.50,0.50]) # A-priori fair coin toss model
number_of_heads = len(np.where(sampling_30 == 1)[0])
num_heads_list.append(number_of_heads)
if number_of_heads == heads_condition:
constraint_satisy = constraint_satisy + 1
num_heads_list = np.array(num_heads_list)
```

```
len(num_heads_list)
```

Defining a normal distribution function from `scipy`

or we could also use the function defined previously.

```
from scipy.stats import norm
x = np.linspace(min(num_heads_list), max(num_heads_list))
std_norm_coin = norm(np.mean(num_heads_list), np.std(num_heads_list))
```

```
quantiles_95_confidence = np.quantile(num_heads_list, [0.025, 0.975])
```

```
fig, ax = plt.subplots(1,1, figsize=(8,5))
# Plot histogram density
sns.histplot(num_heads_list, stat='density', kde=False, ax=ax)
# Plot normal distribution
ax.plot(x, std_norm_coin.pdf(x), color='black', linestyle='--', label='Normal Distribution')
# Plot sample mean
ax.axvline(np.mean(num_heads_list), color='red', linestyle='--', linewidth=2.0, label='Sample Mean')
ax.axvline(heads_condition, color='blue', linestyle='-', linewidth=2.0, label='Experiment condition')
# Plot confidence interval
ax.axvspan(quantiles_95_confidence[0], quantiles_95_confidence[1], alpha=0.15, color='yellow',label='95% confidence interval')
ax.set_xlabel('Number of heads in 30 coin tosses')
plt.title('Visualize distribution of number of heads for 30 coin tosses sampled {} times'.format(num_trials), fontsize=15);
plt.legend(loc="upper left")
plt.tight_layout()
```

**p-value estimate**

```
p_value = constraint_satisy / num_trials
print(p_value)
```

Since p-value is less than 0.05, this means the coin is **not fair**

For most problems, we only care about the order of magnitude: if the p-value is smaller that 1/100, the effect is likely to be real; if it is greater than 1/10, probably not. If you think there is a difference between a 4.8% (significant!) and 5.2% (not significant!), you are taking it too seriously.