Introduction to Data Science

1MS041, 2023

©2023 Raazesh Sainudiin, Benny Avelin. Attribution 4.0 International (CC BY 4.0)

Common discrete random variables

Bernoulli random variable

Single trial with success probability $p$.

In [1]:
from Utils import plotEMF
In [3]:
p = 0.1
plotEMF([(0,1-p),(1,p)])
In [4]:
import numpy as np
In [5]:
np.random.randint(0,2,size=10)
Out[5]:
array([0, 1, 0, 0, 1, 0, 1, 1, 0, 1])
In [7]:
from Utils import plotEDF,emfToEdf
In [8]:
plotEDF(emfToEdf([(0,0),(0,1-p),(1,p)]))

Binomial random variable

If we do $n$ trials with success probability $p$, then the binomial random variable is the number of successes. The PMF is $$ f(x) = {n \choose x} p^x (1-p)^{n-x} $$ Can only produce numbers $0,1,\ldots,n$.

In [9]:
from scipy.special import binom as binomial
n = 20
p = 0.5
plotEMF([(i,binomial(n,i)*(p**i)*((1-p)**(n-i))) for i in range(n)])
In [ ]:
np.random.binomial(20,0.5,size=10)
In [10]:
plotEDF(emfToEdf([(i,binomial(n,i)*(p**i)*((1-p)**(n-i))) for i in range(n)]))

Poisson random variable

Pois($\lambda$) where $\lambda \in (0,\infty)$ is called the rate $$ f(x) = \frac{\lambda^x e^{-\lambda}}{x!} $$

In [11]:
from scipy.special import factorial
from math import exp
l = 2
plotEMF([(i,l**i*exp(-l)/factorial(i)) for i in range(10)])
In [ ]:
In [ ]:
np.random.poisson(2,size=10)
In [12]:
plotEDF(emfToEdf([(i,l**i*exp(-l)/factorial(i)) for i in range(10)]))

Empirical means

In [13]:
from random import randint

def X():
    """Produces a single random number from DeMoivre(1/3,1/3,1/3)"""
    return randint(0,2)

def empirical_mean(n=1):
    """Produces the empirical mean of n experiments of the X above"""
    Z = [X() for i in range(n)]
    return sum(Z)/n
In [20]:
# Run this to get an observation of X and rerun for another
X
Out[20]:
<function __main__.X()>
In [25]:
# Run this to get an observation of the empirical mean of X
# when doing 10 experiments
empirical_mean(10)
Out[25]:
1.3

Common continuous random variables

The uniform [0,1] random variable

In this case we have

$$ f(x) = \begin{cases} 1 & \text{if } 0 \leq x \leq 1 \\ 0 & \text{otherwise} \end{cases} $$

Also, for $x \in [0,1]$ we have

$$ F(x) = \int_{-\infty}^x f(v) dv = \int_0^x dv = x $$
500px-Uniform_Distribution_PDF_SVG.svg.png wikipedia image 500px-Uniform_cdf.svg.png
In [26]:
import numpy as np
In [27]:
np.array([1,2,3])
Out[27]:
array([1, 2, 3])
In [30]:
import matplotlib.pyplot as plt
In [35]:
x = np.random.uniform(0,1,size=100)
In [39]:
_=plt.hist(x,density=False,bins=3)
In [40]:
from Utils import makeEDF,makeEMF
plotEMF(makeEMF(np.random.uniform(size=100)))
In [41]:
import numpy as np
from Utils import makeEDF,makeEMF,plotEDF
plotEDF(makeEDF(np.random.uniform(size=100)))

The Gaussian random variable (Normal)

In this case we have $$ f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{1}{2} \left ( \frac{x-\mu}{\sigma}\right )} $$ here we have two parameters, the mean $\mu$ and the standard deviation $\sigma$.

In [42]:
np.random.normal(size=10)
Out[42]:
array([ 1.01386181, -0.56043332,  0.49921684,  0.88863697, -1.64044673,
        0.37383609, -2.41185133,  1.21347747,  0.21825103,  0.97901204])
In [43]:
_=plt.hist(np.random.normal(size=100000),bins=200)