2) Random Variables¶

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Last updated: 9-16-2020

2.1) What is a random variable?¶

It is a statistical model that describes uncertain outcome of a random process.

Let’s model stock price as a random variable X:

stock = ['Increases', 'Stays same','Decreases']
x = [10, 0, -9]
prob = [0.3, 0.5, 0.2]
table = {'Stock Price':stock, 'x': x, 'P(X=x)':prob }
import pandas as pd
X = pd.DataFrame(table)
X

	Stock Price	x	P(X=x)
0	Increases	10	0.3
1	Stays same	0	0.5
2	Decreases	-9	0.2

The capital “X” stands for the random variable; whereas the lower case “x” indicates the possible outcomes (10, 0, -9). The statement $P(X=x_i)$ means the probability of the outcome $x_i$. For example, $P(X=x_0) = P(X=10) = 0.3$.

2.2) How to calculate the expected value of a random variable X, denoted by E(X)?¶

Expected value is a weighted average that uses probabilities to weight the possible outcomes.

Let’s calculate the mean ($\mu$) or expected value of the previous example random variable (X):

\[ \mu = E(X) = \sum_{i=0}^{n} x_iP(X=x_i) \]

\[ x_0P(X=x_0)+...+x_nP(X=x_n)\]

\[ 10*0.3+0*0.5-9*0.2\]

\[ 3+0-1.8 = 1.2\]

Therefore, the average return of this stock X is $1.2.

Let’s show the calculations step by step:

X['x*P(X=x)'] = X['x']*X['P(X=x)']
X

	Stock Price	x	P(X=x)	x*P(X=x)
0	Increases	10	0.3	3.0
1	Stays same	0	0.5	0.0
2	Decreases	-9	0.2	-1.8

Sum up the rows of ‘x*P(X=x)’:

sum(X['x*P(X=x)'])

1.2

2.3) How to calculate the variance of X, denoted by Var(X)?¶

The variance of a random variable X is the expected value of the squared deviation from its mean $\mu$:

\[ \sigma^2=E[(X-\mu)^2] = Var(X)\]

\[ \sum_{i=0}^{n} (x_i - \mu)^2 P(X=x_i) \]

\[ (x_0 - \mu)^2 P(X=x_0)+...+(x_n - \mu)^2 P(X=x_n) \]

\[ 77.44*0.3+1.44*0.5+104.04*0.2 \]

\[ 23.232 + 0.720 + 20.808 \]

\[ 44.76\]

X['x-mu'] = X['x'] - sum(X['x*P(X=x)'])
X['(x-mu)^2'] = X['x-mu']*X['x-mu']
X['[(x-mu)^2]*P(X=x)'] =  X['(x-mu)^2']*X['P(X=x)']
X

	Stock Price	x	P(X=x)	x*P(X=x)	x-mu	(x-mu)^2	[(x-mu)^2]*P(X=x)
0	Increases	10	0.3	3.0	8.8	77.44	23.232
1	Stays same	0	0.5	0.0	-1.2	1.44	0.720
2	Decreases	-9	0.2	-1.8	-10.2	104.04	20.808

# Round 2 decimals
%precision 2

varX = sum(X['[(x-mu)^2]*P(X=x)'])
varX

44.76

Variance is a measure of variability around the mean. It is hard to interpret 44.76, because the measurement unit is the square of the measurement unit ($) of the random variable.

2.4) How to calculate the standard deviation of X, denoted by SD(X) or $\sigma$?¶

Standard Deviation is the square root of the variance.

\[ \sigma = \sqrt{\sigma^2}=\sqrt{Var(X)}\]

\[ \sqrt{44.76} \]

$$ $6.7 $$

sdX = varX**(1/2)
sdX

6.69

The standard deviation, $\sigma = 6.7$, has the same unit ($) of the random variable. Therefore, it is easy to interpret.

One $\sigma$ above the mean ($\mu$) or below the mean ($\mu$) is a very likely outcome.

The standard deviation is a measure of variability around the mean. Bigger the number, bigger the variation.

In Finance, it is a proxy for risk. You want to minimize risk ($\sigma$) and maximize return ($\mu$).

2.5) Prove that $E(cX) = cE(X)$, where c is a constant.¶

In the previous examples, I was using the index $i$ starting from 0. Let’s start counting from 1 but remember that Python starts counting from 0.

By definition:

\[ E(X) = x_1p_1+x_2p_2+...+x_np_n\]

Then:

\[ E(cX) = cx_1p_1+cx_2p_2+...+cx_np_n\]

\[ c(x_1p_1+x_2p_2+...+x_np_n)\]

\[ cE(X)\]

2.6) Prove that $Var(cX)=c^2Var(X)$, where c is a constant.¶

By definition:

\[ Var(X)=E[(X-\mu_x)^2] \]

Then:

\[ Var(cX)=E[(cX-c\mu_x)^2] \]

\[ E[c^2(X-\mu_x)^2] \]

\[ c^2E[(X-\mu_x)^2] \]

\[ c^2Var(X) \]

2.7) Prove that $E(X+c)=E(X)+c$, where c is a constant.¶

\[ E(X+c) = (x_1+c)p_1+(x_2+c)p_2+...+(x_n+c)p_n\]

\[ x_1p_1+cp_1+x_2p_2+cp_2+...+x_np_n+cp_n\]

\[ (x_1p_1+x_2p_2+...+x_np_n) + c(p_1+p_2+...+p_n)\]

As probability must sum up to 1:

\[ \sum_{i=1}^{n}p_i=1 \]

Then:

\[ E(X) + c\]

2.8) Prove that $E(c)=c$, where c is constant.¶

By definition: $$ E(X) = x_1p_1+x_2p_2+...+x_np_n$$

Then:

\[ E(c) = cp_1+cp_2+...+cp_n\]

\[ c(p_1+p_2+...+p_n)\]

\[c\]

2.9) Prove that $Var(c) = 0$, where c is a constant.¶

By definition:

\[ Var(X)=E[(X-\mu_x)^2] \]

Then:

\[ Var(c)=E[(c-\mu_c)^2] \]

\[ E[(c-c)^2] \]

\[ E[(0)^2] \]

\[0\]

Intuition: By definition a constant has no variation.

2.10) Draw the distribution or probability mass function (PMF) of a random variable $X$, that represents one roll from a fair six-sided die.¶

# Library to plot the chart
import matplotlib.pyplot as plt

# Function to plot the chart
def plot_pmf(xs, probs, rv_name='X'):
    plt.plot(xs, probs, 'ro', ms=12, mec='b', color='r')
    plt.vlines(xs, 0, probs, colors='g', lw=4)
    plt.xlabel('$x$')
    plt.ylabel('$P(X = x)$')
    plt.ylim(0, 1)
    plt.title('Probability Mass Function of $X$');

import numpy as np

# Generate x-axis and y-axis
xk = np.arange(1, 7)
pk = (1/6, 1/6, 1/6, 1/6, 1/6, 1/6)

plot_pmf(np.arange(1, 7), np.repeat(1/6, 6))

plt.yticks(np.linspace(0, 1, 7),
           ('0', r'$\frac{1}{6}$', r'$\frac{2}{6}$', r'$\frac{3}{6}$',
            r'$\frac{4}{6}$', r'$\frac{5}{6}$', '1'));

2.11) Show that $Var(X)=E(X^2) -[E(X)]^2$.¶

\[Var(X)=E[X-E(X)]^2\]

\[=E\{X^2 -2XE(X)+[E(X)]^2\}\]

\[=E(X^2) -2E(X)E(X)+[E(X)]^2\]

\[=E(X^2) -[E(X)]^2\]

2.12) What is the Bernoulli random variable?¶

It is binary random variable, that can take only two values: 1 and 0. For example, we can simulate a biased coin with $P(Tail) = p$. Let $X = 1$ if the coin flip is tail, and $X = 0$ if the coin flip is head.

Outcome	x	P(X=x)
Tail	1	$p$
Head	0	$1-p$

The expected value of $X$ is:

\[E[X] = 1 \times p + 0 \times (1 - p) = p\]

The variance of $X$ is:

\[Var(X) = E[X^2] - E[X]^2\]

\[= [1^2 \times p + 0^2 \times (1 - p)] - p^2\]

\[= p - p^2\]

\[= p(1 - p)\]

The standard deviation of $X$ is:

\[SD(X) = \sqrt{Var(X)}\]

\[= \sqrt{p(1 - p)}\]

Exercises¶

1| Let the random variable $X$ be the number of heads in three tosses.

The outcome space is:

\[\Omega = \{ HHH, HHT, HTH, THH, HTT, THT, TTH, TTT \}\]

The probability distribution of $X$ is:

$\text{Possible value of } x$	$~~0~~$	$~~1~~$	$~~2~~$	$~~3~~$
$P(X = x)$	$1/8$	$3/8$	$3/8$	$1/8$

a) What is the chance of getting more than one head, $P(X>1)$?

b) What is the chance of getting two or less than two heads, $P(X \leq 2)$?

2| Let $Y$ be a random variable that represents a stock.

stock = ['Increases', 'Stays same','Decreases']
y = [5, 0, -3]
prob = [0.4, 0.1, 0.5]
table = {'Stock Price':stock, 'y': y, 'P(Y=y)':prob }
import pandas as pd
Y = pd.DataFrame(table)
Y

	Stock Price	y	P(Y=y)
0	Increases	5	0.4
1	Stays same	0	0.1
2	Decreases	-3	0.5

a) Calculate the $E(Y)$.

b) Calculate $Var(Y)$.

c) Calculate the $SD(Y)$.

3| Let $D$ be a random variable that represents the roll of a single fair six-sided die.

a) Calculate the expected value of $D$.

b) Calculate the standard deviation of $D$.

4| Prove that $Var(X+c)=Var(X)$, where c is a constant.

Reference¶

Adhikari, A., Pitman, J. (2020). Probability for Data Science.

Diez, D. M., Barr, C. D., Çetinkaya-Rundel, M. (2014). Introductory Statistics with Randomization and Simulation.

Lau, S., Gonzalez, J., Nolan, D. (2020). Principles and Techniques of Data Science.

Probability and Statistics with Python

2) Random Variables¶

2.1) What is a random variable?¶

2.2) How to calculate the expected value of a random variable X, denoted by E(X)?¶

2.3) How to calculate the variance of X, denoted by Var(X)?¶

2.4) How to calculate the standard deviation of X, denoted by SD(X) or \(\sigma\)?¶

2.5) Prove that \(E(cX) = cE(X)\), where c is a constant.¶

2.6) Prove that \(Var(cX)=c^2Var(X)\), where c is a constant.¶

2.7) Prove that \(E(X+c)=E(X)+c\), where c is a constant.¶

2.8) Prove that \(E(c)=c\), where c is constant.¶

2.9) Prove that \(Var(c) = 0\), where c is a constant.¶

2.10) Draw the distribution or probability mass function (PMF) of a random variable \(X\), that represents one roll from a fair six-sided die.¶

2.11) Show that \(Var(X)=E(X^2) -[E(X)]^2\).¶

2.12) What is the Bernoulli random variable?¶

Exercises¶

Reference¶