1.x — Chapter 1 summary and quiz

📅 January 15, 2025

1.x — Chapter 1 summary and quiz#

Chapter summary#

In this chapter, we covered the fundamental concepts of basic probability theory:

Key concepts#

Sample Space ( $Ω$ ): The set of all possible outcomes of a random experiment - Sample Space
Event: A subset of the sample space - Event (Probability Theory)
Probability: A number between 0 and 1 quantifying how likely an event is to occur - Probability
Independence: Events that don’t affect each other’s probabilities - Independence (Probability Theory)
Dependence: Events that influence each other’s probabilities

What we learned#

Sample Spaces and Events (1.1):
- Discrete vs. continuous sample spaces
- Simple events (single outcome) vs. compound events (multiple outcomes)
- Set operations: union ( $\cup$ ), intersection ( $\cap$ ), complement ( $^{c}$ )
- Mutually exclusive events (cannot occur simultaneously)
Probability Axioms and Rules (1.2):
- Axiom 1: Non-negativity: $P (A) \geq 0$
- Axiom 2: Normalization: $P (Ω) = 1$
- Axiom 3: Additivity: For mutually exclusive events, $P (A_{1} \cup A_{2} \cup ...) = P (A_{1}) + P (A_{2}) + ...$
- Complement rule: $P (A^{c}) = 1 - P (A)$
- Addition rule: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$
- Equally likely outcomes: $P (A) = \frac{∣ A ∣}{∣Ω∣}$
Independence and Dependence (1.3):
- Events $A$ and $B$ are independent if $P (A \cap B) = P (A) \cdot P (B)$
- Equivalently, $P (B ∣ A) = P (B)$ for independent events
- Conditional probability: $P (B ∣ A) = \frac{P ( A \cap B )}{P ( A )}$
- In finance, events are often dependent (correlated)

Applications in quantitative finance#

Portfolio Analysis: Sample spaces for returns, events for risk scenarios
Risk Management: Probability of losses, joint risk events
Trading Signals: Independent vs. dependent signals
Market Conditions: Dependence between market variables

Important formulas#

Concept	Formula
Complement	$P (A^{c}) = 1 - P (A)$
Union (general)	$P (A \cup B) = P (A) + P (B) - P (A \cap B)$
Union (mutually exclusive)	$P (A \cup B) = P (A) + P (B)$
Independence	$P (A \cap B) = P (A) \cdot P (B)$
Conditional Probability	$P (B ∣ A) = \frac{P ( A \cap B )}{P ( A )}$
Equally Likely	$P(A) = \frac{

Next steps#

In Chapter 2, we’ll explore conditional probability in more depth, including how to update probabilities when new information becomes available. This leads to Bayes’ Theorem, which is fundamental for decision-making under uncertainty in quantitative finance.

Quiz time#

Question #1

A fair die is rolled. Define the following events:

$A$ = “Roll an even number”
$B$ = “Roll a number greater than 4”
$C$ = “Roll a 6”

What are the sets for events $A$ , $B$ , and $C$ ? What is $A \cap B$ ?

Show Solution

$A = {2, 4, 6}$ (even numbers)
$B = {5, 6}$ (numbers greater than 4)
$C = {6}$ (simple event)

$A \cap B = {6}$ (numbers that are both even AND greater than 4)

Question #2

Using the events from Question #1, calculate:

$P (A)$
$P (B)$
$P (A \cup B)$
$P (A \cap B)$

Show Solution

Since all outcomes are equally likely:

$P (A) = \frac{3}{6} = \frac{1}{2}$ (3 even numbers out of 6)
$P (B) = \frac{2}{6} = \frac{1}{3}$ (2 numbers > 4 out of 6)
$P (A \cap B) = \frac{1}{6}$ (only 6 is in both)

Using the addition rule: $P (A \cup B) = P (A) + P (B) - P (A \cap B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6} = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$

Alternatively, we can count: $A \cup B = {2, 4, 5, 6}$ , so $P (A \cup B) = \frac{4}{6} = \frac{2}{3}$ .

Question #3

Two fair coins are tossed. Let:

$A$ = “First coin is heads”
$B$ = “Second coin is heads”

Are events $A$ and $B$ independent? Justify your answer.

Show Solution

Yes, $A$ and $B$ are independent.

Method 1: Check definition

$P (A) = \frac{1}{2}$
$P (B) = \frac{1}{2}$
$P (A \cap B) = P (both heads) = \frac{1}{4}$

Since $P (A \cap B) = \frac{1}{4} = \frac{1}{2} \cdot \frac{1}{2} = P (A) \cdot P (B)$ , the events are independent.

Method 2: Check conditional probability

$P (B ∣ A) = \frac{P ( A \cap B )}{P ( A )} = \frac{1/4}{1/2} = \frac{1}{2} = P (B)$

Since $P (B ∣ A) = P (B)$ , the events are independent.

Intuition: The outcome of the first coin toss doesn’t affect the second coin toss.

Question #4

A portfolio has three possible outcomes:

High return: probability 0.4
Medium return: probability 0.35
Low return: probability 0.25

What is the probability of NOT getting a low return? What is the probability of getting either a high or medium return?

Show Solution

Let $L$ = “Low return”, so $P (L) = 0.25$ .

Using the complement rule: $P (L^{c}) = 1 - P (L) = 1 - 0.25 = 0.75$

The probability of NOT getting a low return is 0.75.

For the second part, let $H$ = “High return” and $M$ = “Medium return”. Since these are mutually exclusive events: $P (H \cup M) = P (H) + P (M) = 0.4 + 0.35 = 0.75$

Note that $P (H \cup M) = P (L^{c})$ , which makes sense since high and medium are the complement of low.

Question #5

Two stocks have the following probabilities:

Stock 1 goes up: $P (A) = 0.6$
Stock 2 goes up: $P (B) = 0.5$
Both go up: $P (A \cap B) = 0.4$

Are the stock movements independent? What is the probability that at least one stock goes up?

Show Solution

Check independence:

If independent: $P (A \cap B) = P (A) \cdot P (B) = 0.6 \cdot 0.5 = 0.3$

Actual: $P (A \cap B) = 0.4 \neq = 0.3$

Since $0.4 \neq = 0.3$ , the events are dependent (correlated). The stocks tend to move together more than if they were independent.

Probability of at least one going up:

Using the addition rule: $P (A \cup B) = P (A) + P (B) - P (A \cap B) = 0.6 + 0.5 - 0.4 = 0.7$

The probability that at least one stock goes up is 0.7.

Question #6

A trading strategy has a 60% win rate. If we assume each trade is independent, what is the probability of:

Winning the first trade?
Losing the first trade?
Winning both the first and second trades?

Show Solution

Let $W$ = “Win a trade”, so $P (W) = 0.6$ and $P (W^{c}) = 1 - 0.6 = 0.4$ .

First trade win: $P (W) = 0.6$
First trade loss: $P (W^{c}) = 0.4$

Since trades are independent:

Both trades win: $P (W_{1} \cap W_{2}) = P (W_{1}) \cdot P (W_{2}) = 0.6 \cdot 0.6 = 0.36$

The probability of winning both trades is 0.36 (36%).

Question #7

Write Python code to:

Check if two events are independent given their probabilities
Calculate the probability of the union of two events
Calculate the probability of the complement of an event

Show Solution

def are_independent(p_a, p_b, p_intersection, tolerance=1e-6):
    """Check if events A and B are independent"""
    expected = p_a * p_b
    return abs(p_intersection - expected) < tolerance

def probability_union(p_a, p_b, p_intersection):
    """Calculate P(A ∪ B) = P(A) + P(B) - P(A ∩ B)"""
    return p_a + p_b - p_intersection

def probability_complement(p_a):
    """Calculate P(A^c) = 1 - P(A)"""
    return 1 - p_a

# Example usage
p_a = 0.6
p_b = 0.5
p_intersection = 0.4

print(f"Are independent? {are_independent(p_a, p_b, p_intersection)}")
print(f"P(A ∪ B) = {probability_union(p_a, p_b, p_intersection):.2f}")
print(f"P(A^c) = {probability_complement(p_a):.2f}")

Output:

Are independent? False
P(A ∪ B) = 0.70
P(A^c) = 0.40