2.x — Chapter 2 summary and quiz

📅 January 15, 2025

2.x — Chapter 2 summary and quiz#

Chapter summary#

In this chapter, we explored conditional probability, one of the most important concepts in probability theory and quantitative finance. Conditional probability allows us to update our beliefs about events when we receive new information.

Key concepts#

Conditional Probability: The probability of an event given that another event has occurred - Conditional Probability
Dependent Events: Events that influence each other’s probabilities
Multiplication Rule: Formula for calculating joint probabilities using conditional probabilities
Chain Rule: Extension to multiple events
Conditional Outcomes: Outcomes whose probabilities depend on other events

What we learned#

Conditional Probability and Dependent Events (2.1):
- Definition: $P (B ∣ A) = \frac{P ( A \cap B )}{P ( A )}$ (provided $P (A) > 0$ )
- Multiplication rule: $P (A \cap B) = P (A) \cdot P (B ∣ A)$
- Relationship to independence: $P (B ∣ A) = P (B)$ if and only if $A$ and $B$ are independent
- Chain rule for multiple events: $P (A_{1} \cap ... \cap A_{n}) = P (A_{1}) \cdot P (A_{2} ∣ A_{1}) \cdot ... \cdot P (A_{n} ∣ A_{1} \cap ... \cap A_{n - 1})$
Conditional Outcomes and Applications (2.2):
- Conditional outcomes in complex scenarios
- Probability trees for visualizing conditional scenarios
- Law of Total Probability (introduction): $P (B) = \sum_{i} P (B ∣ A_{i}) \cdot P (A_{i})$
- Conditional expectations: $E [X] = \sum_{i} E [X ∣ A_{i}] \cdot P (A_{i})$
- Applications to risk assessment, portfolio management, and trading strategies

Applications in quantitative finance#

Risk Assessment: Adjusting default probabilities based on economic conditions
Portfolio Management: Calculating portfolio returns conditional on market regimes
Trading Strategies: Evaluating signal accuracy under different market conditions
Credit Analysis: Understanding default probabilities given borrower characteristics
Market Analysis: Modeling stock movements conditional on market conditions

Important formulas#

Concept	Formula
Conditional Probability	$P (B ∣ A) = \frac{P ( A \cap B )}{P ( A )}$
Multiplication Rule	$P (A \cap B) = P (A) \cdot P (B ∣ A)$
Chain Rule (2 events)	$P (A \cap B) = P (A) \cdot P (B ∣ A)$
Chain Rule (n events)	$P (A_{1} \cap ... \cap A_{n}) = P (A_{1}) \cdot P (A_{2} ∣ A_{1}) \cdot ... \cdot P (A_{n} ∣ A_{1} \cap ... \cap A_{n - 1})$
Independence (conditional)	$P (B ∣ A) = P (B)$ if and only if independent
Law of Total Probability	$P (B) = \sum_{i} P (B ∣ A_{i}) \cdot P (A_{i})$ (for mutually exclusive, exhaustive $A_{i}$ )
Conditional Expectation	$E [X] = \sum_{i} E [X ∣ A_{i}] \cdot P (A_{i})$

Next steps#

In Chapter 3, we’ll formally learn Bayes’ Theorem and the Law of Total Probability, which provide powerful tools for:

Updating probabilities when new information arrives
Reversing conditional probabilities
Making decisions under uncertainty
Understanding how evidence affects our beliefs

These concepts are fundamental for quantitative finance applications like risk modeling, signal processing, and decision-making under uncertainty.

Quiz time#

Question #1

A fair die is rolled. Let:

$A$ = “Roll an even number”
$B$ = “Roll a number greater than 3”

Calculate:

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

Show Solution

Sample space: $Ω = {1, 2, 3, 4, 5, 6}$

$A = {2, 4, 6}$ (even numbers)
$B = {4, 5, 6}$ (numbers > 3)
$A \cap B = {4, 6}$ (even numbers > 3)

Since all outcomes are equally likely:

$P (A) = \frac{3}{6} = \frac{1}{2}$
$P (B) = \frac{3}{6} = \frac{1}{2}$
$P (A \cap B) = \frac{2}{6} = \frac{1}{3}$

Using conditional probability formula:

$P (B ∣ A) = \frac{P ( A \cap B )}{P ( A )} = \frac{1/3}{1/2} = \frac{2}{3}$

$P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )} = \frac{1/3}{1/2} = \frac{2}{3}$

Interpretation:

Given an even number, there’s a $\frac{2}{3}$ chance it’s greater than 3
Given a number greater than 3, there’s a $\frac{2}{3}$ chance it’s even

Question #2

Two cards are drawn from a standard deck without replacement. Calculate:

$P (First card is an ace)$
$P (Second card is an ace ∣ First card is an ace)$
$P (Both cards are aces)$

Show Solution

$P (First card is an ace) = \frac{4}{52} = \frac{1}{13}$
After drawing one ace, 3 aces remain out of 51 cards: $P (Second card is an ace ∣ First card is an ace) = \frac{3}{51} = \frac{1}{17}$
Using multiplication rule: $P (Both cards are aces) = P (First ace) \cdot P (Second ace ∣ First ace) = \frac{4}{52} \cdot \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}$

Question #3

A stock’s movement depends on market conditions:

Bull market (probability 0.6): Stock goes up with probability 0.8
Bear market (probability 0.4): Stock goes up with probability 0.2

Calculate:

$P (Stock goes up)$ (unconditional probability)
$P (Bull market ∣ Stock goes up)$

Show Solution

Let:

$B$ = “Bull market”: $P (B) = 0.6$
$B e a r$ = “Bear market”: $P (B e a r) = 0.4$
$U$ = “Stock goes up”

Part 1: Unconditional probability using Law of Total Probability:

$P (U) = P (U ∣ B) \cdot P (B) + P (U ∣ B e a r) \cdot P (B e a r)$

$P (U) = 0.8 \cdot 0.6 + 0.2 \cdot 0.4 = 0.48 + 0.08 = 0.56$

Part 2: Reversing the conditional probability (using the formula we’ll learn as Bayes’ Theorem in Chapter 3):

$P (B ∣ U) = \frac{P ( U ∣ B ) \cdot P ( B )}{P ( U )} = \frac{0.8 \cdot 0.6}{0.56} = \frac{0.48}{0.56} = \frac{6}{7} \approx 0.857$

Interpretation:

The unconditional probability of the stock going up is 56%
If the stock goes up, there’s an 85.7% chance we’re in a bull market

Question #4

A trading strategy has two signals:

Signal 1 is correct with probability 0.7
Signal 2 is correct with probability 0.6
If Signal 1 is correct, Signal 2 is correct with probability 0.8
If Signal 1 is incorrect, Signal 2 is correct with probability 0.3

Calculate:

$P (Both signals correct)$
$P (At least one signal correct)$
$P (Signal 1 correct ∣ Signal 2 correct)$

Show Solution

Let:

$S_{1}$ = “Signal 1 correct”: $P (S_{1}) = 0.7$
$S_{2}$ = “Signal 2 correct”: $P (S_{2} ∣ S_{1}) = 0.8$ , $P (S_{2} ∣ S_{1}^{c}) = 0.3$

Part 1: Both signals correct

$P (S_{1} \cap S_{2}) = P (S_{1}) \cdot P (S_{2} ∣ S_{1}) = 0.7 \cdot 0.8 = 0.56$

Part 2: At least one signal correct

First, calculate $P (S_{2})$ using Law of Total Probability:

$P (S_{2}) = P (S_{2} ∣ S_{1}) \cdot P (S_{1}) + P (S_{2} ∣ S_{1}^{c}) \cdot P (S_{1}^{c})$

$P (S_{2}) = 0.8 \cdot 0.7 + 0.3 \cdot 0.3 = 0.56 + 0.09 = 0.65$

Now use addition rule:

$P (S_{1} \cup S_{2}) = P (S_{1}) + P (S_{2}) - P (S_{1} \cap S_{2}) = 0.7 + 0.65 - 0.56 = 0.79$

Part 3: Reverse conditional probability

$P (S_{1} ∣ S_{2}) = \frac{P ( S _{1} \cap S _{2} )}{P ( S _{2} )} = \frac{0.56}{0.65} = \frac{56}{65} \approx 0.862$

Interpretation:

Both signals are correct 56% of the time
At least one signal is correct 79% of the time
If Signal 2 is correct, Signal 1 is also correct 86.2% of the time

Question #5

A portfolio contains stocks from two sectors:

Technology (60% of portfolio): Expected return 12%
Finance (40% of portfolio): Expected return 8%

Calculate the unconditional expected return of the portfolio.

Show Solution

Let:

$T$ = “Technology sector”: $P (T) = 0.6$
$F$ = “Finance sector”: $P (F) = 0.4$
$E [R ∣ T] = 0.12$
$E [R ∣ F] = 0.08$

Using Law of Total Expectation:

$E [R] = E [R ∣ T] \cdot P (T) + E [R ∣ F] \cdot P (F)$

$E [R] = 0.12 \cdot 0.6 + 0.08 \cdot 0.4 = 0.072 + 0.032 = 0.104$

Interpretation: The unconditional expected return is 10.4%, which is a weighted average of the sector-specific expected returns.

Question #6

Three cards are drawn from a standard deck without replacement. Calculate the probability that all three are aces.

Show Solution

Using the chain rule:

$P (All three aces) = P (First ace) \cdot P (Second ace ∣ First ace) \cdot P (Third ace ∣ First two aces)$

$P (All three aces) = \frac{4}{52} \cdot \frac{3}{51} \cdot \frac{2}{50} = \frac{24}{132600} = \frac{1}{5525}$

Interpretation: The probability of drawing three aces in a row (without replacement) is approximately 0.018% (1 in 5,525).

Question #7

Write Python code to:

Calculate conditional probability $P (B ∣ A)$
Calculate joint probability using multiplication rule
Check if events are independent using conditional probability
Calculate unconditional probability using Law of Total Probability

Show Solution

def conditional_probability(p_intersection, p_given):
    """Calculate P(B | A) = P(A ∩ B) / P(A)"""
    if p_given == 0:
        return None
    return p_intersection / p_given

def joint_probability(p_a, p_b_given_a):
    """Calculate P(A ∩ B) = P(A) * P(B | A)"""
    return p_a * p_b_given_a

def are_independent_conditional(p_a, p_b, p_b_given_a, tolerance=1e-6):
    """Check independence using conditional probability: P(B | A) = P(B)"""
    if p_a == 0:
        return None
    return abs(p_b_given_a - p_b) < tolerance

def law_of_total_probability(conditional_probs, prior_probs):
    """Calculate P(B) = sum of P(B | A_i) * P(A_i)"""
    if len(conditional_probs) != len(prior_probs):
        raise ValueError("Lists must have same length")
    return sum(c * p for c, p in zip(conditional_probs, prior_probs))

# Example 1: Conditional probability
p_intersection = 0.3
p_given = 0.6
p_conditional = conditional_probability(p_intersection, p_given)
print(f"P(B | A) = {p_conditional:.2f}")

# Example 2: Joint probability
p_a = 0.6
p_b_given_a = 0.5
p_joint = joint_probability(p_a, p_b_given_a)
print(f"P(A ∩ B) = {p_joint:.2f}")

# Example 3: Check independence
p_b = 0.5
is_independent = are_independent_conditional(p_a, p_b, p_b_given_a)
print(f"Are independent? {is_independent}")

# Example 4: Law of Total Probability
conditional_probs = [0.8, 0.2]  # P(up | bull), P(up | bear)
prior_probs = [0.6, 0.4]  # P(bull), P(bear)
p_unconditional = law_of_total_probability(conditional_probs, prior_probs)
print(f"P(up) = {p_unconditional:.2f}")

Output:

P(B | A) = 0.50
P(A ∩ B) = 0.30
Are independent? True
P(up) = 0.56