Mathematical Statistics

Continuous Random Variables and Probability Distributions

Samir Orujov, PhD

ADA University, School of Business

Information Communication Technologies Agency, Statistics Unit

2025-11-30

🎯 Learning Objectives

By the end of this lecture, you will be able to:

• Define continuous random variables and distinguish them from discrete random variables in financial contexts (such as stock price movements)

• Derive and interpret distribution functions \(F(y)\) and probability density functions \(f(y)\) for continuous random variables

• Apply properties of distribution and density functions to calculate probabilities over intervals

• Compute quantiles and percentiles (including the median) for continuous distributions to support risk analysis

• Use real financial data to model continuous random variables and evaluate probabilistic outcomes

📋 Overview

📚 Topics Covered Today

• Introduction – Continuous vs discrete random variables with economic examples

• Distribution Functions – Definitions, properties, and cumulative distribution interpretation

• Probability Density Functions – Derivatives of CDFs and their properties

• Quantiles & Percentiles – Median and other measures for financial risk assessment

• Applications – Real-world financial data analysis and probability calculations

📖 Introduction to Continuous Random Variables

🔍 Motivation

In probability theory and statistics, random variables model real-world phenomena:

• Discrete random variables – Take countable values (e.g., number of defaults in a loan portfolio)

• Continuous random variables – Take any value within an interval (uncountably infinite outcomes)

• Economic example – Daily stock price movements: prices vary continuously and can fluctuate between any two real values

• Key insight – For continuous variables defined over positive real line segments, no single value can be entirely ruled out as a potential outcome

📖 Definition: Distribution Function

📝 Definition 1: Distribution Function

Let \(Y\) denote any random variable. The distribution function of \(Y\), denoted by \(F(y)\), is defined as: \[F(y) = P(Y \leq y) \quad \text{for } -\infty < y < \infty\]

• Property 1 – \(F(y)\) represents the probability that \(Y\) takes on a value less than or equal to \(y\) for all possible values of \(y\) (cumulative probability interpretation)

• Property 2 – \(F(y)\) is also called the cumulative distribution function (CDF)

• Property 3 – The CDF applies to both discrete and continuous random variables

📌 Example 1: Binomial Distribution Function

Problem: Suppose \(Y\) has a binomial distribution with \(n = 2\) and \(p = \frac{1}{2}\). Find the distribution function \(F(y)\).

Solution:

Step 1: For \(y < 0\), no binomial value exists: \[F(y) = P(Y \leq y) = 0\]

Step 2: For \(0 \leq y < 1\), only \(Y = 0\) is included: \[F(y) = P(Y = 0) = \binom{2}{0} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^{2} = \frac{1}{4}\]

Step 3: For \(1 \leq y < 2\), include \(Y = 0\) and \(Y = 1\): \[F(y) = P(Y = 0) + P(Y = 1) = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}\]

Step 4: For \(y \geq 2\), all values included: \[F(y) = P(Y = 0) + P(Y = 1) + P(Y = 2) =\] \[\frac{1}{4} + \frac{1}{2} + \frac{1}{4} = 1\]

📌 Example 1: Piecewise CDF

The distribution function for the binomial random variable \(Y\) with \(n=2\) and \(p=0.5\) is:

\[F(y) = \begin{cases} 0, & \text{if } y < 0 \\ \frac{1}{4}, & \text{if } 0 \leq y < 1 \\ \frac{3}{4}, & \text{if } 1 \leq y < 2 \\ 1, & \text{if } y \geq 2 \end{cases}\]

Key observations:

• This is a step function (discrete distribution)

• The CDF jumps at integer values (0, 1, 2)

• Between jumps, \(F(y)\) remains constant

🧮 Theorem: Properties of Distribution Functions

Theorem 1: Properties of \(F(y)\)

If \(F(y)\) is a distribution function, then it possesses the following properties:

Property 1: \(F(-\infty) \equiv \lim_{y \to -\infty} F(y) = 0\)

• As \(y\) approaches negative infinity, the probability approaches zero (no outcomes below \(-\infty\))

Property 2: \(F(\infty) \equiv \lim_{y \to \infty} F(y) = 1\)

• As \(y\) approaches positive infinity, the cumulative probability approaches one (all outcomes eventually included)

Property 3: \(F(y)\) is a non-decreasing function of \(y\)

• For any \(y_1 < y_2\), we have \(F(y_1) \leq F(y_2)\) (probabilities cannot decrease as we move right on the number line)

📖 Definition: Continuous Random Variable

📝 Definition 2: Continuous Random Variable

A random variable \(Y\) with distribution function \(F(y)\) is said to be continuous if the distribution function \(F(y)\) is continuous for \(-\infty < y < \infty\).

• Interpretation – \(Y\) is continuous if its distribution function forms a smooth, unbroken curve (no jumps or discontinuities)

• Distinction from discrete – Discrete variables have step-function CDFs; continuous variables have smooth CDFs

• Examples – Stock returns, interest rates, asset prices, portfolio values

📖 Property: Zero Point Probability

Property: Point Probability for Continuous Variables

If \(Y\) is a continuous random variable, then for any real number \(y\): \[P(Y = y) = 0\]

Key implications:

• The probability of \(Y\) taking on any exact value is zero

• Continuous random variables can assume an uncountably infinite number of values within an interval

• Probabilities are meaningful only over intervals, not at individual points

• This is why \(P(a \leq Y \leq b) = P(a < Y < b)\) for continuous \(Y\)

📖 Definition: Probability Density Function

📝 Definition 3: Probability Density Function (PDF)

Let \(F(y)\) be the distribution function for a continuous random variable \(Y\). The function \(f(y)\), given by: \[f(y) = \frac{dF(y)}{dy} = F'(y)\] wherever the derivative exists, is called the probability density function (PDF) for \(Y\).

• Property 1 – \(f(y)\) is the derivative of the CDF (rate of change of cumulative probability)

• Property 2 – Conversely, \(F(y) = \int_{-\infty}^{y} f(t) \, dt\) (CDF is the integral of PDF)

• Property 3 – \(f(y)\) itself is not a probability; it is a density (probability per unit length)

🧮 Theorem: Properties of Density Functions

Theorem 2: Properties of PDF

If \(f(y)\) is a probability density function for a continuous random variable, then:

Property 1: \(f(y) \geq 0\) for all \(y\), \(-\infty < y < \infty\)

• The density function is always non-negative (densities cannot be negative)

Property 2: \(\int_{-\infty}^{\infty} f(y) \, dy = 1\)

• The total area under the density curve equals one (total probability is 1)

Verification: To check if \(f(y)\) is a valid PDF, verify both properties above.

📌 Example 2: Finding PDF from CDF

Problem: Given the distribution function: \[F(y) = \begin{cases} 0, & \text{for } y < 0 \\ y, & \text{for } 0 \leq y \leq 1 \\ 1, & \text{for } y > 1 \end{cases}\] Find the probability density function \(f(y)\).

Solution: Take the derivative \(f(y) = \frac{dF(y)}{dy}\):

Region 1: For \(y < 0\), \(F(y) = 0\) (constant), so \(f(y) = 0\)

Region 2: For \(0 \leq y \leq 1\), \(F(y) = y\) (linear), so \(f(y) = 1\)

Region 3: For \(y > 1\), \(F(y) = 1\) (constant), so \(f(y) = 0\)

Result: This is the uniform distribution on \([0,1]\): \[f(y) = \begin{cases} 1, & 0 \leq y \leq 1 \\ 0, & \text{elsewhere} \end{cases}\]

📌 Example 3: Finding CDF from PDF

Problem: Let \(Y\) be a continuous random variable with PDF: \[f(y) = \begin{cases} 3y^2, & 0 \leq y \leq 1 \\ 0, & \text{elsewhere} \end{cases}\] Find the distribution function \(F(y)\).

Solution: Integrate \(F(y) = \int_{-\infty}^{y} f(t) \, dt\):

Region 1: For \(y < 0\): \[F(y) = \int_{-\infty}^{y} 0 \, dt = 0\]

Region 2: For \(0 \leq y \leq 1\): \[F(y) = \int_{-\infty}^{0} 0 \, dt + \int_{0}^{y} 3t^2 \, dt = \left[t^3\right]_0^y = y^3\]

Region 3: For \(y > 1\): \[F(y) = \int_{0}^{1} 3t^2 \, dt = \left[t^3\right]_0^1 = 1\]

📌 Example 3: Result

The distribution function is: \[F(y) = \begin{cases} 0, & y < 0 \\ y^3, & 0 \leq y \leq 1 \\ 1, & y > 1 \end{cases}\]

Key observations:

• The CDF is continuous (smooth curve, no jumps)

• \(F(y)\) is strictly increasing on \([0,1]\)

• The PDF \(f(y) = 3y^2\) is continuous on \((0,1)\), though it jumps to zero at the boundaries

Remark: The distribution function for a continuous random variable must be continuous, but the density function need not be everywhere continuous.

🎮 Interactive: PDF and CDF Visualization

Explore the Relationship: Adjust the parameter \(a\) to see how the PDF \(f(y) = ay^{a-1}\) on \([0,1]\) changes the CDF.

viewof a_param = Inputs.range([0.5, 5], {
  value: 3, 
  step: 0.5, 
  label: "Parameter a:"
})

md`**Current Parameter:**  
a = ${a_param.toFixed(1)}  

**PDF:** f(y) = ${a_param.toFixed(1)} y<sup>${(a_param-1).toFixed(1)}</sup>  

**CDF:** F(y) = y<sup>${a_param.toFixed(1)}</sup>`

Observations:

• When a > 1, density increases toward y=1

• When a < 1, density decreases toward y=1

• When a = 1, uniform distribution

y_values = d3.range(0, 1.01, 0.01)

pdf_data = y_values.map(y => ({
  y: y,
  value: y === 0 && a_param < 1 ? null : a_param * Math.pow(y, a_param - 1)
})).filter(d => d.value !== null)

cdf_data = y_values.map(y => ({
  y: y,
  value: Math.pow(y, a_param)
}))

Plot.plot({
  width: 800,
  height: 450,
  marginLeft: 60,
  marginBottom: 40,
  x: {
    label: "y",
    domain: [0, 1],
    grid: true
  },
  y: {
    label: "Function Value",
    domain: [0, Math.max(5, a_param * 1.2)],
    grid: true
  },
  marks: [
    Plot.line(pdf_data, {x: "y", y: "value", stroke: "steelblue", strokeWidth: 3, tip: true}),
    Plot.line(cdf_data, {x: "y", y: "value", stroke: "orange", strokeWidth: 2.5, strokeDasharray: "5,5", tip: true}),
    Plot.ruleX([0]),
    Plot.ruleY([0]),
    Plot.ruleY([1], {stroke: "red", strokeDasharray: "3,3", strokeWidth: 1})
  ],
  caption: html`<span style="color: steelblue; font-weight: bold;">━━</span> PDF f(y) | <span style="color: orange; font-weight: bold;">━ ━</span> CDF F(y) | <span style="color: red;">- -</span> Probability = 1`
})

📖 Definition: Quantiles and Percentiles

📝 Definition 4: Quantiles

Let \(Y\) denote any random variable. If \(0 < p < 1\), the p-th quantile of \(Y\), denoted by \(\phi_p\), is the smallest value such that: \[P(Y \leq \phi_p) = F(\phi_p) \geq p\]

For continuous \(Y\), \(\phi_p\) satisfies \(F(\phi_p) = p\).

• Percentile notation – Some prefer to call \(\phi_p\) the \(100p\)-th percentile of \(Y\)

• Median – If \(p = 0.5\), then \(\phi_{0.5}\) is the median of \(Y\)

• Financial interpretation – Quantiles are used in Value-at-Risk (VaR) calculations for portfolio risk assessment

📌 Example 4: Computing the Median

Problem: For the random variable \(Y\) from Example 3 with \(f(y) = 3y^2\) on \([0,1]\), find the median.

Solution: The median \(\phi_{0.5}\) satisfies \(F(\phi_{0.5}) = 0.5\).

Step 1: Recall that \(F(y) = y^3\) for \(0 \leq y \leq 1\).

Step 2: Set up the equation: \[(\phi_{0.5})^3 = 0.5\]

Step 3: Solve for \(\phi_{0.5}\): \[\phi_{0.5} = (0.5)^{1/3} = 0.7937\]

Interpretation: The median of \(Y\) is approximately 0.794. This means that \(P(Y \leq 0.794) = 0.5\), so half the probability mass lies below this value (useful for summarizing skewed distributions in finance).

🧮 Theorem: Probability Over Intervals

Theorem 3: Interval Probabilities

If the random variable \(Y\) has density function \(f(y)\) and \(a < b\), then: \[P(a \leq Y \leq b) = \int_{a}^{b} f(y) \, dy\]

Key consequences:

Consequence 1: \(P(a \leq Y \leq b) = F(b) - F(a)\)

• The probability over \([a,b]\) is the difference in CDF values

Consequence 2: For continuous \(Y\): \[P(a \leq Y \leq b) = P(a < Y < b) = P(a \leq Y < b) = P(a < Y \leq b)\]

• Whether endpoints are included or excluded does not matter (since \(P(Y = a) = P(Y = b) = 0\))

📌 Example 5: Finding Valid PDF Constants

Problem: Given \(f(y) = cy^2\) for \(0 \leq y \leq 2\), and \(f(y) = 0\) elsewhere, find the value of \(c\) for which \(f(y)\) is a valid density function.

Solution: For \(f(y)\) to be a valid PDF, it must satisfy: \[\int_{-\infty}^{\infty} f(y) \, dy = 1\]

Step 1: Compute the integral over the support: \[\int_{0}^{2} cy^2 \, dy = 1\]

Step 2: Evaluate: \[c \int_{0}^{2} y^2 \, dy = c \left[\frac{y^3}{3}\right]_0^2 = c \cdot \frac{8}{3} = 1\]

Step 3: Solve for \(c\): \[c = \frac{3}{8}\]

Result: The valid PDF is \(f(y) = \frac{3}{8}y^2\) for \(0 \leq y \leq 2\).

📌 Example 6: Computing Interval Probabilities

Problem: Given \(Y\) with density \(f(y) = \frac{3}{8}y^2\) for \(0 \leq y \leq 2\), find \(P(1 \leq Y \leq 2)\) and \(P(1 < Y < 2)\).

Solution:

Step 1: Compute the integral: \[P(1 \leq Y \leq 2) = \int_{1}^{2} \frac{3}{8}y^2 \, dy \] \[= \frac{3}{8} \int_{1}^{2} y^2 \, dy\]

Step 2: Evaluate: \[\frac{3}{8} \left[\frac{y^3}{3}\right]_1^2 = \frac{3}{8} \left(\frac{8}{3} - \frac{1}{3}\right)\] \[ = \frac{3}{8} \cdot \frac{7}{3} = \frac{7}{8}\]

Step 3: For continuous \(Y\), \(P(Y = 1) = P(Y = 2) = 0\), so: \[P(1 < Y < 2) = P(1 \leq Y \leq 2) = \frac{7}{8}\]

Interpretation: There is an 87.5% probability that \(Y\) falls between 1 and 2.

💰 Case Study: Stock Returns (Real Data)

📈 Stock Return Analysis

Context: Modeling daily log-returns of S&P 500 index as a continuous random variable.

Key Questions:

• Can we model stock returns as continuous distributions?

• What is the empirical CDF and how does it compare to theoretical distributions?

• What are key quantiles (VaR measures) for risk management?

📊 Data Source

We analyze daily S&P 500 log-returns from January 2020 to October 2024.

Source: Yahoo Finance API (via quantmod)

Period: 2020-01-01 to 2024-10-31

Data Quality: Adjusted closing prices

Verification: Cross-checked with multiple financial data providers

💰 Case Study: Data Loading and Processing

library(quantmod)
library(tidyverse)
library(lubridate)

# Load S&P 500 data
getSymbols("^GSPC", from = "2020-01-01", 
           to = "2024-10-31", auto.assign = TRUE)

[1] "GSPC"

# Compute log-returns
sp500 <- data.frame(
  Date = index(GSPC),
  Close = as.numeric(GSPC$GSPC.Adjusted)
) %>%
  mutate(LogReturn = c(NA, diff(log(Close))))

# Remove NA
sp500_clean <- sp500 %>% filter(!is.na(LogReturn))

cat("Number of observations:", nrow(sp500_clean), "
")

Number of observations: 1215 

cat("Date range:", min(sp500_clean$Date), 
    "to", max(sp500_clean$Date), "
")

Date range: 18264 to 20026 

# Summary statistics
summary_stats <- sp500_clean %>%
  summarise(
    Mean = mean(LogReturn),
    StdDev = sd(LogReturn),
    Median = median(LogReturn),
    Min = min(LogReturn),
    Max = max(LogReturn),
    Q5 = quantile(LogReturn, 0.05),
    Q95 = quantile(LogReturn, 0.95)
  )

knitr::kable(summary_stats, digits = 5, 
             caption = "Summary Statistics of S&P 500 Log-Returns")

Summary Statistics of S&P 500 Log-Returns

Mean	StdDev	Median	Min	Max	Q5	Q95
0.00048	0.01363	0.00089	-0.12765	0.08968	-0.01911	0.01828

Key insights:

• Mean return is slightly positive

• Standard deviation shows volatility

• 5th percentile (VaR) indicates risk

💰 Case Study: Empirical CDF and Density

# Empirical CDF
ggplot(sp500_clean, aes(x = LogReturn)) +
  stat_ecdf(geom = "step", color = "steelblue", 
            size = 1.2) +
  labs(title = "Empirical CDF of S&P 500 Log-Returns",
       subtitle = "January 2020 - October 2024",
       x = "Log-Return",
       y = "Cumulative Probability F(y)") +
  theme_minimal(base_size = 13) +
  geom_hline(yintercept = c(0.05, 0.5, 0.95), 
             linetype = "dashed", color = "red", alpha = 0.5)

# Empirical PDF (histogram + density)
ggplot(sp500_clean, aes(x = LogReturn)) +
  geom_histogram(aes(y = ..density..), 
                 bins = 50, fill = "lightblue", 
                 color = "black", alpha = 0.6) +
  geom_density(color = "darkblue", size = 1.2) +
  labs(title = "Empirical PDF of S&P 500 Log-Returns",
       subtitle = "Histogram and Kernel Density Estimate",
       x = "Log-Return",
       y = "Density f(y)") +
  theme_minimal(base_size = 13)

💰 Case Study: Quantile Analysis

# Compute key quantiles
quantiles <- sp500_clean %>%
  summarise(
    `1% (VaR 99%)` = quantile(LogReturn, 0.01),
    `5% (VaR 95%)` = quantile(LogReturn, 0.05),
    `25% (Q1)` = quantile(LogReturn, 0.25),
    `50% (Median)` = quantile(LogReturn, 0.50),
    `75% (Q3)` = quantile(LogReturn, 0.75),
    `95%` = quantile(LogReturn, 0.95),
    `99%` = quantile(LogReturn, 0.99)
  )

knitr::kable(t(quantiles), digits = 5, 
             col.names = c("Log-Return Value"),
             caption = "Key Quantiles of S&P 500 Log-Returns")

Key Quantiles of S&P 500 Log-Returns

	Log-Return Value
1% (VaR 99%)	-0.03687
5% (VaR 95%)	-0.01911
25% (Q1)	-0.00539
50% (Median)	0.00089
75% (Q3)	0.00735
95%	0.01828
99%	0.03043

Financial interpretation:

• VaR 95% (5th percentile): There is a 5% chance daily losses exceed this value (risk measure for portfolio management)

• Median: The middle value of the distribution (50% probability above/below)

• VaR 99% (1st percentile): Extreme loss threshold for stress testing

📝 Quiz #1: Distribution Functions

What is the value of the distribution function \(F(y)\) as \(y\) approaches positive infinity?

• 1
• 0
• 0.5
• Undefined

📝 Quiz #2: Continuous Random Variables

For a continuous random variable Y, what is the probability that Y equals exactly 5?

• 0
• 0.5
• Depends on the distribution
• Cannot be determined

📝 Quiz #3: PDF Properties

Which of the following must be true for a valid probability density function f(y)?

• The integral of f(y) over all y equals 1
• f(y) can be negative for some values of y
• f(y) must be continuous everywhere
• The maximum value of f(y) cannot exceed 1

📝 Summary

✅ Key Takeaways

• Continuous random variables can take any value in an interval; individual point probabilities are zero

• Distribution function \(F(y) = P(Y \leq y)\) is cumulative and non-decreasing, with limits 0 and 1 at negative and positive infinity

• Probability density function \(f(y) = F'(y)\) satisfies \(f(y) \geq 0\) and \(\int f(y) dy = 1\); probabilities are found by integrating \(f(y)\) over intervals

• Quantiles and percentiles (including the median) provide important summary measures for risk assessment in financial applications

• Real financial data (such as S&P 500 returns) can be modeled using continuous distributions to compute risk measures like Value-at-Risk

📚 Practice Problems

📝 Homework Problems

Problem 1: Given \(f(y) = 2y\) for \(0 \leq y \leq 1\) and zero elsewhere, verify this is a valid PDF and find \(F(y)\). Then compute \(P(0.25 \leq Y \leq 0.75)\).

Problem 2: For the PDF \(f(y) = ke^{-y}\) for \(y > 0\) and zero elsewhere, find the constant \(k\) that makes this a valid density, and determine the median.

Problem 3: A stock’s daily log-return is modeled by \(f(y) = \frac{1}{2}\) for \(-0.02 \leq y \leq 0.02\). Find the probability that the return is between -0.01 and 0.01, and find the 5th percentile (VaR 95%).

Problem 4: Using the real S&P 500 data from the case study, compute the interquartile range (IQR) and interpret it in the context of financial risk.

👋 Thank You!

📬 Contact Information:

Samir Orujov

Assistant Professor

School of Business

ADA University

📧 Email: sorujov@ada.edu.az

🏢 Office: D312

⏰ Office Hours: By appointment

📅 Next Class:

Topic: Expected Value and Variance of Continuous Random Variables

Reading: Review properties of integrals and moments

Preparation: Refresh calculus (integration by parts)

⏰ Reminders:

✅ Complete practice problems

✅ Review quantile computations

✅ Work hard and stay curious!

❓ Questions?

💬 Open Discussion (5 minutes)

Discussion Point 1: How do continuous distributions differ from discrete distributions in practical financial modeling?

Discussion Point 2: Why is the median sometimes preferred over the mean for skewed distributions (e.g., stock returns during crises)?

Discussion Point 3: How can quantiles be used to set risk limits in portfolio management?

Discussion Point 4: What are the limitations of using historical data to estimate distribution functions?