Mathematical Statistics

Method of Transformations and Moment-Generating Functions

Samir Orujov, PhD

ADA University, School of Business

Information Communication Technologies Agency, Statistics Unit

2026-02-22

🎯 Learning Objectives

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

• Apply the method of transformations (change of variables) to find pdfs of transformed random variables

• Use the Jacobian for the univariate transformation formula

• Apply the moment-generating function method to identify distributions of sums

• Recognize when the MGF method is more efficient than direct integration

• Apply these techniques to derive distributions of portfolio returns and risk measures

📱 Attendance Check-in

📋 Overview

📚 Topics Covered Today

• Method of Transformations – Direct formula for monotonic functions

• The Jacobian in 1D – Why \(|dy/du|\) appears in the formula

• MGF Method – Using moment-generating functions to identify distributions

• Sums of Random Variables – Convolutions and the MGF approach

• Case Study – Distribution of portfolio returns using MGF

📖 Why Faster Transformation Methods?

🎯 Motivation

The distribution function method from last lecture works but can be tedious. In finance, we need efficient tools:

Direct Formula Needed For:

• Modeling stock prices as exponentials of normal returns (log-normal)
• Simulating asset paths from uniform random numbers
• Pricing derivatives requiring PDF transformations
• Converting between return types (simple vs log)

MGF Method Needed For:

• Finding portfolio return distributions (sums of asset returns)
• Proving normality of diversified portfolios
• Insurance claim aggregation (sum of exponentials)
• Testing portfolio variance with chi-square distributions

📖 Recall: Distribution Function Method

🔗 From Last Lecture

The distribution function method works for any transformation:

1. Find \(F_U(u) = P(U \leq u) = P(g(Y) \leq u)\)
2. Express in terms of \(Y\)
3. Differentiate to get \(f_U(u)\)

Limitation: Can be tedious when the function is monotonic and differentiable.

Today’s Question: Can we get a direct formula for \(f_U(u)\) without going through the CDF?

Answer: Yes! The method of transformations provides exactly this shortcut.

📖 Definition: Method of Transformations

📝 Theorem 6.2: Univariate Transformation

Let \(Y\) be a continuous random variable with pdf \(f_Y(y)\) on support \((a, b)\).

Let \(U = g(Y)\) where \(g\) is strictly monotonic and differentiable.

Let \(h = g^{-1}\) be the inverse function, so \(Y = h(U)\).

Then the pdf of \(U\) is:

\[f_U(u) = f_Y(h(u)) \cdot \left| \frac{dh}{du} \right| = f_Y(h(u)) \cdot |h'(u)|\]

for \(u\) in the range of \(g\).

The term \(|h'(u)| = |dy/du|\) is called the Jacobian of the transformation.

🧮 Why Does the Jacobian Appear?

Intuition: Stretching and Compressing

Consider a small interval \([y, y + dy]\) that maps to \([u, u + du]\).

Probability is conserved: \[f_Y(y) \, dy = f_U(u) \, du\]

Solving for \(f_U(u)\): \[f_U(u) = f_Y(y) \cdot \frac{dy}{du} = f_Y(h(u)) \cdot |h'(u)|\]

The absolute value ensures the density is positive!

If \(g\) is increasing: \(h'(u) > 0\)

• Orientation preserved
• \(|h'(u)| = h'(u)\)

If \(g\) is decreasing: \(h'(u) < 0\)

• Orientation reversed
• \(|h'(u)| = -h'(u)\)

📌 Example 1: Simulating Default Times

Problem: A bank models time-to-default of a loan as \(Y \sim \text{Exponential}(\beta)\). To simulate defaults using uniform random numbers, define \(U = 1 - e^{-Y/\beta}\). Find the distribution of \(U\).

Solution:

Step 1: Identify \(g(y) = 1 - e^{-y/\beta}\). This is strictly increasing on \(y > 0\).

Step 2: Find the inverse. From \(u = 1 - e^{-y/\beta}\):

\[e^{-y/\beta} = 1 - u \implies y = -\beta \ln(1-u) = h(u)\]

Step 3: Compute the Jacobian: \[h'(u) = \frac{d}{du}[-\beta \ln(1-u)] = \frac{\beta}{1-u}\]

Step 4: Apply the formula. Since \(f_Y(y) = \frac{1}{\beta}e^{-y/\beta}\): \[f_U(u) = \frac{1}{\beta}e^{-h(u)/\beta} \cdot \frac{\beta}{1-u} = \frac{1}{\beta}(1-u) \cdot \frac{\beta}{1-u} = 1 \; \text{for} \; 0 < u < 1 \rightarrow \text{Uniform}(0,1)!\]

Finance Application: Any exponential default time can be simulated from Uniform(0,1) – the basis of Monte Carlo credit risk models.

🧮 Theorem: Probability Integral Transform

Theorem 6.3: Probability Integral Transform

Let \(Y\) be a continuous random variable with CDF \(F_Y(y)\).

Part 1: If \(U = F_Y(Y)\), then \(U \sim \text{Uniform}(0, 1)\).

Part 2: Conversely, if \(U \sim \text{Uniform}(0, 1)\), then \(Y = F_Y^{-1}(U)\) has CDF \(F_Y\).

Why This Matters:

• Simulation: Generate any distribution from uniform random numbers!

• Goodness-of-fit: Transform data to uniform for testing

• Copulas: Model dependence using uniform marginals

Finance Application: Generate correlated asset returns by transforming correlated uniforms.

📌 Example 2: Log-Normal Distribution

Problem: If \(Y \sim N(\mu, \sigma^2)\), find the distribution of \(U = e^Y\).

Solution:

Step 1: \(g(y) = e^y\) is strictly increasing. Range: \(U > 0\).

Step 2: Inverse: \(y = \ln(u) = h(u)\)

Step 3: Jacobian: \(h'(u) = \frac{1}{u}\)

Step 4: Apply formula: \[f_U(u) = \frac{1}{\sigma\sqrt{2\pi}} \exp\left[-\frac{(\ln u - \mu)^2}{2\sigma^2}\right] \cdot \frac{1}{u}\]

\[= \frac{1}{u\sigma\sqrt{2\pi}} \exp\left[-\frac{(\ln u - \mu)^2}{2\sigma^2}\right] \quad \text{for } u > 0\]

This is the Log-Normal distribution: \(U \sim \text{LogNormal}(\mu, \sigma^2)\).

📖 The Log-Normal in Finance

📝 Why Log-Normal for Stock Prices?

If log-returns are normal: \(r_t = \ln(P_t/P_{t-1}) \sim N(\mu, \sigma^2)\)

Then: \(\ln(P_t) = \ln(P_0) + \sum_{i=1}^{t} r_i\)

By CLT, \(\ln(P_t)\) is approximately normal, so \(P_t = e^{\ln(P_t)}\) is log-normal.

Properties of Log-Normal:

• Always positive (prices can’t be negative!)
• Right-skewed
• Mean: \(E[U] = e^{\mu + \sigma^2/2}\)
• Variance: \(V(U) = e^{2\mu + \sigma^2}(e^{\sigma^2} - 1)\)

Black-Scholes Model:

Stock price follows geometric Brownian motion: \[dS_t = \mu S_t \, dt + \sigma S_t \, dW_t\]

Solution: \(S_t\) is log-normally distributed!

🎮 Interactive: Log-Normal Parameters

Explore: How do \(\mu\) and \(\sigma\) affect the log-normal stock price distribution?

viewof mu_ln = {
  const input = Inputs.range([-1, 2], {
    value: 0,
    step: 0.1,
    label: "μ (log-scale mean):"
  });

  ['pointerdown', 'touchstart', 'mousedown', 'click', 'wheel', 'pointermove', 'touchmove', 'mousemove'].forEach(e =>
    input.addEventListener(e, ev => ev.stopPropagation())
  );
  return input;
}

viewof sigma_ln = {
  const input = Inputs.range([0.1, 1.5], {
    value: 0.5,
    step: 0.1,
    label: "σ (log-scale std):"
  });

  ['pointerdown', 'touchstart', 'mousedown', 'click', 'wheel', 'pointermove', 'touchmove', 'mousemove'].forEach(e =>
    input.addEventListener(e, ev => ev.stopPropagation())
  );
  return input;
}

// Log-normal statistics
ln_mean = Math.exp(mu_ln + sigma_ln * sigma_ln / 2)
ln_median = Math.exp(mu_ln)
ln_mode = Math.exp(mu_ln - sigma_ln * sigma_ln)

Mean:

Median:

Mode:

lnPdf = {
  const points = [];
  for (let x = 0.01; x <= 8; x += 0.05) {
    const logx = Math.log(x);
    const pdf = Math.exp(-(logx - mu_ln) * (logx - mu_ln) / (2 * sigma_ln * sigma_ln)) / (x * sigma_ln * Math.sqrt(2 * Math.PI));
    points.push({x: x, pdf: pdf});
  }
  return points;
}

Plot.plot({
  width: 620,
  height: 400,
  x: { domain: [0, 8], label: "U = eʸ" },
  y: { domain: [0, 2], label: "Density" },
  marks: [
    Plot.line(lnPdf, {x: "x", y: "pdf", stroke: "steelblue", strokeWidth: 2}),
    Plot.ruleX([ln_mean], {stroke: "red", strokeDasharray: "5,5"}),
    Plot.ruleX([ln_median], {stroke: "green", strokeDasharray: "3,3"}),
    Plot.ruleY([0])
  ]
})

Red dashed: Mean | Green dashed: Median

📖 Method of Moment-Generating Functions

📝 Theorem 6.4: MGF Uniqueness

If two random variables have the same moment-generating function in a neighborhood of \(t = 0\), they have the same distribution.

Method: 1. Find \(M_U(t) = E[e^{tU}]\) for the transformed variable \(U\) 2. Recognize \(M_U(t)\) as the MGF of a known distribution 3. Conclude \(U\) has that distribution

When to Use MGF Method:

• Finding distribution of sums of independent RVs
• When the MGF has a recognizable form
• Avoiding complex integration

🧮 Key MGF Property: Sums of Independent RVs

Theorem 6.5: MGF of Sums

If \(Y_1, Y_2, \ldots, Y_n\) are independent random variables, then:

\[M_{Y_1 + Y_2 + \cdots + Y_n}(t) = M_{Y_1}(t) \cdot M_{Y_2}(t) \cdots M_{Y_n}(t)\]

Proof: \[M_{\sum Y_i}(t) = E[e^{t\sum Y_i}] = E[\prod e^{tY_i}] = \prod E[e^{tY_i}] = \prod M_{Y_i}(t)\]

The last equality uses independence.

Power of This Result: Instead of complex convolution integrals, just multiply MGFs!

📌 Example 3: Two-Asset Portfolio Return

Problem: A portfolio holds two uncorrelated assets with returns \(Y_1 \sim N(\mu_1, \sigma_1^2)\) and \(Y_2 \sim N(\mu_2, \sigma_2^2)\), independent. Find the distribution of the total return \(U = Y_1 + Y_2\).

Solution:

Step 1: Recall the normal MGF: \(M_Y(t) = e^{\mu t + \sigma^2 t^2/2}\)

Step 2: Compute product of MGFs: \[M_U(t) = M_{Y_1}(t) \cdot M_{Y_2}(t) = e^{\mu_1 t + \sigma_1^2 t^2/2} \cdot e^{\mu_2 t + \sigma_2^2 t^2/2}\]

\[= e^{(\mu_1 + \mu_2)t + (\sigma_1^2 + \sigma_2^2)t^2/2}\]

Step 3: Recognize this as the MGF of \(N(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2)\)

\[\boxed{Y_1 + Y_2 \sim N(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2)}\]

Portfolio Insight: Variances add for uncorrelated assets – this is why diversification across truly independent markets reduces risk most effectively!

📌 Example 4: Total Insurance Claim Processing Time

Problem: An insurance company processes \(n\) independent claims, each taking \(Y_i \sim \text{Exponential}(\beta)\) days. Find the distribution of total processing time \(U = \sum_{i=1}^n Y_i\).

Solution:

Step 1: Exponential MGF: \(M_Y(t) = (1 - \beta t)^{-1}\) for \(t < 1/\beta\)

Step 2: Product of MGFs: \[M_U(t) = \prod_{i=1}^n M_{Y_i}(t) = [(1 - \beta t)^{-1}]^n = (1 - \beta t)^{-n}\]

Step 3: Recognize as Gamma MGF with \(\alpha = n\):

\[\boxed{U = \sum_{i=1}^n Y_i \sim \text{Gamma}(n, \beta)}\]

Insurance Insight: Total time to process \(n\) claims follows a Gamma distribution – actuaries use Gamma quantiles to plan staffing and reserves.

📌 Example 5: Testing Portfolio Variance

Problem: A risk analyst standardizes \(n\) independent asset returns to get \(Z_i \sim N(0,1)\). To test portfolio variance, they compute \(U = \sum_{i=1}^n Z_i^2\). What is the distribution of \(U\)?

Solution:

Step 1: We showed \(Z^2 \sim \chi^2(1)\), which has MGF: \(M_{Z^2}(t) = (1 - 2t)^{-1/2}\)

Step 2: Product of MGFs: \[M_U(t) = \prod_{i=1}^n (1 - 2t)^{-1/2} = (1 - 2t)^{-n/2}\]

Step 3: This is the MGF of \(\chi^2(n)\):

\[\boxed{\sum_{i=1}^n Z_i^2 \sim \chi^2(n)}\]

Risk Testing: The \(\chi^2\) distribution underlies variance tests in portfolio risk management – risk managers use it to test if realized volatility exceeds model predictions.

💰 Case Study: Portfolio Return Data

Scenario: 40% AAPL, 35% MSFT, 25% GOOG (daily log returns, 2020-2024)

library(tidyverse)
library(tidyquant)
library(knitr)

# Download data for 3 assets
symbols <- c("AAPL", "MSFT", "GOOG")
prices <- tq_get(symbols, from = "2020-01-01", to = "2024-12-31")

# Calculate log returns
returns <- prices %>%
  group_by(symbol) %>%
  mutate(log_ret = log(adjusted / lag(adjusted))) %>%
  na.omit() %>%
  select(date, symbol, log_ret) %>%
  pivot_wider(names_from = symbol, values_from = log_ret)

# Portfolio weights
w <- c(AAPL = 0.4, MSFT = 0.35, GOOG = 0.25)

# Portfolio return (weighted sum)
returns <- returns %>%
  mutate(portfolio = w["AAPL"]*AAPL + w["MSFT"]*MSFT + w["GOOG"]*GOOG)

# Summary statistics table
summary_stats <- data.frame(
  Asset = c("AAPL", "MSFT", "GOOG", "Portfolio"),
  Weight = c("40%", "35%", "25%", "—"),
  Mean = c(mean(returns$AAPL), mean(returns$MSFT),
           mean(returns$GOOG), mean(returns$portfolio)) * 252 * 100,
  StdDev = c(sd(returns$AAPL), sd(returns$MSFT),
             sd(returns$GOOG), sd(returns$portfolio)) * sqrt(252) * 100
)

kable(summary_stats, digits = 2,
      col.names = c("Asset", "Weight", "Ann. Mean (%)", "Ann. SD (%)"),
      caption = "Daily Log Return Statistics (Annualized)")

Daily Log Return Statistics (Annualized)

Asset	Weight	Ann. Mean (%)	Ann. SD (%)
AAPL	40%	24.93	31.66
MSFT	35%	20.41	30.53
GOOG	25%	20.86	32.42
Portfolio	—	22.33	28.40

💰 Case Study: Portfolio Return Distribution

Does the portfolio return look normal? MGF theory predicts yes, if returns are approximately normal and independent.

# Plot portfolio return distribution
ggplot(returns, aes(x = portfolio)) +
  geom_histogram(aes(y = after_stat(density)),
                 bins = 50, fill = "steelblue", alpha = 0.7) +
  stat_function(fun = dnorm,
                args = list(mean = mean(returns$portfolio),
                           sd = sd(returns$portfolio)),
                color = "red", linewidth = 1.2) +
  labs(title = "Portfolio Return Distribution",
       subtitle = "40% AAPL, 35% MSFT, 25% GOOG | Red = Normal fit",
       x = "Daily Log Return", y = "Density") +
  theme_minimal(base_size = 14)

💰 Case Study: Correlation & Portfolio Math

# Correlation matrix
cor_matrix <- cor(returns %>% select(AAPL, MSFT, GOOG))
kable(cor_matrix, digits = 3, caption = "Correlation Matrix")

Correlation Matrix

	AAPL	MSFT	GOOG
AAPL	1.000	0.751	0.651
MSFT	0.751	1.000	0.745
GOOG	0.651	0.745	1.000

The Math (from MGF perspective):

If returns were independent: \[V(R_p) = w_1^2\sigma_1^2 + w_2^2\sigma_2^2 + w_3^2\sigma_3^2\]

With correlation: \[V(R_p) = \sum_i w_i^2\sigma_i^2 + 2\sum_{i<j}w_i w_j \rho_{ij}\sigma_i\sigma_j\]

Since \(\rho_{ij} < 1\), portfolio variance is less than the case with perfect correlation!

💰 Case Study: Diversification Benefit

# Weighted average of individual SDs
weighted_avg_sd <- sum(w * c(sd(returns$AAPL),
                              sd(returns$MSFT),
                              sd(returns$GOOG)))

# Actual portfolio SD
portfolio_sd <- sd(returns$portfolio)

# Risk reduction table
risk_comparison <- data.frame(
  Metric = c("Weighted Avg SD", "Actual Portfolio SD", "Risk Reduction"),
  Value = c(sprintf("%.4f", weighted_avg_sd),
            sprintf("%.4f", portfolio_sd),
            sprintf("%.1f%%", 100 * (1 - portfolio_sd/weighted_avg_sd)))
)

kable(risk_comparison, caption = "Diversification Benefit")

Diversification Benefit

Metric	Value
Weighted Avg SD	0.0198
Actual Portfolio SD	0.0179
Risk Reduction	9.7%

This is the mathematical foundation of diversification!

💰 Case Study: Key Findings

📊 Analysis Results

MGF Insight:

• Portfolio return = weighted sum of individual returns

• If returns were normal and independent, portfolio would be normal

• MGF method: multiply individual MGFs

Reality Check:

• Returns are not independent (correlated)

• Correlations < 1 enable diversification

• Portfolio SD significantly less than weighted average

Practical Implications:

1. Diversification works because correlations < 1

2. Risk reduction is quantifiable using variance formulas

3. Portfolio optimization balances return vs. risk

🤝 Think-Pair-Share: Choosing the Right

💭 Activity (4 minutes)

A hedge fund analyst needs to find the distribution of a weighted portfolio return \(R_p = 0.5R_1 + 0.3R_2 + 0.2R_3\) where each \(R_i \sim N(\mu_i, \sigma_i^2)\) independently.

🧠 Think (1 min):

• Which method is most efficient: distribution function, transformation, or MGF?
• What makes this problem easy for MGFs?

👫 Pair (2 min):

• Compare your method choices
• Work out \(M_{R_p}(t)\) together
• What distribution do you recognize?

🗣️ Share (1 min):

• What is the final distribution of \(R_p\)?
• How would correlation change your approach?

📝 Quiz #1: Transformation Method

A risk analyst models a stock return \(Y\) with pdf \(f_Y(y)\). To convert to a price ratio \(U = e^Y\) (strictly increasing, inverse \(h(u) = \ln(u)\)), the pdf of \(U\) is:

• \(f_U(u) = f_Y(\ln(u)) \cdot \frac{1}{u}\)
• \(f_U(u) = f_Y(e^u) \cdot e^u\)
• \(f_U(u) = f_Y(\ln(u))\)
• \(f_U(u) = f_Y(\ln(u)) \cdot u\)

📝 Quiz #2: Stock Price Distribution

If daily log-returns of a stock follow \(Y \sim N(\mu, \sigma^2)\), and the price ratio is \(S/S_0 = e^Y\), what distribution does \(S/S_0\) follow?

• Log-Normal\((\mu, \sigma^2)\)
• Normal\((\mu, \sigma^2)\)
• Exponential\((e^\mu)\)
• Gamma\((\mu, \sigma^2)\)

📝 Quiz #3: Portfolio Return MGF

Two uncorrelated assets have return MGFs \(M_1(t)\) and \(M_2(t)\). Total portfolio return is \(R = R_1 + R_2\). What is \(M_R(t)\)?

• \(M_1(t) \cdot M_2(t)\)
• \(M_1(t) + M_2(t)\)
• \(M_1(t) / M_2(t)\)
• \(M_1(M_2(t))\)

📝 Quiz #4: Insurance Claim Aggregation

An insurer processes 5 independent claims, each with processing time Exponential\((\beta)\). Using the MGF method, the distribution of total processing time is:

• Gamma\((5, \beta)\)
• Exponential\((5\beta)\)
• Normal\((5\beta, 5\beta^2)\)
• Chi-square\((5)\)

📝 Summary

✅ Key Takeaways

• Transformation method: \(f_U(u) = f_Y(h(u)) \cdot |h'(u)|\) for monotonic \(g\) with inverse \(h\) — the Jacobian accounts for stretching/compressing

• Probability integral transform: \(F_Y(Y) \sim \text{Uniform}(0,1)\) — foundation for Monte Carlo simulation

• Log-normal distribution: \(e^Y\) where \(Y \sim N(\mu, \sigma^2)\) — models stock prices and other positive quantities

• MGF method: For sums of independent RVs, \(M_{\sum Y_i}(t) = \prod M_{Y_i}(t)\)

• Key results: Normals sum to normal, exponentials to gamma, \(\chi^2(1)\)’s to \(\chi^2(n)\) — proven elegantly via MGFs

📚 Practice Problems

📝 Homework Problems

Problem 1 (Transformation – Risk): An insurance loss is modeled as \(Y \sim \text{Exponential}(2)\). Use the transformation method to find the pdf of the cube-root loss severity \(U = Y^{1/3}\).

Problem 2 (Log-Normal – Stock Pricing): A stock price follows \(S \sim \text{LogNormal}(4, 0.09)\). Find \(E[S]\) and \(P(S > 60)\).

Problem 3 (MGF – Claim Counts): Two branch offices have independent claim counts \(Y_1 \sim \text{Poisson}(\lambda_1)\) and \(Y_2 \sim \text{Poisson}(\lambda_2)\). Use MGFs to show total claims \(Y_1 + Y_2 \sim \text{Poisson}(\lambda_1 + \lambda_2)\).

Problem 4 (Chi-Square – Variance Test): A risk analyst standardizes 10 independent daily returns to get \(Z_i \sim N(0,1)\). Find \(P(\sum Z_i^2 > 18.31)\) to test if realized variance is unexpectedly high.

Problem 5 (Portfolio): A pension fund holds two uncorrelated asset classes with returns \(N(0.08, 0.04)\) and \(N(0.05, 0.01)\). Find the distribution of the 60-40 portfolio return using the MGF method.

📱 Late Check-in

👋 Thank You!

📬 Contact Information:

Samir Orujov, PhD

Assistant Professor

School of Business

ADA University

📧 Email: sorujov@ada.edu.az

🏢 Office: D312

⏰ Office Hours: By appointment

📅 Next Class:

Topic: Multivariate Transformations and Order Statistics

Reading: Chapter 6, Sections 6.6-6.7

Preparation: Review partial derivatives and Jacobian determinants

⏰ Reminders:

✅ Complete Practice Problems 1-5

✅ Review matrix determinants

✅ Think about min/max of random samples

✅ Work hard!

❓ Questions?

💬 Open Discussion

Key Topics for Discussion:

• When is the MGF method preferable to the distribution function method?

• Why do stock prices follow log-normal rather than normal distributions?

• How does the probability integral transform enable Monte Carlo simulation?

• What happens when random variables are not independent?