Mathematical Statistics

The Beta Distribution and Choosing a Continuous Model

Samir Orujov, PhD

ADA University, School of Business

Information Communication Technologies Agency, Statistics Unit

2026-09-24

🎯 Learning Objectives

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

• Recognise a beta density \(f(y) \propto y^{\alpha-1}(1-y)^{\beta-1}\) on \([0, 1]\) and find its constant \(B(\alpha, \beta)\)

• Read the shape of a beta density from \(\alpha\) and \(\beta\), and rescale it to any interval \([c, d]\)

• Compute \(E(Y)\), \(V(Y)\) and beta probabilities by direct integration, by the binomial link, and with pbeta

• Choose a continuous model for an economic quantity from theory, from its support, and from a histogram

🗺️ Where We Are

Wackerly §§4.7–4.8

Saturday closed the gamma family: the exponential and the chi-square as special cases, all living on \((0, \infty)\). Waiting times and claim sizes, with no ceiling.

Many quantities in finance do have a ceiling. A recovery rate, a market share, a utilisation rate, a portfolio weight: each is a proportion, trapped between 0 and 1.

Today, the model built for that: the beta (§4.7). Then §4.8 steps back and asks how we pick among the Chapter 4 densities in the first place.

❓ Motivating Question

The workout desk’s problem

A Baku bank’s workout unit has closed 250 defaulted SME loans. The share of each exposure it recovered averaged 0.40, with a standard deviation of 0.20.

For loss-given-default provisioning it needs \(P(\text{recovery} < 0.20)\).

A normal with that mean and sd gives 0.159, and also puts 2.3% probability on recovering less than nothing. A model for a proportion must live on \([0, 1]\).

📐 Definition 4.12: The Beta Density

Definition 4.12

\(Y\) has a beta distribution with parameters \(\alpha > 0\) and \(\beta > 0\) if and only if its density is \[f(y) = \frac{y^{\alpha-1}(1-y)^{\beta-1}}{B(\alpha, \beta)}, \quad 0 \le y \le 1, \qquad f(y) = 0 \text{ elsewhere,}\] where \[B(\alpha, \beta) = \int_0^1 y^{\alpha-1}(1-y)^{\beta-1}\,dy = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}\]

\(B(\alpha, \beta)\) only makes the area 1. With integers: \(B(2, 3) = 1!\,2!/4! = 1/12\), so \(f(y) = 12\,y(1-y)^2\).

🔬 Interactive: Two Parameters, Many Shapes

viewof a = {
  const input = Inputs.range([0.5, 8], {value: 2, step: 0.5, label: "alpha:"});
  ['pointerdown','touchstart','mousedown','click','wheel','pointermove','touchmove']
    .forEach(e => input.addEventListener(e, ev => ev.stopPropagation()));
  return input;
}
viewof b = {
  const input = Inputs.range([0.5, 8], {value: 3, step: 0.5, label: "beta:"});
  ['pointerdown','touchstart','mousedown','click','wheel','pointermove','touchmove']
    .forEach(e => input.addEventListener(e, ev => ev.stopPropagation()));
  return input;
}

lgam = {                 // Lanczos approximation to log Gamma(x), x > 0
  function f(x) {
  const g = 7, c = [0.99999999999980993, 676.5203681218851, -1259.1392167224028,
    771.32342877765313, -176.61502916214059, 12.507343278686905,
    -0.13857109526572012, 9.9843695780195716e-6, 1.5056327351493116e-7];
  if (x < 0.5) return Math.log(Math.PI / Math.sin(Math.PI * x)) - f(1 - x);
  x -= 1; let s = c[0];
  for (let i = 1; i < g + 2; i++) s += c[i] / (x + i);
  const t = x + g + 0.5;
  return 0.5 * Math.log(2 * Math.PI) + (x + 0.5) * Math.log(t) - t + Math.log(s);
  }
  return f;
}
logB = lgam(a) + lgam(b) - lgam(a + b)
dens = Array.from({length: 399}, (_, i) => {
  const y = (i + 1) / 400;
  return {y, f: Math.exp((a - 1) * Math.log(y) + (b - 1) * Math.log(1 - y) - logB)};
})
mu = a / (a + b)
sd = Math.sqrt(a * b / ((a + b) ** 2 * (a + b + 1)))

md`Beta(**${a}**, **${b}**): mean **${mu.toFixed(3)}**, standard deviation **${sd.toFixed(3)}**`

Plot.plot({
  width: 1150, height: 250, marginLeft: 78, marginBottom: 58,
  style: {fontSize: "18px"},
  x: {label: "y (a proportion)", domain: [0, 1]},
  y: {label: "f(y)", domain: [0, 5]},
  marks: [
    Plot.areaY(dens, {x: "y", y: "f", fill: "#8ba3c7", fillOpacity: 0.5, clip: true}),
    Plot.line(dens, {x: "y", y: "f", stroke: "#14130f", strokeWidth: 2, clip: true}),
    Plot.ruleX([mu], {stroke: "#8b2635", strokeWidth: 2, strokeDasharray: "6 4"}),
    Plot.ruleY([0])
  ]
})

🧭 Reading the Shape

• \(\alpha = \beta = 1\): \(f(y) = 1\), the uniform on \((0, 1)\) (Exercise 4.127)

• \(\alpha = \beta\): symmetric about \(1/2\); larger values pull it tighter (Exercise 4.115)

• \(\alpha < \beta\): mass nearer 0, skewed right; \(\alpha > \beta\): skewed left

• \(\alpha, \beta > 1\): one hump, zero at both ends; setting \(f^{(1)}(y) = 0\) puts the mode at \(\dfrac{\alpha-1}{\alpha+\beta-2}\)

• \(\alpha < 1\) or \(\beta < 1\): the density rises without bound at that edge

• \(1 - Y\) is beta with the parameters swapped, \((\beta, \alpha)\) (a Chapter 6 result, Exercise 4.114)

📏 Beyond Proportions: Rescaling

If \(c \le y \le d\), then \(y^* = (y - c)/(d - c)\) lies in \([0, 1]\). The beta is not restricted to proportions; it is restricted to bounded quantities.

A pension fund’s mandate keeps its equity weight \(Y\) between 30% and 70%. Model \(Y^* = (Y - 0.30)/0.40\) as Beta(2, 2), so \(F(y) = 3y^2 - 2y^3\).

\[E(Y) = 0.30 + 0.40 \times 0.5 = 0.50\] \[P(Y > 0.60) = P(Y^* > 0.75) = 1 - F(0.75) = 1 - 0.84375 = 0.156\]

📐 Theorem 4.11: Mean and Variance

Theorem 4.11

If \(Y\) is beta with parameters \(\alpha > 0\) and \(\beta > 0\), \[\mu = E(Y) = \frac{\alpha}{\alpha + \beta} \qquad \text{and} \qquad \sigma^2 = V(Y) = \frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}\]

Proof of the mean. Multiplying by \(y\) adds one to the exponent, so the integral is another beta constant: \[E(Y) = \frac{B(\alpha+1, \beta)}{B(\alpha, \beta)} = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \cdot \frac{\alpha\,\Gamma(\alpha)\Gamma(\beta)}{(\alpha+\beta)\,\Gamma(\alpha+\beta)} = \frac{\alpha}{\alpha+\beta}\] The variance is Exercise 4.130.

🏦 Worked Example: Recovery Rates

The workout desk models the recovery rate as Beta(2, 3), so \(f(y) = 12\,y(1-y)^2\) on \([0, 1]\).

\[E(Y) = \frac{2}{5} = 0.40, \qquad V(Y) = \frac{2 \times 3}{5^2 \times 6} = 0.04, \qquad \sigma = 0.20\] It matches the desk’s mean and standard deviation.

\[P(Y < 0.2) = \int_0^{0.2} 12\,(y - 2y^2 + y^3)\,dy = \Big[\,6y^2 - 8y^3 + 3y^4\,\Big]_0^{0.2}\] \[= 0.24 - 0.064 + 0.0048 = 0.1808\] The normal said 0.159. About 18% of defaulted loans recover under a fifth, and the beta never predicts a negative recovery.

🔗 The Binomial Link

The beta CDF is the incomplete beta function \(I_y(\alpha, \beta)\). When \(\alpha\) and \(\beta\) are integers, with \(n = \alpha + \beta - 1\), \[F(y) = \int_0^y \frac{t^{\alpha-1}(1-t)^{\beta-1}}{B(\alpha, \beta)}\,dt = \sum_{i=\alpha}^{n} \binom{n}{i} y^i (1-y)^{n-i}\]

The right side is \(P(W \ge \alpha)\) for \(W \sim\) binomial\((n, p = y)\). For the recoveries, \(n = 4\) and \(p = 0.2\): \[P(Y < 0.2) = P(W \ge 2) = 1 - 0.8^4 - 4(0.2)(0.8)^3 = 1 - 0.4096 - 0.4096 = 0.1808\]

The same number, with no integral. Exercise 4.134 uses exactly this.

💻 The Same Numbers in R

pbeta(0.2, shape1 = 2, shape2 = 3)          # P(Y < 0.2)

[1] 0.1808

1 - pbinom(1, size = 4, prob = 0.2)         # binomial link: P(W >= 2)

[1] 0.1808

qbeta(c(0.50, 0.95), 2, 3)                  # median and 95th percentile

[1] 0.3857276 0.7513954

rec <- rbeta(1e5, 2, 3)                     # 100,000 simulated recoveries
round(c(mean = mean(rec), sd = sd(rec), below_0.2 = mean(rec < 0.2)), 4)

     mean        sd below_0.2 
   0.4004    0.2004    0.1803 

The book prints pbeta(y0, α, 1/β), a reciprocal carried over from the gamma; for the beta the call is pbeta(y0, α, β).

⛽ Worked Example: A Pipeline

Example 4.11, moved to an export pipeline. The fraction \(Y\) of weekly capacity actually shipped is Beta(4, 2). Above 90% the operator pays a congestion surcharge. How often?

\[f(y) = \frac{\Gamma(6)}{\Gamma(4)\Gamma(2)}\,y^3(1-y) = 20\,(y^3 - y^4), \qquad 0 \le y \le 1\]

\[P(Y > 0.9) = 20\left[\frac{y^4}{4} - \frac{y^5}{5}\right]_{0.9}^{1} = 20 \times 0.004073 = 0.0815\] Roughly one week in twelve; the book rounds it to 0.08. The mean utilisation is \(4/6 = 0.667\), with \(\sigma = 0.178\).

🧠 Think-Pair-Share

A mobile operator’s share \(Y\) of new subscriptions in a region, month by month, is modelled as Beta(7, 13).

Four minutes, in pairs:

1. Find \(E(Y)\) and the standard deviation of \(Y\).

2. What is the most likely share (the mode)?

3. The rivals together hold \(1 - Y\). What is its distribution, and its mean?

✅ Think-Pair-Share: Solution

1. Theorem 4.11 with \(\alpha + \beta = 20\): \[E(Y) = \frac{7}{20} = 0.35, \qquad V(Y) = \frac{7 \times 13}{20^2 \times 21} = 0.01083, \qquad \sigma = 0.104\]

2. Both parameters exceed 1, so the mode is \(\dfrac{7 - 1}{20 - 2} = \dfrac{1}{3}\), just below the mean: the density is skewed right.

3. Swap the parameters: \(1 - Y\) is Beta(13, 7), with mean \(13/20 = 0.65 = 1 - 0.35\) and the same standard deviation, since the formula for \(V\) is symmetric in \(\alpha, \beta\).

🧭 §4.8: How Do We Choose a Model?

No density is a perfect picture of nature. A good model is one that yields good inferences about the population, not one that fits every bump.

Wackerly names three routes:

1. Theory. Poisson events in time make the gap between them exponential, and the time from the \(a\)th to the \(b\)th event gamma with \(\alpha = b - a\). The central limit theorem (Chapter 7) points to the normal.

2. A histogram (Chapter 1) of sample data, compared with candidate curves.

3. Formal tests of fit, later in the course.

🗂️ Start From the Support

Model	Support	Typical economic quantity
Uniform\((\theta_1, \theta_2)\)	\([\theta_1, \theta_2]\)	arrival time inside a settlement window
Normal\((\mu, \sigma^2)\)	\((-\infty, \infty)\)	a daily return, a forecast error
Exponential, gamma, \(\chi^2\)	\((0, \infty)\)	time between events, a claim amount
Beta\((\alpha, \beta)\)	\([0, 1]\), or \([c, d]\) rescaled	recovery rate, market share, utilisation, a weight

The support rules models out before any data are seen. Within what is left, the shape of a histogram decides.

💻 Fitting the Recoveries

set.seed(2026)
recov <- rbeta(250, 2, 3)          # stand-in for the desk's 250 closed files
m <- mean(recov); s <- sd(recov)
k <- m * (1 - m) / s^2 - 1         # Theorem 4.11 solved for alpha + beta
round(c(mean = m, sd = s, alpha = m * k, beta = (1 - m) * k), 3)

 mean    sd alpha  beta 
0.406 0.198 2.095 3.063 

round(c(normal_below_0 = pnorm(0, m, s),
        beta_below_0.2 = pbeta(0.2, m * k, (1 - m) * k),
        data_below_0.2 = mean(recov < 0.2)), 3)

normal_below_0 beta_below_0.2 data_below_0.2 
         0.020          0.168          0.160 

Matching the sample mean and sd recovers \(\alpha \approx 2.1\), \(\beta \approx 3.1\). The normal still spends 2% on impossible values.

📈 Two Candidates, One Histogram

📝 Quiz #1: The Mean

A telecom regulator models an operator’s market share as Beta(3, 7). What is its expected share?

• \(0.30\)
• \(0.43\)
• \(0.70\)
• \(0.019\)

📝 Quiz #2: The Shape

Recovery rates on unsecured consumer loans follow Beta(2, 8). Which description fits?

• Skewed right: most recoveries are small, with a tail towards 1
• Symmetric about 0.5
• Skewed left: most recoveries are close to 1
• Uniform on [0, 1]

📝 Quiz #3: Choosing a Model

A fund reports the proportion of its assets held in equities each quarter. Which model respects what that number can be?

• Beta, because the proportion lies in [0, 1]
• Normal, because averages tend to be normal
• Exponential, because it is positive
• Poisson, because it is reported at regular intervals

📋 Key Formulas

	Statement
Definition 4.12	\(f(y) = \dfrac{y^{\alpha-1}(1-y)^{\beta-1}}{B(\alpha, \beta)}\), \(\; 0 \le y \le 1\)
beta function	\(B(\alpha, \beta) = \dfrac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}\)
Theorem 4.11, mean	\(E(Y) = \alpha/(\alpha+\beta)\)
Theorem 4.11, variance	\(V(Y) = \dfrac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}\)
binomial link, \(n = \alpha+\beta-1\)	\(F(y) = \sum_{i=\alpha}^{n} \binom{n}{i} y^i(1-y)^{n-i}\)
in R	pbeta(y0, α, β), qbeta(p, α, β)

📋 Summary

• The beta is the Chapter 4 model for a bounded quantity: a proportion, or anything rescaled to \([0, 1]\)

• \(\alpha\) and \(\beta\) set the shape: symmetric, skewed either way, humped or U-shaped; \((1, 1)\) is the uniform

• \(E(Y) = \alpha/(\alpha + \beta)\); with integer parameters the CDF is a binomial tail

• A good model yields good inferences; perfect fit is not the test

• Choose by theory, then by support, then by the histogram

📚 Practice Problems

Wackerly, 7th edition

• §4.7: Exercises 4.124 – 4.134; start with 4.124, 4.125, 4.128, 4.132 and 4.134(a)

• Applet Exercises 4.114 – 4.119: use today’s slider in place of the applet

• Redo the workout desk with Beta(3, 4.5): same mean, smaller spread. What happens to \(P(Y < 0.2)\)?

Week 11, Problem Set 1 is open now and closes Sunday 29 November at 23:59 on WeBWorK, covering §§4.7–4.8.

Next class: 21 November, other expected values, Tchebysheff’s theorem for continuous variables, and Chapter 4 in review (Wackerly §§4.9–4.11).

🙏 Thank You

Dr. Samir Orujov

📧 sorujov@ada.edu.az
🏢 Building D, Room D325
🕓 Office hours: Wednesday, 16:00 – 18:00

Slides and readings: sorujov.net/teaching

❓ Questions

• Beta(2, 3) and Beta(20, 30) have the same mean. How do their standard deviations compare, and what does that say about the evidence behind each?

• Why does the beta density blow up at 0 when \(\alpha < 1\), and could a recovery-rate distribution plausibly look like that?

• A loss rate can be exactly 0 for many loans. Can a continuous beta model that point mass?