```{r} #| label: setup #| include: false set.seed(2026) library(ggplot2) theme_set(theme_minimal(base_size = 18)) ``` ## π― Learning Objectives ::: {style="font-size: 32px"} 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 ::: --- ## πΊοΈ Where We Are ::: {style="font-size: 30px"} **Wackerly §§4.7β4.8** 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. ::: {.fragment} 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. ::: ::: {.fragment} Today, the model built for that: the **beta** (Β§4.7). Then Β§4.8 steps back and asks how we pick among the Chapter 4 densities in the first place. ::: ::: --- ## β Motivating Question ::: {.callout-important} ## 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 $P(\text{recovery} < 0.20)$. ::: ::: {.fragment style="font-size: 30px"} 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 $[0, 1]$. ::: --- ## π Definition 4.12: The Beta Density ::: {style="font-size: 30px"} ::: {.callout-note} ## 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 $$B(\alpha, \beta) = \int_0^1 y^{\alpha-1}(1-y)^{\beta-1}\,dy = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$$ ::: ::: {.fragment} $B(\alpha, \beta)$ only makes the area 1. With integers: $B(2, 3) = 1!\,2!/4! = 1/12$, so $f(y) = 12\,y(1-y)^2$. ::: ::: --- ## π¬ Interactive: Two Parameters, Many Shapes {.smaller} ```{ojs} //| echo: false 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; } ``` ```{ojs} //| echo: false 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]) ] }) ``` --- ## π§ Reading the Shape ::: {style="font-size: 29px"} - $\alpha = \beta = 1$: $f(y) = 1$, 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) ::: --- ## π Beyond Proportions: Rescaling ::: {style="font-size: 30px"} 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. ::: {.fragment} 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 $F(y) = 3y^2 - 2y^3$. ::: ::: {.fragment} $$E(Y) = 0.30 + 0.40 \times 0.5 = 0.50$$ $$P(Y > 0.60) = P(Y^* > 0.75) = 1 - F(0.75) = 1 - 0.84375 = 0.156$$ ::: ::: --- ## π Theorem 4.11: Mean and Variance ::: {style="font-size: 30px"} ::: {.callout-important} ## Theorem 4.11 If $Y$ is beta with parameters $\alpha > 0$ and $\beta > 0$, $$\mu = E(Y) = \frac{\alpha}{\alpha + \beta} \qquad \text{and} \qquad \sigma^2 = V(Y) = \frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}$$ ::: ::: {.fragment} **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. ::: ::: --- ## π¦ Worked Example: Recovery Rates ::: {style="font-size: 30px"} The workout desk models the recovery rate as Beta(2, 3), so $f(y) = 12\,y(1-y)^2$ on $[0, 1]$. ::: {.fragment} $$E(Y) = \frac{2}{5} = 0.40, \qquad V(Y) = \frac{2 \times 3}{5^2 \times 6} = 0.04, \qquad \sigma = 0.20$$ It matches the desk's mean and standard deviation. ::: ::: {.fragment} $$P(Y < 0.2) = \int_0^{0.2} 12\,(y - 2y^2 + y^3)\,dy = \Big[\,6y^2 - 8y^3 + 3y^4\,\Big]_0^{0.2}$$ $$= 0.24 - 0.064 + 0.0048 = 0.1808$$ The normal said 0.159. **About 18% of defaulted loans recover under a fifth**, and the beta never predicts a negative recovery. ::: ::: --- ## π The Binomial Link ::: {style="font-size: 30px"} The beta CDF is the **incomplete beta function** $I_y(\alpha, \beta)$. When $\alpha$ and $\beta$ are integers, with $n = \alpha + \beta - 1$, $$F(y) = \int_0^y \frac{t^{\alpha-1}(1-t)^{\beta-1}}{B(\alpha, \beta)}\,dt = \sum_{i=\alpha}^{n} \binom{n}{i} y^i (1-y)^{n-i}$$ ::: {.fragment} The right side is $P(W \ge \alpha)$ for $W \sim$ binomial$(n, p = y)$. For the recoveries, $n = 4$ and $p = 0.2$: $$P(Y < 0.2) = P(W \ge 2) = 1 - 0.8^4 - 4(0.2)(0.8)^3 = 1 - 0.4096 - 0.4096 = 0.1808$$ ::: ::: {.fragment} The same number, with no integral. Exercise 4.134 uses exactly this. ::: ::: --- ## π» The Same Numbers in R ```{r} #| label: beta-r #| code-fold: false pbeta(0.2, shape1 = 2, shape2 = 3) # P(Y < 0.2) 1 - pbinom(1, size = 4, prob = 0.2) # binomial link: P(W >= 2) qbeta(c(0.50, 0.95), 2, 3) # median and 95th percentile 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) ``` ::: {style="font-size: 28px"} The book prints `pbeta(y0, Ξ±, 1/Ξ²)`, a reciprocal carried over from the gamma; for the beta the call is `pbeta(y0, Ξ±, Ξ²)`. ::: --- ## β½ Worked Example: A Pipeline ::: {style="font-size: 30px"} 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? ::: {.fragment} $$f(y) = \frac{\Gamma(6)}{\Gamma(4)\Gamma(2)}\,y^3(1-y) = 20\,(y^3 - y^4), \qquad 0 \le y \le 1$$ ::: ::: {.fragment} $$P(Y > 0.9) = 20\left[\frac{y^4}{4} - \frac{y^5}{5}\right]_{0.9}^{1} = 20 \times 0.004073 = 0.0815$$ Roughly **one week in twelve**; the book rounds it to 0.08. The mean utilisation is $4/6 = 0.667$, with $\sigma = 0.178$. ::: ::: --- ## π§ Think-Pair-Share ```{r} #| label: tps-timer #| echo: false # The timer is the only thing in this deck that needs a package beyond base R. # Guarded so a machine without it renders the deck anyway, with a static # figure in the same corner, rather than halting the whole build. if (requireNamespace("countdown", quietly = TRUE)) { countdown::countdown(minutes = 4, seconds = 0, top = 0, right = 0, font_size = "2em", warn_when = 30) } else { htmltools::HTML(paste0( '4:00
')) } ``` ::: {style="font-size: 30px"} 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? ::: --- ## β
Think-Pair-Share: Solution ::: {style="font-size: 30px"} 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**. ::: {.fragment} 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 $\alpha, \beta$. ::: ::: --- ## π§ Β§4.8: How Do We Choose a Model? ::: {style="font-size: 29px"} 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. ::: {.fragment} 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. ::: ::: --- ## ποΈ Start From the Support ::: {style="font-size: 28px"} | 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 | ::: {.fragment} The support rules models **out** before any data are seen. Within what is left, the shape of a histogram decides. ::: ::: --- ## π» Fitting the Recoveries ```{r} #| label: fit-recoveries #| code-fold: false 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) 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) ``` ::: {style="font-size: 28px"} Matching the sample mean and sd recovers $\alpha \approx 2.1$, $\beta \approx 3.1$. The normal still spends 2% on impossible values. ::: --- ## π Two Candidates, One Histogram ```{r} #| label: fit-figure #| echo: false #| fig-width: 10 #| fig-height: 4.4 a_hat <- m * k; b_hat <- (1 - m) * k yy <- seq(-0.3, 1.1, by = 0.002) curves <- rbind( data.frame(y = yy, f = ifelse(yy >= 0 & yy <= 1, dbeta(pmin(pmax(yy, 1e-6), 1 - 1e-6), a_hat, b_hat), 0), model = "Beta, mean and sd matched"), data.frame(y = yy, f = dnorm(yy, m, s), model = "Normal, mean and sd matched")) ggplot(data.frame(y = recov), aes(y)) + annotate("rect", xmin = -0.3, xmax = 0, ymin = 0, ymax = Inf, fill = "#8b2635", alpha = 0.10) + geom_histogram(aes(y = after_stat(density)), breaks = seq(0, 1, 0.05), fill = "grey80", colour = "white") + geom_line(data = curves, aes(y, f, colour = model), linewidth = 1.3) + annotate("text", x = -0.15, y = 2.1, label = "recovery < 0:\nimpossible", colour = "#8b2635", size = 6) + scale_colour_manual(values = c("#14130f", "#3d6e8f")) + scale_x_continuous(breaks = seq(-0.25, 1, 0.25)) + labs(x = "Recovery rate on a defaulted loan", y = "Density", colour = NULL) + theme(legend.position = "top", legend.text = element_text(size = 18)) ``` --- ## π Quiz #1: The Mean {.quiz-question} A telecom regulator models an operator's market share as Beta(3, 7). What is its expected share? - [$0.30$]{.correct data-explanation="β
Theorem 4.11: E(Y) = Ξ±/(Ξ± + Ξ²) = 3/10 = 0.30."} - $0.43$ - $0.70$ - $0.019$ --- ## π Quiz #2: The Shape {.quiz-question} Recovery rates on unsecured consumer loans follow Beta(2, 8). Which description fits? - [Skewed right: most recoveries are small, with a tail towards 1]{.correct data-explanation="β
Ξ± < Ξ² puts the mass near 0. With both above 1 the density is 0 at the ends, and the mode is (2 β 1)/(2 + 8 β 2) = 0.125."} - Symmetric about 0.5 - Skewed left: most recoveries are close to 1 - Uniform on [0, 1] --- ## π Quiz #3: Choosing a Model {.quiz-question} 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]]{.correct data-explanation="β
Start from the support (Β§4.8). A weight is bounded by 0 and 1; only the beta lives there. The normal and the exponential put probability outside it, and the Poisson is for counts."} - Normal, because averages tend to be normal - Exponential, because it is positive - Poisson, because it is reported at regular intervals --- ## π Key Formulas ::: {style="font-size: 28px"} | | 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, Ξ±, Ξ²)`, `qbeta(p, Ξ±, Ξ²)` | ::: --- ## π Summary ::: {style="font-size: 30px"} - 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** ::: --- ## π Practice Problems ::: {style="font-size: 28px"} **Wackerly, 7th edition** - Β§4.7: Exercises 4.124 β 4.134; start with 4.124, 4.125, 4.128, 4.132 and 4.134(a) - Applet Exercises 4.114 β 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 §§4.7β4.8. **Next class:** 21 November, other expected values, Tchebysheff's theorem for continuous variables, and Chapter 4 in review (Wackerly §§4.9β4.11). ::: --- ## π Thank You ::: {style="font-size: 34px"} **Dr. Samir Orujov** π§ sorujov@ada.edu.az\ π’ Building D, Room D325\ π Office hours: Wednesday, 16:00 β 18:00 Slides and readings: **sorujov.net/teaching** ::: --- ## β Questions ::: {style="font-size: 32px"} - 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? ::: 