```{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: - **State** the gamma density (Definition 4.9) and use $\Gamma(n) = (n-1)!$ - **Derive** $E(Y) = \alpha\beta$ and $V(Y) = \alpha\beta^2$ by recognising a gamma integral (Theorem 4.8) - **Compute** gamma probabilities and quantiles in R, minding that R wants the rate $1/\beta$ - **Identify** the chi-square and exponential distributions as gamma special cases - **Use** the memoryless property of the exponential, and say when it is a poor model ::: --- ## πΊοΈ Where We Are ::: {style="font-size: 32px"} **Wackerly Β§4.6** Wednesday: the normal distribution, symmetric about $\mu$ and spread over the whole real line. ::: {.fragment} Many quantities in finance are neither. A waiting time, a lifetime, a repair time cannot be negative, and most of their mass sits near zero with a long tail to the right. ::: ::: {.fragment} Today's family is built for exactly that shape, and it contains two special cases we will use for the rest of the course: the **exponential** and the **chi-square**. ::: ::: --- ## β Motivating Question ::: {style="font-size: 30px"} On an interbank AZN/USD desk, the gap between two **large trades** (above 1 million USD) averages 25 minutes, with a standard deviation of about 25 minutes. ::: {.fragment} Model the gap as normal with $\mu = \sigma = 25$ and it gives $$P(Y < 0) = P(Z < -1) = 0.1587$$ **Sixteen percent of gaps would be negative.** The model is not slightly wrong; it is wrong in kind. ::: ::: {.fragment} We need a density that lives on $[0, \infty)$, is skewed right, and still has a mean and a variance we can compute. ::: ::: --- ## π Definition 4.9: The Gamma ::: {style="font-size: 30px"} ::: {.callout-important} ## Definition 4.9 $Y$ has a **gamma distribution** with parameters $\alpha > 0$ and $\beta > 0$ if and only if $$f(y) = \begin{cases} \dfrac{y^{\alpha-1} e^{-y/\beta}}{\beta^{\alpha}\,\Gamma(\alpha)}, & 0 \le y < \infty, \\[4pt] 0, & \text{elsewhere,} \end{cases} \qquad \Gamma(\alpha) = \int_0^\infty y^{\alpha-1} e^{-y}\,dy.$$ ::: ::: {.fragment} $\alpha$ is the **shape** parameter; $\beta$ is the **scale** parameter. Multiplying $Y$ by a positive constant changes $\beta$ and leaves $\alpha$ alone. ::: ::: --- ## π’ The Gamma Function ::: {style="font-size: 30px"} $\Gamma(\alpha)$ is only there to make the density integrate to 1. Three facts do all the work: - $\Gamma(1) = 1$, by direct integration - $\Gamma(\alpha) = (\alpha - 1)\,\Gamma(\alpha - 1)$ for $\alpha > 1$, by parts - $\Gamma(n) = (n-1)!$ for a positive integer $n$ ```{r} #| label: gamma-fn #| code-fold: false gamma(c(1, 2, 4, 7)) # 0!, 1!, 3!, 6! ``` So the gamma function is the factorial, shifted by one and extended to every $\alpha > 0$. ::: --- ## π What the Shape Parameter Does ```{r} #| label: shape-figure #| echo: false #| fig-width: 10 #| fig-height: 4.6 yy <- seq(0, 10, by = 0.02) shape_df <- rbind( data.frame(y = yy, f = dgamma(yy, shape = 1, rate = 1), a = "alpha == 1"), data.frame(y = yy, f = dgamma(yy, shape = 2, rate = 1), a = "alpha == 2"), data.frame(y = yy, f = dgamma(yy, shape = 4, rate = 1), a = "alpha == 4")) ggplot(shape_df, aes(y, f, colour = a)) + geom_line(linewidth = 1.4) + scale_colour_manual(values = c("#14130f", "#3d6e8f", "#b3833a"), labels = scales::label_parse()) + scale_y_continuous(limits = c(0, 1), breaks = seq(0, 1, 0.25)) + scale_x_continuous(breaks = seq(0, 10, 2)) + labs(x = "y", y = "f(y)", colour = NULL, title = expression("Gamma densities with" ~ beta == 1 ~ "(Figure 4.16)")) + theme(legend.position = "top", legend.text = element_text(size = 20)) ``` ::: {style="font-size: 28px"} At $\alpha = 1$ the mode is at zero; as $\alpha$ grows the peak moves right and the tail thins. ::: --- ## π Theorem 4.8: Mean and Variance ::: {style="font-size: 30px"} ::: {.callout-important} ## Theorem 4.8 If $Y$ has a gamma distribution with parameters $\alpha$ and $\beta$, then $\mu = E(Y) = \alpha\beta$ and $\sigma^2 = V(Y) = \alpha\beta^2$. ::: ::: {.fragment} **The trick.** The density integrates to 1, so $\int_0^\infty y^{\alpha-1}e^{-y/\beta}\,dy = \beta^{\alpha}\Gamma(\alpha)$ for **every** $\alpha > 0$. Then $$E(Y) = \frac{1}{\beta^{\alpha}\Gamma(\alpha)}\int_0^\infty y^{\alpha}e^{-y/\beta}\,dy = \frac{\beta^{\alpha+1}\Gamma(\alpha+1)}{\beta^{\alpha}\Gamma(\alpha)} = \alpha\beta.$$ ::: ::: {.fragment} The same step with $\alpha + 2$ gives $E(Y^2) = \alpha(\alpha+1)\beta^2$, so $V(Y) = \alpha\beta^2$. ::: ::: --- ## π’ Worked Example: Waiting for Claims ::: {style="font-size: 30px"} A Baku property insurer tracks the time $Y$, in days, until its **third** large claim of the quarter. $Y$ is gamma with $\alpha = 3$, $\beta = 4$. ::: {.fragment} **Moments.** $E(Y) = 3 \times 4 = 12$ days; $V(Y) = 3 \times 16 = 48$, so $\sigma = 6.93$ days. ::: ::: {.fragment} **A tail probability.** With integer $\alpha$, Exercise 4.99 turns the gamma tail into a Poisson sum with mean $20/\beta = 5$: $$P(Y > 20) = \sum_{x=0}^{2} \frac{5^x e^{-5}}{x!} = 18.5\,e^{-5} = 0.1247$$ ::: ::: {.fragment} One quarter in eight, the claims desk waits more than 20 days for its third large claim. ::: ::: --- ## π» The Same Numbers in R ```{r} #| label: claims-r #| code-fold: false alpha <- 3; beta <- 4 c(mean = alpha * beta, var = alpha * beta^2) 1 - pgamma(20, shape = alpha, rate = 1 / beta) # P(Y > 20) sum(dpois(0:2, lambda = 20 / beta)) # Exercise 4.99 route qgamma(0.95, shape = alpha, rate = 1 / beta) # 95th percentile, days ``` ::: {style="font-size: 28px"} **The trap:** R's third argument is the **rate** $1/\beta$, as in the book's `pgamma(y0, Ξ±, 1/Ξ²)`. Pass $\beta$ there and the answer is silently wrong. ::: --- ## π Definition 4.10: Chi-Square ::: {style="font-size: 30px"} ::: {.callout-important} ## Definition 4.10 Let $\nu$ be a positive integer. $Y$ has a **chi-square distribution with $\nu$ degrees of freedom** if and only if $Y$ is gamma with $\alpha = \nu/2$ and $\beta = 2$. ::: ::: {.callout-important} ## Theorem 4.9 If $Y$ is $\chi^2$ with $\nu$ degrees of freedom, then $E(Y) = \nu$ and $V(Y) = 2\nu$. ::: ::: {.fragment} Proof: Theorem 4.8 with $\alpha = \nu/2$, $\beta = 2$. The name "degrees of freedom" is explained in Theorem 6.4; for now it is just the parameter. ::: ::: --- ## π₯οΈ Gamma to Chi-Square: A Repair Time ::: {style="font-size: 30px"} The time $Y$, in hours, to restore a failed rack in a bank's data centre is gamma with $\alpha = 1.5$, $\beta = 4$. What is $P(Y < 3.5)$? ::: {.fragment} Exercise 6.46 will show: if $\alpha = n/2$, then $2Y/\beta$ is $\chi^2$ with $n$ degrees of freedom. Here $2Y/4 = Y/2$ is $\chi^2_3$, so $$P(Y < 3.5) = P(Y/2 < 1.75) = P(\chi^2_3 < 1.75)$$ ::: ```{r} #| label: chisq-link #| code-fold: false c(gamma = pgamma(3.5, shape = 1.5, rate = 1/4), chisq = pchisq(1.75, df = 3)) ``` About 37% of rack failures are fixed within 3.5 hours. ::: --- ## π Definition 4.11: The Exponential ::: {style="font-size: 30px"} ::: {.callout-important} ## Definition 4.11 $Y$ has an **exponential distribution** with parameter $\beta > 0$ if and only if $$f(y) = \begin{cases} \dfrac{1}{\beta}\, e^{-y/\beta}, & 0 \le y < \infty, \\[4pt] 0, & \text{elsewhere.} \end{cases}$$ ::: ::: {.callout-important} ## Theorem 4.10 If $Y$ is exponential with parameter $\beta$, then $E(Y) = \beta$ and $V(Y) = \beta^2$. ::: ::: {.fragment} It is the gamma with $\alpha = 1$, and the one member whose distribution function has a closed form: $P(Y > y) = e^{-y/\beta}$. ::: ::: --- ## πΎ Worked Example: Drive Lifetimes ::: {style="font-size: 30px"} Solid-state drives in a data centre fail after an exponential lifetime with mean $\beta = 5$ years. ::: {.fragment} **Survival past the warranty.** $P(Y > 3) = e^{-3/5} = 0.5488$. Just over half the drives outlast a 3-year warranty. ::: ::: {.fragment} **An early-replacement threshold.** Which age $t$ do only 10% of drives fail before? $$1 - e^{-t/5} = 0.10 \;\Rightarrow\; t = -5\ln(0.90) = 0.527 \text{ years} \approx 6.3 \text{ months}$$ ::: ::: {.fragment} Because $\sigma = \beta = 5$ years, lifetimes are as variable as they are long: a fleet mean says little about one drive. ::: ::: --- ## π§ Example 4.10: Memoryless ::: {style="font-size: 30px"} If $Y$ is exponential and $a, b > 0$, then $$P(Y > a + b \mid Y > a) = \frac{P(Y > a + b)}{P(Y > a)} = \frac{e^{-(a+b)/\beta}}{e^{-a/\beta}} = e^{-b/\beta} = P(Y > b)$$ ::: {.fragment} The intersection of $(Y > a + b)$ and $(Y > a)$ is just $(Y > a + b)$; that is the whole proof. ::: ::: {.fragment} **Having survived $a$ units tells you nothing about the next $b$.** The geometric distribution in Chapter 3 had the same property (Exercise 4.95 links the two). ::: ::: --- ## βοΈ Memoryless at a Bank Call Centre ::: {style="font-size: 28px"} Hold time at a bank's call centre is exponential with mean 4 minutes, so $P(Y > 2) = e^{-0.5} = 0.6065$. A caller has already held 5 minutes. Simulating 100,000 calls: ::: ```{r} #| label: memoryless-sim #| code-fold: false hold <- rexp(100000, rate = 1/4) c(fresh_caller = mean(hold > 2), waited_5_mins = mean(hold[hold > 5] > 7), exact = exp(-2/4)) ``` ::: {style="font-size: 28px"} The caller who has waited 5 minutes is no closer to being answered. Whether that is **realistic** is a modelling question: a queue that clears in order does remember. ::: --- ## π¬ Same Mean, Different Shape ```{ojs} //| echo: false viewof shape = { const input = Inputs.range([1, 20], {value: 1, step: 0.5, label: "Shape Ξ± (mean fixed at 10):"}); ['pointerdown','touchstart','mousedown','click','wheel','pointermove','touchmove'] .forEach(e => input.addEventListener(e, ev => ev.stopPropagation())); return input; } ``` ```{ojs} //| echo: false lgam = 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]; x -= 1; let a = c[0]; const t = x + g + 0.5; for (let i = 1; i < 9; i++) a += c[i] / (x + i); return 0.5 * Math.log(2 * Math.PI) + (x + 0.5) * Math.log(t) - t + Math.log(a); } scaleB = 10 / shape dens = y => y <= 0 ? (shape === 1 ? 1 / scaleB : 0) : Math.exp((shape - 1) * Math.log(y) - y / scaleB - shape * Math.log(scaleB) - lgam(shape)) grid = Array.from({length: 601}, (_, i) => ({y: i * 0.05, f: dens(i * 0.05)})) tail = { let s = 0; const h = 0.01; for (let y = 15; y < 200; y += h) s += h * (dens(y) + dens(y + h)) / 2; return s; } md`Ξ± = **${shape}**, Ξ² = **${scaleB.toFixed(2)}**: Ο = **${(10 / Math.sqrt(shape)).toFixed(2)}**, and P(Y > 15) = **${tail.toFixed(4)}**` Plot.plot({ width: 1150, height: 310, marginTop: 40, marginLeft: 78, marginBottom: 58, style: {fontSize: "18px"}, x: {label: "Waiting time y (mean 10)", domain: [0, 30]}, y: {label: "f(y)", grid: true}, marks: [ Plot.areaY(grid.filter(d => d.y >= 15), {x: "y", y: "f", fill: "#cbb8a9"}), Plot.line(grid, {x: "y", y: "f", stroke: "#14130f", strokeWidth: 2.5}), Plot.ruleX([10], {stroke: "#8b2635", strokeDasharray: "4 4"}), Plot.ruleY([0]) ] }) ``` --- ## 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"} Back to the AZN/USD desk. Gaps between large trades are now modelled as **exponential with mean 25 minutes**. **Four minutes, in pairs:** 1. What is the probability that a gap exceeds 30 minutes? 2. No large trade for 40 minutes. What is the probability that one arrives in the next 10? 3. The time $W$ until the third large trade is gamma with $\alpha = 3$, $\beta = 25$. Find $E(W)$ and the standard deviation of $W$. ::: --- ## β
Think-Pair-Share: Solution ::: {style="font-size: 30px"} 1. $P(Y > 30) = e^{-30/25} = e^{-1.2} = 0.3012$. 2. Memoryless: the 40 minutes are irrelevant. $$P(Y \le 50 \mid Y > 40) = P(Y \le 10) = 1 - e^{-0.4} = 0.3297$$ 3. Theorem 4.8: $E(W) = 3 \times 25 = 75$ minutes, $V(W) = 3 \times 625 = 1875$, so $\sigma_W = 43.30$ minutes. ::: {.fragment} Compare with the normal model in the motivating question: this one never puts mass below zero, and its $\sigma$ equals its mean for a single gap, as the desk observed. ::: ::: --- ## π Quiz #1: Chi-Square Moments {.quiz-question} $Y$ has a chi-square distribution with 8 degrees of freedom. What are $E(Y)$ and $V(Y)$? - [$E(Y) = 8$, $V(Y) = 16$]{.correct data-explanation="β
Theorem 4.9: E(Y) = Ξ½ and V(Y) = 2Ξ½. Equivalently, gamma with Ξ± = 4 and Ξ² = 2 gives Ξ±Ξ² = 8 and Ξ±Ξ²Β² = 16."} - $E(Y) = 4$, $V(Y) = 8$ - $E(Y) = 8$, $V(Y) = 64$ - $E(Y) = 16$, $V(Y) = 32$ --- ## π Quiz #2: A Server Fan {.quiz-question} A cooling fan's lifetime is exponential with mean 4 years. It has already run 2 years. What is the probability it runs at least 1 more year? - [$e^{-1/4} = 0.7788$]{.correct data-explanation="β
Memoryless: P(Y > 3 | Y > 2) = P(Y > 1) = e^(β1/4). The two years already run do not enter."} - $e^{-3/4} = 0.4724$ - $1 - e^{-1/4} = 0.2212$ - $e^{-2/4} = 0.6065$ --- ## π Quiz #3: Back Out the Parameters {.quiz-question} A payment-processing delay is gamma distributed with mean 6 seconds and variance 12 secondsΒ². What are $\alpha$ and $\beta$? - [$\alpha = 3$, $\beta = 2$]{.correct data-explanation="β
Divide the variance by the mean: Ξ±Ξ²Β²/(Ξ±Ξ²) = Ξ² = 12/6 = 2, then Ξ± = 6/2 = 3."} - $\alpha = 2$, $\beta = 3$ - $\alpha = 6$, $\beta = 1$ - $\alpha = 0.5$, $\beta = 12$ --- ## π Key Formulas ::: {style="font-size: 28px"} | | Density on $y \ge 0$ | $E(Y)$ | $V(Y)$ | |---|---|---|---| | Gamma, Def. 4.9 | $\dfrac{y^{\alpha-1}e^{-y/\beta}}{\beta^{\alpha}\Gamma(\alpha)}$ | $\alpha\beta$ | $\alpha\beta^2$ | | $\chi^2_\nu$, Def. 4.10 | gamma, $\alpha = \nu/2$, $\beta = 2$ | $\nu$ | $2\nu$ | | Exponential, Def. 4.11 | $\dfrac{1}{\beta}e^{-y/\beta}$ | $\beta$ | $\beta^2$ | - $\Gamma(n) = (n-1)!$ and $\Gamma(\alpha) = (\alpha-1)\Gamma(\alpha-1)$ - Exponential: $P(Y > y) = e^{-y/\beta}$ and $P(Y > a+b \mid Y > a) = P(Y > b)$ - In R: `pgamma(y, alpha, 1/beta)`, `qgamma(p, alpha, 1/beta)` ::: --- ## π Summary ::: {style="font-size: 30px"} - The gamma family models quantities that are **nonnegative and skewed right**: waiting times, lifetimes, repair times - $\alpha$ sets the shape, $\beta$ the scale; with the mean fixed, a larger $\alpha$ means a thinner tail - Theorem 4.8 follows from one idea: every gamma integral is $\beta^{\alpha}\Gamma(\alpha)$ - Chi-square ($\alpha = \nu/2$, $\beta = 2$) and exponential ($\alpha = 1$) are gamma special cases - The exponential is **memoryless**, which is both its convenience and its limitation ::: --- ## π Practice Problems ::: {style="font-size: 28px"} **Wackerly, 7th edition** - Exercises at the end of Β§4.6: start with 4.81, 4.82, 4.88, 4.89 and 4.91, then 4.96, 4.104 and 4.106 β 4.109 - Re-set Exercise 4.91 as a bank's cash-demand problem: what vault holding keeps the chance of a shortfall at 1%? **Week 10, Problem Set 2** is open now and closes **Sunday 22 November at 23:59** on WeBWorK, covering Β§4.6. **Next class:** 18 November, the beta distribution and how to choose a continuous model (Wackerly Β§4.7 β 4.8). ::: --- ## 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"} - A drive that has run 4 years is as good as new under the exponential model. What would you look for in real failure data to reject that? - Why does the gamma tail have a Poisson-sum form only when $\alpha$ is an integer? - With the mean held fixed, what happens to the gamma density as $\alpha \to \infty$, and which distribution does it start to resemble? ::: 