```{r} #| label: setup #| include: false set.seed(2026) library(ggplot2) theme_set(theme_minimal(base_size = 18)) ``` ## π¬ The Idea in 3 Minutes ::: {style="text-align:center"} [Watch this short intuition video before (or after) the slides. Captions: CC button.]{style="font-size:22px"} ::: --- ## π― Learning Objectives ::: {style="font-size: 32px"} By the end of this lecture, you will be able to: - **Recognise** sampling without replacement from a finite pool, and model it with the hypergeometric distribution - **Compute** hypergeometric probabilities, mean and variance, and explain the factor $\frac{N-n}{N-1}$ - **Apply** the Poisson distribution to counts of events in time, including a Poisson process over $a$ units - **Use** the Poisson as the limit of the binomial for rare events - **Choose** among the five discrete models of Chapter 3 for Midterm I ::: --- ## πΊοΈ Where We Are ::: {style="font-size: 32px"} **Wackerly Β§3.7β3.8** Saturday ended on one sentence: *the binomial fixes the trials; the geometric and negative binomial fix the successes and count the trials.* All three rest on the same two assumptions β **independent** trials with a **constant** $p$. ::: {.fragment} Today we drop the assumptions themselves, one at a time. Draw **without replacement** from a small pool and independence fails: that is the hypergeometric. Let $n \to \infty$ while $p \to 0$ and there is no $n$ left to count: that is the Poisson. ::: ::: --- ## π§Ύ Why the Binomial Fails in an Audit ::: {style="font-size: 30px"} An auditor receives a batch of **25 invoices** from a Baku construction supplier. Unknown to her, **5** are misstated. She pulls **6** at random for testing. ::: {.fragment} The first draw is misstated with probability $5/25 = 0.20$. If it was, the second is misstated with probability $4/24 = 0.167$; if it was not, $5/24 = 0.208$. ::: ::: {.fragment} **The trials are not independent and $p$ is not constant**, so $Y$, the number of misstated invoices in the sample, is not binomial. It is a counting problem over subsets β Section 2.6 β and that is where the book derives the answer. ::: ::: --- ## π Definition 3.10: Hypergeometric ::: {style="font-size: 30px"} ::: {.callout-important} ## Definition 3.10 A random variable $Y$ has a **hypergeometric probability distribution** if and only if $$p(y) = \frac{\binom{r}{y}\binom{N-r}{n-y}}{\binom{N}{n}},$$ where $y$ is an integer $0, 1, 2, \ldots, n$, subject to $y \le r$ and $n - y \le N - r$. ::: ::: {.fragment} Numerator: choose $y$ of the $r$ "red" elements **and** $n-y$ of the $N-r$ others (the $mn$ rule). Denominator: all $\binom{N}{n}$ equally likely samples. In R: `dhyper(y, r, N - r, n)`. ::: ::: --- ## π Worked Example: The Audit Rule ::: {style="font-size: 30px"} The firm's rule: **escalate to a full audit if two or more** of the 6 sampled invoices are misstated. With $N = 25$, $r = 5$, $n = 6$: ::: {.fragment} $$p(0) = \frac{\binom{5}{0}\binom{20}{6}}{\binom{25}{6}} = \frac{38760}{177100} = 0.2189, \qquad p(1) = \frac{\binom{5}{1}\binom{20}{5}}{\binom{25}{6}} = \frac{77520}{177100} = 0.4377$$ ::: ::: {.fragment} $$P(Y \ge 2) = 1 - p(0) - p(1) = 1 - 0.2189 - 0.4377 = 0.3434$$ ::: ::: {.fragment} A batch with **one invoice in five** misstated still passes the test about two times in three. The complement did the work, exactly as in Chapter 2. ::: ::: --- ## π Theorem 3.10: Mean and Variance ::: {style="font-size: 30px"} ::: {.callout-important} ## Theorem 3.10 If $Y$ has a hypergeometric distribution, $$\mu = E(Y) = \frac{nr}{N} \qquad \sigma^2 = V(Y) = n\left(\frac{r}{N}\right)\left(\frac{N-r}{N}\right)\left(\frac{N-n}{N-1}\right)$$ ::: ::: {.fragment} With $p = r/N$ this is $\mu = np$ and $\sigma^2 = npq\,\frac{N-n}{N-1}$: the binomial, times a **finite-population correction**. ::: ::: {.fragment} Audit: $\mu = 6(5)/25 = 1.2$ and $\sigma^2 = 6(0.2)(0.8)(19/24) = 0.76$. The binomial would say $0.96$ β sampling without replacement removes about a fifth of the variance. ::: ::: --- ## π» When Does the Pool Size Matter? ```{r} #| label: hyper-vs-binom #| fig-height: 3.9 #| fig-width: 10 #| echo: false y <- 0:6 pmf <- rbind( data.frame(y, p = dhyper(y, 5, 20, 6), model = "Hypergeometric, N = 25"), data.frame(y, p = dhyper(y, 50, 200, 6), model = "Hypergeometric, N = 250"), data.frame(y, p = dbinom(y, 6, 0.2), model = "Binomial, p = 0.2")) pmf$model <- factor(pmf$model, levels = unique(pmf$model)) ggplot(pmf, aes(factor(y), p, fill = model)) + geom_col(position = position_dodge(width = 0.85), width = 0.8) + scale_fill_manual(values = c("#8b2635", "#cbb8a9", "#14130f")) + labs(x = "Misstated invoices in a sample of 6", y = "p(y)", fill = NULL) + theme(legend.position = "top", legend.text = element_text(size = 18)) ``` ::: {style="font-size: 28px"} Same $r/N = 0.2$ throughout. At $N = 250$ the sample is 2.4% of the pool and the hypergeometric is indistinguishable from the binomial; at $N = 25$ it is 24% and the distribution is visibly tighter. ::: --- ## π§οΈ From Binomial to Poisson ::: {style="font-size: 30px"} Split a day into $n$ tiny subintervals, each holding at most one event with probability $p$, independently. The count is binomial. Now let $n \to \infty$ with $\lambda = np$ fixed: $$\lim_{n\to\infty}\binom{n}{y}p^y(1-p)^{n-y} = \frac{\lambda^y}{y!}e^{-\lambda}$$ ::: {.callout-important} ## Definition 3.11 $Y$ has a **Poisson probability distribution** if and only if $$p(y) = \frac{\lambda^y}{y!}e^{-\lambda}, \qquad y = 0, 1, 2, \ldots, \quad \lambda > 0.$$ ::: ::: --- ## π¦ Worked Example: A Claims Desk ::: {style="font-size: 30px"} The cargo-insurance desk of a Baku insurer receives on average $\lambda = 3$ claims per day, and can fully process **4** in a day. Let $Y$ be tomorrow's claim count. ::: {.fragment} $$P(Y = 0) = e^{-3} = 0.0498 \quad P(Y = 1) = 3e^{-3} = 0.1494 \quad P(Y = 2) = 4.5e^{-3} = 0.2240$$ ::: ::: {.fragment} The desk falls behind when $Y \ge 5$: $$P(Y \ge 5) = 1 - P(Y \le 4) = 1 - 0.8153 = 0.1847$$ ::: ::: {.fragment} Roughly **one working day in five** leaves a backlog. In R: `1 - ppois(4, 3)`, or Table 3 of Appendix 3. ::: ::: --- ## π Theorem 3.11: Mean = Variance ::: {style="font-size: 30px"} ::: {.callout-important} ## Theorem 3.11 If $Y$ has a Poisson distribution with parameter $\lambda$, then $\mu = E(Y) = \lambda$ and $\sigma^2 = V(Y) = \lambda$. ::: ::: {.fragment} **Proof idea.** In $\sum_{y} y\,\lambda^y e^{-\lambda}/y!$ the $y = 0$ term vanishes; cancel $y$ against $y!$, factor out $\lambda$, and put $z = y - 1$. What remains is a Poisson sum, which equals 1. The variance is Exercise 3.138. ::: ::: {.fragment} A practical test: if claim counts show a sample variance far above their mean, the Poisson model β not just its $\lambda$ β is in doubt. ::: ::: --- ## π§ A Poisson Process: One ATM ::: {style="font-size: 28px"} If events occur as a Poisson process at $\lambda$ per unit, the count in $a$ units is Poisson with mean $a\lambda$. An ATM in Nizami Street averages **12 withdrawals per hour**. ::: ```{r} #| label: atm lambda_hr <- 12 hours <- c(1/6, 1/2, 1, 2) data.frame( window = c("10 min", "30 min", "1 hour", "2 hours"), mean = hours * lambda_hr, P_none = round(exp(-hours * lambda_hr), 4), P_over_30 = round(1 - ppois(30, hours * lambda_hr), 4)) ``` ::: {style="font-size: 28px"} No withdrawal in 10 minutes: $e^{-2} = 0.1353$. A cassette good for 30 withdrawals runs out within 2 hours with probability $0.0958$. ::: --- ## β οΈ The Poisson Limit: Rare Losses ::: {style="font-size: 28px"} A bank's trading desk has a loss event (a failed settlement, a fat-finger trade) on any of **250** trading days with probability **0.008**, independently. Binomial, or Poisson with $\lambda = np = 2$? ::: ```{r} #| label: oprisk y <- 0:5 data.frame(y, binomial = round(dbinom(y, 250, 0.008), 4), poisson = round(dpois(y, 2), 4)) ``` ::: {style="font-size: 28px"} Agreement to the third decimal: $P(Y \ge 4)$ is $0.1421$ exact and $0.1429$ Poisson. The book's rule of thumb: large $n$, small $p$, $\lambda = np$ below about 7. ::: --- ## π¬ Interactive: Binomial to Poisson {.smaller} ```{ojs} //| echo: false viewof n_sub = { const input = Inputs.range([4, 200], {value: 6, step: 1, label: "Number of trials n (Ξ» = np = 2):"}); ['pointerdown','touchstart','mousedown','click','wheel','pointermove','touchmove'] .forEach(e => input.addEventListener(e, ev => ev.stopPropagation())); return input; } ``` ```{ojs} //| echo: false lam = 2 binomPmf = (n, p) => { const out = [Math.pow(1 - p, n)]; for (let k = 0; k < 10; k++) out.push(k + 1 > n ? 0 : out[k] * (n - k) / (k + 1) * p / (1 - p)); return out; } poisPmf = Array.from({length: 11}, (_, k) => { let f = 1; for (let i = 2; i <= k; i++) f *= i; return Math.exp(-lam) * Math.pow(lam, k) / f; }) bin = binomPmf(n_sub, lam / n_sub).map((p, k) => ({y: k, p})) pois = poisPmf.map((p, k) => ({y: k, p})) gap = d3.max(bin, (d, k) => Math.abs(d.p - pois[k].p)) md`With **n = ${n_sub}** and **p = ${(lam / n_sub).toFixed(4)}**, the largest gap between binomial bars and Poisson dots is **${gap.toFixed(4)}**.` Plot.plot({ width: 1150, height: 300, marginLeft: 78, marginBottom: 58, style: {fontSize: "18px"}, x: {label: "y", domain: d3.range(0, 11), padding: 0.25}, y: {label: "p(y)", domain: [0, 0.45], tickFormat: ".2f"}, marks: [ Plot.barY(bin, {x: "y", y: "p", fill: "#cbb8a9"}), Plot.dot(pois, {x: "y", y: "p", r: 8, fill: "#8b2635"}), 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"} A bank's risk desk has recorded operational-loss events as a Poisson process at **2 per month** for years. Last quarter (3 months) there were **11**. **Four minutes, in pairs:** 1. What is the distribution of $Y$, the number of events in a quarter? 2. Where does 11 fall relative to $\mu \pm 2\sigma$? 3. Is 11 alarming enough to report to the risk committee? ::: --- ## β
Think-Pair-Share: Solution ::: {style="font-size: 30px"} 1. A Poisson process over $a = 3$ months: $Y$ is Poisson with $\lambda^\star = 3 \times 2 = 6$. 2. Theorem 3.11: $\mu = 6$, $\sigma = \sqrt{6} = 2.45$, so $\mu + 2\sigma = 6 + 2(2.45) = 10.90$. The observed 11 lies **just outside**. ::: {.fragment} 3. Exactly: $P(Y \ge 11) = 1 - P(Y \le 10) = 1 - 0.9574 = 0.0426$, from `1 - ppois(10, 6)`. Not impossible, but a one-in-25 quarter under the old rate β worth an investigation into whether $\lambda$ has risen, which is the reasoning of Example 3.22. ::: ::: --- ## π Quiz #1: A Licence Lottery {.quiz-question} The Energy Ministry awards 6 solar licences by lottery among 16 bidders, of whom 4 are joint ventures. What is the probability that **no** joint venture wins? - [$\binom{12}{6}\big/\binom{16}{6} = 0.1154$]{.correct data-explanation="β
Licences are drawn without replacement from a finite pool of 16, so Y is hypergeometric with N = 16, r = 4, n = 6: p(0) = 924/8008 = 0.1154."} - $(0.75)^6 = 0.1780$ - $\binom{12}{6}\big/\binom{16}{4} = 0.5077$ - $12/16 = 0.75$ --- ## π Quiz #2: Scaling the Interval {.quiz-question} Withdrawals at an ATM follow a Poisson process at 12 per hour. What is the probability of **no** withdrawal in a 5-minute window? - [$e^{-1} = 0.368$]{.correct data-explanation="β
Five minutes is a = 1/12 of an hour, so the count is Poisson with mean aΞ» = 1, and P(Y = 0) = eβ»ΒΉ = 0.368."} - $e^{-12} = 0.000006$ - $1 - e^{-1} = 0.632$ - $1/12 = 0.083$ --- ## π§ Review: Which Discrete Model? ::: {style="font-size: 28px"} | Model | $Y$ isβ¦ | $E(Y)$ | $V(Y)$ | |--------|--------------|----|------| | Binomial Β§3.4 | successes in $n$ independent trials | $np$ | $npq$ | | Geometric Β§3.5 | the trial of the **first** success | $1/p$ | $q/p^2$ | | Negative binomial Β§3.6 | the trial of the $r$-th success | $r/p$ | $rq/p^2$ | | Hypergeometric Β§3.7 | "red" items in $n$ drawn **without replacement** | $nr/N$ | $npq\frac{N-n}{N-1}$ | | Poisson Β§3.8 | events in a fixed interval | $\lambda$ | $\lambda$ | ::: {.fragment} Ask in order: **is $n$ fixed?** Independent trials, binomial; a finite pool, hypergeometric. **Is the trial count the variable?** Geometric or negative binomial. **Only a rate?** Poisson. ::: ::: --- ## βοΈ Exam-Style Question 1 ::: {style="font-size: 30px"} Of a bank's 12 branches, 3 failed an internal compliance check. The Central Bank inspects 4 branches chosen at random. Find the probability that it finds **at least one** failing branch, and the mean and variance of the number it finds. ::: {.fragment} Without replacement from $N = 12$ with $r = 3$, $n = 4$: $$P(Y \ge 1) = 1 - \frac{\binom{3}{0}\binom{9}{4}}{\binom{12}{4}} = 1 - \frac{126}{495} = 0.7455$$ ::: ::: {.fragment} $\mu = 4(3)/12 = 1$ and $\sigma^2 = 4(0.25)(0.75)(8/11) = 0.5455$. Treating it as binomial would give $1 - 0.75^4 = 0.6836$ β wrong, and by 6 points. ::: ::: --- ## βοΈ Exam-Style Question 2 ::: {style="font-size: 30px"} Large claims (above 50,000 AZN) reach a reinsurer as a Poisson process at 0.5 per week. Call a 4-week month **quiet** if it has at most one large claim. Months are independent. Find the probability that at least 2 of the next 6 months are quiet. ::: {.fragment} **Step 1 (Β§3.8).** A month has $\lambda = 4(0.5) = 2$: $w = P(Y \le 1) = e^{-2} + 2e^{-2} = 3e^{-2} = 0.4060$. ::: ::: {.fragment} **Step 2 (Β§3.4).** Quiet months $X$ are binomial with $n = 6$, $p = w = 0.4060$: $$P(X \ge 2) = 1 - (0.5940)^6 - 6(0.4060)(0.5940)^5 = 1 - 0.0439 - 0.1801 = 0.7759$$ ::: ::: --- ## π Quiz #3: Exam-Style β A Cost Built on a Count {.quiz-question} Claims at a desk are Poisson with $\lambda = 4$ per day. Daily handling cost is $C = 500 + 800Y$ AZN. What is the **standard deviation** of $C$? - [$1600$ AZN]{.correct data-explanation="β
V(C) = 800Β² V(Y) = 640000 Γ 4 = 2,560,000, so Ο = 800 Γ β4 = 1600 AZN. The constant 500 shifts the cost but adds no spread; E(C) = 500 + 800(4) = 3700 AZN."} - $3200$ AZN - $3700$ AZN - $2100$ AZN --- ## π Key Formulas ::: {style="font-size: 30px"} | | Statement | |---|---| | Definition 3.10 | $p(y) = \binom{r}{y}\binom{N-r}{n-y}\big/\binom{N}{n}$ | | Theorem 3.10 | $\mu = \frac{nr}{N}$, $\ \sigma^2 = n\frac{r}{N}\frac{N-r}{N}\frac{N-n}{N-1}$ | | Definition 3.11 | $p(y) = \frac{\lambda^y}{y!}e^{-\lambda}$, $\ y = 0, 1, 2, \ldots$ | | Theorem 3.11 | $\mu = \sigma^2 = \lambda$ | | Poisson process | count in $a$ units is Poisson with mean $a\lambda$ | | Poisson limit | $\binom{n}{y}p^y q^{n-y} \approx \frac{(np)^y}{y!}e^{-np}$, large $n$, small $p$ | ::: --- ## π Summary ::: {style="font-size: 30px"} - Drawing **without replacement** from a small pool breaks independence: the count is hypergeometric - Its mean is the binomial's $np$; its variance is $npq$ shrunk by $\frac{N-n}{N-1}$, which matters when $n/N$ is not small - Counts of rare events in an interval are Poisson, and the single parameter $\lambda$ is both the mean and the variance - Over $a$ units of a Poisson process the mean is $a\lambda$; for large $n$ and small $p$ the binomial is close to Poisson($np$) - For Midterm I, name the model **before** computing: fixed $n$? finite pool? trials to a success? only a rate? ::: --- ## π Practice Problems ::: {style="font-size: 28px"} **Wackerly, 7th edition** - Β§3.7: Exercises 3.103, 3.105, 3.108, 3.112, 3.117 - Β§3.8: Exercises 3.121, 3.122, 3.126, 3.128, 3.134, 3.139 **Week 7, Problem Set 1** is open now and closes **Sunday 1 November at 23:59** on WeBWorK, covering §§3.7β3.8. It is the only set this week. **Midterm Examination I** is on **Saturday 24 October**: Chapters 1β3, §§3.1β3.8. **Next class:** 28 October, after the midterm β moments and moment-generating functions (Wackerly Β§3.9). ::: --- ## 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"} - In the audit, how large must the batch $N$ be before the binomial answer is within 0.005 of the hypergeometric one? - Why can a Poisson count exceed any bound, when a binomial count never exceeds $n$? - Two independent ATMs average 12 and 8 withdrawals an hour. What would you guess the distribution of their total is β and how would you check it? ::: 