```{r} #| label: setup #| include: false set.seed(2026) library(ggplot2) ``` ## ๐ฌ The Idea in 2 Minutes ::: {style="text-align:center"} [Watch this short intuition video before (or after) the slides. Captions: CC button.]{style="font-size:22px"} ::: --- ## ๐ฏ Learning Objectives ::: {style="font-size: 32px"} By the end of this lecture, you will be able to: - **Check** an experiment against the five properties of a binomial experiment (Definition 3.6) - **Derive** $p(y) = \binom{n}{y} p^y q^{n-y}$ from sample points and the counting rule - **Compute** binomial probabilities by hand, from Table 1 and with `dbinom` / `pbinom` - **Use** $E(Y) = np$ and $V(Y) = npq$ (Theorem 3.7) to price and to judge risk - **Read** an unlikely count as evidence against an assumed $p$ ::: --- ## ๐บ๏ธ Where We Are ::: {style="font-size: 30px"} Last class ended on: *ยง3.4 โ the binomial distribution.* On Wednesday every $p(y)$ came as a **table**, and $E(Y)$ and $V(Y)$ were two sums over it. Today one **formula** covers a whole family of tables โ indexed by just two numbers, $n$ and $p$. ::: {.callout-important} ## The question this lecture answers A Baku bank's SME desk holds 20 loans. Each defaults this year with probability 0.05, independently of the others. **How likely are three or more defaults โ and how many should the desk budget for?** ::: ::: --- ## ๐ The Binomial Experiment ::: {style="font-size: 29px"} ::: {.callout-note} ## Definition 3.6 1. The experiment consists of a **fixed number, $n$**, of identical trials. 2. Each trial results in one of **two outcomes**: success, $S$, or failure, $F$. 3. The probability of success on a single trial is equal to some value $p$ and **remains the same** from trial to trial. The probability of a failure is $q = 1 - p$. 4. The trials are **independent**. 5. The random variable of interest is $Y$, the **number of successes** observed during the $n$ trials. ::: "Success" is only a name for one of the two outcomes. For a credit analyst, a *default* is the success. ::: --- ## ๐ณ Checking the Five: Card Declines ::: {style="font-size: 29px"} A payment processor watches the next **8** card transactions at a Baku supermarket. Each is declined with probability $0.03$, independently. $Y$ = number declined. Binomial? ::: {.fragment} 1. **Fixed $n$**: 8 transactions, decided in advance. โ 2. **Two outcomes**: declined ($S$) or approved ($F$). โ 3. **Same $p$**: $P(S) = 0.03$ on every transaction. โ 4. **Independent**: stated. โ 5. **Counts successes**: $Y$ counts declines. โ So $n = 8$, $p = 0.03$. ::: ::: {.fragment} "Transactions **until** the first decline" breaks property 1 โ that is ยง3.5. ::: ::: --- ## ๐ Approximately Binomial ::: {style="font-size: 29px"} An ISP has 250,000 households in Baku; 35% of them reach the advertised 100 Mbps at peak hour. We test **10** households at random. $Y$ = number reaching 100 Mbps. ::: {.fragment} Sampling is **without** replacement, so strictly the trials are dependent. But removing a few households barely changes the 35%: the conditional probabilities stay very close to $0.35$, and $Y$ is **approximately** binomial with $n = 10$, $p = 0.35$. ::: ::: {.fragment} If the sample were a large fraction of the population โ say 10% โ this fails, and the right model is the **hypergeometric** of ยง3.7. ::: ::: --- ## ๐งฎ Deriving $p(y)$ ::: {style="font-size: 29px"} A sample point is an $n$-tuple of letters, e.g. $SSFSF\cdots FS$. Take the one with $y$ successes first, then $n - y$ failures: $$S\,S \cdots S\;F\,F \cdots F \quad\text{has probability}\quad (p \cdot p \cdots p)(q \cdot q \cdots q) = p^y q^{n-y}$$ by independence (Theorem 2.5, applied $n$ times): $y$ factors of $p$, $n - y$ factors of $q$. ::: {.fragment} Every **other** ordering of $y$ $S$'s and $n - y$ $F$'s has the same probability, and by Theorem 2.3 there are $\binom{n}{y} = \frac{n!}{y!(n-y)!}$ of them. ::: ::: {.fragment} So $P(Y = y) = \binom{n}{y} p^y q^{n-y}$ โ a count times a probability. ::: ::: --- ## ๐ Definition 3.7 ::: {style="font-size: 30px"} ::: {.callout-note} ## Definition 3.7 A random variable $Y$ has a **binomial distribution** based on $n$ trials with success probability $p$ if and only if $$p(y) = \binom{n}{y} p^y q^{n-y}, \qquad y = 0, 1, 2, \dots, n \text{ and } 0 \le p \le 1.$$ ::: ::: {.fragment} Theorem 3.1 holds: each term is positive, and they are the terms of the binomial expansion, $$\sum_{y=0}^{n} \binom{n}{y} p^y q^{n-y} = (q + p)^n = 1^n = 1.$$ ::: ::: --- ## ๐ฆ Example: The SME Loan Book ::: {style="font-size: 29px"} $n = 20$ loans, default ($S$) probability $p = 0.05$, so $q = 0.95$. ::: {.fragment} **No defaults:** $\;p(0) = \binom{20}{0}(0.05)^0(0.95)^{20} = 0.3585$ ::: ::: {.fragment} **At least one** (the complement, as in Lecture 6): $\;1 - p(0) = 0.6415$ ::: ::: {.fragment} **Three or more:** $\;P(Y \ge 3) = 1 - [p(0) + p(1) + p(2)]$ $$= 1 - [0.3585 + 0.3774 + 0.1887] = 1 - 0.9245 = 0.0755$$ (Terms rounded; the unrounded sum is 0.92452.) With 50,000 AZN per loan, a year of 150,000 AZN or more in defaulted principal has about a **7.5%** chance. ::: ::: --- ## ๐ป The Same Numbers in R ```{r} #| label: sme-book #| code-fold: false n <- 20; p <- 0.05 round(dbinom(0:4, size = n, prob = p), 4) # p(0), ..., p(4) c(none = dbinom(0, n, p), at_least_one = 1 - dbinom(0, n, p), three_plus = 1 - pbinom(2, n, p), # P(Y >= 3) = 1 - P(Y <= 2) total = sum(dbinom(0:n, n, p))) |> round(4) # Simulate 100,000 years of the loan book years <- rbinom(100000, size = n, prob = p) round(mean(years >= 3), 4) ``` --- ## ๐ The Shape Depends on $p$ and $n$ ```{r} #| label: shapes-figure #| echo: false #| fig-width: 11 #| fig-height: 4.3 shapes <- rbind( data.frame(panel = "n = 10, p = 0.1", y = 0:10, p = dbinom(0:10, 10, 0.1)), data.frame(panel = "n = 10, p = 0.5", y = 0:10, p = dbinom(0:10, 10, 0.5)), data.frame(panel = "n = 20, p = 0.5", y = 0:20, p = dbinom(0:20, 20, 0.5))) ggplot(shapes, aes(y, p)) + geom_col(width = 0.85, fill = "#8ba3c7") + facet_wrap(~ panel, scales = "free_x") + scale_x_continuous(breaks = function(l) if (l[2] > 12) seq(0, 20, 5) else seq(0, 10, 2), minor_breaks = NULL) + scale_y_continuous(breaks = seq(0, 0.4, 0.1)) + labs(x = "Number of successes, y", y = "p(y)") + theme_minimal(base_size = 20) + theme(strip.text = element_text(size = 20, face = "bold")) ``` ::: {style="font-size: 28px"} Small $p$: piled up near 0 and skewed right. $p = 0.5$: symmetric. Larger $n$: wider and lower. ::: --- ## ๐ Table 1 and `pbinom`: Card Declines ::: {style="font-size: 29px"} The acquiring bank's contract says a terminal's decline rate is **5%**. An auditor pulls $n = 25$ transactions and finds **4** declines. How surprising is that, if the contract is right? ::: {.fragment} $$P(Y \ge 4) = 1 - P(Y \le 3) = 1 - \sum_{y=0}^{3} p(y)$$ Table 1, Appendix 3: the table for $n = 25$, column $p = .05$, row $a = 3$ gives $0.966$. In R: `pbinom(3, 25, 0.05)` returns the same $0.966$. ::: ::: {.fragment} $$P(Y \ge 4) = 1 - 0.966 = 0.034$$ Either a 3-in-100 event happened, or the true decline rate is **higher than 5%**. As in Example 3.9, the auditor has grounds to question the reported rate. ::: ::: --- ## ๐ Theorem 3.7: Mean and Variance ::: {style="font-size: 30px"} ::: {.callout-important} ## Theorem 3.7 Let $Y$ be a binomial random variable based on $n$ trials and success probability $p$. Then $$\mu = E(Y) = np \qquad\text{and}\qquad \sigma^2 = V(Y) = npq.$$ ::: ::: {.fragment} **SME book:** $\mu = 20(0.05) = 1$ default a year, $\sigma^2 = 20(0.05)(0.95) = 0.95$, $\sigma = 0.975$. ::: ::: {.fragment} The mean is what intuition says: 5% of 20. The variance is the new information โ and $pq$ is largest at $p = 0.5$, where the outcome is least predictable. ::: ::: --- ## ๐ The Proof: Two Tricks Worth Keeping ::: {style="font-size: 28px"} **Mean.** The $y = 0$ term vanishes; cancel $y$ against $y!$, factor out $np$, set $z = y - 1$: $$E(Y) = np \sum_{z=0}^{n-1} \binom{n-1}{z} p^z q^{n-1-z} = np \cdot 1$$ The sum is a binomial distribution on $n - 1$ trials, so it equals 1. ::: {.fragment} **Variance.** $E(Y^2)$ does not cancel against $y!$, but $E[Y(Y-1)]$ does. The same device gives $E[Y(Y-1)] = n(n-1)p^2$, so $$\sigma^2 = E[Y(Y-1)] + \mu - \mu^2 = n(n-1)p^2 + np - n^2p^2 = npq.$$ ::: ::: {.fragment} Trick 1: $\sum p(y) = 1$ evaluates a sum for free. Trick 2: find $E[Y(Y-1)]$, not $E(Y^2)$. ::: ::: --- ## ๐ Worked Example: A Sales Shift ::: {style="font-size: 28px"} A mobile operator's outbound team makes $n = 40$ calls a shift; each converts to a sale with $p = 0.15$, independently. An agent earns **30 AZN** base plus **12 AZN** per sale. ::: {.fragment} Sales: $E(Y) = 40(0.15) = 6$, $\;V(Y) = 40(0.15)(0.85) = 5.1$, $\;\sigma = 2.26$. ::: ::: {.fragment} Pay $W = 30 + 12Y$ โ Wednesday's linear rule, Theorems 3.3โ3.5 and $V(aY + b) = a^2 V(Y)$: $$E(W) = 30 + 12(6) = 102 \text{ AZN}, \qquad V(W) = 12^2(5.1) = 734.4, \qquad \sigma_W = 27.10 \text{ AZN}$$ ::: ::: {.fragment} A 10-sale shift (150 AZN): $P(Y \ge 10) = 1 -$ `pbinom(9, 40, 0.15)` $= 0.067$. ::: ::: --- ## ๐ฏ Using the Model Backwards ::: {style="font-size: 28px"} The regulator tests 20 households of a new ISP and **6** reach 100 Mbps. Which $p$ makes that result most probable? (Example 3.10) ::: {.fragment} Maximise $\ln P(Y = 6) = \ln\binom{20}{6} + 6\ln p + 14\ln(1-p)$: $$\frac{6}{p} - \frac{14}{1-p} = 0 \;\Rightarrow\; \hat{p} = \frac{6}{20} = 0.30$$ ::: ```{r} #| label: likelihood-figure #| echo: false #| fig-width: 11 #| fig-height: 2.3 lik <- data.frame(p = seq(0, 1, by = 0.005)) lik$L <- dbinom(6, 20, lik$p) ggplot(lik, aes(p, L)) + geom_line(linewidth = 1.2, colour = "#14130f") + geom_vline(xintercept = 0.3, colour = "#8b2635", linewidth = 1.1, linetype = "dashed") + annotate("text", x = 0.42, y = 0.17, hjust = 0, size = 6.5, colour = "#8b2635", label = "p = 0.30, P(Y = 6) = 0.192") + scale_x_continuous(breaks = seq(0, 1, 0.1)) + labs(x = "Candidate p", y = "P(Y = 6)") + theme_minimal(base_size = 19) ``` ::: --- ## ๐ง 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 microfinance lender in Ganja holds **10** independent loans of **8,000 AZN** each. Each defaults with probability **0.10**, and a default loses the full principal. **Four minutes, in pairs:** 1. What is the probability that no loan defaults? 2. What is the probability of **two or more** defaults? 3. What are the mean and standard deviation of the AZN loss? ::: --- ## โ
Think-Pair-Share: Solution ::: {style="font-size: 29px"} $Y \sim$ binomial with $n = 10$, $p = 0.10$, $q = 0.90$. 1. $p(0) = (0.9)^{10} = 0.3487$ 2. $p(1) = \binom{10}{1}(0.1)(0.9)^9 = 0.3874$, so $P(Y \ge 2) = 1 - 0.3487 - 0.3874 = 0.2639$ ::: {.fragment} 3. Loss $L = 8000\,Y$. $E(Y) = np = 1$ and $V(Y) = npq = 0.9$, so $$E(L) = 8{,}000 \text{ AZN}, \qquad \sigma_L = 8000\sqrt{0.9} = 7{,}589 \text{ AZN}$$ The standard deviation is nearly as large as the mean: ten loans are too few to diversify away default risk. ::: ::: --- ## ๐ฌ Interactive: How $p$ Moves the Shape {.smaller} ```{ojs} //| echo: false viewof p_hit = { const input = Inputs.range([0.05, 0.95], {value: 0.35, step: 0.01, label: "p, each household hits 100 Mbps:"}); ['pointerdown','touchstart','mousedown','click','wheel','pointermove','touchmove'] .forEach(e => input.addEventListener(e, ev => ev.stopPropagation())); return input; } ``` ```{ojs} //| echo: false nTest = 20 choose = (n, k) => { let c = 1; for (let i = 1; i <= k; i++) c = c * (n - k + i) / i; return c; } pmfB = Array.from({length: nTest + 1}, (_, y) => ({y: y, p: choose(nTest, y) * Math.pow(p_hit, y) * Math.pow(1 - p_hit, nTest - y)})) md`n = 20 households ยท mean **np = ${(nTest * p_hit).toFixed(1)}** ยท variance **npq = ${(nTest * p_hit * (1 - p_hit)).toFixed(2)}**` Plot.plot({ width: 1150, height: 300, marginTop: 40, marginLeft: 78, marginBottom: 58, style: {fontSize: "18px"}, x: {label: "Households of 20 reaching 100 Mbps, y", domain: d3.range(0, 21), padding: 0.15}, y: {label: "p(y)", domain: [0, 0.4], ticks: [0, 0.1, 0.2, 0.3, 0.4], tickFormat: ".1f", grid: true}, marks: [ Plot.barY(pmfB, {x: "y", y: "p", fill: d => d.y === Math.round(nTest * p_hit) ? "#8b2635" : "#8ba3c7"}), Plot.ruleY([0]) ] }) ``` ::: {style="font-size: 28px"} The red bar is the one nearest $np$. Near $p = 0.05$ or $0.95$ the mass piles against an end; at $p = 0.5$ it is symmetric and widest. ::: --- ## ๐ Quiz #1: Is It Binomial? {.quiz-question} Which of these random variables has a binomial distribution? - [The number of 50 card transactions declined, each declined with probability 0.03, independently]{.correct data-explanation="โ
Fixed n = 50, two outcomes, constant p = 0.03, independent trials, and Y counts successes: all five properties of Definition 3.6 hold."} - The number of cold calls made until the first sale - The total AZN value of 20 independent transactions - The number of bad loans in a file review of 6 drawn from a pool of 10 loans, 3 of them bad --- ## ๐ Quiz #2: The Variance {.quiz-question} A telecom sells to each of $n = 50$ contacted customers with probability $p = 0.2$, independently. What is $V(Y)$? - [$8$]{.correct data-explanation="โ
Theorem 3.7: V(Y) = npq = 50 ร 0.2 ร 0.8 = 8. The mean is np = 10; the standard deviation is โ8 = 2.83."} - $10$ - $2.83$ - $40$ --- ## ๐ Quiz #3: Which Outcome Is the Success? {.quiz-question} Each of 8 independent SME loans **defaults** with probability 0.1. $Y$ = number of loans that **repay**. What is $P(Y = 8)$? - [$0.9^8 = 0.4305$]{.correct data-explanation="โ
Y counts repayments, so the success probability is p = 0.9, not 0.1. P(Y = 8) = 0.9โธ = 0.4305. Define the success first, then make sure p is its probability."} - $0.1^8 = 10^{-8}$ - $1 - 0.9^8 = 0.5695$ - $8 \times 0.9 = 7.2$ --- ## ๐ Key Formulas ::: {style="font-size: 30px"} | | Statement | |---|---| | Definition 3.7 | $p(y) = \binom{n}{y} p^y q^{n-y}, \quad y = 0, 1, \dots, n$ | | validity | $\sum_{y=0}^{n} p(y) = (q + p)^n = 1$ | | at least one | $P(Y \ge 1) = 1 - q^n$ | | Theorem 3.7 | $E(Y) = np, \qquad V(Y) = npq$ | | $p(y)$ in R | `dbinom(y, n, p)` | | $P(Y \le a)$ in R | `pbinom(a, n, p)` | | estimate of $p$ (Ex. 3.10) | $\hat{p} = y/n$ | ::: --- ## ๐ Summary ::: {style="font-size: 30px"} - A binomial experiment: fixed $n$, two outcomes, constant $p$, independent trials, $Y$ counts successes - Sampling a small part of a large population is approximately binomial; a large part is not (ยง3.7) - $p(y)$ is a count, $\binom{n}{y}$, times the probability of one ordering, $p^y q^{n-y}$ - $E(Y) = np$ and $V(Y) = npq$; with Wednesday's linear rules they price a loss or a pay scheme - An improbable count under an assumed $p$ is evidence against that $p$ - Name the success first โ then make sure $p$ is **its** probability ::: --- ## ๐ Practice Problems ::: {style="font-size: 28px"} **Wackerly, 7th edition**, exercises at the end of ยง3.4: - 3.38, 3.39, 3.40, 3.43, 3.48, 3.54, 3.56, 3.57, 3.59 **Week 6, Problem Set 1** is open now and closes **Sunday 25 October at 23:59** on WeBWorK, covering ยง3.4. **Quiz I** is this **Saturday, 17 October**, in the first 30 minutes of class, on **Chapters 1โ2**. **Next class:** Saturday 17 October โ the geometric and negative binomial distributions, Wackerly ยงยง3.5โ3.6: what happens when the number of trials is the random quantity. ::: --- ## ๐ Thank You ::: {style="font-size: 34px"} **Dr. Samir Orujov** ๐ง sorujov@ada.edu.az\ ๐ข Building D, Room D325\ ๐ Office hours: Wednesday, 16:00 โ 18:00 Slides and readings: **sorujov.net/teaching** ::: --- ## โ Questions ::: {style="font-size: 32px"} - Loan defaults tend to rise together in a recession. Which property of Definition 3.6 does that break, and in which direction does it push $P(Y \ge 3)$? - If $Y$ is binomial$(n, p)$, what is the distribution of $n - Y$? - For fixed $n$, which $p$ makes $V(Y)$ largest, and why is that the hardest book to reserve for? ::: 