```{r} #| label: setup #| include: false set.seed(2026) library(ggplot2) ``` ## π¬ The Idea in 5 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** a partition of $S$ (Definition 2.11) and choose one that makes the conditionals easy - **Compute** $P(A)$ by the law of total probability (Theorem 2.8), branch by branch - **Invert** a conditional with Bayes' rule (Theorem 2.9), reading the answer off a tree diagram - **Explain** why an accurate screen for a rare event produces mostly false alarms - **Update** a probability twice, using two conditionally independent signals in sequence ::: --- ## πΊοΈ Where We Are ::: {style="font-size: 32px"} **Wackerly Β§2.10** Last class ended on this: *what to do when the event depends on which state the world is in* β the law of total probability and Bayes' rule. ::: {.fragment} A loan's chance of default depends on its rating grade. A flagged card payment is fraud more often if fraud is common. In both cases $P(A \mid \text{state})$ is easy; $P(A)$ is not. ::: ::: {.fragment} Today: **add up over the states** to get $P(A)$, then **run the conditional backwards** to learn the state from $A$. ::: ::: --- ## π³ A Question Before Any Notation ::: {style="font-size: 32px"} A Baku bank screens every card payment with a fraud model. - It flags **95%** of fraudulent payments - It clears **97%** of legitimate ones - About **1 payment in 500** is fraudulent ::: {.fragment} A payment has just been flagged. **Is it probably fraud?** Write down a number now. We will come back to it. ::: ::: --- ## π Definition 2.11: A Partition ::: {style="font-size: 30px"} ::: {.callout-important} ## Definition 2.11 For some positive integer $k$, let the sets $B_1, B_2, \ldots, B_k$ be such that 1. $S = B_1 \cup B_2 \cup \cdots \cup B_k$ 2. $B_i \cap B_j = \emptyset$, for $i \neq j$. Then the collection $\{B_1, B_2, \ldots, B_k\}$ is said to be a **partition** of $S$. ::: ::: {.fragment} Every loan has **exactly one** rating grade: the grades cover the book and never overlap. "Rated A", "rated B" and "in arrears" are not a partition β a loan can be B and in arrears. ::: ::: --- ## π Theorem 2.8: Total Probability ::: {style="font-size: 30px"} ::: {.callout-important} ## Theorem 2.8 If $\{B_1, \ldots, B_k\}$ is a partition of $S$ with $P(B_i) > 0$ for every $i$, then for any event $A$ $$P(A) = \sum_{i=1}^{k} P(A \mid B_i)\,P(B_i)$$ ::: ::: {.fragment} **Proof in one line.** $A = (A \cap B_1) \cup \cdots \cup (A \cap B_k)$, and these pieces are mutually exclusive, so Axiom 3 adds them. Theorem 2.5 writes each piece as $P(A \mid B_i)P(B_i)$. ::: ::: --- ## π¦ Default Across Three Grades ::: {style="font-size: 28px"} A bank's SME loan book is split by internal rating grade. $D$ is the event that a loan defaults within the year. | Grade | Share of book $P(B_i)$ | Default rate $P(D \mid B_i)$ | Product | |---|---|---|---| | A | 0.50 | 0.01 | 0.005 | | B | 0.35 | 0.04 | 0.014 | | C | 0.15 | 0.12 | 0.018 | | | **1.00** | | **0.037** | ::: {.fragment} $$P(D) = 0.005 + 0.014 + 0.018 = 0.037$$ The book-wide default rate, 3.7%, is a **weighted average** of the grade rates, with the grade shares as weights. ::: ::: --- ## π Theorem 2.9: Bayes' Rule ::: {style="font-size: 30px"} ::: {.callout-important} ## Theorem 2.9 (Bayes' Rule) If $\{B_1, \ldots, B_k\}$ is a partition of $S$ with $P(B_i) > 0$ for every $i$, then $$P(B_j \mid A) = \frac{P(A \mid B_j)\,P(B_j)}{\sum_{i=1}^{k} P(A \mid B_i)\,P(B_i)}$$ ::: ::: {.fragment} The numerator is **one row** of the table: $P(A \cap B_j)$. The denominator is **the total**: $P(A)$ from Theorem 2.8. Bayes' rule is Definition 2.9 with both pieces filled in. ::: ::: --- ## π Running the Loan Book Backwards ::: {style="font-size: 30px"} A loan has defaulted. Which grade did it come from? $$P(C \mid D) = \frac{0.018}{0.037} = 0.486, \quad P(B \mid D) = \frac{0.014}{0.037} = 0.378, \quad P(A \mid D) = \frac{0.005}{0.037} = 0.135$$ ::: {.fragment} Grade C is **15%** of the book but **49%** of the defaults. The three posteriors sum to 1 because they divide one partition's products by their own total. ::: ::: {.fragment} **Prior** $P(B_j)$: what you believed before the default. **Posterior** $P(B_j \mid D)$: what you believe after. ::: ::: --- ## π³ Example 2.23: A Card Batch ::: {style="font-size: 29px"} A card manufacturer personalises chip cards on **five lines** at equal rates. Normally 2% of chips are defective. In March **line 1** malfunctioned and produced 5% defectives. A bank receives a March batch of 100 cards, tests three, and **exactly one fails**. What is the probability the batch came from line 1? ::: {.fragment} Let $B$ = "batch from line 1" and $A$ = "exactly one of three fails". Then $P(B) = 0.2$, $P(\bar{B}) = 0.8$, and $$P(A \mid B) = 3(0.05)(0.95)^2 = 0.135375, \qquad P(A \mid \bar{B}) = 3(0.02)(0.98)^2 = 0.057624$$ ::: ::: --- ## π³ The Tree ```{r} #| label: card-tree #| echo: false #| fig-width: 11 #| fig-height: 4.6 pB <- 0.2 a1 <- 3 * 0.05 * 0.95^2 a0 <- 3 * 0.02 * 0.98^2 nodes <- data.frame( x = c(0, 1, 1, 2, 2, 2, 2), y = c(0, 1.5, -1.5, 2.25, 0.75, -0.75, -2.25), lab = c("", "B: line 1", "~B: lines 2-5", "A", "~A", "A", "~A")) edges <- data.frame( x = c(0, 0, 1, 1, 1, 1), y = c(0, 0, 1.5, 1.5, -1.5, -1.5), xend = c(1, 1, 2, 2, 2, 2), yend = c(1.5, -1.5, 2.25, 0.75, -0.75, -2.25), p = sprintf("%.4f", c(pB, 1 - pB, a1, 1 - a1, a0, 1 - a0))) joint <- c(pB * a1, pB * (1 - a1), (1 - pB) * a0, (1 - pB) * (1 - a0)) leaves <- data.frame(x = 2.55, y = c(2.25, 0.75, -0.75, -2.25), lab = sprintf("%.4f", joint), hit = c(TRUE, FALSE, TRUE, FALSE)) ggplot() + geom_segment(data = edges, aes(x, y, xend = xend, yend = yend), linewidth = 1, colour = "grey45") + geom_label(data = edges, aes((x + xend) / 2, (y + yend) / 2, label = p), size = 6.5, label.size = 0, fill = "white") + geom_label(data = nodes[-1, ], aes(x, y, label = lab), size = 7, fill = "#eef2f8") + geom_text(data = leaves, aes(x, y, label = lab, colour = hit, fontface = ifelse(hit, "bold", "plain")), size = 7) + scale_colour_manual(values = c(`TRUE` = "#8b2635", `FALSE` = "grey35"), guide = "none") + annotate("text", x = 2.55, y = 3, label = "P(branch)", size = 6.5) + xlim(-0.1, 2.85) + theme_void() ``` ::: {style="font-size: 28px"} Multiply **along** a path (Thm 2.5); add **across** the paths ending in $A$ (Thm 2.8). ::: --- ## π³ Reading the Answer Off the Tree ::: {style="font-size: 30px"} The two red leaves are the paths that end in $A$: $$P(A) = 0.0271 + 0.0461 = 0.0732$$ ::: {.fragment} Bayes' rule is **one red leaf over both red leaves**: $$P(B \mid A) = \frac{(0.135375)(0.2)}{0.0731742} = 0.37, \qquad P(\bar{B} \mid A) = 1 - 0.37 = 0.63$$ ::: ::: {.fragment} One failure in three nearly doubles the chance of line 1, from 0.20 to 0.37 β and still leaves the other four lines the likelier source. ::: ::: --- ## β οΈ Back to the Fraud Screen ::: {style="font-size: 30px"} $F$ = fraud, $T$ = flagged. $P(F) = 0.002$, $P(T \mid F) = 0.95$, $P(T \mid \bar{F}) = 0.03$. $$P(F \mid T) = \frac{0.95 \times 0.002}{0.95 \times 0.002 + 0.03 \times 0.998} = \frac{0.0019}{0.03184} = 0.060$$ ::: {.fragment} **Six percent.** In 100,000 payments: 200 are fraud and 190 of them are flagged; 99,800 are legitimate and **2,994** of them are flagged. Of 3,184 flags, 190 are fraud. ::: ::: {.fragment} This is the **base-rate paradox**: the screen is accurate, but the legitimate pool is so large that its 3% error outnumbers all the fraud there is. ::: ::: --- ## π¬ Interactive: Posterior vs Base Rate {.smaller} ```{ojs} //| echo: false viewof base_pct = { const input = Inputs.range([0.1, 50], {value: 0.2, step: 0.1, label: "Base rate of fraud (%):"}); ['pointerdown','touchstart','mousedown','click','wheel','pointermove','touchmove'] .forEach(e => input.addEventListener(e, ev => ev.stopPropagation())); return input; } ``` ```{ojs} //| echo: false sens = 0.95 fpr = 0.03 posterior = p => p * sens / (p * sens + (1 - p) * fpr) grid = Array.from({length: 121}, (_, i) => { const pct = 0.1 * Math.pow(500, i / 120); return {pct: pct, post: posterior(pct / 100)}; }) current = [{pct: base_pct, post: posterior(base_pct / 100)}] md`Screen fixed at 95% / 97%. At a base rate of **${base_pct.toFixed(1)}%**, a flag means fraud with probability **${(100 * current[0].post).toFixed(1)}%**.` Plot.plot({ width: 1150, height: 310, marginLeft: 78, marginTop: 40, marginBottom: 58, style: {fontSize: "18px"}, x: {type: "log", domain: [0.1, 50], label: "Base rate of fraud (%, log scale)", ticks: [0.1, 0.2, 0.5, 1, 2, 5, 10, 20, 50], tickFormat: d => String(d)}, y: {label: "P(fraud | flagged)", domain: [0, 1], tickFormat: ".1f"}, marks: [ Plot.ruleY([0.5], {stroke: "#cbb8a9", strokeDasharray: "4 4"}), Plot.line(grid, {x: "pct", y: "post", stroke: "#14130f", strokeWidth: 2}), Plot.dot(current, {x: "pct", y: "post", r: 9, fill: "#8b2635"}), Plot.ruleY([0]) ] }) ``` --- ## π§ What Would Actually Help? ```{r} #| label: screen-designs posterior <- function(base, sens, fpr) base * sens / (base * sens + (1 - base) * fpr) data.frame( design = c("current", "sensitivity 99.9%", "false alarms 1%", "false alarms 0.1%", "base rate 2%"), base = c(0.002, 0.002, 0.002, 0.002, 0.02), sens = c(0.95, 0.999, 0.95, 0.95, 0.95), fpr = c(0.03, 0.03, 0.01, 0.001, 0.03) ) |> transform(posterior = round(posterior(base, sens, fpr), 3)) ``` ::: {style="font-size: 28px"} Catching more fraud barely moves it. **Cutting false alarms** does β the denominator is where the legitimate pool lives. The posterior crosses 50% at a base rate of $0.03/0.98 = 3.1\%$. ::: --- ## π Sequential Updating ::: {style="font-size: 29px"} Flagged payments go to a **device check**: it fails for 90% of fraudulent payments and 5% of legitimate ones. Given fraud status, the two signals are **conditionally independent**. ::: {.fragment} **Today's posterior is tomorrow's prior.** After the flag, $P(F) = 0.0597$. Apply Bayes' rule again with the device check: $$P(F \mid T \cap V) = \frac{0.90 \times 0.0597}{0.90 \times 0.0597 + 0.05 \times 0.9403} = 0.533$$ ::: ::: {.fragment} Or do it in one step from the start, multiplying the conditionals (independence given $F$): $$\frac{0.002 \times 0.95 \times 0.90}{0.002 \times 0.95 \times 0.90 + 0.998 \times 0.03 \times 0.05} = \frac{0.00171}{0.003207} = 0.533$$ ::: ::: --- ## π» Two Routes, One Answer ```{r} #| label: sequential bayes <- function(prior, hit, false_alarm) { prior * hit / (prior * hit + (1 - prior) * false_alarm) } p0 <- 0.002 # base rate p1 <- bayes(p0, 0.95, 0.03) # after the model flags it p2 <- bayes(p1, 0.90, 0.05) # after the device check fails too one_step <- bayes(p0, 0.95 * 0.90, 0.03 * 0.05) round(c(prior = p0, after_flag = p1, after_both = p2, one_step = one_step), 4) ``` ::: {style="font-size: 28px"} The order of the checks does not matter, and neither does whether you stop in between. Without conditional independence the one-step route would need $P(V \mid T \cap F)$, not $P(V \mid F)$. ::: --- ## 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 subscribers are **60% prepaid**, **30% postpaid** and **10% business**. Monthly churn rates are 6%, 2% and 1%. **Four minutes, in pairs:** 1. What share of all subscribers churn this month? 2. A subscriber has just churned. What is the probability they were prepaid? 3. A subscriber stayed. What is the probability they are business? ::: --- ## β
Think-Pair-Share: Solution ::: {style="font-size: 30px"} 1. Let $L$ = the subscriber leaves. Theorem 2.8 over the three plans: $$P(L) = 0.6(0.06) + 0.3(0.02) + 0.1(0.01) = 0.036 + 0.006 + 0.001 = 0.043$$ 2. Theorem 2.9: $P(\text{prepaid} \mid L) = 0.036 / 0.043 = 0.837$. ::: {.fragment} 3. The same partition, with $\bar{L}$ in place of $L$. $P(\bar{L}) = 1 - 0.043 = 0.957$ and $$P(\text{business} \mid \bar{L}) = \frac{0.1 \times 0.99}{0.957} = 0.103$$ Staying tells you almost nothing β nearly everyone stays β so the posterior barely moves off the 10% prior. ::: ::: --- ## π Quiz #1: Total Probability {.quiz-question} $\{B_1, B_2\}$ is a partition with $P(B_1) = 0.7$. $P(A \mid B_1) = 0.1$ and $P(A \mid B_2) = 0.4$. What is $P(A)$? - [$0.19$]{.correct data-explanation="β
Theorem 2.8: 0.7 Γ 0.1 + 0.3 Γ 0.4 = 0.07 + 0.12 = 0.19. The conditionals are weighted by the partition probabilities, not averaged equally."} - $0.25$ - $0.50$ - $0.28$ --- ## π Quiz #2: Bayes' Rule {.quiz-question} Same partition as Quiz #1. Given that $A$ occurred, what is $P(B_2 \mid A)$? - [$0.63$]{.correct data-explanation="β
Theorem 2.9: P(Bβ | A) = 0.12 / 0.19 = 0.63. The numerator is Bβ's branch of the tree; the denominator is P(A) from Quiz #1."} - $0.40$ - $0.12$ - $0.30$ --- ## π Quiz #3: The Base-Rate Paradox {.quiz-question} A screen for a rare event (base rate 0.2%) has 95% sensitivity and a 3% false-alarm rate. Which change raises $P(\text{event} \mid \text{flag})$ the most? - Raising sensitivity to 99.9% - [Cutting the false-alarm rate to 0.1%]{.correct data-explanation="β
The posterior rises from 0.06 to 0.66. False alarms come from the huge non-event pool, which is the denominator's dominant term; sensitivity only acts on the tiny event pool."} - Screening twice as many payments - Raising sensitivity to 100% --- ## π Key Formulas ::: {style="font-size: 30px"} | | Statement | |---|---| | Definition 2.11 | $S = B_1 \cup \cdots \cup B_k$, $\; B_i \cap B_j = \emptyset$ for $i \neq j$ | | decomposition | $A = (A \cap B_1) \cup \cdots \cup (A \cap B_k)$ | | Theorem 2.8 | $P(A) = \sum_{i=1}^k P(A \mid B_i)P(B_i)$ | | Theorem 2.9 | $P(B_j \mid A) = \dfrac{P(A \mid B_j)P(B_j)}{\sum_{i=1}^k P(A \mid B_i)P(B_i)}$ | | two signals, independent given $B_j$ | $P(B_j \mid A_1 \cap A_2) \propto P(A_1 \mid B_j)P(A_2 \mid B_j)P(B_j)$ | ::: --- ## π Summary ::: {style="font-size: 30px"} - A partition splits $S$ into states that cover everything and never overlap - Total probability is a **weighted average** of the conditionals, weighted by the states - Bayes' rule is one branch of the tree divided by all the branches that end in $A$ - For a rare event, even an accurate screen gives a small posterior: **the base rate dominates** - False alarms, not missed detections, are what a rare-event screen must cut - Updating is sequential: yesterday's posterior is today's prior ::: --- ## π Practice Problems ::: {style="font-size: 28px"} **Wackerly, 7th edition** - Exercises at the end of Β§2.10 β start with 2.124, 2.125 and 2.129, then 2.126 β 2.127 on sensitivity and specificity, and 2.132 - Re-draw today's card-batch tree with $P(A \mid B)$ lowered to 0.12, and find $P(B \mid A)$ (compare Exercise 2.123(b)) **Week 4, Problem Set 2** is open now and closes **Sunday 11 October at 23:59** on WeBWorK, covering Β§2.10. **Next class:** turning outcomes into numbers β random variables, random sampling, and a synthesis of Chapter 2 (§§2.11β2.13). ::: --- ## 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 flag on the fraud screen raised the probability from 0.2% to 6%. Is that a failure of the screen, or a success? - If the two fraud signals were positively correlated among legitimate payments, would the one-step answer be too high or too low? - Why must every $P(B_i)$ be strictly positive in Theorems 2.8 and 2.9? ::: 