```{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: - **Recover** the marginal distribution of one variable from a joint table by summing, and from a joint density by integrating (Definition 5.4) - **Read** the limits of that integral off a sketch of the support - **Compute** a conditional probability function $p(y_1 \mid y_2)$ and a conditional density $f(y_1 \mid y_2)$ (Definitions 5.5 and 5.7) - **Build** a joint table from a marginal and a set of conditionals, and reverse the conditioning - **Condition** a claim-size density on the event that the claim exceeds a deductible ::: --- ## πΊοΈ Where We Are ::: {style="font-size: 30px"} **Wackerly Β§5.3** Wednesday opened Chapter 5 with **bivariate and multivariate probability distributions**: a joint probability function $p(y_1, y_2)$ or a joint density $f(y_1, y_2)$ describes two random variables *at once*, and we found probabilities of events about the pair. ::: {.fragment} A joint distribution holds more than we usually ask of it. Today we take two things out of it: the distribution of **one variable alone** (the *marginal*), and the distribution of one variable **once the other is known** (the *conditional*). ::: ::: --- ## β Motivating Question ::: {.callout-important} ## The retail desk's question A Baku bank's household survey records monthly **income** $Y_1$ and monthly **consumption spending** $Y_2$, in AZN, as a joint table. The card team wants the distribution of spending across all customers. The credit team wants it **for customers earning 600 AZN**. Same table, two different questions. ::: ::: {.fragment style="font-size: 30px"} The first is a **marginal** distribution; the second is a **conditional** one. Both are read out of the joint table without any new data. ::: --- ## π Definition 5.4: Marginal Distributions ::: {style="font-size: 30px"} ::: {.callout-note} ## Definition 5.4 **a.** If $Y_1, Y_2$ are jointly discrete with probability function $p(y_1, y_2)$, the marginal probability functions are $$p_1(y_1) = \sum_{\text{all } y_2} p(y_1, y_2), \qquad p_2(y_2) = \sum_{\text{all } y_1} p(y_1, y_2).$$ **b.** If $Y_1, Y_2$ are jointly continuous with density $f(y_1, y_2)$, the marginal densities are $$f_1(y_1) = \int_{-\infty}^{\infty} f(y_1, y_2)\,dy_2, \qquad f_2(y_2) = \int_{-\infty}^{\infty} f(y_1, y_2)\,dy_1.$$ ::: The events $(Y_1 = y_1, Y_2 = y_2)$ are mutually exclusive, so $(Y_1 = y_1)$ is their union over $y_2$: sum out the variable you do not want. ::: --- ## π Income and Consumption ::: {style="font-size: 30px"} | $p(y_1, y_2)$ | $y_2 = 500$ | $y_2 = 900$ | $y_2 = 1500$ | $p_1(y_1)$ | |---|---|---|---|---| | $y_1 = 600$ | 0.20 | 0.08 | 0.02 | **0.30** | | $y_1 = 1200$ | 0.10 | 0.25 | 0.10 | **0.45** | | $y_1 = 2400$ | 0.02 | 0.08 | 0.15 | **0.25** | | $p_2(y_2)$ | **0.32** | **0.41** | **0.27** | 1 | ::: {.fragment} Row totals give income alone; column totals give spending alone. The card team's answer is the bottom row: $p_2(500) = 0.20 + 0.10 + 0.02 = 0.32$. ::: ::: {.fragment} The totals sit in the **margins** of the table, which is where the name comes from. ::: ::: --- ## β½ A Fuel Station's Day ::: {style="font-size: 30px"} A Sumgait filling station starts the day with $Y_2$ thousand litres of diesel and sells $Y_1$. With no delivery during the day, $Y_1 \le Y_2$, and $$f(y_1, y_2) = \tfrac{1}{8}, \qquad 0 \le y_1 \le y_2 \le 4 \quad (0 \text{ elsewhere}).$$ ::: {.fragment} **Sales alone.** For fixed $y_1$, the density is positive only for $y_1 \le y_2 \le 4$: $$f_1(y_1) = \int_{y_1}^{4} \tfrac{1}{8}\,dy_2 = \frac{4 - y_1}{8}, \qquad 0 \le y_1 \le 4.$$ ::: ::: {.fragment} **Stock alone.** For fixed $y_2$, it is positive only for $0 \le y_1 \le y_2$: $$f_2(y_2) = \int_{0}^{y_2} \tfrac{1}{8}\,dy_1 = \frac{y_2}{8}, \qquad 0 \le y_2 \le 4.$$ ::: ::: --- ## π The Limits Come From the Picture ```{r} #| label: triangle-figure #| echo: false #| fig-width: 10 #| fig-height: 4.6 tri <- data.frame(y1 = c(0, 4, 0), y2 = c(0, 4, 4)) ggplot(tri, aes(y1, y2)) + geom_polygon(fill = "#8ba3c7", alpha = 0.45, colour = "#14130f") + annotate("segment", x = 1, xend = 1, y = 1, yend = 4, colour = "#8b2635", linewidth = 2) + annotate("segment", x = 0, xend = 3, y = 3, yend = 3, colour = "#3d6e8f", linewidth = 2) + annotate("text", x = 1.15, y = 1.6, hjust = 0, size = 6.5, colour = "#8b2635", label = "f[1](1):~y[2]~runs~from~1~to~4", parse = TRUE) + annotate("text", x = 3.15, y = 3, hjust = 0, size = 6.5, colour = "#3d6e8f", label = "f[2](3):~y[1]~runs~from~0~to~3", parse = TRUE) + coord_cartesian(xlim = c(0, 6.8), ylim = c(0, 4.1)) + scale_x_continuous(breaks = 0:4) + scale_y_continuous(breaks = 0:4) + labs(x = expression(y[1]~"= sales (thousand litres)"), y = expression(y[2]~"= opening stock")) ``` ::: {style="font-size: 28px"} Integrating out $y_2$ walks **up** a vertical slice; integrating out $y_1$ walks **across** a horizontal one. Sketch the support first, every time. ::: --- ## π Definition 5.5: Discrete Conditionals ::: {style="font-size: 30px"} The multiplicative law, applied to the events $(Y_1 = y_1)$ and $(Y_2 = y_2)$, gives $$p(y_1, y_2) = p_1(y_1)\,p(y_2 \mid y_1) = p_2(y_2)\,p(y_1 \mid y_2).$$ ::: {.callout-note} ## Definition 5.5 If $Y_1, Y_2$ are jointly discrete with joint probability function $p(y_1, y_2)$ and marginals $p_1(y_1)$, $p_2(y_2)$, the conditional discrete probability function of $Y_1$ given $Y_2$ is $$p(y_1 \mid y_2) = P(Y_1 = y_1 \mid Y_2 = y_2) = \frac{p(y_1, y_2)}{p_2(y_2)}, \quad \text{provided } p_2(y_2) > 0.$$ ::: ::: --- ## π The Credit Team's Answer ::: {style="font-size: 30px"} Condition on income $Y_1 = 600$: keep that **row** and divide it by its total, $p_1(600) = 0.30$. $$\begin{aligned} p(500 \mid 600) &= 0.20 / 0.30 = 0.667 \\ p(900 \mid 600) &= 0.08 / 0.30 = 0.267 \\ p(1500 \mid 600) &= 0.02 / 0.30 = 0.067 \end{aligned}$$ ::: {.fragment} The three add to 1: a conditional distribution **is** a distribution, just over a smaller world. ::: ::: {.fragment} Across all customers, $P(Y_2 = 500) = 0.32$. Among the 600-AZN earners it is **0.667**. Knowing income more than doubles the chance of the lowest spending band. ::: ::: --- ## π» Every Conditional at Once ```{r} #| label: conditional-table #| code-fold: false p <- matrix(c(0.20, 0.08, 0.02, 0.10, 0.25, 0.10, 0.02, 0.08, 0.15), nrow = 3, byrow = TRUE, dimnames = list(income = c(600, 1200, 2400), spend = c(500, 900, 1500))) rowSums(p) # p1(y1): income alone round(p / rowSums(p), 3) # p(y2 | y1): each row divided by its total ``` --- ## π Spending Shifts With Income ```{r} #| label: conditional-figure #| echo: false #| fig-width: 10.5 #| fig-height: 4.6 cond <- p / rowSums(p) df <- rbind( data.frame(spend = colnames(p), prob = colSums(p), given = "All customers (marginal)"), data.frame(spend = colnames(p), prob = cond["600", ], given = "Income 600 AZN"), data.frame(spend = colnames(p), prob = cond["2400", ], given = "Income 2400 AZN")) df$spend <- factor(df$spend, levels = colnames(p)) df$given <- factor(df$given, levels = unique(df$given)) ggplot(df, aes(spend, prob, fill = given)) + geom_col(position = position_dodge(width = 0.8), width = 0.75) + geom_text(aes(label = sprintf("%.2f", prob)), vjust = -0.3, size = 5.5, position = position_dodge(width = 0.8)) + scale_fill_manual(values = c("grey70", "#8b2635", "#3d6e8f")) + scale_y_continuous(limits = c(0, 0.75), breaks = seq(0, 0.75, 0.25)) + labs(x = "Monthly consumption spending (AZN)", y = "Probability", fill = NULL) + theme(legend.position = "top", legend.text = element_text(size = 18)) ``` --- ## π¦ Loan Size and Default ::: {style="font-size: 29px"} A lender's book is 50% small loans ($Y_1 = 1$), 30% medium ($Y_1 = 2$), 20% large ($Y_1 = 3$). Annual default rates $p(1 \mid y_1)$ are 6%, 5% and 10%. So $p(y_1, 1) = p_1(y_1)\,p(1 \mid y_1)$: | | small | medium | large | $p_2(y_2)$ | |---|---|---|---|---| | $y_2 = 1$ (default) | 0.030 | 0.015 | 0.020 | **0.065** | | $y_2 = 0$ (repays) | 0.470 | 0.285 | 0.180 | **0.935** | ::: {.fragment} The risk committee asks the **other** conditional: what share of defaults are large? $$p(3 \mid 1) = \frac{p(3, 1)}{p_2(1)} = \frac{0.020}{0.065} = 0.308$$ Large loans are 20% of the book but 31% of the defaults. ::: ::: --- ## π Definition 5.7: Conditional Densities ::: {style="font-size: 29px"} For continuous variables $P(Y_2 = y_2) = 0$, so Definition 5.5 would divide by zero. Wackerly instead starts from the **conditional distribution function** $F(y_1 \mid y_2) = P(Y_1 \le y_1 \mid Y_2 = y_2)$ (Definition 5.6) and its integrand. ::: {.callout-note} ## Definition 5.7 For any $y_2$ with $f_2(y_2) > 0$, the conditional density of $Y_1$ given $Y_2 = y_2$ is $$f(y_1 \mid y_2) = \frac{f(y_1, y_2)}{f_2(y_2)},$$ and for any $y_1$ with $f_1(y_1) > 0$, $\;f(y_2 \mid y_1) = f(y_1, y_2)/f_1(y_1)$. ::: Same recipe as the table: take a **slice** of the joint, divide by its total. ::: --- ## β½ Back at the Fuel Station ::: {style="font-size: 30px"} Given an opening stock $y_2$, with $0 < y_2 \le 4$: $$f(y_1 \mid y_2) = \frac{1/8}{y_2/8} = \frac{1}{y_2}, \qquad 0 \le y_1 \le y_2.$$ Given the stock, sales are **uniform** on $[0, y_2]$. ::: {.fragment} Probability of selling at most 1 thousand litres: | | $P(Y_1 \le 1 \mid \cdot)$ | |---|---| | stock $Y_2 = 3$ | $\int_0^1 \frac{1}{3}\,dy_1 = 0.333$ | | stock $Y_2 = 1.5$ | $\int_0^1 \frac{1}{1.5}\,dy_1 = 0.667$ | | stock unknown, $f_1$ | $\int_0^1 \frac{4 - y_1}{8}\,dy_1 = 0.4375$ | ::: ::: --- ## π A Claim Above the Deductible ::: {style="font-size: 28px"} Motor claims $Y$ (thousand AZN) are gamma, $\alpha = 2$, $\beta = 0.4$: mean 800 AZN. With a 500 AZN deductible the insurer only **sees** claims with $Y > 0.5$. By Definition 2.9, $P(Y \le y \mid Y > d) = [F(y) - F(d)]/[1 - F(d)]$; differentiate: $$f(y \mid Y > d) = \frac{f(y)}{P(Y > d)}, \quad y > d \qquad (0 \text{ for } y \le d).$$ ```{r} #| label: deductible S <- function(y) pgamma(y, shape = 2, scale = 0.4, lower.tail = FALSE) round(c(P_above_d = S(0.5), scale_up = 1 / S(0.5), P_over_1.5 = S(1.5), P_over_1.5_given_d = S(1.5) / S(0.5)), 3) ``` Cut the density at $d$, then stretch what is left so it again has area 1. ::: --- ## π¬ Interactive: Move the Deductible {.smaller} ```{ojs} //| echo: false viewof d = { const input = Inputs.range([0, 2], {value: 0.5, step: 0.05, label: "Deductible d (thousand AZN):"}); ['pointerdown','touchstart','mousedown','click','wheel','pointermove','touchmove'] .forEach(e => input.addEventListener(e, ev => ev.stopPropagation())); return input; } ``` ```{ojs} //| echo: false fy = y => y * Math.exp(-y / 0.4) / 0.16 // gamma(2, 0.4) density Sd = Math.exp(-d / 0.4) * (1 + d / 0.4) // P(Y > d) grid = Array.from({length: 400}, (_, i) => { const y = (i + 1) / 100; return {y, f: fy(y), c: y > d ? fy(y) / Sd : 0}; }) md`P(Y > d) = **${Sd.toFixed(3)}**, so every surviving density value is multiplied by **${(1 / Sd).toFixed(2)}**. Dashed: f(y). Shaded: f(y | Y > d).` Plot.plot({ width: 1150, height: 290, marginLeft: 78, marginBottom: 58, style: {fontSize: "18px"}, x: {label: "Claim size y (thousand AZN)", domain: [0, 4], ticks: 8}, y: {label: "Density", domain: [0, 2.6]}, marks: [ Plot.areaY(grid, {x: "y", y: "c", fill: "#8b2635", fillOpacity: 0.35}), Plot.line(grid, {x: "y", y: "c", stroke: "#8b2635", strokeWidth: 2.5}), Plot.line(grid, {x: "y", y: "f", stroke: "#14130f", strokeWidth: 2, strokeDasharray: "6 4"}), Plot.ruleX([d], {stroke: "#3d6e8f", strokeWidth: 2}), 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: 29px"} An internet provider links $Y_1$, a household's broadband outages last month, to $Y_2 = 1$ if it then switched provider. | | $y_1 = 0$ | $y_1 = 1$ | $y_1 = 2$ | |---|---|---|---| | $y_2 = 1$ (switched) | 0.06 | 0.08 | 0.06 | | $y_2 = 0$ (stayed) | 0.54 | 0.22 | 0.04 | **Four minutes, in pairs:** (1) Find $p_1(y_1)$ and $p_2(1)$. (2) Find $p(1 \mid y_1)$ for each $y_1$. (3) Of the households that switched, what share had two outages? ::: --- ## β
Think-Pair-Share: Solution ::: {style="font-size: 30px"} 1. Column totals: $p_1(0) = 0.60$, $p_1(1) = 0.30$, $p_1(2) = 0.10$. Row total: $p_2(1) = 0.20$. 2. Divide each switching entry by its column total: $$p(1 \mid 0) = \tfrac{0.06}{0.60} = 0.10, \quad p(1 \mid 1) = \tfrac{0.08}{0.30} = 0.267, \quad p(1 \mid 2) = \tfrac{0.06}{0.10} = 0.60$$ ::: {.fragment} 3. Now condition the other way: $p(2 \mid 1) = 0.06/0.20 = 0.30$. Only 10% of households had two outages, but they are 30% of the churn. ::: ::: {.fragment} Part 2 is the engineer's number; part 3 is the marketing team's. **Name which variable is given before you divide.** ::: ::: --- ## π Quiz #1: Condition on the Column {.quiz-question} $Y_1 = 1$ if a household holds a term deposit, $Y_2 = 1$ if it holds a card loan: $p(0,0) = 0.30$, $p(0,1) = 0.20$, $p(1,0) = 0.10$, $p(1,1) = 0.40$. What is $P(Y_1 = 1 \mid Y_2 = 1)$? - [$0.667$]{.correct data-explanation="β
First the marginal: p2(1) = 0.20 + 0.40 = 0.60. Then Definition 5.5: 0.40 / 0.60 = 0.667."} - $0.40$ - $0.80$ - $0.50$ --- ## π Quiz #2: A Continuous Slice {.quiz-question} At the fuel station, $f(y_1, y_2) = 1/8$ on $0 \le y_1 \le y_2 \le 4$. Given an opening stock of $Y_2 = 2$, the conditional density of sales $Y_1$ is - [$1/2$ for $0 \le y_1 \le 2$]{.correct data-explanation="β
f2(2) = 2/8, so f(y1 | 2) = (1/8)/(2/8) = 1/2 on [0, 2]: uniform over what could be sold."} - $1/8$ for $0 \le y_1 \le 2$ - $(4 - y_1)/8$ for $0 \le y_1 \le 4$ - $1/4$ for $0 \le y_1 \le 4$ --- ## π Quiz #3: The Deductible {.quiz-question} A claim density has $P(Y > d) = 0.64$ at the deductible $d$. For a claim size $y > d$, the conditional density $f(y \mid Y > d)$ equals - [$f(y) / 0.64$]{.correct data-explanation="β
Cut at d, then rescale the surviving area 0.64 back to 1: divide by P(Y > d). The conditional density is larger than f(y) wherever it is not zero."} - $f(y)$, unchanged - $0.64 \, f(y)$ - $f(y) - f(d)$ --- ## π Key Formulas ::: {style="font-size: 28px"} | | Statement | |---|---| | Definition 5.4(a) | $p_1(y_1) = \sum_{\text{all } y_2} p(y_1, y_2)$ | | Definition 5.4(b) | $f_1(y_1) = \int_{-\infty}^{\infty} f(y_1, y_2)\,dy_2$ | | multiplicative law | $p(y_1, y_2) = p_1(y_1)\,p(y_2 \mid y_1) = p_2(y_2)\,p(y_1 \mid y_2)$ | | Definition 5.5 | $p(y_1 \mid y_2) = p(y_1, y_2)/p_2(y_2)$, $\; p_2(y_2) > 0$ | | Definition 5.6 | $F(y_1 \mid y_2) = P(Y_1 \le y_1 \mid Y_2 = y_2)$ | | Definition 5.7 | $f(y_1 \mid y_2) = f(y_1, y_2)/f_2(y_2)$, $\; f_2(y_2) > 0$ | | given an event | $f(y \mid Y > d) = f(y)/P(Y > d)$, $\; y > d$ | ::: --- ## π Summary ::: {style="font-size: 30px"} - A **marginal** removes a variable: sum it out of a table, integrate it out of a density - The integration limits come from the **support**; sketch it before integrating - A **conditional** keeps one slice of the joint and divides by that slice's total - $p(y_2 \mid y_1)$ and $p(y_1 \mid y_2)$ answer different questions; name what is given first - Conditioning on an event such as $Y > d$ truncates the density and rescales it by $1/P(Y > d)$ ::: --- ## π Practice Problems ::: {style="font-size: 28px"} **Wackerly, 7th edition** - Β§5.3: Exercises 5.19 β 5.38; start with 5.19, 5.22, 5.23, 5.25, 5.27, 5.33 and 5.38 - Exercise 5.25(f) and 5.33(d) ask whether a conditional equals a marginal. Hold on to your answers - Redo the deductible slide with $d = 1$ thousand AZN: what is $P(Y > 1.5 \mid Y > 1)$? **Week 12, Problem Set 2** is open now and closes **Sunday 6 December at 23:59** on WeBWorK, covering Β§5.3. **Next class:** 2 December, independent random variables (Wackerly Β§5.4): what it means when conditioning on one variable leaves the other's distribution unchanged. ::: --- ## 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"} - Two different joint tables can share the same two marginals. Can you build one for the loan book? - In the fuel example, why is $f(y_1 \mid y_2)$ undefined for $y_2 > 4$? - If $p(y_2 \mid y_1)$ were the same for every $y_1$, what would the income and consumption table look like? ::: 