```{r} #| label: setup #| include: false set.seed(2026) library(ggplot2) theme_set(theme_minimal(base_size = 22)) ``` ## π― Learning Objectives ::: {style="font-size: 32px"} By the end of this lecture, you will be able to: - **Explain** why two marginal probabilities do not determine the probability of a joint event - **Construct** a joint probability function from a sample space, and check it against Theorem 5.1 - **Evaluate** a joint distribution function $F(y_1, y_2)$ from a table, including the rectangle rule of Theorem 5.2 - **Compute** a joint probability as a double integral of a joint density over a sketched region - **Extend** the definitions from two random variables to $n$ ::: --- ## πΊοΈ Where We Are ::: {style="font-size: 32px"} **Wackerly Β§5.1β5.2 β the start of Chapter 5** Saturday closed Chapter 4: one continuous random variable at a time β its density, its expected values, its MGF, and Tchebysheff's bound on how far it strays. ::: {.fragment} Every model so far has described **one** quantity. A loan book, a portfolio, a payment system is never one quantity. ::: ::: {.fragment} Today the object becomes a **pair** $(Y_1, Y_2)$, and the question becomes how two random variables behave *together*. This chapter carries the rest of the course. ::: ::: --- ## β The Question This Lecture Answers ::: {style="font-size: 30px"} Two SME borrowers of a Baku bank, both in construction, each default within the year with probability $0.07$. What is the probability that **both** default? ::: {.fragment} - If they are unrelated: $0.07 \times 0.07 = 0.0049$ - If they always move together: $0.07$ ::: ::: {.fragment} A fourteen-fold range, and the two marginals cannot choose between them. The answer lives in the **joint** distribution, which Β§5.2 defines. ::: ::: {.fragment} Β§5.1: every sample $(y_1, \ldots, y_n)$ is also such an intersection, $(Y_1 = y_1, \ldots, Y_n = y_n)$. We build the tools with $n = 2$, then extend. ::: ::: --- ## π Definition 5.1 and Theorem 5.1 ::: {style="font-size: 29px"} ::: {.callout-note} ## Definition 5.1 Let $Y_1$ and $Y_2$ be discrete random variables. The **joint (or bivariate) probability function** for $Y_1$ and $Y_2$ is $$p(y_1, y_2) = P(Y_1 = y_1, Y_2 = y_2), \quad -\infty < y_1 < \infty,\ -\infty < y_2 < \infty.$$ ::: ::: {.callout-important} ## Theorem 5.1 1. $p(y_1, y_2) \ge 0$ for all $y_1, y_2$. 2. $\sum_{y_1, y_2} p(y_1, y_2) = 1$, summing over all pairs given nonzero probability. ::: ::: --- ## π¦ Routing Two Loan Applications ::: {style="font-size: 28px"} Two applications are each sent **at random, independently**, to committee 1 (Corporate), 2 (SME) or 3 (Retail). $Y_1$ = number sent to Corporate, $Y_2$ = number to SME. All $3 \times 3 = 9$ routings $\{i, j\}$ are equally likely. | | $y_1 = 0$ | $y_1 = 1$ | $y_1 = 2$ | |---|:---:|:---:|:---:| | $y_2 = 0$ | $1/9$ | $2/9$ | $1/9$ | | $y_2 = 1$ | $2/9$ | $2/9$ | $0$ | | $y_2 = 2$ | $1/9$ | $0$ | $0$ | ::: {.fragment} $p(1,1) = 2/9$: both $\{1,2\}$ and $\{2,1\}$ give it. The zeros say $Y_1 + Y_2 \le 2$: **the support is a region, not a rectangle**. ::: ::: --- ## π Joint Returns of Two Assets ::: {style="font-size: 28px"} Monthly returns (%) on a Baku-listed bank share, $Y_1$, and an oil-sector bond fund, $Y_2$: | | $y_2 = -1$ | $y_2 = 1$ | $y_2 = 3$ | |---|:---:|:---:|:---:| | $y_1 = -2$ | $0.10$ | $0.08$ | $0.02$ | | $y_1 = 0$ | $0.06$ | $0.30$ | $0.09$ | | $y_1 = 4$ | $0.02$ | $0.12$ | $0.21$ | ::: {.fragment} - Both gain: $p(4,1) + p(4,3) = 0.33$. Both lose: $p(-2,-1) = 0.10$. - An equal-weight portfolio gains when $y_1 + y_2 > 0$: six cells, total $0.76$. ::: ::: --- ## π» The Table in R ```{r} #| label: joint-returns p <- matrix(c(0.10, 0.08, 0.02, 0.06, 0.30, 0.09, 0.02, 0.12, 0.21), nrow = 3, byrow = TRUE, dimnames = list(y1 = c(-2, 0, 4), y2 = c(-1, 1, 3))) y1 <- as.numeric(rownames(p)); y2 <- as.numeric(colnames(p)) Fjoint <- function(a, b) sum(p[y1 <= a, y2 <= b]) # Definition 5.2 c(min = min(p), total = sum(p), # Theorem 5.1 both_gain = sum(p[y1 > 0, y2 > 0]), port_gain = sum(p[outer(y1, y2, "+") > 0]), F01 = Fjoint(0, 1), F4m1 = Fjoint(4, -1)) ``` --- ## πΌοΈ Mass Sits on the Diagonal ```{r} #| label: joint-returns-fig #| echo: false #| fig-width: 10 #| fig-height: 4.6 d <- expand.grid(y1 = y1, y2 = y2) d$p <- as.vector(p) ggplot(d, aes(y1, y2)) + geom_point(aes(size = p), colour = "#8b2635", alpha = 0.8) + geom_text(aes(label = sprintf("%.2f", p)), nudge_x = 0.75, size = 7) + scale_size_area(max_size = 26, guide = "none") + scale_x_continuous(breaks = y1, limits = c(-2.7, 5.2)) + scale_y_continuous(breaks = y2, limits = c(-1.8, 3.8)) + labs(x = "Bank share return (%)", y = "Bond fund return (%)") ``` ::: {style="font-size: 28px"} The large masses run from bottom left to top right: the two assets tend to move together. ::: --- ## π Definition 5.2: The Joint CDF ::: {style="font-size: 29px"} ::: {.callout-note} ## Definition 5.2 For any random variables $Y_1$ and $Y_2$, the **joint distribution function** is $F(y_1, y_2) = P(Y_1 \le y_1, Y_2 \le y_2)$. For discrete variables, $F(y_1, y_2) = \sum_{t_1 \le y_1} \sum_{t_2 \le y_2} p(t_1, t_2)$. ::: On the returns table: - $F(0, 1) = 0.10 + 0.08 + 0.06 + 0.30 = 0.54$ - $F(0.5, 1.5) = 0.54$ as well β no new mass between the grid points - $F(-3, 5) = 0$ and $F(5, 5) = 1$ ::: --- ## π Theorem 5.2: Properties of $F$ ::: {style="font-size: 28px"} ::: {.callout-important} ## Theorem 5.2 1. $F(-\infty, -\infty) = F(-\infty, y_2) = F(y_1, -\infty) = 0$. 2. $F(\infty, \infty) = 1$. 3. If $y_1^* \ge y_1$ and $y_2^* \ge y_2$, then $F(y_1^*, y_2^*) - F(y_1^*, y_2) - F(y_1, y_2^*) + F(y_1, y_2) \ge 0$. ::: ::: {.fragment} Part 3 is $P(y_1 < Y_1 \le y_1^*,\ y_2 < Y_2 \le y_2^*)$: the rectangle rule. On the returns table, $$P(-2 < Y_1 \le 4,\ -1 < Y_2 \le 3) = 1 - 0.18 - 0.20 + 0.10 = 0.72,$$ which is the four cells $0.30 + 0.09 + 0.12 + 0.21$. ::: ::: --- ## π Quiz #1: Reading $F$ From a Table {.quiz-question} In the joint returns table, what is $F(4, -1) = P(Y_1 \le 4, Y_2 \le -1)$? - [$0.18$]{.correct data-explanation="β
Every value of Y1 is at most 4, so only the condition Y2 β€ β1 bites: the whole y2 = β1 column, 0.10 + 0.06 + 0.02 = 0.18."} - $0.02$ - $0.10$ - $1$ --- ## π Definition 5.3 and Theorem 5.3 ::: {style="font-size: 28px"} ::: {.callout-note} ## Definition 5.3 If there is a nonnegative $f(y_1, y_2)$ with $F(y_1, y_2) = \int_{-\infty}^{y_1} \int_{-\infty}^{y_2} f(t_1, t_2)\, dt_2\, dt_1$ for all $y_1, y_2$, then $Y_1, Y_2$ are **jointly continuous**, and $f$ is the **joint probability density function**. ::: ::: {.callout-important} ## Theorem 5.3 1. $f(y_1, y_2) \ge 0$ for all $y_1, y_2$. 2. $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(y_1, y_2)\, dy_1\, dy_2 = 1$. ::: Probabilities are now **volumes**: $P(a_1 \le Y_1 \le a_2,\ b_1 \le Y_2 \le b_2) = \int_{b_1}^{b_2} \int_{a_1}^{a_2} f(y_1, y_2)\, dy_1\, dy_2$. ::: --- ## β±οΈ Two Settlement Times ::: {style="font-size: 29px"} An AZN/USD foreign-exchange trade settles in two legs. $Y_1$ = hours until the **first** leg settles, $Y_2$ = hours until the **second**. By construction $Y_1 \le Y_2$. Suppose $$f(y_1, y_2) = \begin{cases} 2e^{-(y_1 + y_2)}, & 0 \le y_1 \le y_2 < \infty,\\ 0, & \text{elsewhere.} \end{cases}$$ ::: {.fragment} **Check Theorem 5.3.** $f \ge 0$, and, integrating $y_1$ first over $0 \le y_1 \le y_2$, $$\int_0^\infty 2e^{-y_2}\left(1 - e^{-y_2}\right) dy_2 = 2\left(1 - \tfrac{1}{2}\right) = 1.$$ ::: ::: --- ## βοΈ Worked Example: Two Probabilities ::: {style="font-size: 28px"} **Both legs settled within one hour.** Only $Y_2 \le 1$ bites, because $Y_1 \le Y_2$: $$F(1, 1) = \int_0^1 \int_0^{y_2} 2e^{-(y_1 + y_2)}\, dy_1\, dy_2 = 2(1 - e^{-1}) - (1 - e^{-2}) = (1 - e^{-1})^2 = 0.3996$$ ::: {.fragment} **First leg within 30 minutes, second after an hour.** The rectangle $0 \le y_1 \le 0.5,\ y_2 > 1$ lies wholly inside the support: $$\int_0^{0.5} \int_1^{\infty} 2e^{-(y_1 + y_2)}\, dy_2\, dy_1 = 2e^{-1}\left(1 - e^{-0.5}\right) = 0.2895$$ ::: ::: --- ## β οΈ Settlement Risk: The Gap ::: {style="font-size: 28px"} While one leg has settled and the other has not, the bank is exposed to its counterparty. What is $P(Y_2 - Y_1 > 1)$? ::: {.fragment} **Sketch the region first.** It is the part of the support above the line $y_2 = y_1 + 1$. For each $y_1$, $y_2$ runs from $y_1 + 1$ to $\infty$: $$\int_0^{\infty} \int_{y_1 + 1}^{\infty} 2e^{-(y_1 + y_2)}\, dy_2\, dy_1 = \int_0^{\infty} 2e^{-1}e^{-2y_1}\, dy_1 = e^{-1} = 0.3679$$ ::: ::: {.fragment} More than a third of trades leave the bank exposed for over an hour. ::: ::: --- ## π» Checking the Integrals by Simulation ```{r} #| label: settlement-sim # Two independent exponential(1) legs; the earlier is Y1, the later Y2. # This pair has exactly the joint density 2exp(-(y1 + y2)) on 0 <= y1 <= y2. n <- 100000 a <- rexp(n); b <- rexp(n) Y1 <- pmin(a, b); Y2 <- pmax(a, b) rbind(simulated = c(F_1_1 = mean(Y1 <= 1 & Y2 <= 1), rect = mean(Y1 <= 0.5 & Y2 > 1), gap = mean(Y2 - Y1 > 1)), exact = c((1 - exp(-1))^2, 2 * exp(-1) * (1 - exp(-0.5)), exp(-1))) |> round(4) ``` ::: {style="font-size: 28px"} A hundred thousand simulated trades land within $0.003$ of all three integrals β the region and the limits were right. ::: --- ## πΌοΈ Where the Gap Lives ```{r} #| label: settlement-fig #| echo: false #| fig-width: 10 #| fig-height: 4.8 s <- data.frame(y1 = Y1[1:3000], y2 = Y2[1:3000]) s$gap <- ifelse(s$y2 - s$y1 > 1, "Gap over 1 hour", "Gap under 1 hour") ggplot(s, aes(y1, y2, colour = gap)) + geom_point(alpha = 0.45, size = 1.6) + geom_abline(intercept = 0, slope = 1, linewidth = 0.9) + geom_abline(intercept = 1, slope = 1, linetype = "dashed", linewidth = 0.9) + scale_colour_manual(values = c("Gap over 1 hour" = "#8b2635", "Gap under 1 hour" = "#8ba3c7"), name = NULL) + coord_cartesian(xlim = c(0, 3), ylim = c(0, 5)) + labs(x = expression("First leg " * Y[1] * " (hours)"), y = expression("Second leg " * Y[2] * " (hours)")) + theme(legend.position = "top", legend.text = element_text(size = 18)) ``` ::: {style="font-size: 28px"} No point falls below the solid line $y_2 = y_1$; the red points above the dashed line are the event. ::: --- ## 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"} Back to the two construction borrowers. $Y_i = 1$ if borrower $i$ defaults, $0$ if not: | | $y_2 = 0$ | $y_2 = 1$ | |---|:---:|:---:| | $y_1 = 0$ | $0.88$ | $0.05$ | | $y_1 = 1$ | $0.05$ | $c$ | **Four minutes, in pairs:** (1) find $c$; (2) find the probability that at least one defaults; (3) compare $P(\text{both})$ with $0.07 \times 0.07$. ::: --- ## β
Think-Pair-Share: Solution ::: {style="font-size: 29px"} 1. Theorem 5.1: the entries sum to 1, so $c = 1 - 0.88 - 0.05 - 0.05 = 0.02$. Each borrower still defaults with probability $0.05 + 0.02 = 0.07$. 2. "At least one" fails only at $(0, 0)$: $1 - p(0,0) = 1 - 0.88 = 0.12$. ::: {.fragment} 3. $P(Y_1 = 1, Y_2 = 1) = 0.02$, about **four times** the $0.0049$ that unrelated borrowers would give. ::: ::: {.fragment} Same marginals, very different risk of a double loss. That is the opening question answered: **only the joint table could say.** ::: ::: --- ## π Quiz #2: Making a Density Valid {.quiz-question} A credit line's limit $Y_1$ and the amount drawn $Y_2$ (million AZN) have constant density $f(y_1, y_2) = k$ on the triangle $0 \le y_2 \le y_1 \le 2$. What is $k$? - [$1/2$]{.correct data-explanation="β
Theorem 5.3 needs the volume to be 1. The triangle has area (1/2)(2)(2) = 2, so k Γ 2 = 1 and k = 1/2."} - $1$ - $1/4$ - $2$ --- ## π Key Formulas ::: {style="font-size: 28px"} | | Statement | |---|---| | Definition 5.1 | $p(y_1, y_2) = P(Y_1 = y_1, Y_2 = y_2)$ | | Theorem 5.1 | $p \ge 0$, $\ \sum_{y_1, y_2} p(y_1, y_2) = 1$ | | Definition 5.2 | $F(y_1, y_2) = P(Y_1 \le y_1, Y_2 \le y_2)$ | | Theorem 5.2, part 3 | $P(y_1 < Y_1 \le y_1^*, y_2 < Y_2 \le y_2^*) = F(y_1^*, y_2^*) - F(y_1^*, y_2) - F(y_1, y_2^*) + F(y_1, y_2)$ | | Theorem 5.3 | $f \ge 0$, $\ \iint f(y_1, y_2)\, dy_1\, dy_2 = 1$ | | $n$ variables | $F(y_1, \ldots, y_n) = \int_{-\infty}^{y_1} \cdots \int_{-\infty}^{y_n} f(t_1, \ldots, t_n)\, dt_n \cdots dt_1$ | ::: --- ## π Summary ::: {style="font-size: 30px"} - Marginal probabilities alone cannot price a joint event; the joint distribution can - A joint probability function is a table of nonnegative entries summing to 1 - $F(y_1, y_2)$ accumulates mass below and to the left; differences of $F$ give rectangles - For jointly continuous variables, probabilities are volumes under $f(y_1, y_2)$ - **Sketch the support and the event before writing the limits** - Everything extends to $(Y_1, \ldots, Y_n)$ β the setting of every sample ::: --- ## π Practice Problems ::: {style="font-size: 28px"} **Wackerly, 7th edition** - Exercises at the end of Β§5.2 β start with 5.1, 5.3 and 5.4 (tables), then 5.5, 5.7, 5.9 and 5.11 (densities: sketch every region first) - On the settlement model, find $P(Y_1 + Y_2 < 2)$ and check it with the simulation code above **Week 12, Problem Set 1** is open now and closes **Sunday 6 December at 23:59** on WeBWorK, covering Β§5.1β5.2. **Next class:** 28 November β marginal and conditional probability distributions (Wackerly Β§5.3): how to recover each borrower's 7% from the joint table, and what one default says about the other. ::: --- ## 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"} - Given two marginal default probabilities of $0.07$, what are the largest and smallest values $P(\text{both default})$ can take? - Why must a joint density's support be sketched before the limits are written, when a single density's never had to be? - In the settlement model, is $P(Y_2 - Y_1 > h)$ a shape worth knowing for every $h$? Guess its form from $h = 1$. ::: 