```{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: - **Place** every tool of Chapters 1β5 on one map, and say which kind of question each one answers - **Choose** a probability model by reading what the random variable counts or measures - **Move** between joint, marginal and conditional distributions, and test independence - **Combine** Chapter 5's covariance and linear-function results with Chapter 4's normal model - **Solve** multi-part, exam-style problems that cross chapter boundaries ::: --- ## πΊοΈ Where We Are ::: {style="font-size: 30px"} **Wackerly Β§5.12** Wednesday: the bivariate normal distribution and conditional expectations. $E(Y_1) = E[E(Y_1 \mid Y_2)]$ (Theorem 5.14) and $V(Y_1) = E[V(Y_1 \mid Y_2)] + V[E(Y_1 \mid Y_2)]$ (Theorem 5.15), the last new theorems of the course. ::: {.fragment} Today there is **no new theory**. We put the five chapters on one page, then work four exam-style problems that each cross at least two chapters. ::: ::: {.fragment} **Midterm Examination II** is on **23 December**: comprehensive over Chapters 1β5, with the emphasis on Chapters 4 and 5. ::: ::: --- ## π§ What Chapter 5 Was For ::: {style="font-size: 30px"} The book's summary names the multinomial experiment (Β§5.9) as the chapter's theme: most experiments yield **several** measurements $y_1, y_2, \ldots, y_k$, observations on $k$ random variables. ::: {.fragment} To say how likely a whole sample is, we need **joint, marginal and conditional** distributions. ::: ::: {.fragment} A sample $Y_1, \ldots, Y_n$ is usually **independent** with one common distribution, and independence is what lets the joint probability factor. ::: ::: {.fragment} Wackerly's warning: do not drown in the details. They exist to prepare **inference**, the subject of Math Stat II. ::: ::: --- ## πΊοΈ The Course on One Page ::: {style="font-size: 28px"} | Ch. | The tools | The question they answer | |---|---|---| | 1 | $\bar{y}$, $s^2$, $s$; the empirical rule | What does this data set look like? | | 2 | axioms; counting; $P(A \mid B)$; Theorems 2.5β2.9 | How likely is this **event**? | | 3 | $p(y)$, $E(Y)$, $V(Y)$; five discrete models; $m(t)$, $P(t)$; Tchebysheff | How is this **count** distributed? | | 4 | $F(y)$, $f(y)$; uniform, normal, gamma, beta; $m(t)$; Tchebysheff | How is this **measurement** distributed? | | 5 | joint, marginal, conditional; Cov, $\rho$; Theorems 5.12β5.15 | How do **several** variables behave together? | : {tbl-colwidths="[7,56,37]"} Each chapter reuses the last: Chapter 5 is Chapters 3 and 4 in two dimensions. ::: --- ## π Which Model? Read the Variable ::: {style="font-size: 28px"} | The variable is | Model | Mean | |---|---|---| | successes in $n$ independent trials | binomial | $np$ | | the trial of the first (or $r$-th) success | geometric (negative binomial) | $1/p$ ($r/p$) | | successes drawn **without** replacement | hypergeometric | $nr/N$ | | events in a fixed time or space | Poisson | $\lambda$ | | a time to the next (or $\alpha$-th) event | exponential (gamma) | $\beta$ ($\alpha\beta$) | | a sum of many small effects | normal | $\mu$ | | a proportion on $[0, 1]$ | beta | $\alpha/(\alpha + \beta)$ | | counts in $k$ categories | multinomial | $np_i$ | : {tbl-colwidths="[45,35,20]"} ::: --- ## βοΈ Chapter 5 as One Chain ::: {style="font-size: 28px"} | Step | Discrete form | Result | |---|---|---| | joint | $p(y_1, y_2)$ | Definition 5.1 | | marginal | $p_1(y_1) = \sum_{y_2} p(y_1, y_2)$ | Definition 5.4 | | conditional | $p(y_1 \mid y_2) = p(y_1, y_2)/p_2(y_2)$ | Definition 5.5 | | independence | $p(y_1, y_2) = p_1(y_1)p_2(y_2)$ for **all** pairs | Theorem 5.4 | | covariance | $\text{Cov}(Y_1, Y_2) = E(Y_1Y_2) - \mu_1\mu_2$ | Theorem 5.10 | | linear functions | $V(\sum a_iY_i) = \sum a_i^2V(Y_i) + 2\sum_{i= 9), binomial", "P(Y >= 9), Poisson(4)", "Tchebysheff bound, k = 2", "exact P(1 <= Y <= 7)"), value = round(c(n * p, n * p * (1 - p), 1 - pbinom(8, n, p), 1 - ppois(8, 4), 1 - 1/2^2, pbinom(7, n, p) - pbinom(0, n, p)), 4)) ``` ::: {style="font-size: 28px"} Binomial, $\mu = 4$, $\sigma^2 = 3.96$. The treaty is triggered with probability about **0.021**; Poisson ($\lambda = np = 4$) misses by less than 0.001. Tchebysheff promises **at least 0.75**; the truth is **0.932**. ::: --- ## π Problem 2: Two Loans, One Sector ::: {style="font-size: 30px"} A Baku bank lends to two construction firms. $D_i = 1$ if firm $i$ defaults within the year: | | $D_2 = 0$ | $D_2 = 1$ | |---|---|---| | $D_1 = 0$ | 0.90 | 0.04 | | $D_1 = 1$ | 0.04 | 0.02 | a. Find the marginal default probabilities and $P(D_2 = 1 \mid D_1 = 1)$. Are $D_1$ and $D_2$ independent? b. Find $\text{Cov}(D_1, D_2)$ and $\rho$. c. Given that at least one firm defaults, what is the probability that firm 1 did? ::: --- ## β
Problem 2: Solution ::: {style="font-size: 30px"} a. $P(D_1 = 1) = 0.04 + 0.02 = 0.06$, and likewise $P(D_2 = 1) = 0.06$. $$P(D_2 = 1 \mid D_1 = 1) = \frac{0.02}{0.06} = 0.333 \ne 0.06,$$ so they are **dependent** (Theorem 5.4 fails at the cell $(1, 1)$: $0.02 \ne 0.06^2$). b. $E(D_1D_2) = P(D_1 = 1, D_2 = 1) = 0.02$, so $\text{Cov} = 0.02 - 0.06^2 = 0.0164$, and $$\rho = \frac{0.0164}{\sqrt{0.06 \times 0.94}\sqrt{0.06 \times 0.94}} = \frac{0.0164}{0.0564} = 0.291.$$ c. $P(\text{at least one}) = 1 - 0.90 = 0.10$, so $P(D_1 = 1 \mid \text{at least one}) = 0.06/0.10 = 0.6$. ::: --- ## π» Problem 2, Continued: The Loss ::: {style="font-size: 28px"} Exposures are 80,000 and 120,000 AZN, with 45% lost on default, so $L = 36{,}000D_1 + 54{,}000D_2$. ::: ```{r} #| label: loan-loss a <- 36000; b <- 54000 v <- 0.06 * 0.94; cv <- 0.02 - 0.06^2 EL <- (a + b) * 0.06 VL <- a^2 * v + b^2 * v + 2 * a * b * cv # Theorem 5.12 c(E_L = EL, sd_L = round(sqrt(VL)), sd_if_independent = round(sqrt(a^2 * v + b^2 * v))) ``` ::: {style="font-size: 28px"} The expected loss, 5,400 AZN, needs only the marginals. The spread does not: the covariance term raises the standard deviation from 15,413 to **17,359 AZN**. Same-sector loans default together, and Theorem 5.12 prices that in. ::: --- ## π Problem 3: A Two-Fund Portfolio ::: {style="font-size: 28px"} A client puts **60%** of 50,000 AZN in a bank-share fund $A$ and **40%** in an AZN bond fund $B$. Monthly returns (%): $\mu_A = 1.2$, $\sigma_A = 5$; $\mu_B = 0.8$, $\sigma_B = 1.5$; $\rho = -0.2$. Assume the portfolio return $R$ is normal. ::: {.fragment} **Mean** (Theorem 5.12): $E(R) = 0.6(1.2) + 0.4(0.8) = 1.04$. **Variance**: $V(R) = 0.36(25) + 0.16(2.25) + 2(0.6)(0.4)(-0.2)(5)(1.5) = 9 + 0.36 - 0.72 = 8.64$, so $\sigma_R = 2.939$. ::: ::: {.fragment} **Loss of more than 3%** (Chapter 4): $P(R < -3) = P\left(Z < \frac{-3 - 1.04}{2.939}\right) = P(Z < -1.37) = 0.085$. **5% worst case**: $1.04 - 1.645(2.939) = -3.79\%$, a loss of about **1,897 AZN**. ::: ::: --- ## π Problem 3: The Picture ```{r} #| label: portfolio-figure #| echo: false #| fig-width: 10 #| fig-height: 3.7 mp <- 1.04; sp <- sqrt(8.64) x <- seq(-10, 12, by = 0.02) df <- data.frame(x = x, port = dnorm(x, mp, sp), fundA = dnorm(x, 1.2, 5)) tail_df <- subset(df, x <= -3) ggplot(df, aes(x)) + geom_area(data = tail_df, aes(y = port), fill = "#8b2635", alpha = 0.35) + geom_line(aes(y = port, colour = "60/40 portfolio"), linewidth = 1.4) + geom_line(aes(y = fundA, colour = "fund A alone"), linewidth = 1.2, linetype = "dashed") + geom_vline(xintercept = -3, colour = "grey40") + scale_colour_manual(values = c("60/40 portfolio" = "#14130f", "fund A alone" = "#3d6e8f")) + labs(x = "Monthly return (%)", y = "Density", colour = NULL) + theme(legend.position = "top", legend.text = element_text(size = 18)) ``` ::: {style="font-size: 28px"} Shaded: $P(R < -3) = 0.085$. Holding fund $A$ alone, the same loss has probability $0.200$. The negative $\rho$ and the bond fund's small $\sigma$ do the work. ::: --- ## π§ Think-Pair-Share: Waiting Times ```{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 call centre has a **retail** line and a **business** line. The time to the next call is exponential, mean 4 minutes on retail ($Y_1$) and 6 minutes on business ($Y_2$), independently. **Four minutes, in pairs:** 1. Find $P(Y_1 > 5)$. The retail line has been silent for 3 minutes: now what is the chance of 5 more? 2. Find $E(Y_1 + Y_2)$ and $V(Y_1 + Y_2)$. 3. Find $P(Y_1 < Y_2)$, the chance that retail rings first. ::: --- ## β
Think-Pair-Share: Solution ::: {style="font-size: 28px"} 1. $P(Y_1 > 5) = e^{-5/4} = 0.287$, and by memorylessness $P(Y_1 > 8 \mid Y_1 > 3) = e^{-5/4} = 0.287$ as well. 2. Theorem 4.10 and Theorem 5.12 with $\text{Cov} = 0$ (independence, Theorem 5.11): $$E(Y_1 + Y_2) = 4 + 6 = 10, \qquad V(Y_1 + Y_2) = 4^2 + 6^2 = 52.$$ 3. Independence makes the joint density the product (Chapter 5), then integrate over $y_1 < y_2$: $$P(Y_1 < Y_2) = \int_0^\infty \frac{1}{4}e^{-y_1/4}\int_{y_1}^\infty \frac{1}{6}e^{-y_2/6}\,dy_2\,dy_1 = \int_0^\infty \frac{1}{4}e^{-5y_1/12}\,dy_1 = \frac{1/4}{5/12} = 0.6.$$ ::: --- ## π» Checking Part 3 by Simulation ```{r} #| label: waiting-sim y1 <- rexp(100000, rate = 1/4); y2 <- rexp(100000, rate = 1/6) round(c(retail_first = mean(y1 < y2), mean_sum = mean(y1 + y2), var_sum = var(y1 + y2)), 3) ``` ```{r} #| label: waiting-figure #| echo: false #| fig-width: 10 #| fig-height: 3.5 pts <- data.frame(y1 = y1[1:1500], y2 = y2[1:1500]) ggplot(pts, aes(y1, y2, colour = y1 < y2)) + geom_point(size = 1.8, alpha = 0.6) + geom_abline(slope = 1, intercept = 0) + coord_cartesian(xlim = c(0, 20), ylim = c(0, 20)) + scale_colour_manual(values = c("TRUE" = "#8b2635", "FALSE" = "#3d6e8f"), labels = c("TRUE" = "retail first", "FALSE" = "business first")) + labs(x = expression(y[1] ~ "(minutes)"), y = expression(y[2] ~ "(minutes)"), colour = NULL) + theme(legend.position = "right", legend.text = element_text(size = 18)) ``` ::: {style="font-size: 28px"} 100,000 simulated pairs: retail rings first about 60% of the time, as the integral said. ::: --- ## β οΈ Where Marks Are Lost ::: {style="font-size: 28px"} - **Variance of a sum without the covariance.** $V(Y_1 + Y_2) = V(Y_1) + V(Y_2)$ only when $\text{Cov} = 0$. - **$\rho$ divided by variances.** It is $\text{Cov}/(\sigma_1\sigma_2)$: standard deviations. - **Independence from one cell.** One cell with $p(y_1, y_2) = p_1(y_1)p_2(y_2)$ proves nothing; one cell without it proves dependence. - **Zero covariance read as independence.** Theorem 5.11 runs one way only. - **Limits of integration.** On a triangular support such as $0 \le y_2 \le y_1 \le 1$, the inner limits depend on the outer variable. - **$\beta$ read as a rate.** In Wackerly, the exponential with parameter $\beta$ has **mean** $\beta$. ::: --- ## π Quiz #1: Default Across the Cycle {.quiz-question} A borrower's default probability is 0.02 in an expansion and 0.08 in a recession. The economy is in expansion with probability 0.7. What is the unconditional default probability? - [$0.038$]{.correct data-explanation="β
Theorem 5.14 with D an indicator, which is Theorem 2.8 in Chapter 5 dress: E(D) = 0.7(0.02) + 0.3(0.08) = 0.014 + 0.024 = 0.038."} - $0.050$ - $0.080$ - $0.0016$ --- ## π Quiz #2: A Hedged Position {.quiz-question} A trader is long $Y_1$ and short $Y_2$, with $V(Y_1) = 4$, $V(Y_2) = 9$ and $\text{Cov}(Y_1, Y_2) = 3$. What is $V(Y_1 - Y_2)$? - [$7$]{.correct data-explanation="β
Theorem 5.12 with a1 = 1, a2 = β1: V = 4 + 9 + 2(1)(β1)(3) = 13 β 6 = 7. Positive covariance makes the hedge less risky than either leg would suggest added together."} - $13$ - $19$ - $-5$ --- ## π Quiz #3: Which Model? {.quiz-question} On the retail line above (exponential, mean 4 minutes), what is the distribution of the time until the **third** call? - [Gamma with $\alpha = 3$, $\beta = 4$: mean 12 minutes]{.correct data-explanation="β
A sum of three independent exponentials with mean Ξ² = 4 is gamma(Ξ± = 3, Ξ² = 4), with mean Ξ±Ξ² = 12 and variance Ξ±Ξ²Β² = 48."} - Exponential with mean 12 minutes - Poisson with $\lambda = 3$ - Geometric with $p = 1/4$ --- ## π Key Formulas ::: {style="font-size: 28px"} | | Statement | |---|---| | Theorem 2.9 (Bayes) | $P(B_j \mid A) = P(A \mid B_j)P(B_j) / \sum_i P(A \mid B_i)P(B_i)$ | | Tchebysheff | $P(\lvert Y - \mu \rvert < k\sigma) \ge 1 - 1/k^2$ | | Theorem 5.4 | independent iff $p(y_1, y_2) = p_1(y_1)p_2(y_2)$ for all pairs | | Theorem 5.10 | $\text{Cov}(Y_1, Y_2) = E(Y_1Y_2) - \mu_1\mu_2$, $\quad \rho = \text{Cov}/(\sigma_1\sigma_2)$ | | Theorem 5.12 | $V(\sum a_iY_i) = \sum a_i^2V(Y_i) + 2\sum_{i 