```{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: - **Describe** the bivariate normal by its five parameters, and say what $\rho$ does to its shape - **Compute** a conditional expectation $E(Y_1 \mid Y_2 = y_2)$ from a conditional density (Definition 5.13) - **Apply** Theorem 5.14, $E(Y_1) = E[E(Y_1 \mid Y_2)]$, to a hierarchical model - **Split** a variance with Theorem 5.15 into a within part and a between part - **Read** each piece of that split in economic terms ::: --- ## πΊοΈ Where We Are ::: {style="font-size: 30px"} **Wackerly §§5.10β5.11** Friday: the multinomial, the joint distribution of the counts in several categories at once. ::: {.fragment} Today closes Chapter 5 with two ideas. **Β§5.10** is the book's optional section: we meet the bivariate normal descriptively, as the picture behind "two correlated returns". ::: ::: {.fragment} **Β§5.11** is the load-bearing one. Once we know the distribution of $Y_1$ *given* $Y_2$, its mean and variance are ordinary means and variances, and Theorems 5.14β5.15 put them back together. ::: ::: --- ## β Motivating Question ::: {style="font-size: 30px"} ::: {.callout-important} ## The question this lecture answers A household in Baku earns 1,500 AZN this month. How much do we expect it to spend, and what does averaging that forecast over **all** incomes tell us about average spending? ::: ::: {.fragment} The first answer is a number that **depends on the income**: a regression function. The second is Theorem 5.14. ::: ::: --- ## π Β§5.10: The Bivariate Normal Density ::: {style="font-size: 28px"} $$f(y_1, y_2) = \frac{e^{-Q/2}}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}, \qquad -\infty < y_1, y_2 < \infty,$$ $$Q = \frac{1}{1-\rho^2}\left[\frac{(y_1-\mu_1)^2}{\sigma_1^2} - 2\rho\frac{(y_1-\mu_1)(y_2-\mu_2)}{\sigma_1\sigma_2} + \frac{(y_2-\mu_2)^2}{\sigma_2^2}\right]$$ ::: {.fragment} Five parameters: $\mu_1, \mu_2, \sigma_1^2, \sigma_2^2, \rho$. Nobody integrates this in class; we read off what it **implies**. ::: ::: --- ## π What the Five Parameters Mean ::: {style="font-size: 28px"} - **Marginals** are normal: $Y_1 \sim N(\mu_1, \sigma_1^2)$ and $Y_2 \sim N(\mu_2, \sigma_2^2)$ (Exercise 5.128) - **Covariance**: $\text{Cov}(Y_1, Y_2) = \rho\sigma_1\sigma_2$, so $\rho$ is the correlation - **Independence**: $\rho = 0$ makes $f$ factor, so $Y_1, Y_2$ are independent **if and only if** $\text{Cov} = 0$ - **Conditional** (Exercise 5.129): given $Y_2 = y_2$, $Y_1$ is normal with $$\text{mean } \mu_1 + \rho\frac{\sigma_1}{\sigma_2}(y_2 - \mu_2), \qquad \text{variance } \sigma_1^2(1-\rho^2)$$ ::: ::: {style="font-size: 28px"} ::: {.fragment} Zero covariance does **not** imply independence in general (recall the Brent straddle from Β§5.7). The bivariate normal is the exception. ::: ::: --- ## π Two Correlated Indices ```{r} #| label: bvn-figure #| echo: false #| fig-width: 11 #| fig-height: 3.9 mu1 <- 0.8; s1 <- 4.5; mu2 <- 0.6; s2 <- 3.8 z1 <- rnorm(300); z2 <- rnorm(300) panels <- do.call(rbind, lapply(c(0, 0.5, 0.9), function(r) { y2 <- mu2 + s2 * z1 y1 <- mu1 + s1 * (r * z1 + sqrt(1 - r^2) * z2) data.frame(rho = r, y1 = y1, y2 = y2) })) lines <- data.frame(rho = c(0, 0.5, 0.9)) lines$slope <- lines$rho * s1 / s2 lines$int <- mu1 - lines$slope * mu2 ggplot(panels, aes(y2, y1)) + geom_point(colour = "#3d6e8f", alpha = 0.55, size = 1.8) + geom_abline(data = lines, aes(slope = slope, intercept = int), colour = "#8b2635", linewidth = 1.3) + facet_wrap(~ rho, labeller = label_bquote(rho == .(rho))) + scale_x_continuous(breaks = c(-8, -4, 0, 4, 8)) + labs(x = "Emerging-markets index, monthly return (%)", y = "Regional index (%)") + theme(strip.text = element_text(size = 20), panel.spacing = unit(2, "lines")) ``` ::: {style="font-size: 28px"} Same marginals in all three panels. The red line is $E(Y_1 \mid Y_2 = y_2)$; as $\rho$ rises the cloud tilts onto it and the spread about it shrinks from 4.5% to 1.96%. ::: --- ## π Definition 5.13 ::: {style="font-size: 30px"} ::: {.callout-note} ## Definition 5.13 If $Y_1$ and $Y_2$ are any two random variables, the conditional expectation of $g(Y_1)$, given that $Y_2 = y_2$, is $$E(g(Y_1) \mid Y_2 = y_2) = \int_{-\infty}^{\infty} g(y_1) f(y_1 \mid y_2)\, dy_1$$ if jointly continuous, and $\sum_{\text{all } y_1} g(y_1)\,p(y_1 \mid y_2)$ if jointly discrete. ::: ::: {.fragment} An ordinary expectation, computed with the **conditional** density in place of the marginal. ::: ::: --- ## π Consumption Given Income ::: {style="font-size: 28px"} $Y_2$ = a household's monthly income and $Y_1$ = its consumption, both in thousand AZN, with spending never above income: $$f(y_1, y_2) = 1/2, \qquad 0 \le y_1 \le y_2 \le 2$$ ::: {.fragment} **Step 1.** $f_2(y_2) = \int_0^{y_2} \tfrac12\, dy_1 = y_2/2$, so $f(y_1 \mid y_2) = \dfrac{1/2}{y_2/2} = \dfrac{1}{y_2}$ on $0 < y_1 \le y_2$. ::: ::: {.fragment} **Step 2.** $E(Y_1 \mid Y_2 = y_2) = \int_0^{y_2} y_1 \dfrac{1}{y_2}\, dy_1 = \dfrac{y_2}{2}$. ::: ::: {.fragment} **Step 3.** At $y_2 = 1.5$: expected spending is $0.75$, i.e. **750 AZN**. The slope $1/2$ is a marginal propensity to consume. ::: ::: --- ## π A Random Variable of Its Own ::: {style="font-size: 30px"} $E(Y_1 \mid Y_2 = y_2) = y_2/2$ is a **number** for each income level. ::: {.fragment} Let the income vary and it becomes $E(Y_1 \mid Y_2) = Y_2/2$: a function of the random variable $Y_2$, so **itself a random variable**, with a mean and a variance of its own. ::: ::: {.fragment} Its mean is Theorem 5.14. Its variance is half of Theorem 5.15. ::: ::: --- ## π Theorem 5.14: Average the Averages ::: {style="font-size: 28px"} ::: {.callout-important} ## Theorem 5.14 $$E(Y_1) = E[E(Y_1 \mid Y_2)]$$ the inner expectation over the conditional distribution of $Y_1$ given $Y_2$, the outer over the distribution of $Y_2$. ::: ::: {.fragment} Consumption check: $E(Y_2) = \int_0^2 y_2 \cdot \tfrac{y_2}{2}\, dy_2 = \tfrac43$, so $$E(Y_1) = E(Y_2/2) = \tfrac23 \approx 667 \text{ AZN}$$ Directly, $f_1(y_1) = (2 - y_1)/2$ gives $\int_0^2 y_1 \tfrac{2-y_1}{2}\, dy_1 = \tfrac23$ as well. ::: ::: --- ## π Claims per Fleet Policy ::: {style="font-size: 28px"} An insurer writes fleet policies on 10 delivery vans each. Within a fleet, each van has a claim in the year independently with probability $p$, so $Y \mid p \sim \text{Bin}(10, p)$. But fleets differ: $p$ is uniform on $(0, 1/4)$ across the book (Example 5.32). ::: {.fragment} $$E(Y) = E[E(Y \mid p)] = E(10p) = 10 \cdot \frac{0 + 1/4}{2} = \frac{10}{8} = 1.25$$ ::: ::: {.fragment} 1.25 claims per policy per year, found **without** the marginal distribution of $Y$. ::: ::: --- ## π Theorem 5.15: Splitting a Variance ::: {style="font-size: 28px"} The conditional variance: $V(Y_1 \mid Y_2 = y_2) = E(Y_1^2 \mid Y_2 = y_2) - [E(Y_1 \mid Y_2 = y_2)]^2$. ::: {.callout-important} ## Theorem 5.15 $$V(Y_1) = E[V(Y_1 \mid Y_2)] + V[E(Y_1 \mid Y_2)]$$ ::: ::: {.fragment} - $E[V(Y_1 \mid Y_2)]$: the **within** part, average spread around each conditional mean - $V[E(Y_1 \mid Y_2)]$: the **between** part, how much the conditional means themselves move ::: ::: --- ## π The Variance of Claims ::: {style="font-size: 28px"} $E(Y \mid p) = 10p$ and $V(Y \mid p) = 10p(1-p)$. With $E(p) = \tfrac18$, $V(p) = \tfrac{(1/4)^2}{12} = \tfrac{1}{192}$, $E(p^2) = \tfrac{1}{192} + \tfrac{1}{64} = \tfrac{1}{48}$: ::: {.fragment} $$E[V(Y \mid p)] = 10\left(\tfrac18 - \tfrac{1}{48}\right) = 1.0417, \qquad V[E(Y \mid p)] = 100 \cdot \tfrac{1}{192} = 0.5208$$ $$V(Y) = 1.5625, \qquad \sigma = 1.25$$ ::: ::: {.fragment} A binomial with $p$ fixed at $1/8$ would give only $10 \cdot \tfrac18 \cdot \tfrac78 = 1.09$. **Not knowing the fleet adds risk**, and the between part is where it shows. ::: ::: --- ## π» The Fleet Book, Simulated ```{r} #| label: fleet-sim #| code-fold: false set.seed(2026) p <- runif(1e5, 0, 1/4) # one risk level per fleet y <- rbinom(1e5, 10, p) # claims on that fleet c(mean = mean(y), var = var(y), within = mean(10 * p * (1 - p)), between = var(10 * p)) ``` ::: {style="font-size: 28px"} Theory: 1.25, 1.5625, 1.0417, 0.5208. The simulation agrees to two decimals; within + between recovers the total. ::: --- ## ποΈ Risk Across Market Regimes ::: {style="font-size: 28px"} A pension fund's monthly return $R$ (%) depends on the regime $M$: | Regime | $P(M)$ | $E(R \mid M)$ | $\text{SD}(R \mid M)$ | |---|---|---|---| | calm | 0.8 | 1.2 | 3 | | oil-price stress | 0.2 | $-2.5$ | 7 | ::: {.fragment} $E(R) = 0.8(1.2) + 0.2(-2.5) = 0.46\%$ by Theorem 5.14. ::: ::: {.fragment} Within: $E[V(R \mid M)] = 0.8(9) + 0.2(49) = 17.00$ ::: ::: --- ## ποΈ The Split, Read as Risk ::: {style="font-size: 28px"} Between: $V[E(R \mid M)] = 0.8(1.2)^2 + 0.2(2.5)^2 - 0.46^2 = 2.40 - 0.21 = 2.19$ ::: {.fragment} $$V(R) = 17.00 + 2.19 = 19.19, \qquad \text{SD}(R) = 4.38\%$$ ::: ::: {.fragment} - **89%** of the variance is ordinary month-to-month noise inside a regime - **11%** comes from not knowing which regime we are in ::: ::: {.fragment} Quoting only the calm-regime 3% understates the fund's risk by almost a third. ::: ::: --- ## 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: 28px"} Monthly household income $Y_2$ and consumption $Y_1$ (AZN) are bivariate normal: $\mu_2 = 1400$, $\sigma_2 = 400$, $\mu_1 = 1100$, $\sigma_1 = 250$, $\rho = 0.8$. **Four minutes, in pairs:** 1. Expected consumption of a household earning 1,800 AZN? Earning 1,000? 2. The standard deviation of consumption **given** income? 3. Check Theorem 5.15: do within and between add up to $\sigma_1^2$? ::: --- ## β
Think-Pair-Share: Solution ::: {style="font-size: 28px"} 1. Slope $\rho\,\sigma_1/\sigma_2 = 0.8 \times 250/400 = 0.5$, so $$E(Y_1 \mid Y_2 = 1800) = 1100 + 0.5(400) = 1300, \qquad E(Y_1 \mid Y_2 = 1000) = 900$$ ::: {.fragment} 2. $V(Y_1 \mid Y_2) = 250^2(1 - 0.64) = 22{,}500$, so the conditional SD is **150 AZN**, the same at every income. ::: ::: {.fragment} 3. Within $= 22{,}500$. Between $= V(0.5\,Y_2) = 0.25 \times 400^2 = 40{,}000$. Sum $= 62{,}500 = 250^2$. β ::: ::: --- ## π Quiz #1: Claims per Policy {.quiz-question} 70% of a motor book is standard, with mean 0.10 claims per policy; 30% is high-risk, with mean 0.40. What is the expected number of claims per policy? - [$0.19$]{.correct data-explanation="β
Theorem 5.14: E(Y) = 0.7(0.10) + 0.3(0.40) = 0.07 + 0.12 = 0.19. The conditional means are weighted by how often each type occurs."} - $0.25$ - $0.40$ - $0.50$ --- ## π Quiz #2: Within Plus Between {.quiz-question} For a loan book, $E[V(Y \mid X)] = 9$ and $V[E(Y \mid X)] = 4$. What is $V(Y)$? - [$13$]{.correct data-explanation="β
Theorem 5.15 adds the two parts: V(Y) = 9 + 4 = 13. Variances add here, not standard deviations."} - $5$ - $9$ - $25$ --- ## π Quiz #3: Zero Covariance {.quiz-question} $Y_1$ and $Y_2$ are bivariate normal with $\text{Cov}(Y_1, Y_2) = 0$. What follows? - [They are independent]{.correct data-explanation="β
With Ο = 0 the bivariate normal density factors into g(y1)h(y2), so Theorem 5.5 gives independence. For the bivariate normal, zero covariance and independence are equivalent."} - They may still be dependent - Their marginals need not be normal - $E(Y_1 \mid Y_2) = Y_2$ --- ## π Key Formulas ::: {style="font-size: 28px"} | | Statement | |---|---| | Bivariate normal | $\text{Cov}(Y_1, Y_2) = \rho\sigma_1\sigma_2$; independent iff $\rho = 0$ | | its conditional | $E(Y_1 \mid y_2) = \mu_1 + \rho\frac{\sigma_1}{\sigma_2}(y_2 - \mu_2)$, $V = \sigma_1^2(1-\rho^2)$ | | Definition 5.13 | $E(g(Y_1) \mid Y_2 = y_2) = \int g(y_1) f(y_1 \mid y_2)\, dy_1$ | | conditional variance | $V(Y_1 \mid y_2) = E(Y_1^2 \mid y_2) - [E(Y_1 \mid y_2)]^2$ | | Theorem 5.14 | $E(Y_1) = E[E(Y_1 \mid Y_2)]$ | | Theorem 5.15 | $V(Y_1) = E[V(Y_1 \mid Y_2)] + V[E(Y_1 \mid Y_2)]$ | ::: --- ## π Summary ::: {style="font-size: 30px"} - The bivariate normal is fixed by two means, two variances and $\rho$; its marginals and conditionals are normal - For it alone among our models, $\rho = 0$ means independence - $E(Y_1 \mid Y_2)$ is a regression function, and a random variable - Theorem 5.14: average the conditional means to get the mean - Theorem 5.15: total risk = average within-group risk + spread of the group means - Hierarchical models (a rate that itself varies) are where both theorems pay off ::: --- ## π Practice Problems ::: {style="font-size: 28px"} **Wackerly, 7th edition** - Β§5.11 exercises: start with 5.133, 5.135 and 5.137, then 5.136 and 5.138 - Optional, Β§5.10: the starred 5.128 and 5.129 derive the marginal and conditional used today **Week 15, Problem Set 1** is open now and closes **Sunday 27 December at 23:59** on WeBWorK, covering §§5.10β5.11. Midterm Examination II is on 23 December. **Next class:** 19 December, Chapter 5 in review and a comprehensive review of Chapters 1β5 (Wackerly Β§5.12). ::: --- ## 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: 30px"} - In the fleet book, what would $p$'s distribution have to be for the between part to vanish? - Why is the conditional SD of consumption the same at every income level in the bivariate normal, but not in the triangle example? - Theorem 5.15 says conditioning never increases variance on average. Where have you already relied on that in finance? ::: 