```{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: - **Compute** $E[g(Y_1, Y_2)]$ from a joint probability table or a joint density (Definition 5.9) - **Use** Theorems 5.6β5.8 to take expectations of sums term by term, with or without independence - **Apply** Theorem 5.9 to factor $E[g(Y_1)h(Y_2)]$ when $Y_1$ and $Y_2$ are independent - **Compute** $\text{Cov}(Y_1, Y_2)$ and $\rho$ by Theorem 5.10, and read their sign - **Explain** why independence implies zero covariance, but zero covariance does not imply independence ::: --- ## πΊοΈ Where We Are ::: {style="font-size: 30px"} **Wackerly §§5.5β5.7** Wednesday (Β§5.4): independence means the joint distribution is the product of the marginals, everywhere. It ended on co-movement: at correlation 0.6, $P(\text{both fall}) = 0.352$, not $0.5 \times 0.5 = 0.25$. ::: {.fragment} We used that word, **correlation**, before defining it. Today we define it, as one expected value of a function of two variables: $E[(Y_1 - \mu_1)(Y_2 - \mu_2)]$. ::: ::: {.fragment} First, the general tool: $E[g(Y_1, Y_2)]$ for total revenue, a product of shares, a spread. ::: ::: --- ## β Motivating Question ::: {.callout-important} ## The appliance retailer's peak hour A Baku home-appliance store sells air conditioners at **1,200 AZN** and refrigerators at **900 AZN**. In the Saturday peak hour it sells $Y_1$ air conditioners and $Y_2$ refrigerators. What revenue should it **expect** in that hour, and do the two product lines sell **together**? ::: ::: {.fragment style="font-size: 30px"} Revenue $R = 1200Y_1 + 900Y_2$ is a function of **two** random variables. Chapter 4 told us how to find $E[g(Y)]$ for one. ::: --- ## π Definition 5.9 ::: {style="font-size: 30px"} ::: {.callout-note} ## Definition 5.9 Let $g(Y_1, \ldots, Y_k)$ be a function of discrete random variables with probability function $p(y_1, \ldots, y_k)$. Then $$E[g(Y_1, \ldots, Y_k)] = \sum_{\text{all } y_k} \cdots \sum_{\text{all } y_1} g(y_1, \ldots, y_k)\, p(y_1, \ldots, y_k).$$ For continuous variables with joint density $f$, the sums become integrals over $(-\infty, \infty)$. ::: ::: {.fragment} In words: **weight each value of $g$ by the probability of the cell that produces it**, and add. Taking $g(Y_1, Y_2) = Y_1$ gives back Definition 4.5 through the marginal density. ::: ::: --- ## π¬ Worked Example: The Peak Hour ::: {style="font-size: 28px"} :::: {.columns} ::: {.column width="50%"} | $y_1 \backslash y_2$ | 0 | 1 | 2 | $p_1$ | |---|---|---|---|---| | **0** | .30 | .15 | .05 | .50 | | **1** | .10 | .15 | .10 | .35 | | **2** | .02 | .05 | .08 | .15 | | $p_2$ | .42 | .35 | .23 | 1 | ::: ::: {.column width="50%"} Each cell carries a revenue $g(y_1, y_2) = 1200y_1 + 900y_2$: from 0 AZN (nothing sold) to 4200 AZN (two of each). ::: :::: ::: {.fragment} Definition 5.9, cell by cell: $$E(R) = 0(.30) + 900(.15) + 1800(.05) + \cdots + 4200(.08) = \mathbf{1509} \text{ AZN}$$ ::: ::: --- ## π Theorems 5.6β5.8: Special Theorems ::: {style="font-size: 30px"} ::: {.callout-important} ## Theorems 5.6, 5.7, 5.8 For a constant $c$ and functions $g, g_1, \ldots, g_k$ of $Y_1$ and $Y_2$: $$E(c) = c, \qquad E[c\,g(Y_1, Y_2)] = c\,E[g(Y_1, Y_2)],$$ $$E[g_1(Y_1, Y_2) + \cdots + g_k(Y_1, Y_2)] = E[g_1(Y_1, Y_2)] + \cdots + E[g_k(Y_1, Y_2)].$$ ::: ::: {.fragment} The proofs are the univariate ones: a sum or an integral is linear. **Nothing here assumes independence.** The expectation of a sum is the sum of the expectations, however the terms are related. ::: ::: --- ## β‘ Expected Revenue the Short Way ::: {style="font-size: 28px"} With $g_1 = 1200Y_1$ and $g_2 = 900Y_2$, Theorems 5.7 and 5.8 give $$E(R) = 1200E(Y_1) + 900E(Y_2) = 1200(0.65) + 900(0.81) = 780 + 729 = 1509 \text{ AZN}.$$ Only the **marginals** were needed. The same 1509 as nine cells of arithmetic: ::: ```{r} #| label: peak-hour p <- matrix(c(.30, .15, .05, .10, .15, .10, .02, .05, .08), nrow = 3, byrow = TRUE) y <- 0:2 R <- outer(1200 * y, 900 * y, "+") # revenue in each cell c(by_definition_5.9 = sum(R * p), by_linearity = 1200 * sum(y * rowSums(p)) + 900 * sum(y * colSums(p))) ``` --- ## β‘ Worked Example: A Power Plant ::: {style="font-size: 28px"} A gas-fired plant sells a share $Y_1$ of its monthly output on the spot market (the rest under contract), and a share $Y_2$ of those spot sales clears at peak-hour prices. Model: $$f(y_1, y_2) = 2(1 - y_1), \qquad 0 \le y_1 \le 1,\; 0 \le y_2 \le 1.$$ ::: {.fragment} $Y_1 Y_2$ is the share of **total** output sold at the peak spot price. By Definition 5.9, $$E(Y_1Y_2) = \int_0^1\!\!\int_0^1 2y_1y_2(1 - y_1)\,dy_2\,dy_1 = \int_0^1 (y_1 - y_1^2)\,dy_1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}.$$ ::: ::: {.fragment} On 180 GWh of monthly output, the plant should expect **30 GWh** sold at the peak spot price. ::: ::: --- ## π Theorem 5.9: Products Split ::: {style="font-size: 30px"} ::: {.callout-important} ## Theorem 5.9 If $Y_1$ and $Y_2$ are independent and $g(Y_1)$, $h(Y_2)$ are functions of $Y_1$ only and $Y_2$ only, then $$E[g(Y_1)h(Y_2)] = E[g(Y_1)]\,E[h(Y_2)],$$ provided the expectations exist. ::: ::: {.fragment} The plant's density is $2(1 - y_1) \times 1$ on a rectangle, so by Wednesday's Theorem 5.5 $Y_1$ and $Y_2$ are independent, with $E(Y_1) = 1/3$ and $E(Y_2) = 1/2$: $$E(Y_1Y_2) = \tfrac{1}{3} \cdot \tfrac{1}{2} = \tfrac{1}{6}, \text{ as before, with no double integral.}$$ ::: ::: --- ## π The Sign of the Deviations ```{r} #| label: deviation-figure #| echo: false #| fig-width: 10 #| fig-height: 4.3 n <- 120 z1 <- rnorm(n); z2 <- rnorm(n) bankA <- 1.0 + 2.12 * z1 bankB <- 1.2 + 2.96 * (0.516 * z1 + sqrt(1 - 0.516^2) * z2) dev <- data.frame(a = bankA, b = bankB, sign = ifelse((bankA - 1.0) * (bankB - 1.2) > 0, "product of deviations > 0", "product of deviations < 0")) share_pos <- mean(dev$sign == "product of deviations > 0") ggplot(dev, aes(a, b, colour = sign)) + geom_vline(xintercept = 1.0, linetype = "dashed", colour = "grey40") + geom_hline(yintercept = 1.2, linetype = "dashed", colour = "grey40") + geom_point(size = 2.6, alpha = 0.85) + annotate("text", x = 1.0, y = max(bankB) + 0.8, label = "mu[1]", parse = TRUE, size = 7) + annotate("text", x = min(bankA) - 0.3, y = 1.2 + 0.9, label = "mu[2]", parse = TRUE, size = 7) + scale_colour_manual(values = c("#8b2635", "#3d6e8f")) + labs(x = "Bank A monthly return (%)", y = "Bank B monthly return (%)", colour = NULL) + theme(legend.position = "top", legend.text = element_text(size = 18)) ``` ::: {style="font-size: 28px"} `r round(100 * share_pos)`% of 120 simulated months land where $(y_1 - \mu_1)(y_2 - \mu_2) > 0$, so the **average** product is positive. ::: --- ## π Definition 5.10 and Theorem 5.10 ::: {style="font-size: 28px"} ::: {.callout-note} ## Definition 5.10 If $Y_1$ and $Y_2$ have means $\mu_1$ and $\mu_2$, the **covariance** of $Y_1$ and $Y_2$ is $$\text{Cov}(Y_1, Y_2) = E[(Y_1 - \mu_1)(Y_2 - \mu_2)].$$ Its scale-free version is the **correlation coefficient** $\rho = \dfrac{\text{Cov}(Y_1, Y_2)}{\sigma_1 \sigma_2}$, with $-1 \le \rho \le 1$. ::: ::: {.callout-important} ## Theorem 5.10 $$\text{Cov}(Y_1, Y_2) = E(Y_1Y_2) - E(Y_1)E(Y_2).$$ ::: Proof: expand the product, then apply Theorems 5.7 and 5.8 term by term. ::: --- ## π¦ Worked Example: Two Bank Shares ::: {style="font-size: 28px"} Monthly returns (%) of two bank shares on the Baku Stock Exchange, $Y_1$ for bank A and $Y_2$ for bank B: | $y_1 \backslash y_2$ | $-3$ | $1$ | $5$ | $p_1$ | |---|---|---|---|---| | $-2$ | .15 | .08 | .02 | .25 | | $1$ | .08 | .30 | .12 | .50 | | $4$ | .02 | .07 | .16 | .25 | | $p_2$ | .25 | .45 | .30 | 1 | ::: {.fragment} $E(Y_1) = 1.0$, $E(Y_2) = 1.2$, and summing $y_1y_2\,p(y_1, y_2)$ over the nine cells gives $E(Y_1Y_2) = 4.44$. $$\text{Cov}(Y_1, Y_2) = 4.44 - (1.0)(1.2) = 3.24, \qquad \rho = \frac{3.24}{\sqrt{4.5}\,\sqrt{8.76}} = 0.516.$$ ::: ::: --- ## π» Covariance and Correlation in R ```{r} #| label: bank-cov #| code-fold: false y1 <- c(-2, 1, 4); y2 <- c(-3, 1, 5) p <- matrix(c(.15, .08, .02, .08, .30, .12, .02, .07, .16), nrow = 3, byrow = TRUE) mu1 <- sum(y1 * rowSums(p)); mu2 <- sum(y2 * colSums(p)) cov_def <- sum(outer(y1 - mu1, y2 - mu2) * p) # Definition 5.10 cov_5.10 <- sum(outer(y1, y2) * p) - mu1 * mu2 # Theorem 5.10 s1 <- sqrt(sum(y1^2 * rowSums(p)) - mu1^2); s2 <- sqrt(sum(y2^2 * colSums(p)) - mu2^2) round(c(cov_def = cov_def, cov_5.10 = cov_5.10, rho = cov_5.10 / (s1 * s2)), 3) ``` ::: {style="font-size: 28px"} The covariance of 3.24 is in "percent squared", hard to judge. $\rho = 0.52$ says: **moderate** positive linear dependence. ::: --- ## π Theorem 5.11 ::: {style="font-size: 30px"} ::: {.callout-important} ## Theorem 5.11 If $Y_1$ and $Y_2$ are independent, then $\text{Cov}(Y_1, Y_2) = 0$. ::: Proof: Theorem 5.9 gives $E(Y_1Y_2) = \mu_1\mu_2$, and Theorem 5.10 finishes it. For the power plant, $\text{Cov} = \tfrac{1}{6} - \tfrac{1}{3}\cdot\tfrac{1}{2} = 0$. ::: {.fragment} Read it backwards as a test: the bank shares have $\text{Cov} = 3.24 \ne 0$, so they **cannot** be independent. **The converse is false.** Zero covariance does not make two variables independent. ::: ::: --- ## π’οΈ Zero Covariance, Still Dependent ::: {style="font-size: 28px"} An energy desk holds $Y_1$ Brent and $Y_2$ natural-gas contracts each day ($-1$ short, $0$ flat, $+1$ long). The desk mandate forbids being **flat in both** (Wackerly Table 5.3): | $y_1 \backslash y_2$ | $-1$ | $0$ | $+1$ | |---|---|---|---| | $-1$ | 1/16 | 3/16 | 1/16 | | $0$ | 3/16 | **0** | 3/16 | | $+1$ | 1/16 | 3/16 | 1/16 | ::: {.fragment} Dependent: $p(0, 0) = 0 \ne p_1(0)p_2(0) = (6/16)^2$. Yet $E(Y_1) = E(Y_2) = 0$, and the four corners give $E(Y_1Y_2) = \tfrac{1}{16} - \tfrac{1}{16} - \tfrac{1}{16} + \tfrac{1}{16} = 0$, so **Cov = 0**. ::: ::: --- ## π Dependent but Uncorrelated ```{r} #| label: straddle-figure #| echo: false #| fig-width: 10 #| fig-height: 4.3 brent <- rnorm(250, mean = 0, sd = 2) # daily Brent move, % payoff <- abs(brent) # at-the-money straddle payoff r_str <- cor(brent, payoff) ggplot(data.frame(x = brent, y = payoff), aes(x, y)) + geom_point(colour = "#3d6e8f", size = 2.4, alpha = 0.8) + geom_smooth(method = "lm", se = FALSE, colour = "#8b2635", linewidth = 1.3) + labs(x = "Daily Brent move (%)", y = "Straddle payoff (% of spot)") ``` ::: {style="font-size: 28px"} The payoff is **completely determined** by the move, yet over 250 simulated days the sample correlation is $r = `r sprintf("%.2f", r_str)`$ and the fitted line (red) is almost flat. Covariance sees only **linear** dependence; a V shape cancels itself out. ::: --- ## 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"} :::: {.columns} ::: {.column width="42%"} | $y_1 \backslash y_2$ | 0 | 1 | 2 | |---|---|---|---| | **0** | .30 | .15 | .05 | | **1** | .10 | .15 | .10 | | **2** | .02 | .05 | .08 | ::: ::: {.column width="58%"} The peak hour again: $E(Y_1) = 0.65$ air conditioners, $E(Y_2) = 0.81$ refrigerators. ::: :::: **Four minutes, in pairs:** 1. Find $E(Y_1Y_2)$. 2. Find $\text{Cov}(Y_1, Y_2)$. What sign did you expect, and why? 3. Are $Y_1$ and $Y_2$ independent? Give **two** ways to tell. ::: --- ## β
Think-Pair-Share: Solution ::: {style="font-size: 28px"} 1. Only cells with $y_1, y_2 \ge 1$ contribute: $$E(Y_1Y_2) = 1(.15) + 2(.10) + 2(.05) + 4(.08) = 0.77$$ 2. Theorem 5.10: $\text{Cov}(Y_1, Y_2) = 0.77 - (0.65)(0.81) = 0.2435 > 0$. A customer refitting a kitchen or a flat buys both. ::: {.fragment} 3. Not independent. By Theorem 5.11, a nonzero covariance rules independence out. Directly: $p(0, 0) = 0.30 \ne p_1(0)p_2(0) = 0.50 \times 0.42 = 0.21$. Note that $E(R) = 1509$ AZN did **not** need any of this: linearity ignores dependence. ::: ::: --- ## π Quiz #1: Linearity {.quiz-question} A telecom operator's daily new mobile contracts $Y_1$ and fibre contracts $Y_2$ are **dependent**, with $E(Y_1) = 2$ and $E(Y_2) = 3$ (hundreds). What is $E(4Y_1 - Y_2 + 10)$? - [$15$]{.correct data-explanation="β
Theorems 5.6β5.8: 4(2) β 3 + 10 = 15. Linearity holds whatever the dependence between Y1 and Y2."} - $21$ - $9$ - Cannot be found without the joint distribution --- ## π Quiz #2: From Moments to Ο {.quiz-question} $E(Y_1Y_2) = 5$, $E(Y_1) = 1$, $E(Y_2) = 2$, $V(Y_1) = 4$, $V(Y_2) = 9$. What is $\rho$? - [$0.5$]{.correct data-explanation="β
Theorem 5.10: Cov = 5 β (1)(2) = 3. Then Ο = 3 / (2 Γ 3) = 0.5, dividing by the standard deviations, not the variances."} - $3$ - $0.083$ - $0.833$ --- ## π Quiz #3: Reading a Zero {.quiz-question} Two exchange-rate changes are found to have $\text{Cov}(Y_1, Y_2) = 0$. Which conclusion is justified? - They are independent - [There is no linear dependence; they may still be dependent]{.correct data-explanation="β
Theorem 5.11 runs one way only. Example 5.24 (the energy desk) and the straddle both have zero covariance and strong dependence."} - $E(Y_1Y_2) = 0$ - Their correlation could still be positive --- ## π Key Formulas ::: {style="font-size: 28px"} | | Statement | |---|---| | Definition 5.9 | $E[g(Y_1, Y_2)] = \sum\sum g(y_1, y_2)\,p(y_1, y_2)$ or $\iint g\,f\,dy_1\,dy_2$ | | Theorems 5.6β5.8 | $E(c) = c$, $E(cg) = cE(g)$, $E(g_1 + g_2) = E(g_1) + E(g_2)$ | | Theorem 5.9 | independent: $E[g(Y_1)h(Y_2)] = E[g(Y_1)]E[h(Y_2)]$ | | Definition 5.10 | $\text{Cov}(Y_1, Y_2) = E[(Y_1 - \mu_1)(Y_2 - \mu_2)]$ | | Theorem 5.10 | $\text{Cov}(Y_1, Y_2) = E(Y_1Y_2) - \mu_1\mu_2$ | | correlation | $\rho = \text{Cov}(Y_1, Y_2)/(\sigma_1\sigma_2)$, $\;-1 \le \rho \le 1$ | | Theorem 5.11 | independent $\Rightarrow$ Cov $= 0$ (not conversely) | ::: --- ## π Summary ::: {style="font-size: 30px"} - Definition 5.9: weight each value of $g$ by its joint probability or density, and add - The expectation of a sum is the sum of the expectations, with **no** independence needed - Under independence, expectations of products split into products of expectations (Theorem 5.9) - Covariance is the average product of deviations; its sign says whether two variables move together - $\rho$ removes the units, so it can be read as weak, moderate or strong - Independence forces zero covariance; zero covariance allows dependence that is not linear ::: --- ## π Practice Problems ::: {style="font-size: 28px"} **Wackerly, 7th edition** - §§5.5β5.6: Exercises 5.72 β 5.88; start with 5.72, 5.74, 5.77, 5.80 and 5.84 - Β§5.7: Exercises 5.89 β 5.97; start with 5.89, 5.91, 5.93, 5.95 and 5.97 - Change the bank-share table so that $\rho < 0$ while keeping both marginals. Which cells must move? **Week 13, Problem Set 2** is open now and closes **Sunday 13 December at 23:59** on WeBWorK, covering §§5.5β5.7. **Next class:** 9 December, the expected value and variance of linear functions of random variables, and portfolio variance (Wackerly Β§5.8). ::: --- ## 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"} - The retailer's revenue needed only the marginals. Which quantity about $R$ will need the covariance as well? - If you double every return of bank A, what happens to the covariance, and what happens to $\rho$? - Can you build a joint table where $Y_2 = Y_1^2$ exactly and the covariance is zero? What must the distribution of $Y_1$ look like? ::: 