```{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: - **State** Definition 5.8: $Y_1, Y_2$ are independent when $F(y_1, y_2) = F_1(y_1)F_2(y_2)$ - **Test** a joint table cell by cell with Theorem 5.4, and **stop** at the first cell that fails - **Apply** the factorisation criterion (Theorem 5.5) to a joint density without finding its marginals - **Recognise** a support that is not a rectangle as a proof of dependence - **Explain** why co-moving asset returns cannot be priced as independent ::: --- ## πΊοΈ Where We Are ::: {style="font-size: 32px"} **Wackerly Β§5.4** Saturday (Β§5.3): a **marginal** removes a variable, $p_1(y_1) = \sum_{y_2} p(y_1, y_2)$; a **conditional** keeps one slice of the joint and divides by its total, $p(y_1 \mid y_2) = p(y_1, y_2)/p_2(y_2)$. ::: {.fragment} We ended on the question for today: what does it mean when conditioning on one variable **leaves the other's distribution unchanged**? Then learning $Y_2$ tells you nothing about $Y_1$, and the joint is just the two marginals multiplied. ::: ::: --- ## β The Question ::: {.callout-important} ## Two branches, one cash van A Baku bank refills the ATMs of two branches, one in Nizami and one in Khatai, from a single cash centre. Does a heavy day of withdrawals in Nizami make a heavy day in Khatai more likely? ::: ::: {style="font-size: 30px"} If **no**, the treasury can plan each branch from its own history and multiply. If **yes**, it has to plan from the joint distribution, and both branches running dry on the same day is more likely than the product suggests. ::: --- ## π Definition 5.8 ::: {style="font-size: 30px"} ::: {.callout-note} ## Definition 5.8 (Wackerly Β§5.4) Let $Y_1$ have distribution function $F_1(y_1)$, $Y_2$ have distribution function $F_2(y_2)$, and $Y_1$ and $Y_2$ have joint distribution function $F(y_1, y_2)$. Then $Y_1$ and $Y_2$ are **independent** if and only if $$F(y_1, y_2) = F_1(y_1)F_2(y_2)$$ for every pair of real numbers $(y_1, y_2)$. Otherwise they are **dependent**. ::: It is the Chapter 2 rule $P(A \cap B) = P(A)P(B)$, applied to every pair of events $\{Y_1 \le y_1\}$ and $\{Y_2 \le y_2\}$ at once. ::: --- ## π Theorem 5.4: The Working Test ::: {style="font-size: 30px"} ::: {.callout-important} ## Theorem 5.4 **Discrete:** $Y_1$ and $Y_2$ are independent if and only if $$p(y_1, y_2) = p_1(y_1)\,p_2(y_2) \quad \text{for all pairs } (y_1, y_2).$$ **Continuous:** $Y_1$ and $Y_2$ are independent if and only if $$f(y_1, y_2) = f_1(y_1)\,f_2(y_2) \quad \text{for all pairs } (y_1, y_2).$$ ::: ::: {.fragment} "For **all** pairs" cuts both ways: to prove independence you check every cell; to prove dependence **one** failing cell is enough. ::: ::: --- ## π¦ Two Branches: Large Withdrawals ::: {style="font-size: 28px"} $Y_1$, $Y_2$ = number of withdrawals above 20,000 AZN before noon at Nizami and at Khatai. | | $y_2 = 0$ | $y_2 = 1$ | $y_2 = 2$ | $p_1(y_1)$ | |---|---|---|---|---| | $y_1 = 0$ | 0.30 | 0.15 | 0.05 | 0.50 | | $y_1 = 1$ | 0.18 | 0.09 | 0.03 | 0.30 | | $y_1 = 2$ | 0.12 | 0.06 | 0.02 | 0.20 | | $p_2(y_2)$ | 0.60 | 0.30 | 0.10 | 1.00 | ::: {.fragment} Cell $(1, 1)$: $p_1(1)\,p_2(1) = 0.30 \times 0.30 = 0.09 = p(1, 1)$. The same holds in all nine cells, so $Y_1$ and $Y_2$ are **independent**. ::: ::: --- ## π‘ Tower Outages After a Storm ::: {style="font-size: 28px"} Daily cell-tower outages: $Y_1$ in Baku, $Y_2$ in Sumgait. One storm hits both. | | $y_2 = 0$ | $y_2 = 1$ | $y_2 = 2$ | $p_1(y_1)$ | |---|---|---|---|---| | $y_1 = 0$ | 0.40 | 0.08 | 0.02 | 0.50 | | $y_1 = 1$ | 0.08 | 0.14 | 0.08 | 0.30 | | $y_1 = 2$ | 0.02 | 0.08 | 0.10 | 0.20 | | $p_2(y_2)$ | 0.50 | 0.30 | 0.20 | 1.00 | ::: {.fragment} Cell $(0, 0)$: $p(0, 0) = 0.40$ but $p_1(0)\,p_2(0) = 0.25$. **Dependent.** Also $P(Y_2 = 0 \mid Y_1 = 0) = 0.80$ but $P(Y_2 = 0 \mid Y_1 = 2) = 0.10$. ::: ::: --- ## π» Checking Every Cell in R ```{r} #| label: cell-check branches <- matrix(c(.30, .15, .05, .18, .09, .03, .12, .06, .02), nrow = 3, byrow = TRUE) towers <- matrix(c(.40, .08, .02, .08, .14, .08, .02, .08, .10), nrow = 3, byrow = TRUE) # ratio p(y1, y2) / (p1(y1) p2(y2)): every entry is 1 exactly when independent ratio <- function(p) p / outer(rowSums(p), colSums(p)) round(ratio(branches), 3) round(ratio(towers), 3) ``` ::: {style="font-size: 28px"} Ratios above 1 on the diagonal: outages **cluster** on the same days. ::: --- ## π Continuous: A Density That Splits ::: {style="font-size: 28px"} $Y_1$, $Y_2$ = fraction of each branch's ATM cassette dispensed by closing time, with $$f(y_1, y_2) = 6y_1y_2^2, \qquad 0 \le y_1 \le 1,\ 0 \le y_2 \le 1.$$ ::: {.fragment} Marginals (limits of integration are **constants**): $$f_1(y_1) = \int_0^1 6y_1y_2^2\,dy_2 = 2y_1, \qquad f_2(y_2) = \int_0^1 6y_1y_2^2\,dy_1 = 3y_2^2$$ ::: ::: {.fragment} $f_1(y_1)f_2(y_2) = 6y_1y_2^2 = f(y_1, y_2)$ everywhere: **independent** (compare Example 5.11). So $$P(Y_1 > 0.5,\ Y_2 > 0.5) = (1 - 0.5^2)(1 - 0.5^3) = 0.75 \times 0.875 = 0.656$$ ::: ::: --- ## πΊ A Triangle Is Never Independent ::: {style="font-size: 28px"} $Y_1$ = fraction of the vault's capacity stocked at opening, $Y_2$ = fraction withdrawn during the day. You cannot withdraw more than was stocked: $$f(y_1, y_2) = 2, \qquad 0 \le y_2 \le y_1 \le 1.$$ ::: {.fragment} $f_1(y_1) = \int_0^{y_1} 2\,dy_2 = 2y_1$ and $f_2(y_2) = \int_{y_2}^{1} 2\,dy_1 = 2(1 - y_2)$ (compare Example 5.12). ::: ::: {.fragment} At $(0.25, 0.75)$: $f = 0$, but $f_1(0.25)f_2(0.75) = 0.5 \times 0.5 = 0.25$. **Dependent.** The limits of integration depended on the other variable, and that was the warning. ::: ::: --- ## π Theorem 5.5: Factorisation ::: {style="font-size: 30px"} ::: {.callout-important} ## Theorem 5.5 Let $Y_1$ and $Y_2$ have a joint density $f(y_1, y_2)$ that is positive if and only if $a \le y_1 \le b$ and $c \le y_2 \le d$, for constants $a, b, c, d$; and $f(y_1, y_2) = 0$ otherwise. Then $Y_1$ and $Y_2$ are independent if and only if $$f(y_1, y_2) = g(y_1)\,h(y_2),$$ where $g$ is a nonnegative function of $y_1$ alone and $h$ of $y_2$ alone. ::: No marginals needed, and $g$, $h$ need not be densities. Two conditions: a **rectangle** support, and a **product** formula. ::: --- ## π Factor or Not? ::: {style="font-size: 28px"} | Joint density | Support | Verdict | |---|---|---| | $12\,y_1\,y_2(1 - y_2)$ | $[0,1] \times [0,1]$ | $g = y_1$, $h = 12y_2(1-y_2)$: **independent** | | $\tfrac32(y_1^2 + y_2^2)$ | $[0,1] \times [0,1]$ | a sum, no split: **dependent** | | $8\,y_1y_2$ | $0 \le y_2 \le y_1 \le 1$ | a product, but a triangle: **dependent** | ::: {.fragment} Row 2: $f_1(y_1) = \tfrac32 y_1^2 + \tfrac12$, so at $(0,0)$ we have $f = 0$ but $f_1 f_2 = 0.25$. Row 3 is the trap. The formula factors, the support does not; at $(0.25, 0.75)$, $f = 0$ but $f_1 f_2 = 0.0625 \times 1.3125 = 0.082$. ::: ::: --- ## π΅ Worked Example: Cash Demand ::: {style="font-size: 28px"} Daily cash withdrawn (units of 100,000 AZN): $Y_1$ at Nizami, $Y_2$ at Khatai, with $$f(y_1, y_2) = \tfrac12\, y_1 e^{-y_1} e^{-y_2/2}, \qquad y_1 > 0,\ y_2 > 0.$$ ::: {.fragment} **Step 1.** Support $(0, \infty) \times (0, \infty)$ is a rectangle; $g(y_1) = y_1e^{-y_1}$, $h(y_2) = \tfrac12 e^{-y_2/2}$. Independent by Theorem 5.5: $Y_1$ is gamma$(2, 1)$, $Y_2$ exponential with mean 2. ::: ::: {.fragment} **Step 2.** Both branches above 200,000 AZN: one double integral becomes two single ones, $$P(Y_1 > 2,\ Y_2 > 2) = 3e^{-2} \times e^{-1} = 0.406 \times 0.368 = 3e^{-3} = 0.149.$$ ::: ::: --- ## π» Cash Demand in R ```{r} #| label: cash-sim n <- 100000 y1 <- rgamma(n, shape = 2, rate = 1) # Nizami, mean 2 (200,000 AZN) y2 <- rexp(n, rate = 1/2) # Khatai, mean 2 c(P_Y1_gt2 = pgamma(2, 2, 1, lower.tail = FALSE), P_Y2_gt2 = pexp(2, 1/2, lower.tail = FALSE), product = 3 * exp(-3), simulated = mean(y1 > 2 & y2 > 2)) |> round(4) ``` ::: {style="font-size: 28px"} About one day in seven, both branches need a large top-up: the cash centre plans for that directly from the two marginals. ::: --- ## π’οΈ Co-Moving Returns ```{r} #| label: comoving #| echo: false #| fig-width: 11 #| fig-height: 4.2 set.seed(2026) n_days <- 750 z1 <- rnorm(n_days); z2 <- rnorm(n_days) make <- function(rho, lab) data.frame( brent = 2.0 * z1, # Brent daily return, % bond = 0.8 * (rho * z1 + sqrt(1 - rho^2) * z2), # energy-issuer bond, % panel = lab) df <- rbind(make(0, "independent"), make(0.6, "co-moving")) df$panel <- factor(df$panel, levels = c("independent", "co-moving")) ggplot(df, aes(brent, bond)) + geom_point(alpha = 0.35, size = 1.6, colour = "#2f4b7c") + geom_vline(xintercept = -2, linetype = "dashed", colour = "#8b2635") + facet_wrap(~ panel) + labs(x = "Brent daily return (%)", y = "Bond daily return (%)") + theme(strip.text = element_text(size = 20, face = "bold")) ``` ::: {style="font-size: 28px"} Same marginals in both panels. When Brent falls more than 2% (left of the dashes), the bond falls with probability 0.50 on the left but **0.86** on the right. ::: --- ## 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"} $Y_1$ = tanker loadings delayed at an oil terminal today (0 or 1); $Y_2$ = pipeline pumping-station alarms (0, 1, 2). | | $y_2 = 0$ | $y_2 = 1$ | $y_2 = 2$ | |---|---|---|---| | $y_1 = 0$ | 0.42 | 0.21 | 0.07 | | $y_1 = 1$ | 0.18 | ? | ? | 1. If $Y_1$ and $Y_2$ are independent, fill in the two missing cells. 2. A colleague enters $0.12$ and $0$ instead. Independent? ::: --- ## β
Think-Pair-Share: Solution ::: {style="font-size: 28px"} 1. Row 0 gives $p_1(0) = 0.70$, so $p_1(1) = 0.30$. Independence in row 0 fixes $p_2$: $$p_2(0) = \tfrac{0.42}{0.70} = 0.6,\quad p_2(1) = \tfrac{0.21}{0.70} = 0.3,\quad p_2(2) = 0.1$$ Check $p(1,0) = 0.3 \times 0.6 = 0.18$ β. Then $p(1,1) = 0.09$ and $p(1,2) = 0.03$. ::: {.fragment} 2. The table still sums to 1, but $p(1, 2) = 0$ while $p_1(1)\,p_2(2) = 0.30 \times 0.07 = 0.021$. **Dependent.** A zero cell under two positive marginals always is (compare Example 5.10). ::: ::: --- ## π Quiz #1: Factorisation {.quiz-question} $f(y_1, y_2) = \tfrac19\, y_1^2 e^{-y_2}$ for $0 \le y_1 \le 3$, $y_2 > 0$, and 0 elsewhere. Are $Y_1$, $Y_2$ independent? - [Yes: the support is a rectangle and $f = g(y_1)h(y_2)$]{.correct data-explanation="β
Theorem 5.5: the support [0,3] Γ (0,β) is a rectangle, and f splits as g(y1) = y1Β²/9 times h(y2) = e^(βy2)."} - No: $y_1$ has a finite range but $y_2$ does not - Cannot say without computing both marginals - No: $g(y_1) = y_1^2$ is not a density --- ## π Quiz #2: One Cell {.quiz-question} In a joint table of two exporters' monthly shipment counts, $p(2, 0) = 0$, while $p_1(2) = 0.2$ and $p_2(0) = 0.4$. What follows? - [$Y_1$ and $Y_2$ are dependent]{.correct data-explanation="β
Theorem 5.4 needs p(y1, y2) = p1(y1)p2(y2) for every pair. Here 0 β 0.2 Γ 0.4 = 0.08, and one failing cell settles it."} - Nothing, until the other cells are checked - They are independent, since $0 \le 0.08$ - The table is invalid --- ## π Quiz #3: Using Definition 5.8 {.quiz-question} Daily losses $Y_1$, $Y_2$ (in million AZN) on two unrelated trading books are independent, with $P(Y_1 \le 1) = 0.7$ and $P(Y_2 \le 1) = 0.4$. What is $P(Y_1 \le 1,\ Y_2 \le 1)$? - [$0.28$]{.correct data-explanation="β
Definition 5.8: F(1, 1) = F1(1)F2(1) = 0.7 Γ 0.4 = 0.28."} - $0.82$ - $1.10$ - $0.40$ --- ## π Key Formulas ::: {style="font-size: 28px"} | | Independent if and only if | |---|---| | Definition 5.8 | $F(y_1, y_2) = F_1(y_1)F_2(y_2)$ for all $(y_1, y_2)$ | | Theorem 5.4, discrete | $p(y_1, y_2) = p_1(y_1)p_2(y_2)$ for all pairs | | Theorem 5.4, continuous | $f(y_1, y_2) = f_1(y_1)f_2(y_2)$ for all pairs | | Theorem 5.5 (rectangle support) | $f(y_1, y_2) = g(y_1)h(y_2)$ | | Exercise 5.43 | $f(y_1 \mid y_2) = f_1(y_1)$ whenever $f_2(y_2) > 0$ | | $n$ variables | $F(y_1, \ldots, y_n) = F_1(y_1)\cdots F_n(y_n)$ | ::: --- ## π Summary ::: {style="font-size: 30px"} - Independence means the joint distribution is the product of the marginals, **everywhere** - Tables: check every cell to prove it, find one cell to refute it; a zero cell is the quickest refutation - Densities: a rectangle support plus a product formula proves it (Theorem 5.5), no integration needed - A support that is not a rectangle proves dependence, however neatly the formula factors - A common factor (oil, a storm) makes variables co-move; at correlation 0.6, $P(\text{both fall}) = 0.352$, not $0.5 \times 0.5 = 0.25$ ::: --- ## π Practice Problems ::: {style="font-size: 28px"} **Wackerly, 7th edition** - Exercises at the end of Β§5.4: start with 5.45, 5.48, 5.51 and 5.53, then 5.56, 5.60, 5.61, and 5.63 β 5.64 - Return to the cash-demand example and find $P(Y_1 < 1,\ Y_2 < 1)$ as a product of two marginal probabilities **Week 13, Problem Set 1** is open now and closes **Sunday 13 December at 23:59** on WeBWorK, covering Β§5.4. **Next class (5 December):** Wackerly §§5.5β5.7, expected values of functions of random variables and covariance. ::: --- ## 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"} - Is a random variable independent of itself? When? - Can two variables have a triangular support and still have $F(y_1, y_2) = F_1(y_1)F_2(y_2)$ at some points? - If Brent and the bond were independent, what would the right-hand scatter plot look like? ::: 