```{r} #| label: setup #| include: false set.seed(2026) library(ggplot2) theme_set(theme_minimal(base_size = 18)) ``` ## π¬ The Idea in 2 Minutes ::: {style="text-align:center"} [Watch this short intuition video before (or after) the slides. Captions: CC button.]{style="font-size:22px"} ::: --- ## π― Learning Objectives ::: {style="font-size: 32px"} By the end of this lecture, you will be able to: - **Write** the distribution function $F(y) = P(Y \le y)$ of any random variable, discrete or continuous - **Explain** why $P(Y = y) = 0$ for a continuous $Y$, and what replaces a probability function - **Move** between a density $f(y)$ and its distribution function $F(y)$ by integrating or differentiating - **Compute** $P(a \le Y \le b)$ as an area under $f(y)$, and find the constant that makes $f(y)$ a density - **Find** a quantile $\phi_p$, such as the median settlement time of a payment ::: --- ## πΊοΈ Where We Are ::: {style="font-size: 30px"} **Wackerly §§4.1β4.2** Friday closed Chapter 3 with a map of the discrete models: name the counting mechanism, pick the model, read off $E(Y)$ and $V(Y)$. Every one of them assigned a probability $p(y)$ to each value. ::: {.fragment} Today we leave counts behind. The time a payment takes to settle, the share of a credit line a firm draws, a day's move in the exchange rate: these take **any value in an interval**. ::: ::: {.fragment} For those, $p(y)$ cannot work. Chapter 4 replaces it with two functions, $F(y)$ and $f(y)$, and today is about both. ::: ::: --- ## β Motivating Question ::: {.callout-important} ## The payments desk's problem A Baku bank sends cross-border transfers that settle anywhere between 0 and 4 hours after release. The desk promises corporate clients settlement **within 3 hours**. *What is the probability a transfer settles in exactly 2 hours? And in at most 3?* ::: ::: {.fragment style="font-size: 30px"} The first question has the surprising answer **zero**, and yet the second has a perfectly good answer. By the end of today we will compute it: $27/32 \approx 0.844$. ::: --- ## π Definition 4.1 and Theorem 4.1 ::: {style="font-size: 30px"} ::: {.callout-note} ## Definition 4.1 Let $Y$ denote any random variable. The **distribution function** of $Y$, denoted by $F(y)$, is such that $F(y) = P(Y \le y)$ for $-\infty < y < \infty$. ::: ::: {.callout-important} ## Theorem 4.1: Properties of a Distribution Function 1. $F(-\infty) \equiv \lim_{y \to -\infty} F(y) = 0$ 2. $F(\infty) \equiv \lim_{y \to \infty} F(y) = 1$ 3. $F(y)$ is nondecreasing: if $y_1 < y_2$ then $F(y_1) \le F(y_2)$ ::: Note the word **any**: $F(y)$ exists for discrete variables too. ::: --- ## πͺ A Discrete F Is a Staircase ::: {style="font-size: 28px"} Example 4.1, in a loan office: two SME applications, each approved independently with probability $1/2$. $Y$ = number approved, so $p(0) = 1/4$, $p(1) = 1/2$, $p(2) = 1/4$ and, for instance, $F(1.5) = 1/4 + 1/2 = 3/4$. ::: ```{r} #| label: staircase #| echo: false #| fig-width: 10 #| fig-height: 3.9 steps <- data.frame(x0 = c(-1, 0, 1, 2), x1 = c(0, 1, 2, 3), F = c(0, 0.25, 0.75, 1)) ggplot(steps) + geom_segment(aes(x = x0, xend = x1, y = F, yend = F), linewidth = 1.4) + geom_point(aes(x = x0, y = F), data = steps[-1, ], size = 4) + geom_point(aes(x = x1, y = F), data = steps[-4, ], size = 4, shape = 21, fill = "white", stroke = 1.3) + scale_x_continuous(breaks = -1:3) + scale_y_continuous(breaks = c(0, 0.25, 0.75, 1), labels = c("0", "1/4", "3/4", "1")) + labs(x = "y (applications approved)", y = "F(y)") + theme(panel.grid.minor = element_blank()) ``` --- ## π Definition 4.2: When F Has No Jumps ::: {style="font-size: 30px"} The staircase jumps at 0, 1, 2 by exactly $p(0)$, $p(1)$, $p(2)$. It is flat in between because those values carry no probability. ::: {.callout-note} ## Definition 4.2 A random variable $Y$ with distribution function $F(y)$ is said to be **continuous** if $F(y)$ is continuous, for $-\infty < y < \infty$. ::: ::: {.fragment} **Consequence.** If $Y$ is continuous, $P(Y = y) = 0$ for every real $y$. A point with $P(Y = y_0) = p_0 > 0$ would put a jump of size $p_0$ into $F$. So a transfer settling in *exactly* 2.000... hours has probability zero, while one settling between 2 and 3 hours does not. ::: ::: --- ## π Definition 4.3: The Density ::: {style="font-size: 29px"} ::: {.callout-note} ## Definition 4.3 Let $F(y)$ be the distribution function for a continuous random variable $Y$. Then $f(y) = \dfrac{dF(y)}{dy}$, wherever the derivative exists, is called the **probability density function** for $Y$. Equivalently, $F(y) = \int_{-\infty}^{y} f(t)\, dt$. ::: ::: {.callout-important} ## Theorem 4.2: Properties of a Density Function 1. $f(y) \ge 0$ for all $y$, $-\infty < y < \infty$ 2. $\int_{-\infty}^{\infty} f(y)\, dy = 1$ ::: ::: --- ## π³ Worked Example: A Credit Line ::: {style="font-size: 29px"} Example 4.3, re-set. $Y$ is the **fraction** of its approved credit line a corporate client has drawn at month-end: $$f(y) = \begin{cases} 3y^2, & 0 \le y \le 1, \\ 0, & \text{elsewhere.} \end{cases}$$ ::: {.fragment} Integrate from $-\infty$ up to $y$. Below 0 nothing accumulates; above 1 everything already has: $$F(y) = \begin{cases} 0, & y < 0, \\ \int_0^y 3t^2\, dt = y^3, & 0 \le y \le 1, \\ 1, & y > 1. \end{cases}$$ ::: ::: {.fragment} So $P(Y \le 0.5) = 0.5^3 = 0.125$: only one client in eight uses half the line or less. Clients draw heavily. ::: ::: --- ## π f and F Side by Side ```{r} #| label: credit-line-figure #| echo: false #| fig-width: 11 #| fig-height: 4.6 yy <- seq(0, 1, by = 0.005) cl <- rbind( data.frame(y = yy, v = 3 * yy^2, panel = "Density~f(y)==3*y^2"), data.frame(y = yy, v = yy^3, panel = "Distribution~'function'~F(y)==y^3")) med <- 0.5^(1/3) marks <- data.frame(panel = c("Density~f(y)==3*y^2", "Distribution~'function'~F(y)==y^3"), v = c(3 * med^2, 0.5)) ggplot(cl, aes(y, v)) + geom_area(data = subset(cl, panel == "Density~f(y)==3*y^2" & y <= med), fill = "#8ba3c7", alpha = 0.6) + geom_line(linewidth = 1.3) + geom_segment(data = marks, aes(x = med, xend = med, y = 0, yend = v), linetype = "dashed", colour = "#8b2635", linewidth = 1) + facet_wrap(~ panel, scales = "free_y", labeller = label_parsed) + labs(x = "y (fraction of credit line drawn)", y = NULL) + theme(strip.text = element_text(size = 19, face = "bold")) ``` ::: {style="font-size: 28px"} The shaded area on the left, 0.5, is the **height** of $F$ on the right at the same $y$. And $f(1) = 3$: a density may exceed 1; $F$ never does. ::: --- ## π Definition 4.4: Quantiles ::: {style="font-size: 30px"} ::: {.callout-note} ## Definition 4.4 Let $Y$ denote any random variable. If $0 < p < 1$, the $p$th **quantile** of $Y$, denoted by $\phi_p$, is the smallest value such that $F(\phi_p) = P(Y \le \phi_p) \ge p$. If $Y$ is continuous, $\phi_p$ is the smallest value such that $F(\phi_p) = p$. ::: ::: {.fragment} Credit line: solve $\phi_p^3 = p$. - Median: $\phi_{.5} = 0.5^{1/3} = 0.7937$. On a 500,000 AZN line, half of clients have drawn more than 396,850 AZN. - $\phi_{.90} = 0.9^{1/3} = 0.9655$: the bank's liquidity desk plans for 96.5% utilisation in the top decile. ::: ::: --- ## π Theorem 4.3: Probability Is Area ::: {style="font-size: 30px"} ::: {.callout-important} ## Theorem 4.3 If the random variable $Y$ has density function $f(y)$ and $a < b$, then the probability that $Y$ falls in the interval $[a, b]$ is $$P(a \le Y \le b) = \int_a^b f(y)\, dy = F(b) - F(a)$$ ::: ::: {.fragment} Because $P(Y = a) = P(Y = b) = 0$, the endpoints do not matter: $$P(a < Y < b) = P(a \le Y < b) = P(a < Y \le b) = P(a \le Y \le b)$$ **This is false for discrete variables.** In the loan office, $P(0 < Y < 2) = 1/2$ but $P(0 \le Y \le 2) = 1$. ::: ::: --- ## π¦ Worked Example: Settlement Time ::: {style="font-size: 29px"} Examples 4.4β4.5, re-set. Settlement time $Y$ (hours) has $f(y) = c\,y(4 - y)$ for $0 \le y \le 4$, zero elsewhere. **Step 1: find $c$.** $$\int_0^4 c(4y - y^2)\, dy = c\left[2y^2 - \frac{y^3}{3}\right]_0^4 = c\left(32 - \frac{64}{3}\right) = \frac{32}{3}c = 1 \;\Rightarrow\; c = \frac{3}{32}$$ ::: {.fragment} **Step 2: the distribution function.** For $0 \le y \le 4$, $F(y) = \dfrac{6y^2 - y^3}{32}$. ::: ::: {.fragment} **Step 3: read off the desk's answers.** $$P(Y \le 3) = \frac{54 - 27}{32} = \frac{27}{32} = 0.8438, \qquad P(1 \le Y \le 3) = \frac{27}{32} - \frac{5}{32} = 0.6875$$ The 3-hour promise is broken for about **one transfer in six**, $5/32 = 0.156$. ::: ::: --- ## π» Checking It in R ```{r} #| label: settlement-check #| code-fold: false f <- function(y) ifelse(y >= 0 & y <= 4, y * (4 - y), 0) c_const <- 1 / integrate(f, 0, 4)$value # should be 3/32 dens <- function(y) c_const * f(y) Fy <- function(y) integrate(dens, 0, y)$value c(c = c_const, P_le_3 = Fy(3), P_1_to_3 = Fy(3) - Fy(1)) # quantiles: solve F(phi) = p numerically q <- function(p) uniroot(function(y) Fy(y) - p, c(0, 4))$root round(c(median = q(0.5), phi_90 = q(0.9)), 4) ``` ::: {style="font-size: 28px"} The median is 2 hours, by symmetry. Nine transfers in ten settle within 3.22 hours. ::: --- ## π¬ Interactive: Area Under f Is the Height of F {.smaller} ```{ojs} //| echo: false viewof b = { const input = Inputs.range([0, 4], {value: 3, step: 0.05, label: "Settlement deadline b (hours):"}); ['pointerdown','touchstart','mousedown','click','wheel','pointermove','touchmove'] .forEach(e => input.addEventListener(e, ev => ev.stopPropagation())); return input; } ``` ```{ojs} //| echo: false dens = y => (3 / 32) * y * (4 - y) cdf = y => (6 * y * y - y * y * y) / 32 grid = Array.from({length: 161}, (_, i) => ({y: i * 0.025, f: dens(i * 0.025)})) shaded = grid.filter(d => d.y <= b + 1e-9) md`Shaded area = F(${b.toFixed(2)}) = P(Y β€ ${b.toFixed(2)}) = **${cdf(b).toFixed(4)}**` Plot.plot({ width: 1150, height: 320, marginLeft: 78, marginBottom: 58, style: {fontSize: "18px"}, x: {label: "Settlement time y (hours)", domain: [0, 4], ticks: [0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4]}, y: {label: "Density f(y)", domain: [0, 0.4], ticks: [0, 0.1, 0.2, 0.3, 0.4], tickFormat: ".1f"}, marks: [ Plot.areaY(shaded, {x: "y", y: "f", fill: "#8ba3c7", fillOpacity: 0.7}), Plot.line(grid, {x: "y", y: "f", stroke: "#14130f", strokeWidth: 2.5}), Plot.ruleX([b], {stroke: "#8b2635", strokeWidth: 2, strokeDasharray: "5 4"}), Plot.ruleY([0]) ] }) ``` --- ## π§ From F to f: Service Time at a Branch ::: {style="font-size: 29px"} Example 4.2 runs the other way: start from $F$ and differentiate. A branch's teller service time $Y$ (minutes) has $$F(y) = 1 - e^{-y^2/50}, \quad y \ge 0 \qquad \Rightarrow \qquad f(y) = \frac{dF(y)}{dy} = \frac{y}{25}\,e^{-y^2/50}, \quad y \ge 0$$ ::: {.fragment} - A customer takes longer than 10 minutes: $1 - F(10) = e^{-2} = 0.1353$ - Between 5 and 10 minutes: $F(10) - F(5) = e^{-0.5} - e^{-2} = 0.4712$ - Median: $1 - e^{-y^2/50} = 0.5$ gives $\phi_{.5} = \sqrt{50 \ln 2} = 5.89$ minutes ::: ::: {.fragment} No integral was needed: when $F$ is given, **every probability is a subtraction**. ::: ::: --- ## 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"} The daily move $Y$ (in %) of the EUR/AZN rate is modelled by $$f(y) = 1 - |y|, \quad -1 \le y \le 1, \qquad 0 \text{ elsewhere.}$$ **Four minutes, in pairs:** 1. Check that $f$ is a density. (Sketch it first.) 2. Find $P(|Y| > 0.5)$, the chance of a move of more than half a percent. 3. Find $\phi_{.95}$. And what is $P(Y = 0)$, where $f$ is highest? ::: --- ## β
Think-Pair-Share: Solution ::: {style="font-size: 29px"} 1. $f \ge 0$, and the graph is a triangle with base 2 and height 1: area $= \tfrac{1}{2} \times 2 \times 1 = 1$. 2. Each tail beyond $\pm 0.5$ is a triangle with base and height $0.5$, area $0.125$. So $P(|Y| > 0.5) = 0.25$. ::: {.fragment} 3. For $0 \le y \le 1$, $F(y) = 1 - \tfrac{(1-y)^2}{2}$. Setting $F(\phi) = 0.95$ gives $(1 - \phi)^2 = 0.1$, so $$\phi_{.95} = 1 - \sqrt{0.1} = 0.6838\%$$ A move above 0.68% happens on about one trading day in twenty. ::: ::: {.fragment} And $P(Y = 0) = 0$. **$f(0) = 1$ is a height, not a probability**; probability lives only in areas. ::: ::: --- ## π Quiz #1: Height or Probability? {.quiz-question} The settlement density has $f(2) = 0.375$. What is $P(Y = 2)$? - [$0$]{.correct data-explanation="β
Y is continuous, so F has no jumps and P(Y = 2) = 0. The value f(2) = 0.375 is a density height; probability is area, and a single point has none."} - $0.375$ - $0.5$, because 2 is the median - It cannot be found without $F(y)$ --- ## π Quiz #2: Making a Density {.quiz-question} A density is $f(y) = cy$ for $0 \le y \le 4$ and $0$ elsewhere. What is $c$? - $1/4$ - [$1/8$]{.correct data-explanation="β
Theorem 4.2 needs the total area to be 1: the integral of cy from 0 to 4 is 8c, so c = 1/8. (It is a triangle of base 4 and height 4c.)"} - $1/16$ - $1/2$ --- ## π Quiz #3: Reading F {.quiz-question} The credit-line fraction has $F(y) = y^3$ on $[0, 1]$. What is the probability a client has drawn **more than half** the line? - $0.125$ - $0.5$ - [$0.875$]{.correct data-explanation="β
P(Y > 0.5) = 1 β F(0.5) = 1 β 0.125 = 0.875. The 0.125 is the complementary event, Y β€ 0.5."} - $0.75$ --- ## π Key Formulas ::: {style="font-size: 30px"} | | Statement | |---|---| | Definition 4.1 | $F(y) = P(Y \le y)$, for any random variable | | Theorem 4.1 | $F(-\infty) = 0$, $F(\infty) = 1$, $F$ nondecreasing | | continuous $Y$ | $P(Y = y) = 0$ for every $y$ | | Definition 4.3 | $f(y) = dF(y)/dy$, $\;F(y) = \int_{-\infty}^{y} f(t)\,dt$ | | Theorem 4.2 | $f(y) \ge 0$, $\;\int_{-\infty}^{\infty} f(y)\,dy = 1$ | | Theorem 4.3 | $P(a \le Y \le b) = \int_a^b f(y)\,dy = F(b) - F(a)$ | | Definition 4.4 | $F(\phi_p) = p$, $Y$ continuous | ::: --- ## π Summary ::: {style="font-size: 30px"} - $F(y) = P(Y \le y)$ describes **every** random variable; a discrete one has a staircase, a continuous one has none - For continuous $Y$, single points carry probability zero, so endpoints never matter - The density $f$ is the derivative of $F$; $F$ is the accumulated area under $f$ - A density is a height, not a probability, and may exceed 1 - Probabilities are areas, $F(b) - F(a)$; quantiles invert $F$ ::: --- ## π Practice Problems ::: {style="font-size: 28px"} **Wackerly, 7th edition** - Β§4.2: Exercises 4.1 β 4.19; start with 4.8, 4.11, 4.12 and 4.14 - Exercise 4.7 is the discrete-endpoints warning from Theorem 4.3, worked from the binomial table - Redo the settlement example with the deadline moved to 2.5 hours: how often is the promise broken? **Week 9, Problem Set 1** is open now and closes **Sunday 15 November at 23:59** on WeBWorK, covering §§4.1β4.2. **Next class:** 7 November, expected values for continuous random variables, and our first named continuous model, the uniform distribution (Wackerly §§4.3β4.4). ::: --- ## 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"} - A distribution function can have jumps **and** rise smoothly in between. What kind of variable would that describe in insurance? - If $f(y) = 3y^2$ on $[0,1]$, why is $f(1) = 3$ not a contradiction of $P \le 1$? - The bank reports the 90th percentile of settlement time rather than the mean. Why might a client prefer that number? ::: 