```{r} #| label: setup #| include: false set.seed(2026) library(ggplot2) ``` ## π¬ The Idea in 3 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: - **Identify** a discrete random variable and write its probability distribution as a formula, a table or a graph - **Check** that a proposed $p(y)$ is a valid distribution using Theorem 3.1 - **Compute** $E(Y)$, $E[g(Y)]$, $V(Y)$ and $\sigma$ for a discrete distribution - **Use** linearity (Theorems 3.3β3.5) and $V(Y) = E(Y^2) - \mu^2$ to shorten the arithmetic - **Price** a premium or compare two projects by their mean and their variance ::: --- ## πΊοΈ Where We Are ::: {style="font-size: 30px"} Last class ended on: *Chapter 3 β discrete random variables, their probability distributions, and expected value (§§3.1β3.3).* Chapter 2 gave us a random variable as a **real-valued function on a sample space**. Today we stop listing sample points and start working with the numbers directly. ::: {.callout-important} ## The question this lecture answers A motor insurer sells hail cover. Most policies never claim; a few claim a lot. **What should one policy cost, and how far can a year's claims stray from that price?** ::: ::: --- ## π Discrete Random Variables ::: {style="font-size: 30px"} ::: {.callout-note} ## Definition 3.1 A random variable $Y$ is said to be **discrete** if it can assume only a finite or countably infinite number of distinct values. ::: Discrete random variables are usually **counts**: - complaints per hour at a mobile operator's call centre: $0, 1, 2, \dots$ - outages per week on an ISP's business network: $0, 1, 2, 3$ - hail claims filed on a motor portfolio in a season Notation: $Y$ is the random variable; $y$ is a particular value it takes. After the hour is over, $y$ is not random. ::: --- ## π Definitions 3.2 and 3.3, Theorem 3.1 ::: {style="font-size: 29px"} ::: {.callout-note} ## Definitions 3.2 and 3.3 $P(Y = y)$ is the **sum of the probabilities of all sample points in $S$ assigned the value $y$**, written $p(y)$ β the *probability function* for $Y$. The **probability distribution** of $Y$ is a formula, a table or a graph giving $p(y)$ for all $y$. ::: ::: {.callout-important} ## Theorem 3.1 For any discrete probability distribution: **(1)** $0 \le p(y) \le 1$ for all $y$; **(2)** $\sum_y p(y) = 1$, summed over all $y$ with nonzero probability. ::: Any $y$ not assigned a positive probability has $p(y) = 0$. ::: --- ## π Example 3.1: An Index Draw ::: {style="font-size: 28px"} A new index will add **2 of 6** shortlisted Baku-listed companies, chosen at random: 3 are banks, 3 are energy firms. Let $Y$ = number of banks added. ::: {.fragment} $S$ has $\binom{6}{2} = 15$ equally likely points. Counting the points assigned each value: $$p(y) = \frac{\binom{3}{y}\binom{3}{2-y}}{\binom{6}{2}}, \quad y = 0, 1, 2$$ ::: ::: {.fragment} | $y$ | 0 | 1 | 2 | |---|---|---|---| | $p(y)$ | $3/15 = 1/5$ | $9/15 = 3/5$ | $3/15 = 1/5$ | Formula, table β and each $p(y) \in [0,1]$, summing to 1, as Theorem 3.1 requires. ::: ::: --- ## π Mean and Variance ::: {style="font-size: 29px"} ::: {.callout-note} ## Definition 3.4 The **expected value** of $Y$ is $\;E(Y) = \sum_y y\,p(y).$ If $p(y)$ describes the population, $E(Y) = \mu$. ::: ::: {.callout-note} ## Definition 3.5 The **variance** of $Y$ is $\;V(Y) = E[(Y - \mu)^2]$. The **standard deviation** $\sigma$ is its positive square root. ::: ::: {.fragment} $E(Y)$ is a **long-run average**: run the experiment $n$ times and about $n\,p(y)$ results equal $y$, so the average of the results is close to $\sum_y y\,p(y)$. ::: ::: --- ## π Example 3.2: ISP Outages ::: {style="font-size: 28px"} $Y$ = outages per week on an ISP's business network in Baku. | $y$ | 0 | 1 | 2 | 3 | |---|---|---|---|---| | $p(y)$ | $1/8$ | $1/4$ | $3/8$ | $1/4$ | ::: {.fragment} $$\mu = 0\left(\tfrac18\right) + 1\left(\tfrac14\right) + 2\left(\tfrac38\right) + 3\left(\tfrac14\right) = 1.75$$ ::: ::: {.fragment} $$\sigma^2 = (0-1.75)^2\tfrac18 + (1-1.75)^2\tfrac14 + (2-1.75)^2\tfrac38 + (3-1.75)^2\tfrac14 = 0.9375$$ $\sigma = \sqrt{0.9375} = 0.97$. The interval $\mu \pm \sigma = (0.78, 2.72)$ holds $y = 1, 2$: probability $5/8$. ::: ::: --- ## π» The Same Numbers in R ```{r} #| label: isp-moments #| code-fold: false y <- 0:3 p <- c(1, 2, 3, 2) / 8 mu <- sum(y * p) # Definition 3.4 sigma2 <- sum((y - mu)^2 * p) # Definition 3.5 round(c(total = sum(p), mu = mu, var = sigma2, sd = sqrt(sigma2)), 4) # A long run of simulated weeks: the average settles on mu weeks <- sample(y, 100000, replace = TRUE, prob = p) round(c(avg = mean(weeks), var = mean((weeks - mean(weeks))^2)), 4) ``` --- ## π The Probability Histogram ```{r} #| label: isp-figure #| echo: false #| fig-width: 10 #| fig-height: 4.4 s <- sqrt(sigma2) ggplot(data.frame(y = y, p = p), aes(y, p)) + annotate("rect", xmin = mu - s, xmax = mu + s, ymin = 0, ymax = Inf, fill = "#cbb8a9", alpha = 0.35) + geom_col(width = 0.9, fill = "#8ba3c7") + geom_vline(xintercept = mu, colour = "#8b2635", linewidth = 1.2) + annotate("text", x = mu + 0.05, y = 0.41, hjust = 0, size = 6, label = "mu == 1.75", parse = TRUE, colour = "#8b2635") + scale_x_continuous(breaks = 0:3, minor_breaks = NULL) + scale_y_continuous(limits = c(0, 0.42), breaks = seq(0, 0.4, 0.1)) + labs(x = "Outages per week, y", y = "p(y)") + theme_minimal(base_size = 20) ``` ::: {style="font-size: 28px"} $\mu$ sits between bars, at a value $Y$ can never take. The shaded band is $\mu \pm \sigma$. ::: --- ## π Theorem 3.2: $E[g(Y)]$ ::: {style="font-size: 28px"} ::: {.callout-important} ## Theorem 3.2 $E[g(Y)] = \sum_{\text{all } y} g(y)\,p(y)$ ::: The ISP's contract credits a corporate client **0 AZN** for 0 or 1 outages, **500 AZN** for 2 and **1,500 AZN** for 3. $$E[g(Y)] = 0\left(\tfrac18\right) + 0\left(\tfrac14\right) + 500\left(\tfrac38\right) + 1500\left(\tfrac14\right) = 562.5 \text{ AZN per week}$$ ::: {.fragment} No need to derive the distribution of $g(Y)$ first. The proof does exactly that grouping β $P(g = 0) = 3/8$, $P(g = 500) = 3/8$, $P(g = 1500) = 1/4$ β and gets the same sum. ::: ::: --- ## π Theorems 3.3β3.5: Linearity ::: {style="font-size: 30px"} ::: {.callout-important} ## Theorems 3.3, 3.4 and 3.5 For a constant $c$ and functions $g, g_1, \dots, g_k$ of $Y$: $$E(c) = c, \qquad E[c\,g(Y)] = c\,E[g(Y)],$$ $$E[g_1(Y) + \cdots + g_k(Y)] = E[g_1(Y)] + \cdots + E[g_k(Y)]$$ ::: ::: {.fragment} Each proof is one line: apply Theorem 3.2, then pull the constant or split the sum. $E(c) = c$ uses $\sum_y p(y) = 1$ from Theorem 3.1. ::: ::: --- ## π Theorem 3.6: A Shortcut ::: {style="font-size: 29px"} ::: {.callout-important} ## Theorem 3.6 $V(Y) = \sigma^2 = E[(Y - \mu)^2] = E(Y^2) - \mu^2$ ::: ::: {.fragment} Expand and use 3.3β3.5: $E(Y^2 - 2\mu Y + \mu^2) = E(Y^2) - 2\mu^2 + \mu^2$. ::: ::: {.fragment} **Example 3.3, ISP outages:** $E(Y^2) = 0 + 1\left(\tfrac14\right) + 4\left(\tfrac38\right) + 9\left(\tfrac14\right) = 4$, so $$\sigma^2 = 4 - 1.75^2 = 0.9375$$ Note $E(Y^2) = 4 \ne \mu^2 = 3.0625$: **the mean of a function is not the function of the mean.** ::: ::: --- ## πΈ A Remittance Fee, $aY + b$ ::: {style="font-size: 28px"} A Baku transfer office charges **1 AZN plus 2%** of the amount. With $Y$ = amount in hundreds of AZN, the fee is $F = 2Y + 1$. | $y$ | 1 | 2 | 3 | 5 | |---|---|---|---|---| | $p(y)$ | 0.4 | 0.3 | 0.2 | 0.1 | $E(Y) = 2.1$, $E(Y^2) = 5.9$, so $V(Y) = 5.9 - 2.1^2 = 1.49$. ::: {.fragment} From Theorems 3.3β3.6: $E(aY + b) = a\mu + b$ and $V(aY + b) = a^2\sigma^2$. $$E(F) = 2(2.1) + 1 = 5.2 \text{ AZN}, \qquad V(F) = 4(1.49) = 5.96, \quad \sigma_F = 2.44 \text{ AZN}$$ The fixed 1 AZN moves the mean and leaves the spread alone. ::: ::: --- ## π¨οΈ Worked Example: Pricing Hail Cover ::: {style="font-size: 28px"} One car's hail claim $X$ in a season: **0** (prob. 0.95), **2,000 AZN** (0.04), **8,000 AZN** (0.01). Admin costs 40 AZN per policy; the insurer targets an expected profit of 60 AZN. What premium $C$? ::: {.fragment} $E(X) = 2000(0.04) + 8000(0.01) = 160$ AZN. Profit is $C - 40 - X$, so by linearity $$E(\text{profit}) = C - 40 - 160 = 60 \;\Rightarrow\; C = 260 \text{ AZN}$$ ::: ::: {.fragment} $E(X^2) = 0.04(2000^2) + 0.01(8000^2) = 800{,}000$, so $\sigma_X = \sqrt{800{,}000 - 160^2} = 880$ AZN β more than three premiums. One policy is a gamble; the business is in **pooling many**. ::: ::: --- ## π Example 3.4: Two Van Models ::: {style="font-size: 28px"} A logistics firm runs a van $t$ hours a day. Daily breakdowns $Y_A$ have mean and variance $0.10t$; $Y_B$ has $0.12t$. Daily cost (AZN): $C_A = 10t + 30Y_A^2$, $C_B = 8t + 30Y_B^2$. ::: {.fragment} The key step is Theorem 3.6 read backwards, $E(Y^2) = V(Y) + \mu^2$: $$E(C_A) = 10t + 30[0.10t + (0.10t)^2] = 13t + 0.3t^2, \quad E(C_B) = 11.6t + 0.432t^2$$ ::: ::: {.fragment} | $t$ | $E(C_A)$ | $E(C_B)$ | cheaper | |---|---|---|---| | 10 h | 160.0 | 159.2 | B | | 20 h | 380.0 | 404.8 | A | The two cross at $t = 1.4/0.132 \approx 10.6$ hours. ::: ::: --- ## 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: 29px"} A bank can finance one of two projects. NPV in millions of AZN: | Project | Outcome 1 | Outcome 2 | Outcome 3 | |---|---|---|---| | P | $-1$ (prob. 0.2) | $2$ (prob. 0.5) | $4$ (prob. 0.3) | | Q | $1$ (prob. 0.25) | $2$ (prob. 0.5) | $3$ (prob. 0.25) | **Four minutes, in pairs:** 1. Find $E(Y)$ for each project. 2. Find $V(Y)$ for each, using Theorem 3.6. 3. Which would a lender who fears losses choose, and what does the mean alone hide? ::: --- ## β
Think-Pair-Share: Solution ::: {style="font-size: 29px"} 1. $E(P) = -0.2 + 1.0 + 1.2 = 2.0$ and $E(Q) = 0.25 + 1.0 + 0.75 = 2.0$. **Same mean.** 2. $E(P^2) = 0.2 + 2.0 + 4.8 = 7.0$, so $V(P) = 7.0 - 4 = 3.0$, $\sigma_P = 1.73$. $E(Q^2) = 0.25 + 2.0 + 2.25 = 4.5$, so $V(Q) = 0.5$, $\sigma_Q = 0.71$. ::: {.fragment} 3. The mean cannot tell them apart. P carries a **20% chance of losing 1 million**; Q never loses. The variance is the first number that sees the difference β which is why Β§1.3 paired every mean with a spread. ::: ::: --- ## π¬ Interactive: Moving Mass to the Tail {.smaller} ```{ojs} //| echo: false viewof shift = { const input = Inputs.range([0, 0.30], {value: 0, step: 0.01, label: "Mass moved from 1 to 5 complaints:"}); ['pointerdown','touchstart','mousedown','click','wheel','pointermove','touchmove'] .forEach(e => input.addEventListener(e, ev => ev.stopPropagation())); return input; } ``` ```{ojs} //| echo: false base = [0.15, 0.35, 0.25, 0.15, 0.05, 0.05] pmf = base.map((b, y) => ({y: y, p: y === 1 ? b - shift : y === 5 ? b + shift : b})) mu = d3.sum(pmf, d => d.y * d.p) ey2 = d3.sum(pmf, d => d.y * d.y * d.p) v = ey2 - mu * mu md`Total **${d3.sum(pmf, d => d.p).toFixed(2)}** Β· mean **ΞΌ = ${mu.toFixed(2)}** Β· variance **ΟΒ² = ${v.toFixed(3)}** Β· Ο = ${Math.sqrt(v).toFixed(3)}` Plot.plot({ width: 1150, height: 300, marginTop: 40, marginLeft: 78, marginBottom: 58, style: {fontSize: "18px"}, x: {label: "Complaints per hour, y", domain: [0, 1, 2, 3, 4, 5], padding: 0.15}, y: {label: "p(y)", domain: [0, 0.4], ticks: [0, 0.1, 0.2, 0.3, 0.4], tickFormat: ".1f", grid: true}, marks: [ Plot.barY(pmf, {x: "y", y: "p", fill: d => d.y === 1 || d.y === 5 ? "#8b2635" : "#8ba3c7"}), Plot.text(pmf, {x: "y", y: "p", text: d => d.p.toFixed(2), dy: -14, fontSize: 18}), Plot.ruleY([0]) ] }) ``` ::: {style="font-size: 28px"} Moving 0.30 of mass four steps raises ΞΌ by $4(0.30) = 1.2$ β but nearly **doubles** ΟΒ². ::: --- ## π Quiz #1: Is It a Distribution? {.quiz-question} Which of these can be the probability function of complaints per hour, $y = 0, 1, 2, 3$? - $p(y) = 0.3, 0.3, 0.3, 0.3$ - [$p(y) = 0.1, 0.4, 0.3, 0.2$]{.correct data-explanation="β
Every value lies in [0, 1] and they sum to 1, which is all Theorem 3.1 asks. The first sums to 1.2; the third has a negative entry; the fourth sums to 0.9."} - $p(y) = 0.5, 0.4, 0.2, -0.1$ - $p(y) = 0.2, 0.2, 0.2, 0.3$ --- ## π Quiz #2: The Mean {.quiz-question} A small fleet files $Y$ = 0, 1 or 2 hail claims in a season, with probabilities 0.5, 0.3, 0.2. What is $E(Y)$? - $1$, the middle value - [$0.7$]{.correct data-explanation="β
Definition 3.4: 0(0.5) + 1(0.3) + 2(0.2) = 0.7. The mean weights each value by its probability; it need not be a value Y can take."} - $0.33$, the average probability - $1.1$ --- ## π Quiz #3: A Fee Rule {.quiz-question} $V(Y) = 4$ for a transfer amount $Y$. The fee is $F = 3Y + 5$ AZN. What is $V(F)$? - $17$ - $12$ - [$36$]{.correct data-explanation="β
V(aY + b) = aΒ²V(Y) = 9 Γ 4 = 36. The constant 5 shifts every fee equally and adds no spread; the multiplier enters squared."} - $41$ --- ## π Key Formulas ::: {style="font-size: 29px"} | | Statement | |---|---| | Theorem 3.1 | $0 \le p(y) \le 1$, $\;\sum_y p(y) = 1$ | | Definition 3.4 | $E(Y) = \sum_y y\,p(y)$ | | Theorem 3.2 | $E[g(Y)] = \sum_y g(y)\,p(y)$ | | Definition 3.5 | $V(Y) = E[(Y - \mu)^2]$, $\;\sigma = \sqrt{V(Y)}$ | | Theorems 3.3β3.5 | $E(c) = c$, $\;E[cg(Y)] = cE[g(Y)]$, $\;E$ of a sum is the sum of $E$ | | Theorem 3.6 | $V(Y) = E(Y^2) - \mu^2$ | | consequence | $E(aY + b) = a\mu + b$, $\;V(aY + b) = a^2\sigma^2$ | ::: --- ## π Summary ::: {style="font-size: 30px"} - A discrete random variable takes countably many values; its distribution is $p(y)$ as a formula, a table or a graph - Theorem 3.1 is the validity check: every $p(y)$ in $[0, 1]$, total 1 - $E(Y)$ is a probability-weighted average β a premium is built on it - $V(Y)$ measures what the mean hides: two projects with the same mean can carry very different risk - Linearity and $E(Y^2) - \mu^2$ turn most calculations into two sums ::: --- ## π Practice Problems ::: {style="font-size: 28px"} **Wackerly, 7th edition** - Exercises at the end of Β§3.2: 3.1, 3.2, 3.9 - Exercises at the end of Β§3.3: 3.12, 3.13, 3.14, 3.19, 3.23, 3.27, 3.30 **Week 5, Problem Set 2** is open now and closes **Sunday 18 October at 23:59** on WeBWorK, covering §§3.1β3.3. The TA-led tutorial runs this week. **Quiz I** is on **17 October**, first 30 minutes of class, on Chapters 1β2 β today's material is not on it, but it is on Midterm I (24 October). **Next class:** Β§3.4 β the binomial distribution. ::: --- ## 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"} - Can $E(Y)$ be a value that $Y$ never takes? Can $V(Y)$ ever be negative? - A premium is set at $E(X)$ plus costs. Why does an insurer with 50 policies charge more than one with 50,000? - When is $E[g(Y)] = g(E(Y))$ exactly true? ::: 