```{r} #| label: setup #| include: false set.seed(2026) library(ggplot2) theme_set(theme_minimal(base_size = 18)) ``` ## 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: 30px"} By the end of this lecture, you will be able to: - **Define** a random variable as a real-valued function on $S$ (Definition 2.12) β *load-bearing* - **Compute** $P(Y = y)$ by summing the probabilities of the sample points in $\{Y = y\}$ β *load-bearing* - **Count** the $\binom{N}{n}$ possible simple random samples and show each unit is included with probability $n/N$ - **Choose** the right Chapter 2 tool for a problem, and combine several in one calculation The first two carry all of Chapter 3. Sampling and the summary we meet more briefly. ::: --- ## πΊοΈ Where We Are ::: {style="font-size: 30px"} **Wackerly §§2.11β2.13** Last class ended: *turning outcomes into numbers β random variables, random sampling, and a synthesis of Chapter 2.* ::: {.fragment} A bank's risk committee holding three SME loans does not ask *which* loans default. It asks **how many**, and **how much money** is lost. ::: ::: {.fragment} Those are numbers, not lists of outcomes. Today we build the bridge from a sample space to a number β and from there, in Chapter 3, to a distribution. ::: ::: --- ## π Definition 2.12: Random Variable ::: {style="font-size: 30px"} ::: {.callout-note} ## Definition 2.12 (Wackerly Β§2.11) A **random variable** is a real-valued function for which the domain is a sample space. ::: ::: {.fragment} So $Y$ takes each sample point $E_i \in S$ to a number $Y(E_i)$. Several points may share a value; no point gets two. ::: ::: {.fragment} The **numerical event** $\{Y = y\}$ is the set of all sample points assigned the value $y$. These sets **partition** $S$: they are mutually exclusive, and together they cover it. ::: ::: --- ## π¦ Three SME Loans ::: {style="font-size: 28px"} A Baku bank holds three loans that default independently: construction, $P = 0.10$; retail, $P = 0.05$; agriculture, $P = 0.20$. Let $Y$ = number of defaults. | Sample point | Probability | $Y$ | |---|---|---| | NNN | $0.9 \times 0.95 \times 0.8 = 0.684$ | 0 | | DNN, NDN, NND | $0.076,\ 0.036,\ 0.171$ | 1 | | DDN, DND, NDD | $0.004,\ 0.019,\ 0.009$ | 2 | | DDD | $0.1 \times 0.05 \times 0.2 = 0.001$ | 3 | The points are **not** equally likely, so we cannot count β we must sum. ::: --- ## β From Sample Points to $P(Y = y)$ ::: {style="font-size: 30px"} $P(Y = y)$ is the sum of the probabilities of the sample points assigned the value $y$. ::: {.fragment} $$P(Y = 1) = P(\text{DNN}) + P(\text{NDN}) + P(\text{NND}) = 0.076 + 0.036 + 0.171 = 0.283$$ ::: ::: {.fragment} $$P(Y = 0) = 0.684, \quad P(Y = 2) = 0.032, \quad P(Y = 3) = 0.001$$ ::: ::: {.fragment} The four values sum to 1, because the events $\{Y = y\}$ partition $S$. **That check is free β always make it.** ::: ::: --- ## π» The Same Calculation in R ```{r} #| label: loans-enumerate #| code-fold: false p <- c(construction = 0.10, retail = 0.05, agriculture = 0.20) S <- expand.grid(c1 = 0:1, c2 = 0:1, c3 = 0:1) # 8 sample points S$prob <- apply(S, 1, function(d) prod(ifelse(d == 1, p, 1 - p))) S$Y <- rowSums(S[, 1:3]) # the function Y(E_i) pY <- tapply(S$prob, S$Y, sum) # sum within {Y = y} round(pY, 3); sum(pY) ``` --- ## π The Picture: $P(Y = y)$ ```{r} #| label: loans-figure #| echo: false #| fig-width: 9 #| fig-height: 4.0 df <- data.frame(y = as.integer(names(pY)), p = as.numeric(pY)) ggplot(df, aes(factor(y), p)) + geom_col(fill = "#8ba3c7", width = 0.6) + geom_text(aes(label = sprintf("%.3f", p)), vjust = -0.4, size = 7) + scale_y_continuous(limits = c(0, 0.78), breaks = seq(0, 0.7, 0.1)) + labs(x = "Number of defaults, y", y = "P(Y = y)") ``` ::: {style="font-size: 28px"} Eight sample points collapsed into four numbers. In Chapter 3 this list of values gets a name: the **probability distribution** of $Y$. ::: --- ## π° A Second Function on the Same $S$ ::: {style="font-size: 28px"} Exposures are 200, 150 and 100 thousand AZN. Let $L$ = loss in default (thousand AZN). | NNN | NND | NDN | DNN | NDD | DND | DDN | DDD | |---|---|---|---|---|---|---|---| | 0 | 100 | 150 | 200 | 250 | 300 | 350 | 450 | ::: {.fragment} Same experiment, same $S$, a **different** function β and a different partition: every point gets its own value. ::: ::: {.fragment} $P(L \ge 250) = 0.009 + 0.019 + 0.004 + 0.001 = 0.033$. Here $\{L \ge 250\} = \{Y \ge 2\}$ exactly: it is one set of sample points, described by two random variables. ::: ::: --- ## π² Β§2.12: The Design Matters ::: {style="font-size: 28px"} A statistical experiment observes a **sample** from a **population**, and uses the probability of the observed sample to make an inference. So the sampling method β the **design** β must be stated. An auditor picks $n = 2$ of a bank's $N = 5$ branches. Probability of drawing one specific pair $\{a, b\}$: ::: {.fragment} - **Without replacement:** $\binom{5}{2} = 10$ equally likely pairs, so $1/10$. ::: ::: {.fragment} - **With replacement:** 25 equally likely ordered draws; $\{a, b\}$ arises as $(a, b)$ or $(b, a)$, so $2/25$. ::: ::: --- ## π Definition 2.13: Random Sample ::: {style="font-size: 30px"} ::: {.callout-note} ## Definition 2.13 (Wackerly Β§2.12) Let $N$ and $n$ be the numbers of elements in the population and sample. If the sampling is conducted so that each of the $\binom{N}{n}$ samples has an equal probability of being selected, the sampling is **random**, and the result is a **random sample**. ::: ::: {.fragment} Each sample therefore has probability $1 / \binom{N}{n}$. In practice the draw is made by software β or, in the book, from a random number table (Table 12, Appendix 3). ::: ::: --- ## ποΈ Worked Example: A Branch Audit ::: {style="font-size: 28px"} The Central Bank audits a simple random sample of $n = 6$ of a commercial bank's $N = 30$ branches. ::: {.fragment} **Number of possible samples:** $\binom{30}{6} = 593{,}775$, each with probability $1/593{,}775$. ::: ::: {.fragment} **Is the Ganjlik branch audited?** Count the samples containing it: fix it, choose the other 5 from 29. $$P(\text{included}) = \frac{\binom{29}{5}}{\binom{30}{6}} = \frac{6}{30} = 0.2 = \frac{n}{N}$$ ::: ::: {.fragment} Every branch faces the same $n/N$. That is the sense in which an SRS is "fair". ::: ::: --- ## π» Checking $n/N$ by Simulation ```{r} #| label: srs-sim #| code-fold: false N <- 30; n <- 6 choose(N, n) # number of possible samples audits <- replicate(20000, sample(1:N, n)) # 20,000 simple random samples incl <- sapply(1:N, function(b) mean(colSums(audits == b))) # share containing b round(range(incl), 3) # every branch near n/N = 0.2 ``` ::: {style="font-size: 28px"} Each of the 30 branches was audited in about 20% of the simulated samples. ::: --- ## ποΈ Not Every Survey Is One SRS ::: {style="font-size: 30px"} A household budget survey needs 800 households across Azerbaijan. One SRS could, by pure chance, draw them mostly from Baku. ::: {.fragment} Instead: allot a number of households to each region, and draw an SRS **within** each region. The pieces are then combined. ::: ::: {.fragment} Such designs are only partially random, and are a course in themselves. For us, Definition 2.13 is the building block β and the random variables of Chapter 3 will be defined on its sample space. ::: ::: --- ## 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"} A survey firm interviews an SRS of $n = 3$ of the $N = 10$ households on one street. Four of the ten hold a bank loan. **Four minutes, in pairs:** 1. How many different samples are possible? 2. What is the probability that a given household is interviewed? 3. Let $Y$ = number of indebted households in the sample. Find $P(Y = 0)$ and $P(Y = 2)$. ::: --- ## β
Think-Pair-Share: Solution ::: {style="font-size: 30px"} 1. $\binom{10}{3} = 120$ samples, each with probability $1/120$. 2. $\binom{9}{2}/\binom{10}{3} = 36/120 = 3/10 = n/N$. ::: {.fragment} 3. $Y$ is a random variable on the 120-point sample space. Count the points in each event: $$P(Y = 0) = \frac{\binom{6}{3}}{120} = \frac{20}{120} = 0.167, \qquad P(Y = 2) = \frac{\binom{4}{2}\binom{6}{1}}{120} = \frac{36}{120} = 0.3$$ ::: ::: {.fragment} Equally likely points, so here summing **is** counting. ::: ::: --- ## π§ Β§2.13: Which Tool, When? ::: {style="font-size: 28px"} | The problem gives you | Reach for | |---|---| | A small $S$ you can list | Sample-point method (Β§2.5) | | Equally likely points, too many to list | Counting rules (Β§2.6) | | "Given that β¦", or a two-way table | Conditional probability (Β§2.7) | | "Unrelated", "separately" | Independence: multiply (Β§2.7) | | "Both" / "either" / "at least one" | Multiplicative, additive laws, complement (Β§2.8) | | A compound event to decompose | Event-composition method (Β§2.9) | | Stages or states, then a reversal | Total probability, Bayes (Β§2.10) | ::: --- ## π Synthesis: Up-Days on an Index ::: {style="font-size: 28px"} A stock index is in a **calm** regime with probability 0.7, **volatile** with 0.3. Given the regime, days are independent: $P(\text{up}) = 0.6$ if calm, $0.4$ if volatile. Let $Y$ = number of up-days in the next 3 trading days. ::: {.fragment} **Given calm** (independence, then counting the $\binom{3}{y}$ orderings): $$P(Y = 3 \mid \text{calm}) = 0.6^3 = 0.216, \qquad P(Y = 3 \mid \text{volatile}) = 0.4^3 = 0.064$$ ::: ::: {.fragment} **Total probability** over the two regimes: $$P(Y = 3) = 0.7(0.216) + 0.3(0.064) = 0.1512 + 0.0192 = 0.1704$$ ::: ::: --- ## π Synthesis: Reading the Regime Back ::: {style="font-size: 28px"} **Bayes' rule:** three up-days in a row β how sure are we the market is calm? $$P(\text{calm} \mid Y = 3) = \frac{0.1512}{0.1704} = 0.887$$ ```{r} #| label: synthesis #| code-fold: false y <- 0:3 calm <- choose(3, y) * 0.6^y * 0.4^(3 - y) # P(Y = y | calm): orderings x product vol <- choose(3, y) * 0.4^y * 0.6^(3 - y) # P(Y = y | volatile) pY <- 0.7 * calm + 0.3 * vol # total probability out <- rbind(P_Y = round(pY, 4), P_calm_given_Y = round(0.7 * calm / pY, 3)) colnames(out) <- paste0("y=", y); out ``` Prior 0.7, posterior 0.887 after three up-days β and 0.409 after none. ::: --- ## π Quiz #1: Summing Sample Points {.quiz-question} Two bond coupons are due; each is paid with probability 0.9, independently. $Y$ = number of missed coupons. What is $P(Y = 1)$? - [$0.18$]{.correct data-explanation="β
{Y = 1} = {paid-missed, missed-paid}, two sample points of 0.9 Γ 0.1 = 0.09 each, so P(Y = 1) = 0.18."} - $0.09$ - $0.10$ - $0.81$ --- ## π Quiz #2: Inclusion Probability {.quiz-question} A compliance officer draws an SRS of 5 client files from 40. What is the probability that a particular client's file is drawn? - $1/40$ - $1/\binom{40}{5}$ - [$5/40 = 0.125$]{.correct data-explanation="β
Samples containing the file: choose(39, 4); all samples: choose(40, 5). The ratio is 5/40 = n/N."} - $5/39$ --- ## π Quiz #3: What Is a Random Variable? {.quiz-question} A loan is observed for one year. Which of these is a **random variable**? - The event that the loan defaults - The sample space $\{\text{default}, \text{no default}\}$ - [The number of defaults: 1 if it defaults, 0 if not]{.correct data-explanation="β
Definition 2.12: a real-valued function on S. The event 'defaults' is a subset of S, not a number; the 0/1 count assigns a number to each sample point."} - The probability of default, 0.08 --- ## π Key Formulas ::: {style="font-size: 30px"} | | Statement | |---|---| | Definition 2.12 | $Y: S \to$ the real line | | numerical event | $\{Y = y\} = \{E_i \in S : Y(E_i) = y\}$ | | probability of a value | $P(Y = y) = \sum_{E_i:\, Y(E_i) = y} P(E_i)$ | | the check | $\sum_y P(Y = y) = 1$ | | Definition 2.13 | each of $\binom{N}{n}$ samples has probability $1/\binom{N}{n}$ | | inclusion probability | $\binom{N-1}{n-1} / \binom{N}{n} = n/N$ | ::: --- ## π Summary ::: {style="font-size: 30px"} - A random variable is a **function** on $S$; the events $\{Y = y\}$ partition $S$ - $P(Y = y)$ is a sum over sample points β a count only when they are equally likely - One $S$ carries many random variables: defaults, losses, up-days - An SRS makes all $\binom{N}{n}$ samples equally likely, so each unit is included with probability $n/N$ - Chapter 2 is a toolbox; hard problems use several tools in sequence ::: --- ## π Practice Problems ::: {style="font-size: 28px"} **Wackerly, 7th edition** - Β§2.11: exercises 2.139 β 2.142; supplementary 2.150, 2.167, 2.170 - Chapter 2 supplementary exercises for review: 2.146, 2.149, 2.152, 2.157, 2.173, 2.174 **Week 5, Problem Set 1** is open now and closes **Sunday 18 October at 23:59** on WeBWorK, covering §§2.11β2.13. **Quiz I (Chapters 1β2)** is on **17 October**, first 30 minutes of class, on WeBWorK. This is our last session on Chapter 2 β the supplementary exercises are your revision list. **Next class:** Chapter 3 β discrete random variables, their probability distributions, and expected value [(§§3.1β3.3).]{style="white-space: nowrap"} ::: --- ## 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"} - If $Y$ and $L$ are two random variables on the same $S$, is $Y + L$ one too? Is $Y^2$? - Sampling with replacement gave $2/25$ for a specific pair, not $1/10$. Which design carries more information about the population? - In the synthesis example, why are the up-days **not** independent once the regime is unknown? ::: 