```{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: - **Recognise** a multinomial experiment from the five properties of Definition 5.11 - **Evaluate** the multinomial probability function for a given set of cell counts - **State** Theorem 5.13: the mean, variance and covariance of the cell counts - **Explain** why every covariance between two cells is negative - **Combine** Theorem 5.13 with Theorem 5.12 to find the mean and variance of a revenue total ::: --- ## πΊοΈ Where We Are ::: {style="font-size: 30px"} **Wackerly Β§5.9** Β· Quiz II is behind us. Wednesday ended on Theorem 5.12: the variance of $\sum a_i Y_i$ is every weight applied to a variance **and to a covariance**. Portfolio risk lives in those cross terms. ::: {.fragment} Today, one distribution in which every cross term is known in advance, from the design of the experiment alone. The binomial counted two outcomes per trial; the **multinomial** counts $k$. ::: ::: {.fragment} A rating review ends in *up*, *stable*, *down* or *default*, not in two classes. How are the counts distributed, and how do they move together? ::: ::: --- ## π A Multinomial Experiment ::: {style="font-size: 30px"} ::: {.callout-important} ## Definition 5.11 1. The experiment consists of $n$ identical trials. 2. The outcome of each trial falls into one of $k$ classes or cells. 3. The probability that a trial falls into cell $i$ is $p_i$, the same from trial to trial, with $p_1 + p_2 + \cdots + p_k = 1$. 4. The trials are independent. 5. $Y_i$ is the number of trials falling into cell $i$, so $Y_1 + Y_2 + \cdots + Y_k = n$. ::: With $k = 2$ this is exactly the binomial experiment of Chapter 3. ::: --- ## π The Probability Function ::: {style="font-size: 30px"} ::: {.callout-important} ## Definition 5.12 $Y_1, \ldots, Y_k$ have a multinomial distribution with parameters $n$ and $p_1, \ldots, p_k$ if $$p(y_1, \ldots, y_k) = \frac{n!}{y_1!\,y_2! \cdots y_k!}\; p_1^{y_1} p_2^{y_2} \cdots p_k^{y_k},$$ where each $y_i = 0, 1, \ldots, n$ and $\sum_{i=1}^{k} y_i = n$. ::: ::: {.fragment} It is Chapter 2 in two factors: **one** ordered sequence with these counts has probability $p_1^{y_1} \cdots p_k^{y_k}$, and the coefficient is the number of ways to partition $n$ trials into groups of sizes $y_1, \ldots, y_k$. ::: ::: --- ## π Worked Example: Rating Transitions ::: {style="font-size: 28px"} A bank's loan book holds 8 firms rated BB. Over one year each firm, independently, is **upgraded** (0.10), stays **stable** (0.75), is **downgraded** (0.12) or **defaults** (0.03). Find the probability of 1 upgrade, 6 stable, 1 downgrade and no default. ::: {.fragment} $$p(1, 6, 1, 0) = \frac{8!}{1!\,6!\,1!\,0!}\,(0.10)^1 (0.75)^6 (0.12)^1 (0.03)^0 = 56 \times 0.10 \times 0.17798 \times 0.12 = 0.1196$$ ::: ::: {.fragment} Even the single most likely year, $(0, 7, 1, 0)$, has probability only 0.1281: with four cells, probability is spread over **165** possible outcomes. ::: ::: --- ## π Moments of the Counts ::: {style="font-size: 30px"} ::: {.callout-important} ## Theorem 5.13 If $Y_1, \ldots, Y_k$ have a multinomial distribution with parameters $n$ and $p_1, \ldots, p_k$, then 1. $E(Y_i) = np_i$ and $V(Y_i) = np_iq_i$, where $q_i = 1 - p_i$. 2. $\text{Cov}(Y_s, Y_t) = -np_sp_t$, if $s \neq t$. ::: ::: {.fragment} **Part 1.** Merge every cell except $i$ into one: each trial lands in cell $i$ or not, so $Y_i$ is **binomial** $(n, p_i)$. ::: ::: {.fragment} **Part 2.** On a single trial, landing in cell $s$ rules out cell $t$, so the two indicators have covariance $0 - p_sp_t$. Different trials are independent, and Theorem 5.12 adds the $n$ same-trial terms: $-np_sp_t$. ::: ::: --- ## π± Worked Example: Tariff Choice ::: {style="font-size: 28px"} A Baku mobile operator signs 200 new subscribers. Each independently picks **Basic** (10 AZN a month, $p_1 = 0.50$), **Standard** (20 AZN, $p_2 = 0.35$) or **Premium** (35 AZN, $p_3 = 0.15$). | | Basic | Standard | Premium | |---|---|---|---| | $E(Y_i) = np_i$ | 100 | 70 | 30 | | $V(Y_i) = np_iq_i$ | 50 | 45.5 | 25.5 | $\text{Cov}(Y_1, Y_2) = -35$, $\ \text{Cov}(Y_1, Y_3) = -15$, $\ \text{Cov}(Y_2, Y_3) = -10.5$. ::: --- ## π° Monthly Revenue ::: {style="font-size: 28px"} Revenue is $R = 10Y_1 + 20Y_2 + 35Y_3$ AZN. $$E(R) = 10(100) + 20(70) + 35(30) = 3450 \text{ AZN}$$ ::: {.fragment} $$\begin{aligned} V(R) &= 100(50) + 400(45.5) + 1225(25.5) \\ &\quad + 2\big[200(-35) + 350(-15) + 700(-10.5)\big] \end{aligned}$$ ::: ::: {.fragment} $V(R) = 54\,437.5 - 39\,200 = 15\,237.5$, so the standard deviation is **123.4 AZN**. Dropping the covariances would report 233.3 AZN: a subscriber who picks Premium is one who did not pick Basic. ::: ::: --- ## π» Checking It by Simulation ```{r} #| label: tariff-sim n <- 200; p <- c(0.50, 0.35, 0.15); fee <- c(10, 20, 35) Y <- t(rmultinom(10000, size = n, prob = p)) # 10,000 cohorts R <- Y %*% fee round(c(cov_basic_premium = cov(Y[, 1], Y[, 3]), theory = -n * p[1] * p[3], sd_revenue = sd(R), theory = sqrt(15237.5)), 1) ``` ```{r} #| label: tariff-fig #| echo: false #| fig-width: 10 #| fig-height: 2.9 cells <- as.data.frame(table(basic = Y[, 1], premium = Y[, 3])) cells <- subset(cells, Freq > 0) cells$basic <- as.integer(as.character(cells$basic)) cells$premium <- as.integer(as.character(cells$premium)) ggplot(cells, aes(basic, premium)) + geom_tile(aes(fill = Freq)) + geom_abline(intercept = 60, slope = -0.3, colour = "#8b2635", linewidth = 1.2) + scale_fill_gradient(low = "grey88", high = "#14130f", name = "Cohorts") + labs(x = "Basic subscribers", y = "Premium subscribers") + theme(legend.title = element_text(size = 16), legend.text = element_text(size = 14)) ``` ::: {style="font-size: 28px"} Each tile is a count of cohorts; the red line is $0.3(200 - y_1)$, the average Premium count given $y_1$ Basic. Its slope is the negative covariance made visible. ::: --- ## 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 trader classifies each of 20 trading days on the Baku exchange as **up** (0.45), **flat** (0.20) or **down** (0.35), independently. **Four minutes, in pairs:** 1. What is the probability of exactly 10 up, 4 flat and 6 down days? 2. Let $D = Y_{\text{up}} - Y_{\text{down}}$. Find $E(D)$ and $V(D)$. 3. What is $\text{Cov}(Y_{\text{up}}, Y_{\text{down}})$, and what does its sign say? ::: --- ## β
Think-Pair-Share: Solution ::: {style="font-size: 28px"} 1. Definition 5.12: $\dfrac{20!}{10!\,4!\,6!}(0.45)^{10}(0.20)^4(0.35)^6 = 38\,798\,760 \times \cdots = 0.0389$ ::: {.fragment} 2. $E(D) = 20(0.45) - 20(0.35) = 2$. By Theorem 5.12, with the covariance from part 3, $$V(D) = 20(0.45)(0.55) + 20(0.35)(0.65) - 2(-3.15) = 4.95 + 4.55 + 6.30 = 15.8$$ ::: ::: {.fragment} 3. $\text{Cov} = -20(0.45)(0.35) = -3.15$. Every up day is a day that was not down. The negative covariance **widens** the spread of a difference: $-2\,\text{Cov}$ adds to $V(D)$. ::: ::: --- ## π Quiz #1: The Sign of the Covariance {.quiz-question} Twelve customers each choose Basic (0.50), Standard (0.35) or Premium (0.15), independently. What is $\text{Cov}(Y_1, Y_2)$? - [$-2.1$]{.correct data-explanation="β
Theorem 5.13: Cov(Y1, Y2) = βn p1 p2 = β12 Γ 0.50 Γ 0.35 = β2.1. It is negative because the counts share a fixed total of 12."} - $2.1$ - $0$, because the customers choose independently - $-0.175$ --- ## π Quiz #2: Is It Multinomial? {.quiz-question} Which of these is **not** a multinomial experiment? - Classifying 30 independent trading days as up, flat or down - Recording which of three tariffs each of 200 independent new subscribers picks - [Drawing 5 firms without replacement from a list of 12, four in each of three rating grades]{.correct data-explanation="β
Without replacement from a small list, each draw changes the probabilities for the next: property 3 (constant p_i) and property 4 (independence) both fail. This is a hypergeometric setting."} - Rating 50 independent loan applicants as approve, refer or decline --- ## π Key Formulas ::: {style="font-size: 30px"} | | Statement | |---|---| | Definition 5.12 | $p(y_1, \ldots, y_k) = \dfrac{n!}{y_1! \cdots y_k!}\,p_1^{y_1} \cdots p_k^{y_k}$ | | constraints | $\sum p_i = 1$, $\ \sum y_i = n$ | | Theorem 5.13, part 1 | $E(Y_i) = np_i$, $\ V(Y_i) = np_iq_i$ | | Theorem 5.13, part 2 | $\text{Cov}(Y_s, Y_t) = -np_sp_t$, $\ s \neq t$ | | each marginal | $Y_i \sim$ binomial $(n, p_i)$ | ::: --- ## π Summary ::: {style="font-size: 30px"} - The multinomial is the binomial with $k$ cells instead of two - Its probability function is one sequence's probability times the number of sequences - Merge the other cells, and any single count is binomial - Two counts always covary negatively: they share the fixed total $n$ - For a total such as revenue, Theorem 5.13 supplies the covariances and Theorem 5.12 combines them ::: --- ## π Practice Problems ::: {style="font-size: 28px"} **Wackerly, 7th edition** - Β§5.9: Exercises 5.119 β 5.127; start with 5.119, 5.123, 5.124, 5.125 and 5.126 - In the tariff example, what premium share $p_3$ maximises $V(Y_3)$ for fixed $n$? **Week 14, Problem Set 2** is open now and closes **Sunday 20 December at 23:59** on WeBWorK, covering Β§5.9. **Next class:** 16 December, the bivariate normal distribution and conditional expectations (§§5.10β5.11). ::: --- ## 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"} - Why can the four counts in the loan book never be independent of one another? - Given $Y_1 = y_1$, what is the distribution of the remaining counts? - The correlation of $Y_s$ and $Y_t$ does not depend on $n$. Why? ::: 