Mathematical Statistics

The Multinomial Probability Distribution

Samir Orujov, PhD

ADA University, School of Business

Information Communication Technologies Agency, Statistics Unit

2026-09-27

🎬 The Idea in 3 Minutes

⬅ Previous lecture: 27 · The Expected Value and Variance of Linear Functions

Watch this short intuition video before (or after) the slides. Captions (English / Azərbaycanca): CC button. ▶ Watch on YouTube

🎯 Learning Objectives

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

🔁 Warm-up from last time

Last class: for a two-asset portfolio with correlation \(\rho < 1\), how does the portfolio’s risk compare to the weighted average of the two assets’ individual risks?

  • It is strictly lower — diversification reduces risk whenever \(\rho < 1\)
  • It equals the weighted average, regardless of \(\rho\)
  • It is higher, since combining two risky assets adds risk
  • It depends only on the two variances, not on \(\rho\)

🗺️ Where We Are

Wackerly §5.9 · 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.

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 \(k\).

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?

📐 A Multinomial Experiment

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.

📐 The Probability Function

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 \(\sum_{i=1}^{k} y_i = n\).

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 \(y_1, \ldots, y_k\).

📊 Worked Example: Rating Transitions

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.

\[p(1, 6, 1, 0) = \frac{8!}{1!\,6!\,1!\,0!}\,(0.10)^1 (0.75)^6 (0.12)^1 (0.03)^0 = 56 \times 0.10 \times 0.17798 \times 0.12 = 0.1196\]

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.

📐 Moments of the Counts

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 \(s \neq t\).

Part 1. Merge every cell except \(i\) into one: each trial lands in cell \(i\) or not, so \(Y_i\) is binomial \((n, p_i)\).

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: \(-np_sp_t\).

📱 Worked Example: Tariff Choice

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

\(\text{Cov}(Y_1, Y_2) = -35\), \(\ \text{Cov}(Y_1, Y_3) = -15\), \(\ \text{Cov}(Y_2, Y_3) = -10.5\).

💰 Monthly Revenue

Revenue is \(R = 10Y_1 + 20Y_2 + 35Y_3\) AZN.

\[E(R) = 10(100) + 20(70) + 35(30) = 3450 \text{ AZN}\]

\[\begin{aligned} V(R) &= 100(50) + 400(45.5) + 1225(25.5) \\ &\quad + 2\big[200(-35) + 350(-15) + 700(-10.5)\big] \end{aligned}\]

\(V(R) = 54\,437.5 - 39\,200 = 15\,237.5\), 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.

💻 Checking It by Simulation

Code 
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)
cov_basic_premium            theory        sd_revenue            theory 
            -15.6             -15.0             125.1             123.4

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.

🧠 Think-Pair-Share

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?

✅ Think-Pair-Share: Solution

  1. Definition 5.12: \(\dfrac{20!}{10!\,4!\,6!}(0.45)^{10}(0.20)^4(0.35)^6 = 38\,798\,760 \times \cdots = 0.0389\)
  1. \(E(D) = 20(0.45) - 20(0.35) = 2\). By Theorem 5.12, with the covariance from part 3, \[V(D) = 20(0.45)(0.55) + 20(0.35)(0.65) - 2(-3.15) = 4.95 + 4.55 + 6.30 = 15.8\]
  1. \(\text{Cov} = -20(0.45)(0.35) = -3.15\). Every up day is a day that was not down. The negative covariance widens the spread of a difference: \(-2\,\text{Cov}\) adds to \(V(D)\).

💰 Case Study: A Month of Trading Days as Multinomial Cells

📈 The question

Sort each trading day into one of four cells by its S&P 500 return.

  1. Do monthly cell counts match a multinomial mean \(np_j\)?
  2. Does the variance match \(np_j(1-p_j)\)?
  3. Does the covariance between two cells stay negative, as Theorem 5.13 requires?

📊 The data

Series: S&P 500 daily close (FRED SP500)

Period: Sep 2016 – Sep 2026, about 2,500 days, 120 full months

Cells: big down (\(< -1\%\)), small down, small up, big up (\(\geq 1\%\))

💰 Case Study: Means Match, Variances Don’t

📦 Cell counts per month vs. the multinomial prediction 
sp <- read.csv("data/sp500.csv", col.names = c("date", "close"))
sp$close <- suppressWarnings(as.numeric(sp$close))    # FRED marks holidays "."
sp <- na.omit(sp)
ret <- diff(log(sp$close))
month <- substr(sp$date[-1], 1, 7)
cell <- cut(ret, breaks = c(-Inf, -0.01, 0, 0.01, Inf),
            labels = c("big_down", "small_down", "small_up", "big_up"))

tab   <- table(month, cell)
full  <- tab[rowSums(tab) >= 18, ]                    # full trading months only
n_bar <- mean(rowSums(full))
p_hat <- colSums(full) / sum(full)

rbind(mean_count = round(colMeans(full), 2),
      np         = round(n_bar * p_hat, 2),
      var_count  = round(apply(full, 2, var), 2),
      npq        = round(n_bar * p_hat * (1 - p_hat), 2))
           big_down small_down small_up big_up
mean_count     2.30       7.18     8.67   2.77
np             2.30       7.18     8.67   2.77
var_count      5.54       5.42    11.96   5.66
npq            2.05       4.71     5.07   2.40

The mean counts track \(np_j\) closely; every cell’s actual variance runs well above \(np_j(1-p_j)\).

💰 Case Study: Cells Move Together, Not Apart

What the data say: the bars line up, so the multinomial mean is a good model of an average month. It fails on the covariances: Theorem 5.13 says any two cell counts move oppositely, but big-down and big-up counts here rise together across months (\(\text{Cov} > 0\)) — calm months have few of both, volatile months have many, because trials are not independent across a month the way Definition 5.11 assumes.

📝 Quiz #1: The Sign of the Covariance

Twelve customers each choose Basic (0.50), Standard (0.35) or Premium (0.15), independently. What is \(\text{Cov}(Y_1, Y_2)\)?

  • \(-2.1\)
  • \(2.1\)
  • \(0\), because the customers choose independently
  • \(-0.175\)

📝 Quiz #2: Is It Multinomial?

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
  • Rating 50 independent loan applicants as approve, refer or decline

📋 Key Formulas

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 \((n, p_i)\)

📋 Summary

  • 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

📚 Practice Problems

Wackerly, 7th edition

  • §5.9: Exercises 5.119 – 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 §5.9.

Next class: 16 December, the bivariate normal distribution and conditional expectations (§§5.10–5.11).

🙏 Thank You

Dr. Samir Orujov

📧 sorujov@ada.edu.az
🏢 Building D, Room D325
🕓 Office hours: Wednesday, 16:00 – 18:00

Slides and readings: sorujov.net/teaching

❓ Questions

  • 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?