# Multinomial

The multinomial is an example of a parametric distribution for multiple random variables. The multinomial is a gentle introduction to joint distributions. It is a extension of the binomial. In both cases, you have $n$ independent experiments. In a binomial each outcome is a "success" or "not success". In a multinomial there can be more than two outcomes (multi). A great analogy for the multinomial is: we are going to roll an $m$ sided dice $n$ times. We care about reporting the number of outcomes of each side of your dice.

Here is the formal definition of the multinomial. Say you perform $n$ independent trials of an experiment where each trial results in one of $m$ outcomes, with respective probabilities: $p_1, p_2, \dots , p_m$ (constrained so that $\sum_i p_i = 1$). Define $X_i$ to be the number of trials with outcome $i$. A multinomial distribution is a closed form function that answers the question: What is the probability that there are $c_i$ trials with outcome $i$. Mathematically: \begin{align*} P(X_1=c_1,X_2 = c_2, \dots , X_m = c_m) &= { {n} \choose {c_1,c_2,\dots , c_m} }\cdot p_1^{c_1} \cdot p_2^{c_2}\dots p_m^{c_m} \\ &= { {n} \choose {c_1,c_2,\dots , c_m} }\cdot \prod_i p_i^{c_i} \end{align*}

This is our first joint random variable model! We can express it in a card, much like we would for random variables:

Multinomial Joint Distribution

Description: Number of outcomes of each possible outcome type in $n$ identical, independent experiments. Each experiment can result in one of $m$ different outcomes. $p_1, \dots, p_m$ where each $p_i \in [0,1]$ is the probability of outcome type $i$ in one experiment.$n \in \{0, 1, \dots\}$, the number of experiments $c_i \in \{0, 1, \dots, n\}$, for each outcome $i$. It must be the case that $\sum_i c_i = n$ \begin{align*} P(X_1=c_1,X_2 = c_2, \dots , X_m = c_m) = { {n} \choose {c_1,c_2,\dots , c_m} } \prod_i p_i^{c_i} \end{align*}

### Examples

Standard Dice Example:
A 6-sided die is rolled 7 times. What is the probability that you roll: 1 one, 1 two, 0 threes, 2 fours, 0 fives, 3 sixes (disregarding order). \begin{align*} \P(X_1=1,X_2 = 1&, X_3 = 0,X_4 = 2,X_5 = 0,X_6 = 3) \\&= \frac{7!}{2!3!}\left(\frac{1}{6}\right)^1\left(\frac{1}{6}\right)^1\left(\frac{1}{6}\right)^0\left(\frac{1}{6}\right)^2\left(\frac{1}{6}\right)^0\left(\frac{1}{6}\right)^3\\ &=420\left(\frac{1}{6}\right)^7 \end{align*}
Weather Example:

Each day the weather in Bayeslandia can be {Sunny, Cloudy, Rainy} where $p_\text{sunny} = 0.7$, $p_\text{cloudy} = 0.2$ and $p_\text{rainy} = 0.1$. Assume each day is independent of one another. What is the probability that over the next 7 days we have 5 sunny days, 1 cloudy day and 1 rainy days? \begin{align*} \P(X_{\text{sunny}}=6,X_{\text{rainy}} = 1&, X_{\text{cloudy}} = 0) \\ &= \frac{7!}{5!1!1!} (0.7)^5 \cdot (0.2)^1 \cdot (0.1) ^1 \\ &\approx 0.14 \end{align*}

How does that compare to the probability that every day is sunny? \begin{align*} \P(X_{\text{sunny}}=7,X_{\text{rainy}} = 0&, X_{\text{cloudy}} = 0) \\ &= \frac{7!}{7!1!} (0.7)^7 \cdot (0.2)^0 \cdot (0.1) ^0 \\ &\approx 0.08 \end{align*}

The multinomial is especially popular because of its use as a model of language. For a full example see the Federalist Paper Authorship example.

### Deriving Joint Probability

A way to deeper understand the multinomial is to derive the joint probability function for a particular multinomial. Consider the multinomial from the previous example. In that multinomial with $n = 7$ outcomes where each outcome can be one of three values $\{S,C,R\}$ where S stands for Sunny, C stands for Cloudy and R stands for Rainy, and days are independent. $p_s = 0.7$, $p_c=0.2$, $p_r = 0.1$. We are going to derive the probability that out of the $n = 7$ days, 5 are sunny, 1 is cloudy and 1 is rainy.

Like our derivation for the binomial, we are going to consider all of the possible weeks with 5 sunny days, 1 rainy day and 1 cloudy day.

  ('S', 'S', 'S', 'S', 'S', 'C', 'R')
('S', 'S', 'S', 'S', 'S', 'R', 'C')
('S', 'S', 'S', 'S', 'C', 'S', 'R')
('S', 'S', 'S', 'S', 'C', 'R', 'S')
('S', 'S', 'S', 'S', 'R', 'S', 'C')
('S', 'S', 'S', 'S', 'R', 'C', 'S')
('S', 'S', 'S', 'C', 'S', 'S', 'R')
('S', 'S', 'S', 'C', 'S', 'R', 'S')
('S', 'S', 'S', 'C', 'R', 'S', 'S')
('S', 'S', 'S', 'R', 'S', 'S', 'C')
('S', 'S', 'S', 'R', 'S', 'C', 'S')
('S', 'S', 'S', 'R', 'C', 'S', 'S')
('S', 'S', 'C', 'S', 'S', 'S', 'R')
('S', 'S', 'C', 'S', 'S', 'R', 'S')
('S', 'S', 'C', 'S', 'R', 'S', 'S')
('S', 'S', 'C', 'R', 'S', 'S', 'S')
('S', 'S', 'R', 'S', 'S', 'S', 'C')
('S', 'S', 'R', 'S', 'S', 'C', 'S')
('S', 'S', 'R', 'S', 'C', 'S', 'S')
('S', 'S', 'R', 'C', 'S', 'S', 'S')
('S', 'C', 'S', 'S', 'S', 'S', 'R')
('S', 'C', 'S', 'S', 'S', 'R', 'S')
('S', 'C', 'S', 'S', 'R', 'S', 'S')
('S', 'C', 'S', 'R', 'S', 'S', 'S')
('S', 'C', 'R', 'S', 'S', 'S', 'S')
('S', 'R', 'S', 'S', 'S', 'S', 'C')
('S', 'R', 'S', 'S', 'S', 'C', 'S')
('S', 'R', 'S', 'S', 'C', 'S', 'S')
('S', 'R', 'S', 'C', 'S', 'S', 'S')
('S', 'R', 'C', 'S', 'S', 'S', 'S')
('C', 'S', 'S', 'S', 'S', 'S', 'R')
('C', 'S', 'S', 'S', 'S', 'R', 'S')
('C', 'S', 'S', 'S', 'R', 'S', 'S')
('C', 'S', 'S', 'R', 'S', 'S', 'S')
('C', 'S', 'R', 'S', 'S', 'S', 'S')
('C', 'R', 'S', 'S', 'S', 'S', 'S')
('R', 'S', 'S', 'S', 'S', 'S', 'C')
('R', 'S', 'S', 'S', 'S', 'C', 'S')
('R', 'S', 'S', 'S', 'C', 'S', 'S')
('R', 'S', 'S', 'C', 'S', 'S', 'S')
('R', 'S', 'C', 'S', 'S', 'S', 'S')
('R', 'C', 'S', 'S', 'S', 'S', 'S')

First, note that each outcome for assignments to the weeks are mutually exclusive. Then note that the probability of any one outcome will be $(p_S)^5 \cdot p_C \cdot p_R$. The number of unique weeks with the chosen count of outcomes can be derived using the rule for Permutations with Indistinct Objects. There are 7 objects, 5 are indistinct from one another. The number of distinct outcomes is: $${ {7} \choose {5,1,1} } = \frac{7!}{5!1!1!} = 7 \cdot 6 = 42$$

Since the outcomes are mutually exclusive, we are going to be adding the probability of each case to itself $\frac{7!}{5!1!1!}$ times. Putting this all together we get the multinomial joint function for this particular case: \begin{align*} \P(X_{\text{sunny}}=5,X_{\text{rainy}} = 1&, X_{\text{cloudy}} = 1) \\ &= \frac{7!}{5!1!1!} (0.7)^5 \cdot (0.2)^1 \cdot (0.1) ^1 \\ &\approx 0.14 \end{align*}