Variance and Covariance of Categorical Distribution

This post is inspired by the lecture given by David Blei on Thursday, 17 September 2020. One of the topics he explained was related to a categorical variable and a categorical distribution. This post will elaborate those two concepts. Let’s get started.


$\pmb{\text{Figure 1}}$: A categorical variable ( $\text{Color}$) and its values ( $\text{Red}$, $\text{Yellow}$, and $\text{Green}$ ). Image taken from Kaggle, some rights reserved.

$\text{Figure 1}$ shows an example of categorical values stored in a categorical variable, $\text{Color}$. Basically, a categorical variable takes one of $K$ values and each categorical value is represented by a $K$-vector with a single $1$ and otherwise $0$s.

Let’s denote a categorical variable as $x^{(k)}$ which means that the $k$th component of it has a a single $1$ and otherwise $0$s.

For example, the categorical variable in $\text{Figure 1}$ has $3$ values ( $\text{Red}$, $\text{Yellow}$, and $\text{Green}$ ) and each value is represented by $3$-vector with a single $1$ and otherwise $0$s as follows:

\[\begin{align} \text{Red} &= x^{(1)} = (1, 0, 0 ) \\ \text{Yellow} &= x^{(2)} = (0, 1, 0 ) \tag{1}\label{eq:yellow} \\ \text{Green} &= x^{(3)} = (0, 0, 1 ). \end{align}\]

The $K$-vector with single $1$ and otherwise $0$s is commonly named one-hot vector.

A categorical distribution is parameterized by $\theta$. Moreover, $\pmb{\theta}$ specifies the probability of each categorical value. Suppose we have $K$ categorical values; therefore,

\[\begin{equation} \theta = (\theta_1, \theta_2, \ldots, \theta_K) \tag{2}\label{eq:theta} \end{equation}\]


\[\begin{equation} \sum_{k=1}^{K}{\theta_k} = 1 \text{ and } 0 \leq \theta_k \leq 1 \text{ for }k=1, \ldots, K. \tag{3}\label{eq:theta-constraints} \end{equation}\]

Consider that $X^{(k)}$ is a random categorical variable which takes one of $K$ values. Moreover, since $X^{(k)}$ is random variable, it has categorical distribution that is described by a discrete probability distribution,

\(\begin{equation} \text{p}(x^{(k)}) = \prod_{l=1}^{K}{\theta_{l}^{x^{(l)}}} \tag{4}\label{eq:pdf-categorical} \end{equation}\)
with $x^{(l)}$ is the $l$th component of $x^{(k)}$. Additionally, we elaborate Equation \eqref{eq:pdf-categorical} into

\[\begin{align} \text{p}(x^{(k)}) &= \prod_{l=1}^{K}{\theta_{l}^{x^{(l)}}} \\ &= \theta_{1}^{x^{(1)}} \times \theta_{2}^{x^{(2)}} \times \cdots \times \theta_{k}^{x^{(k)}} \times \cdots \times \theta_{K}^{x^{(K)}} \\ &= \theta_{1}^{0} \times \theta_{2}^{0} \times \cdots \times \theta_{k}^{1} \times \cdots \times \theta_{K}^{0} \\ &= \theta_{k}. \tag{5}\label{eq:pdf-categorical-simplified} \end{align}\]

Let’s put Equation \eqref{eq:pdf-categorical-simplified} into practice and demonstrate it in one example. Suppose we want to compute $\text{p}(\text{Yellow})$ in Equation \eqref{eq:yellow},

\[\begin{align} \text{p}(\text{Yellow}) &= \text{p}(x^{(2)}) \\ &= \text{p}((0,1,0)) & \Rightarrow \text{1st} = 0, \text{2nd} = 1, \text{3rd} = 0 \\ &= \prod_{l=1}^{3}{\theta_{l}^{x^{(l)}}} \\ &= \theta_{1}^{x^{(1)}} \times \theta_{2}^{x^{(2)}} \times \theta_{3}^{x^{(3)}} \\ &= \theta_{1}^{0} \times \theta_{2}^{1} \times \theta_{3}^{0} \\ &= \theta_{2}. \end{align}\]

With Equation \eqref{eq:pdf-categorical-simplified} in hand, we are now ready to compute the expectation of $X^{(k)}$ as

\[\begin{align} \text{E}(X^{(k)}) &= \sum_{l=1}^{K}{x^{(l)} \text{p}(x^{(l)})} \\ &= \underbrace{0 \times \text{p}(x^{(1)})}_{1\text{st}} + \underbrace{0 \times \text{p}(x^{(2)})}_{2\text{nd}} + \cdots + \underbrace{1 \times \text{p}(x^{(k)})}_{k\text{th}} + \cdots + \underbrace{0 \times \text{p}(x^{(K)}}_{K\text{th}}) \\ &= \text{p}(x^{(k)}) \\ &= \theta_k. \tag{6}\label{eq:expectation} \end{align}\]

Next, we compute the Variance, $\text{Var}$, as follows:

\[\begin{align} \text{Var}(X^{(k)}) &= \underbrace{\text{E}((X^{(k)})^2)}_{\text{Part I}} - \underbrace{(\text{E}(X^{(k)}))^2}_{\text{Part II}}. & \text{the definition of variance} \tag{7}\label{eq:variance-definition} \\ \end{align}\]

Next, we compute $\text{Part I}$, $\text{E}((X^{(k)})^2)$, as follows:

\[\begin{align} \text{E}((X^{(k)})^2) &= \sum_{l=1}^{K}{(x^{(l))^2} \text{p}(x^{(l)})} \\ &= \underbrace{0^2 \times \text{p}(x^{(1)})}_{1\text{st}} + \underbrace{0^2 \times \text{p}(x^{(2)})}_{2\text{nd}} + \cdots + \underbrace{1^2 \times \text{p}(x^{(k)})}_{k\text{th}} + \cdots + \underbrace{0^2 \times \text{p}(x^{(K)}}_{K\text{th}}) \\ &= \text{p}(x^{(k)}) \\ &= \theta_k. \tag{8}\label{eq:expectation-x-square} \end{align}\]

Now, we can finalize computing the Variance in Equation \eqref{eq:variance-definition},

\[\begin{align} \text{Var}(X^{(k)}) &= \text{E}((X^{(k)})^2) - (\text{E}(X^{(k)}))^2 & \text{the definition of variance} \\ &= \theta_k - (\theta_k)^2 & \text{using Equation }\eqref{eq:expectation} \text{ and }\eqref{eq:expectation-x-square} \\ &= \theta_k (1 - \theta_k). & \text{using distributive property} \tag{9}\label{eq:variance} \end{align}\]

Last but not least, we shall compute the Covariance, $\text{Cov}(X^{(j)}, X^{(k)})$. We start by the definition of Covariance,

\[\begin{align} \text{Cov}(X^{(j)}, X^{(k)}) &= \underbrace{\text{E}(X^{(j)} X^{(k)})}_{\text{Part I}} - \underbrace{(\text{E}(X^{(j)}) E(X^{(k)}))}_{\text{Part II}}. & \text{by definition} \tag{10}\label{eq:covariance} \end{align}\]

Let’s compute the $\text{Part I}$ as follows:

\[\begin{align} \text{E}(X^{(j)} X^{(k)}) &= (0)(0) \theta_1 + \cdots + \underbrace{(1)(0) \theta_j}_{j\text{th}} + \cdots + \underbrace{(0)(1) \theta_k}_{k\text{th}} + \cdots + (0)(0) \theta_K \\ &= 0. \tag{11}\label{eq:covariance-zero} \end{align}\]

Eventually, we can finalize Equation \eqref{eq:covariance} as

\[\begin{align} \text{Cov}(X^{(j)}, X^{(k)}) &= \text{E}(X^{(j)} X^{(k)}) - (\text{E}(X^{(j)}) E(X^{(k)})) \\ &= 0 - \theta_j \theta_k & \text{using Equation } \eqref{eq:expectation} \text{ and }\eqref{eq:covariance-zero} \\ &= - \theta_j \theta_k. \end{align}\]

To conclude, we have shown how to derive the expectation, variance, and covariance of a categorical distribution. We hope this post helps anyone who wants to understand a categorical distribution.

Written on September 18, 2020