I know that the following is true for one categorical variable $X \in \{1,...,k\}$

$cov(X=i, X=j) = -p_i p_j$

However, I don't have intuition behind this and cannot find a proof for this. Is there a proof behind this calculation?

To illustrate exactly why I'm interested in this, I'd like to understand the calculation behind the covariance of multinomial random variables. Given two multinomial random variables $Y_a$ and $Y_b$ from the same multinomial distribution with $k$ categories, I know that the covariance can be calculated as follows for $n$ trials

$cov(Y_a, Y_b) = \sum\limits_{s=1}^{n} \sum\limits_{t=1}^{n} cov(1[X=a], 1[X=b]) = -n p_a p_b$

And this depends on the covariance of categorical variables.

  • 2
    $\begingroup$ You refer only to a single variable $X$ rather than two. I suspect the covariance you want to ask about concerns a multinomial distribution with parameters $n=1$ and $(p_1,p_2,\ldots, p_k)$. $\endgroup$
    – whuber
    Commented Jun 11, 2015 at 19:41
  • $\begingroup$ That was my original intention. $\endgroup$
    – mortonjt
    Commented Jun 11, 2015 at 19:42
  • 1
    $\begingroup$ But the covariance between components within a multinomial distribution is given by $-np_i p_j$, which is derived from the covariance from the categorical distribution $\endgroup$
    – mortonjt
    Commented Jun 11, 2015 at 19:43
  • 1
    $\begingroup$ First off, I really appreciate your pedantic attention whuber - I'm still in the process of understanding the problem. In the end, I just want to get a better theoretical understanding behind the covariance calculation behind multinomial random variables. $\endgroup$
    – mortonjt
    Commented Jun 11, 2015 at 19:52
  • 1
    $\begingroup$ @mortonjt you're right about that, but he's talking about something a little earlier. Pay attention to his advice and start with a multinomial with $n=1$ trial. This means that you're modelling a random vector. This random vector has length $k$. Once you write this down, the expression for the covariance just falls out. $\endgroup$
    – Taylor
    Commented Jun 11, 2015 at 20:11

3 Answers 3


Use your crayons!

That's all you need to know. The rest of this answer elaborates on it, for those who have not read the link, and then it supplies a formal demonstration of the claim in that link: coloring rectangles in a scatterplot really does give the correct covariance in all cases.


The figure shows two indicator variables $X$ and $Y$: $(X,Y)=(1,0)$ with probability $p_1$, $(X,Y)=(0,1)$ with probability $p_2$, and otherwise $(X,Y)=(0,0)$. The probabilities are indicated by sets of points, where a proportion $p_1$ of them all are located close to $(1,0)$ (but spread about so you can see each of them), $p_2$ are located close to $(0,1)$, and the remaining fraction $1-p_1-p_2$ around $(0,0)$. All possible rectangles that use some two of these points have been drawn. As explained in the linked post, rectangles are positive (and drawn in red) when the points are at the upper right and lower left and otherwise they are negative (and drawn in cyan).

It is always the case that many of the rectangles cannot be seen because their width, their height, or both are zero. In the present situation, many of the rest are extremely slender because of the slight spreads of the points: they really should be invisible, too. The ones that can be seen all use one point near $(1,0)$ and one point near $(0,1)$. That makes them all negative, explaining the overall cyan cast to the picture.


A fraction $p_1$ of all rectangles have a corner at $(1,0)$. Independently of that, the proportion of those with another corner at $(0,1)$ is $p_2$. When the locations are not spread out, all such rectangles have unit width $1-0$ and unit height $1-0$ and they are negative. Therefore the covariance is

$$\text{Cov}(X,Y) = p_1 p_2 (1-0)(1-0)(-1) = -p_1p_2,$$


Mathematical Proof

The question asks for a proof.

To get started, let's establish the notation. Suppose $X$ is a discrete random variable that takes on the values $x_1,x_2,\ldots,x_k$ with probabilities $p_1,p_2,\ldots,p_k$, respectively. Let $Y_i$ be the indicator of $x_i$; that is, $Y_i = 1$ when $X = x_i$ and otherwise $Y_i=0$. Let $i\ne j$. The chance that $(Y_i,Y_j)=(1,0)$, which corresponds to $X=x_i$, is $p_i$; and the chance that $(Y_i,Y_j)=(0,1)$, which corresponds to $X=x_j$, is $p_j$. Since it is impossible for $(Y_i,Y_j)=(1,1)$, the chance that $(Y_i,Y_j)=(0,0)$ must be $1-p_i-p_j$, corresponding to $X\ne x_i$ and $X\ne x_j$. (The vector-valued random variable $(Y_1,Y_2,\ldots,Y_k)$ has a Multinomial$(1;p_1,p_2,\ldots,p_k)$ distribution.)

The question asks for the covariances of $Y_i$ and $Y_j$ for any indexes $i$ and $j$ in $1,2,\ldots, k$.

The proof uses two ideas. Their demonstrations are simple and easy.

  1. Let $(X,Y)$ be any bivariate random variable. Suppose $(X^\prime, Y^\prime)$ is another random variable with the same distribution but is independent of $(X,Y)$. Then

    $$\text{Cov}(X,Y) = \frac{1}{2}\mathbb{E}((X-X^\prime)(Y-Y^\prime)).$$

    To see why this is so, note that the right hand side remains unchanged when $X$ and $X^\prime$ are shifted by the same amount and also when $Y$ and $Y^\prime$ are shifted by some common amount. We may therefore apply suitable shifts to make the expectations all zero. In this situation

    $$\eqalign{\text{Cov}(X,Y) &= \mathbb{E}(XY)\\ & = \frac{1}{2}\mathbb{E}(XY + X^\prime Y^\prime)\\ & = \frac{1}{2}\mathbb{E}(XY + X^\prime Y^\prime) + \frac{1}{2}\mathbb{E}(X)\mathbb{E}(Y^\prime) + \frac{1}{2}\mathbb{E}(X^\prime)\mathbb{E}(Y) \\ & = \frac{1}{2}\mathbb{E}(XY + X^\prime Y^\prime) + \frac{1}{2}\mathbb{E}(XY^\prime) + \frac{1}{2}\mathbb{E}(X^\prime Y) \\ &= \frac{1}{2}\mathbb{E}((X-X^\prime)(Y-Y^\prime)). }$$

    Those extra terms like $\mathbb{E}(X)\mathbb{E}(Y^\prime)$ could be freely added in the middle step because they are all zero. The equalities of the form $\mathbb{E}(X)\mathbb{E}(Y^\prime) = \mathbb{E}(X Y^\prime)$ in the following step result from the independence of $X$ and $Y^\prime$ and of $X^\prime$ and $Y$.

  2. Where did that factor of $1/2$ go in the crayon calculation? When $(X,Y)$ has a discrete distribution with values $(x_i,y_i)$ and associated probabilities $\pi_{i}$,

    $$\eqalign{ \frac{1}{2}\mathbb{E}((X-X^\prime)\mathbb{E}(Y-Y^\prime)) &=\sum_{i,j=1}^k (x_i-x_j^\prime)(y_i-y_j^\prime)\pi_{i}\pi_{j} \\ &= \frac{1}{2}\left(\sum_{i\gt j} + \sum_{i \lt j} + \sum_{i=j}\right)(x_i-x_j^\prime)(y_i-y_j^\prime)\pi_{i}\pi_{j} \\ &= \frac{1}{2}\left(2\sum_{i\gt j} (x_i-x_j^\prime)(y_i-y_j^\prime)\pi_{i}\pi_{j}\right) + \sum_{i=j} 0 \\ &= \sum_{i \gt j} (x_i-x_j^\prime)(y_i-y_j^\prime)\pi_{i}\pi_{j}. }$$

    In words: the expectation averages over ordered pairs of indices, causing each non-empty rectangle to be counted twice. That's why the factor of $1/2$ is needed in the formula but does not need to be used in the crayon calculation, which counts each distinct rectangle just once.

Applying these two ideas to the bivariate $(Y_i,Y_j)$ in the question, which takes on only four possible values $(0,0),(1,0),(0,1),(1,1)$ with probabilities $1-p_ip_j, p_i, p_j$, and $0$, gives a sum that has only one nonzero term arising from $(1,0)$ and $(0,1)$ equal to

$$\text{Cov}(Y_i,Y_j) = p_i p_j (1-0)(0-1) = -p_i p_j,$$



Consider a single trial from a multinomial, so $n=1$. This will give a random vector $x$ with $k$ components. The $ith$ and $jth$ coordinates of the covariance matrix is given by

$cov(x_i, x_j) = E[(x_i - u_i)(x_j - u_j)] $

$ = E[x_i x_j - p_i x_j -p_j x_i + p_i p_j]$

$ = E[x_i x_j] - E[x_j]p_i - E[x_i]p_j + p_i p_j$

Since $x_i$ and $x_j$ cannot occur simultaneously, $E[x_i x_j]=0$

So $cov(x_i, x_j) = 0 - p_j p_i - p_i p_j + p_i p_j = -p_i p_j$

This definitely wasn't immediately obvious to me. Thanks @whuber, @Taylor!

  • 1
    $\begingroup$ Setting up the probability space to make this argument more formal: the sample space is $\Omega = \{1, 2, ..., k\}$, the law is $\mathbb{P}({ \omega}) = p_\omega, \forall \omega \in \Omega$, and the random variables in question are $x_i (\omega) = \mathbb{I}(\omega = i)$ and $x_j (\omega) = \mathbb{I}(\omega = j)$ for $i \neq j$. Then clearly $\mathbb{E}[x_i] = \mathbb{P}(x_i = 1) = p_i$. $\endgroup$
    – Yibo Yang
    Commented Sep 6, 2023 at 1:19

Plz see https://arxiv.org/abs/1605.05087 This article describes more detail derivation about the covariance of categorical variables.

The derivation show: Defining the variance of categorical variables = Gini-index Defining the covariance of categorical variables = Correspondence Analysis

And also It showed application to natural language processing( word as a category). word2vec is well-known tool in natural language processing. We can show

the covariance of categorical variables ( Correspondence Analysis ) = word2vec.

  • 1
    $\begingroup$ Welcome to the site. We are trying to build a permanent repository of high-quality statistical information in the form of questions & answers. Thus, we're wary of link-only answers, due to linkrot. Can you post a full citation & a summary of the information at the link, in case it goes dead? $\endgroup$ Commented Feb 23, 2019 at 1:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.