Let $n, k \in \mathbb N^*$ such that $p_1, \ldots, p_k \ge 0$ and $p_1 + \cdots + p_k = 1$. Let $p := (p_1, \ldots, p_k)$ and $X \sim \mathrm{Multinomial} (n, p)$. This means $X$ follows a multinomial distribution consisting of $n$ independent trials in which the probability of outcome $i$ is $p_i$.

Then $\frac{X}{n}$ is a vector of probabilities. I would like to ask if the distribution of $\frac{X}{n}$ is already investigated and has some simple form.

  • 1
    $\begingroup$ Define $Y = X/n$, Then $nY$ has a Multinomial($n,p$) distribution... this may help you find the solution. As a note: $X/n$ isn't a vector of probabilities, it's a vector of observed frequencies. $\endgroup$
    – jbowman
    Jul 10, 2022 at 19:01
  • $\begingroup$ Answered for the Poisson distribution at stats.stackexchange.com/questions/240720/…. The answer here will be essentially the same. $\endgroup$
    – whuber
    Jul 10, 2022 at 19:51

1 Answer 1


The pmf of $\mathbf{X}$ is

$$P(\mathbf{X} = \mathbf{x}) = n! \prod_{i=1}^k \frac{p_i^{x_i}}{x_i!}$$

From this you can deduce that the pmf of $\mathbf{Y} = \mathbf{X}/n$ is

$$P(\mathbf{Y} = \mathbf{y}) = P(\mathbf{X} = n\mathbf{y}) = n! \prod_{i=1}^k \frac{p_i^{(ny_i)}}{(ny_i)!}$$


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