When $X=(X_1,X_2,\ldots,X_K)$ follows a multinomial distribution with the size parameter $N$ and probability vector $p=(p_1,p_2,\ldots,p_K)$, what is the expected number of zero outcomes in $X$?

In R code, when $p_i=1/K$ for all $i$, zero outcomes can be simulated as

N <- 100
K <- 50

I would like to be able to predict the number of zeros from $N$, $K$, and $p=(p_1,p_2,\ldots,p_K)$. Or I want to know the distribution of the number of zeros.


The number of zeros in the vector:

$$W \equiv \sum_{i=1}^K \mathbb{I}(X_i = 0).$$

Thus, since each $X_i \sim \text{Bin}(N,p_i)$, the expected number of zeros is:

$$\begin{equation} \begin{aligned} \mathbb{E}(W) = \mathbb{E} \Bigg( \sum_{i=1}^K \mathbb{I}(X_i = 0) \Bigg) &= \sum_{i=1}^K \mathbb{E} ( \mathbb{I}(X_i = 0) ) \\[6pt] &= \sum_{i=1}^K \mathbb{P} (X_i = 0) \\[6pt] &= \sum_{i=1}^K (1-p_i)^N. \\[6pt] \end{aligned} \end{equation}$$

In the special case where each $p_i = 1/K$ you have:

$$\mathbb{E}(W) = K (1-1/K)^N = (K-1)^N.$$

This gives you the expected number of zeros in the general case, and the special case of uniform allocation. The full distribution of the number of zeros is more complicated, owing to non-independence of the counts in the vector.

  • 2
    $\begingroup$ +1 Regarding the distribution, it's a case where simulation should be both fast and straightforward (so a million simulations should be quite feasible for example), and would often be sufficient for a wide variety of purposes. $\endgroup$ – Glen_b -Reinstate Monica Aug 6 at 0:21

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