Non-zero counts in multinomial distribution

Tobias Madsen

What is the distribution of the number of non-zero counts in a multinomial distribution? I recently answered this question on mathematics stackexchange. This question comes up in the field of ecology: When a researcher wants to know the number of different species - the diversity - in a given habitat, she can make a field trip and count the number of individuals she can find from each species. There is however the problem that with large enough habitats, it is imposible to account for every single individual. Thus, there may be species present in the habitat that has not been counted. A reasonable model for the number of individuals we find from each species is a multinomial model, and the total number of different species discovered is exactly the number of non-zero counts in the multinomial distribution. The distribution (and particular the variance) on this number gives us an idea of how certain we are on the diversity in the habitat. A great amount of the theory on this subject has been developed by Anne Chao with the application in ecology in mind, my own interest in the problem stems from the similar problem of estimating the diversity of T-cell receptor repertoires.

To state the problem mathematically, let (n1,n2,,nM)multinomial(p1,p2,,pM,N)(n_1,n_2,\ldots, n_M) \sim \operatorname{multinomial}(p_1,p_2,\ldots, p_M, N) we are interested in the distribution of the sum I=i=1MI(ni>0)I = \sum_{i=1}^M \mathbb{I}(n_i > 0). I doubt that a closed form solution exists for the general problem, but you can find the exact solution using dynamic programming. We can also compute the mean and variance of II and then use a normal approximation. And finally for the special case where p1=p2=pM=1/Mp_1 = p_2 = \ldots p_M = 1/M you can find a closed form solution using Stirling numbers of the second kind. I will go through each in turn below.

Exact solution using dynamic programming

Define Sk=i=1kniS_k = \sum_{i=1}^k n_i, Ik=i=1kI(ni>0)I_k = \sum_{i=1}^k \mathbb{I}(n_i > 0). Using dynamic programming we will iteratively compute the probability that the kk first variables sum to nn and that mm of the first kk are non-zero,

d(k,n,m)=P(Sk=n,Ik=m)d(k, n, m) = P(S_k = n, I_k = m)

Initialize the dynamic programming table with,

d(0,n,m)={1n=0,m=00otherwised(0, n, m) = \begin{cases} 1 & n=0, m=0 \\ 0 & \text{otherwise} \end{cases}

Then recurse over k=1,,Mk=1,\ldots,M,

d(k,n,m)=binomial(0;Nn,p~k)d(k1,n,m)+r=1nbinomial(r;Nn+r,p~k)d(k1,nr,m1),mkd(k, n, m) = \operatorname{binomial}(0; N-n, \tilde{p}_k)d(k-1, n, m) + \sum_{r=1}^{n}\operatorname{binomial}(r; N-n+r,\tilde{p}_k)d(k-1, n-r, m-1), \quad m \leq k

where p~k=pk/i=k+1Mpi\tilde{p}_{k} = p_{k} / \sum_{i=k+1}^M p_i. The first term accounts for the case where the k’th variable is zero and therefore does not contribute new non-zero variables, and the second term (the sum) accounts for all cases where the k’th variable is non-zero. Below is a naive R implementation. The loops can be rearranged to have the final loop being over rr rather than mm to avoid computing the same binomial densities multiple times, still I don’t think this can be done better than O(M2N2)\mathcal{O}(M^2N^2).

p <- c(0.1,0.1,0.2,0.2,0.4)
N <- 6
M <- 5

p_tilde <- p / rev(cumsum(rev(p)))
d <- array(0, dim = c(M+1,N+1,M+1))
d[1,1,1] = 1
for (k in 1:M) {
  for (n in 0:N) {
    for (m in 0:k) {
      d[k+1, n+1, m+1] <- dbinom(0, N-n, prob = p_tilde[k]) * d[k, n+1, m+1]
      if (m > 0 && n > 0) {
        r <- 1:n
        d[k+1, n+1, m+1] <- d[k+1, n+1, m+1] + 
          sum(dbinom(r, r+N-n, prob = p_tilde[k]) * d[k, n-r+1, m])

Verify solution by simulation

> table(colSums(rmultinom(1e6, size = N, prob = p) > 0 )) / 1e6

       1        2        3        4        5 
0.004245 0.114157 0.449484 0.374784 0.057330 
> d[M+1,N+1,]
[1] 0.000000 0.004226 0.114734 0.449280 0.374160 0.057600

Approximation solution

To use a normal approximation we must compute the mean and variance. First the mean can be computed as,

E[m]=E[i=1MI(ni>0)]=i=1ME[I(ni>0)]=i=1M1(1pi)N.\mathbb{E}[m]= \mathbb{E}\left[\sum_{i=1}^M\mathbb{I}(n_i>0)\right] = \sum_{i=1}^M\mathbb{E}[\mathbb{I}(n_i>0)] = \sum_{i=1}^M 1-(1-p_i)^N.

Next the variance,

V[m]=V[i=1MI(ni>0)]=i=1ME[I(ni>0)]=i=1MV[I(ni>0)]+21i<jMCov(I(ni>0),I(nj>0))=i=1M(1(1pi)N)(1pi)N+21i<jMCov(1I(ni>0),1I(nj>0))=i=1M(1(1pi)N)(1pi)N+21i<jM[(1pipj)N(1pi)N(1pj)N]\begin{aligned} \mathbb{V}[m] &= \mathbb{V}\left[\sum_{i=1}^M\mathbb{I}(n_i>0)\right] \\ &= \sum_{i=1}^M\mathbb{E}[\mathbb{I}(n_i>0)] \\ &= \sum_{i=1}^M\mathbb{V}[\mathbb{I}(n_i>0)] + 2 \sum_{1\leq i < j \leq M} \operatorname{Cov}(\mathbb{I}(n_i>0), \mathbb{I}(n_j>0)) \\ &= \sum_{i=1}^M (1-(1-p_i)^N)(1-p_i)^N + 2 \sum_{1\leq i < j \leq M} \operatorname{Cov}(1-\mathbb{I}(n_i>0), 1-\mathbb{I}(n_j>0)) \\ &= \sum_{i=1}^M (1-(1-p_i)^N)(1-p_i)^N + 2 \sum_{1\leq i < j \leq M} \left[(1-p_i-p_j)^N - (1-p_i)^N(1-p_j)^N\right] \end{aligned}

This is an O(N2)\mathcal{O}(N^2) algorithm. If MM is very large you may omit the covariance terms which you should generally expect to be small and you can compute the approximation in O(N)\mathcal{O}(N).

computed_variance <- sum((1-p)^N*(1-(1-p)^N))
for (i in 1:(M-1)) {
  for (j in (i+1):M) {
    computed_variance <- computed_variance + 2*((1-p[i]-p[j])^N-(1-p[i])^N*(1-p[j])^N)

> computed_variance
[1] 0.602115
> var(colSums(rmultinom(1e6, size = N, prob = p) > 0 ))
[1] 0.602461

Uniform case

Finally, in the uniform case pi=pip_i=p\,\forall i, we must count how many ways we can distribute NN numbered objects in MM number using exactly mm of the bins. This can be done in a two step procedure, first choose mm of the MM bins, this can be done in (Mm)M \choose m ways. Then distribute the NN objects into each of the mm bins with at least one object in each bin, this can be done in N!S(N,m)N!S(N,m) where S(N,m)S(N,m) is the Stirling number of the second kind. We end up with the following formula

p(m)=(Mm)S(N,m)/MNp(m) = {M \choose m} S(N,m) / M^N

Example, N=4N=4, M=5M=5, m=3m=3 (S(4,3)=6S(4,3) = 6):

> table(colSums(rmultinom(1e6, size = 4, prob = rep(1/5, 5)) > 0)) / 1e6

       1        2        3        4 
0.008013 0.224771 0.575393 0.191823 
> choose(5, 3) * factorial(3) * 6 / 5^4
[1] 0.576