Number of unique numbers in a vector with replacement

Question

If I get a vector $x$ of length $\ell\geq n$ that contains random draws of $n$ numbers and know that all $n$ numbers appear at least once in such vector, how many numbers appear exactly once in this vector in expectation?

To be more precise: $x$ is drawn with replacement but I know that the vector has at least one copy of each of the $n$ numbers.

The question is related to the existing questions Probability distribution of the number of unique elements when picking $m$ items from $n$ with replacement

Expected number of unique items when drawing with replacement

but it is not the same. I wonder if the solution using Stirling numbers of the second kind is also useful in this case.

Answer 1

Using combinatorial classes as in Analytic Combinatorics by Flajolet and Sedgewick, we have the following class $\mathcal{V}$ of vectors drawn from $[n]$ where each value occurs at least once

$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \mathcal{V} = \textsc{SEQ}_{=n}(\textsc{SET}_{\ge 1}(\mathcal{Z})).$$

This gives the EGF

$$F(z) = (\exp(z)-1)^n.$$

In particular the count of these vectors is

$$\ell! [z^\ell] (\exp(z)-1)^n = n! {\ell \brace n}.$$

Now mark singletons to get

$$\mathcal{V} = \textsc{SEQ}_{=n}(\mathcal{U}\times \textsc{SET}_{=1}(\mathcal{Z}) + \textsc{SET}_{\ge 2}(\mathcal{Z})).$$

We get the bivariate GF

$$Q(z,u) = (\exp(z)-z-1+uz)^n.$$

For the count of singletons we find the EGF

$$\left. \frac{\partial}{\partial u} Q(z,u) \right|_{u=1} = nz (\exp(z)-1)^{n-1}.$$

Extract the coefficient on $z^\ell$

$$\ell! [z^\ell] n z (\exp(z)-1)^{n-1} = \ell (\ell-1)! [z^{\ell-1}] n! \frac{1}{(n-1)!} (\exp(z)-1)^{n-1} \\ = \ell \times n! {\ell-1\brace n-1}.$$

It follows that the expectation is

$$\bbox[5px,border:2px solid #00A000]{ \mathrm{E}[S] = \ell {\ell-1\brace n-1} {\ell\brace n}^{-1}.}$$

Using the asymptotic

$${\ell\brace n} \underset{\ell\rightarrow\infty}{\sim} \frac{n^\ell}{n!}$$

we get with the pair of Stirling numbers

$$\ell \frac{(n-1)^{\ell-1}}{(n-1)!} \frac{n!}{n^\ell}$$

which means that

$$\bbox[5px,border:2px solid #00A000]{ \mathrm{E}[S] \sim \ell \left(1-\frac{1}{n}\right)^{\ell-1}.}$$

We can in fact compute the $r$th factorial moment by differentiating $Q(z,u)$ $r$ times. We get

$$\ell! [z^\ell] n(n-1)\cdots(n-(r-1)) z^r (\exp(z)-1)^{n-r} \\ = \ell(\ell-1)\cdots (\ell-(r-1)) (\ell-r)! n! [z^{\ell-r}] \frac{(\exp(z)-1)^{n-r}}{(n-r)!} \\ = r! {\ell\choose r} n! {\ell-r\brace n-r}.$$

This means that

$$\bbox[5px,border:2px solid #00A000]{ \mathrm{E}[S(S-1)\cdots (S-(r-1))] = r! {\ell\choose r} {\ell-r\brace n-r} {\ell\brace n}^{-1}.}$$

As a sanity check $S$ is of course at most $n$, when $r\gt n$ we thus get zero for the moment in question which agrees with the first Stirling number. When $r=n$ we get zero when $\ell\gt r$ which is $\ell\gt n$ and indeed the number of singletons is at most $n-1$ in that case which gives zero for the moment. When $r=n$ and $\ell=r$ we get $n!$ contributing vectors (only contribution is when $S=n$ so we permute the $n$ unique values) with the statistic equal to $n!$ which is indeed $n!{n\choose n}n!{n-r\brace n-r},$ giving $n!$ as the value of the moment.

Answer 2

Additional statistics as per request. We ask about the conditional expectation of the number of singletons given that we know the vector has at least one singleton. Counting the vectors with no singletons we get the EGF

$$A(z) = (\exp(z)-z)^n.$$

Extracting coefficients we find with $\ell\ge n$

$$\ell! [z^\ell] \sum_{q=0}^n {n\choose q} (-1)^{n-q} z^{n-q} \exp(qz) = \ell! \sum_{q=0}^n {n\choose q} (-1)^{n-q} [z^{\ell+q-n}] \exp(qz) \\ = \ell! \sum_{q=0}^n {n\choose q} (-1)^{n-q} \frac{q^{\ell+q-n}}{(\ell+q-n)!}.$$

Therefore the count of the admissible vectors is

$$n^\ell - \ell! \sum_{q=0}^n {n\choose q} (-1)^{n-q} \frac{q^{\ell+q-n}}{(\ell+q-n)!} = \ell! \sum_{q=0}^{n-1} {n\choose q} (-1)^{n-1-q} \frac{q^{\ell+q-n}}{(\ell+q-n)!}.$$

The contribution from the number of singletons is found by differentiation of the marked mixed GF (this will give a zero contribution from the vectors containing no singletons)

We have

$$B(z) = (\exp(z)-z+uz)^n.$$

Differentiate and evaluate to get

$$\ell! [z^\ell] n z \exp(z(n-1)) = n \ell! [z^{\ell-1}] \exp(z(n-1)) = n \ell (n-1)^{\ell-1}.$$

We thus have for the desired expectation

$$\bbox[5px,border:2px solid #00A000]{ \mathrm{E}[S] = \frac{n (n-1)^{\ell-1}}{(\ell-1)!} \left[ \sum_{q=0}^{n-1} {n\choose q} (-1)^{n-1-q} \frac{q^{\ell+q-n}}{(\ell+q-n)!} \right]^{-1}.}$$

Using the dominant term of the sum we find for $\ell\to\infty$

$$\mathrm{E}[S] \sim \frac{n (n-1)^{\ell-1}}{(\ell-1)!} \left[n \frac{(n-1)^{\ell-1}}{(\ell-1)!}\right]^{-1}$$

This says that

$$\bbox[5px,border:2px solid #00A000]{\mathrm{E}[S] \sim 1.}$$

The formula is exact when $n=2.$

We can also evaluate the count of vectors with no singletons in terms of Stirling set numbers, we get

$$\ell! [z^\ell] \sum_{q=0}^n {n\choose q} (-1)^{n-q} (z-1)^{n-q} (\exp(z)-1)^q \\ = \ell! [z^\ell] \sum_{q=0}^n {n\choose q} q! (-1)^{n-q} \sum_{p=0}^{n-q} (-1)^{n-q-p} {n-q\choose p} z^p (\exp(z)-1)^q/q! \\ = \ell! \sum_{q=0}^n {n\choose q} q! \sum_{p=0}^{n-q} (-1)^p {n-q\choose p} \frac{1}{(\ell-p)!} {\ell-p\brace q} \\ = n! \sum_{q=0}^n \sum_{p=0}^{n-q} (-1)^p \frac{1}{(n-q-p)!} {\ell\choose p} {\ell-p\brace q}.$$