1
$\begingroup$

If we pick a random number from $[1, n]$ with repetition $k$ times. What is the probability distribution of number of distinct numbers picked for a given $k$? The number of distinct numbers picked is $\in [1, min(k, n)]$.

$\endgroup$
  • $\begingroup$ Could you precisely define the indicator random variable in question? Also are you picking one number randomly, or $k$ numbers? $\endgroup$ – Akay Mar 8 '17 at 10:19
  • $\begingroup$ We are picking k random numbers with repetition from $[1, n]$. $\endgroup$ – vamsikal Mar 9 '17 at 11:51
2
$\begingroup$

Sppose we draw $m$ times with $n$ possible values and ask about the number $r$ of distinct values that appeared. The classification by $r$ is given from first principles by

$$\frac{1}{n^m} \sum_{r=1}^n {n\choose r} \times {m\brace r} \times r!.$$

We may include $r=0$ because the Stirling number is zero there. This being a sum of probabilities it should evaluate to one. We get

$$\frac{1}{n^m} \sum_{r=0}^n {n\choose r} \times m! [z^m] (\exp(z)-1)^r = m! [z^m] \frac{1}{n^m} \sum_{r=0}^n {n\choose r} (\exp(z)-1)^r \\ = m! [z^m] \frac{1}{n^m} \exp(nz) = \frac{1}{n^m} n^m = 1$$

and the sanity check goes through. We get for the expected number of distinct values

$$\frac{1}{n^m} \sum_{r=1}^n r {n\choose r} \times m! [z^m] (\exp(z)-1)^r \\ = \frac{1}{n^{m-1}} \sum_{r=1}^n {n-1\choose r-1} \times m! [z^m] (\exp(z)-1)^r \\ = \frac{1}{n^{m-1}} m! [z^m] (\exp(z)-1) \sum_{r=1}^n {n-1\choose r-1} \times (\exp(z)-1)^{r-1} \\ = \frac{1}{n^{m-1}} m! [z^m] (\exp(z)-1) \sum_{r=0}^{n-1} {n-1\choose r} \times (\exp(z)-1)^{r} \\ = \frac{1}{n^{m-1}} m! [z^m] (\exp(z)-1) \exp((n-1)z) = \frac{1}{n^{m-1}} (n^m - (n-1)^m) \\ = n \left(1 - \left(1-\frac{1}{n}\right)^m\right).$$

The species for labeled set partitions is

$$\mathfrak{P}(\mathcal{U}\mathfrak{P}_{\ge 1}(\mathcal{Z}))$$

which yields the generating function

$$G(z, u) = \exp(u(\exp(z)-1)).$$

We verified these with the following script.

ENUM :=
proc(n, m)
    option remember;
    local ind, d, res;

    res := 0;
    for ind from n^m to 2*n^m-1 do
        d := convert(ind, base, n);

        res := res +
        nops(convert(d[1..m], `multiset`));
    od;

    res/n^m;
end;

X := (n,m)-> n*(1-(1-1/n)^m);

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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