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Consider an urn containing $N$ balls of $P$ different colors, with $p_i$ being the proportion of balls of color $i$ among the $N$ balls ($\sum_i p_i = 1$). I draw $n \leq N$ balls from the urn without replacement and look at the number $\gamma$ of different colors among the balls that were drawn. What is the expectation of $\gamma$ as a function of $n/N$, depending on suitable properties of the distribution $\mathbf{p}$?

To give more insight: if $N = P$ and $p_i = 1/P$ for all $i$, then I will always see exactly $n$ colors, that is, $\gamma = P (n/N)$. Otherwise, it can be shown that the expectation of $\gamma$ is $>P(n/N)$. For fixed $P$ and $N$, it would seem that the factor by which to multiply $n/N$ would be maximal when $\mathbf{p}$ is uniform; maybe the expected number of different colors seen be bounded as a function of $n/N$ and, e.g., the entropy of $\mathbf{p}$?

This seems related to the coupon collector's problem, except that sampling is performed without replacement, and the distribution of the coupons is not uniform.

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    $\begingroup$ I think this problem can be stated as: what is the expected number of nonzero entries in a sample from a multivariate hypergeometric distribution? $\endgroup$ Jun 25, 2016 at 19:07
  • $\begingroup$ This is a specific case of the skewed general distribution. See solution here adellera.it/static_html/investigations/distinct_balls/… $\endgroup$
    – RM-
    Aug 23, 2021 at 11:57
  • $\begingroup$ There is nothing generally known as a "skewed general distribution:" that phrase is used in your reference to classify certain kinds of multinomial distributions. $\endgroup$
    – whuber
    Aug 23, 2021 at 13:26

1 Answer 1

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Suppose you have $k$ colors where $k \leq N$. Let $b_i$ denote the number of balls color $i$ so $\sum b_i = N$. Let $B = \{b_1, \ldots, b_k\}$ and let $E_i(B)$ notate the set which consists of the $i$ element subsets of $B$. Let $Q_{n, c}$ denote the number of ways we can choose $n$ elements from the above set such that the number of different colors in the chosen set is $c$. For $c = 1$ the formula is simple:

$$ Q_{n, 1} = \sum_{E \in E_{1}(B)}\binom{\sum_{e \in E}e}{n} $$

For $c = 2$ we can count sets of balls of size $n$ which has at most 2 colors minus the number of sets which have exactly $1$ color:

$$ Q_{n,2} = \sum_{E \in E_{2}(B)}\binom{\sum_{e \in E}e}{n} - \binom{k - 1}{1}Q_{n, 1} $$

$\binom{k - 1}{1}$ is the number of ways you can add a color to a fixed color such that you will have 2 colors if you have $k$ colors in total. The generic formula is if you have $c_1$ fixed colors and you want to make $c_2$ colors out of it while having $k$ colors in total($c_1 \leq c_2 \leq k$) is $\binom{k - c_1}{c_2 - c_1}$. Now we have everything to derive the generic formula for $Q_{n, c}$:

$$ Q_{n, c} = \sum_{E \in E_{c}(B)}\binom{\sum_{e \in E}e}{n} - \sum_{i = 1}^{c - 1}\binom{k - i}{c - i}Q_{n, i} $$

The probability that you will have exactly $c$ colors if you draw $n$ balls is:

$$ P_{n, c} = Q_{n, c} / \binom{N}{n} $$

Also note that $\binom{x}{y} = 0$ if $y > x$.

Probably there are special cases where the formula can be simplified. I didn't bother to find those simplifications this time.

The expected value you're looking for the number of colors dependent on $n$ is the following:

$$ \gamma_{n} = \sum_{i = 1}^{k} P_{n, i} * i $$

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    $\begingroup$ You call $P_{n,c}$ a probability, but you seem to have defined it as a sum of integers. Did you forget to divide by something? $\endgroup$ Jun 27, 2016 at 14:57
  • $\begingroup$ Yes, I guess you're right. You need to divide by $\binom{N}{n}$, but unfortunately it's still not right that way. If $E, F \in E_{c}(B)$ and $E \cap F \neq \emptyset$ I do doublecounting in the above formula. $\endgroup$
    – jakab922
    Jun 29, 2016 at 8:21
  • $\begingroup$ Seems like the formula can be fixed by using the sieve method. I will post a fix later today. $\endgroup$
    – jakab922
    Jun 29, 2016 at 8:38

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