How to minimize $-\sum p_b \ln{p_b}$?

Question

Consider multisets of the form $A = \{a_1,\dots,a_n\}$ of integers. Let $q = P(a_i = a_j)$ when $i$ and $j$ are chosen independently and uniformly from $\{1,\dots, n\}$. Let $B$ be the set of integers in $A$. We know that $|B| \leq n$. Finally let $p_b = P(a_i = b)$ when $i$ is chosen uniformly from $\{1,\dots, n\}$.

How small can $-\sum_{b \in B} p_b \ln{p_b}$ be as a function of $q$ and $n$?

In this setup we can vary $A$ as long as we maintain the constraints that $q = P(a_i = a_j)$ and the size of $A$ is $n$.

We can immediately infer that $|B| \geq 1/q$ . If $n = 100$ and $q=1/2$, say, it seems that $|B|$ can be as large as $20$. I imagine an exact answer may be hard to get so I would be happy with an estimate or bound.

Also asked at https://math.stackexchange.com/questions/748304/how-minimize-sum-p-b-lnp-b .

Answer 1

First note that $q = \sum_b p_b^2$. We're going to use an instance of the AM-GM inequality: $$ \sum_b p_b\cdot p_b \geq \prod_b p_b^{p_b}. $$ This gives $$ -\sum_b p_b \log p_b = \log \Bigl( \prod_b p_b^{-p_b} \Bigr) \geq \log \Bigl( 1\Big/\sum_b p_b^2\Bigr) = -\log q. $$ This gives a lower bound.

Equality holds iff all the $p_b$s are equal. So, for a given $n$ and $q$, we can achieve this lower bound if and only if we can choose $A$ and $B$ in such a way that all $p_b$ are equal. Now if all the $p_b$ are equal then for each $b$, we have $q = |B| p_b^2$ and so $p_b = \sqrt{q/|B|}$. On the other hand, for each $b$, $$ p_b = \frac{1}{n} \cdot \bigl|\{ i : a_i = b\}\bigr|, $$ so $n p_b$ must be an integer. So if we're to achieve this minimum then $n\sqrt{q/|B|}$ must be an integer. Obviously this won't often be achievable for given $n$ and $q$. But I imagine it's the case that when $n$ is large, you can come close to this lower bound by a suitable choice of $A$ and $B$.

Incidentally, nothing about this question requires $A$ to be a multiset of integers. The elements of $A$ could be anything.