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When does the following hold?

$\sum_{(i_1,\ldots,i_k)\in E} \frac{n!}{i_1! \ldots i_k!} \le \exp(n H^*)$

Where

$H^*=\max_{(i_1,\ldots,i_k)\in E} -(\frac{i_1}{n}\log \frac{i_1}{n}+\ldots +\frac{i_k}{n}\log \frac{i_k}{n})$ and E is some subset of {$ \{( i_1,\ldots,i_k):i_1+\ldots+i_k=n \}$}

Motivation: this is a generalization of Chernoff's bound to n tosses of fair k-sided dice where E represents the hypothesis we make about that sample. Another motivation is reconciling tight special-case Chernoff bound with looser but more general bound given by Sanov's theorem

Examples: when k=2, it can be proven to hold for sets of coefficients where first component of the coefficient is less than n/2 (ie here).

When k=3, it seems (empirically) to hold for sets of coefficients where sum of first two components is ≤n/2. For instance, for n=10, highest entropy term gives upper bound of (2/3)^3 *10^5 whereas exact sum is 12585. Since k=3 multinomial coefficients lie in a 2-simplex, the 21 multinomial coefficients in this set can be visualized below. Top vertex represents coefficient (0,0,10)

For higher k, we can look at similar sets, ie corners of the (k-1) simplex. I tried few values and it seems to hold for coefficients where sum of first k-1 components is below n/(k-1)

Here's how you'd check it in Mathematica

getit[n_, k_, c_] := (
   all = Select[Tuples[Range[0, n], k], Total[#] == n &];
   e = Select[all, Total[Most[#]] <= c &];
   hterm[x_] := If[0 < x < 1, x Log[x], 0];
   H[event_] := -Total[hterm /@ (event/n)];
   exact = Total[Multinomial @@@ e];
   upper = Exp[n Max[H /@ e]];
   exact < upper
);
(* Check bound for k=3, n=10, with i1+i2<=5 *)
getit[10, 3, 5]

Update 8/18 Leandro gives a bound on a single multinomial coefficient which gives Sanov's theorem if we consider that there's at most $(n+1)^k$ multinomial coefficients in any set E. It seems that to generalize the proof of the tighter binomial bound to, say, trinomial coefficients, one would need to prove the following inequality first

$$p_1 \log q_1 + p_2 \log q_2 + p_3 \log q_3 \ge q_1 \log q_1 + q_2 \log q_2 + q_3 \log q_3$$

Where p and q add up to 1. For each q, the set of p's for which the above bound holds also gives us the hypothesis for which we can give tight Chernoff-like bound. Empirically, this bound seems to hold for p's "bounded away" from the uniform distribution. Black circle below represents q, blue region is the set of distributions p where the bound above holds. My Mathematica notebook

Update 8/24: the bound holds for sets of coefficients of the form $i_1 a_1 + \ldots + i_n a_n \le C$ where $a_1\ldots a_n$ are arbitrary non-negative numbers, details in answer

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Hi Yaroslav, I guess there is a small typo in the definition of $H^*$. –  Leandro Aug 16 '10 at 22:33
    
thanks, fixed . –  Yaroslav Bulatov Aug 16 '10 at 22:35
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I've no answer to your question. but a minor note on your Mathematica code: you can use this definition for hterm instead hterm[x_] := If[0 <= x <= 1, x Log[x], 0]; , and change the formula for exact to Total[Multinomial @@@ e] . –  J. M. Aug 18 '10 at 2:33
    
Good point, @@@ notation is way more legible –  Yaroslav Bulatov Aug 18 '10 at 3:53
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2 Answers 2

Hi Yaroslav, With the additional hypothesis you are considering, I guess that the inequality can be proved following the text that you linked.

Fix any probability vector $(p_1,\ldots,p_k)$ and consider $E\subset \{1,\ldots, n\}^k$ such that $$ E\subset\left\{(i_1,\ldots,i_k); \sum_{j}^ki_j=n\ \text{and }\ p_j\geq\frac{i_j}{n} \right\} $$

For any element of $E$, we have

$$ \sum_{j}^k p_j\log p_j\leq \sum_{j=1}^k \frac{i_j}{n}\log p_j. $$ Therefore $$ -n\left(\sum_{j=1}^k-p_j\log p_j \right)\leq \sum_{j=1}^k i_j\log p_j. $$ Exponentiating both sides we get $$ \exp\left(-n\left(\sum_{j=1}^k-p_j\log p_j \right)\right)\leq p_1^{i_1}\ldots p_k^{i_k}. $$ For the other hand, we have that $$ 1=(p_1+\ldots+p_k)^n=\sum_{i_1,\ldots,i_k}\frac{n!}{i_1!\ldots i_k!}p_1^{i_1}\ldots p_k^{i_k}\geq \sum_{(i_1,\ldots,i_k)\in E}\frac{n!}{i_1!\ldots i_k!}\exp\left(-n\left(\sum_{j=1}^k-p_j\log p_j \right)\right), $$ where the first summation is taken over all sequences of nonnegative integer indices $i_1$ through $i_k$ such the sum of all $i_j$ is $n$.
So we get that $$ \sum_{(i_1,\ldots,i_k)\in E}\frac{n!}{i_1!\ldots i_k!}\leq\exp\left(n\left(\sum_{j=1}^k-p_j\log p_j \right)\right)\leq e^{n H^*}. $$

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Is it essential to disallow 0's in the coefficients? Allowing 0's would allow to relate to existing bounds, like the one I linked –  Yaroslav Bulatov Aug 18 '10 at 2:51
    
Also, it seems your set E contains at most 1 element –  Yaroslav Bulatov Aug 18 '10 at 3:24
    
The difference between you proof and the one I linked is that they base their proof on an entropy inequality that holds for a range of coefficients. In your case, your starting inequality holds for just one coefficient, so it can be replaced by equality. It gives a bound on a multinomial coefficient in terms of it's entropy. There's at most (n+1)^k coefficients in any set E (+1 comes if we allow 0's), so we can bound the sum over these coefficients over the set by (n+1)^k Exp[n H^*] .... that's the same as Sanov's theorem –  Yaroslav Bulatov Aug 18 '10 at 16:32
    
You right. I did not reply before, because I am investigating the generalization of the case two inequality. I have one, but it also seem stupid as my previous argument. If I have time I will think about it little bit more. –  Leandro Aug 18 '10 at 16:52
    
I added "proof by picture" that the binomial proof can be generalized...what's missing is an explicit characterization of the set of p's where H(p,q)>H(q) –  Yaroslav Bulatov Aug 18 '10 at 19:37
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up vote 0 down vote accepted

Take arbitrary non-negative reals $a_1,\ldots,a_n$ consider and consider set $E$ of n-tuples $(i_1,\ldots,i_n)$ satisfying the following

$i_1 a_1 + \ldots + i_n a_n \le n(a_1 \exp - a_1 + \ldots +a_n \exp -a_n), \sum_k i_k=n, i_k\ge 0$

Sum of multinomial coefficients in this set is bounded by the highest entropy one.

Proof, define $q_i=-\log a_i$ and observe that ${q_1}^{i_1} \cdots {q_n}^{i_n}\ge \exp(-n H(q))$. Then proceed in the same manner as the the analogous proof for binomial coefficients, substituting this inequality in the last step of derivation.

Note that if any $a_i$ is 0, the bound is vacuous. Here are some examples of sets of trinomial coefficients defined in this way for random $a_i$'s. Black dot is the highest entropy coefficient.

One way of to think of these sets is the following -- view multinomial coefficients as multinomial probability distributions. Then I pick my "highest entropy" distribution q and consider the set of distributions $p$ for which $p_1 \log q_1 + \ldots p_n \log q_n \ge q_1 \log q_1 + \ldots + q_n \log q_n$. That includes q, and all distributions "further away" from the uniform distribution than q

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