2
$\begingroup$

Consider the $[n]!_q = \prod\limits_{k = 1}^{n} \frac{q^k - 1}{q - 1} = \sum\limits_{k = 0}^{\binom n 2} c_k q^k$ and let $\{f_n\}_{n \in \mathbb{N}}$ be the sequence of the functions on $[0; 1]$ defined by the following $$f_n(x) = \frac{c_{\lfloor \binom n 2 x \rfloor}}{n!}$$ Is there a formula for $\lim\limits_{n \rightarrow \infty} f_n(x)$? Roughly speaking, what is the limit distribution of its coefficients?

$\endgroup$

2 Answers 2

4
$\begingroup$

Let $Z_n$ be the number of inversions of a random permutation in $S_n$. Then for all $x\in\mathbb{R}$, $$ \mathrm{Prob}\left(Z_n<\frac 14 n^2+\frac 16xn^{3/2}\right)\to \mathcal{N}(x), $$ the standard normal distribution. This goes back to Feller, 1945. See for instance Theorem 3.3.4 of https://www.routledgehandbooks.com/doi/10.1201/b18255-6.

$\endgroup$
0
2
$\begingroup$

Choose $\xi_i\in \{0,1,\dots,i-1\} $ uniformly at random. Then $c_k/n! $ is a probability that $\sum \xi_i=k$. The law of large numbers and central limit theorem work nicely for the distribution of $\sum \xi_i$. The limits you are talking about exist and are equal to zero.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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