Someone commented under my question Calculation of the moments using Hypergeometric distribution that $$ \sum_{k=0}^l\frac{{l \choose k}{2n-l \choose n-k}(2k-l)^q}{{2n\choose n}}\sim \sum_{k=0}^l (2k-l)^q {l \choose k}. $$

I've tried to use the Stirling's approximation formula, but I did not get the result. Maybe I have mistake... So is this approximation is true?

Thank you for your help.

  • $\begingroup$ Well, it doesn't seem to hold. What are the limits of $l, n$ that it is supposed to? $\endgroup$ – Eelvex Jun 12 '12 at 13:43

It's only true when $l = 0$ or $q$ is odd (since both sides are $0$), otherwise it is not true.

(I'll just do the case when $l$ is even, the case when $l$ is odd is pretty much identical.)

Since $\binom{a}{b}$ is at a maximum when $b = a/2$ (or $b = (a\pm 1)/2$ if $a$ is odd) and all the terms in the summation are non-negative ($q$ is even), we have

$$0 \le \sum_{k=0}^l \frac{\binom{2n-l}{n-k}}{\binom{2n}{n}} (2k-l)^q\binom{l}{k} \le \frac{\binom{2n-l}{n-l/2}}{\binom{2n}{n}} \sum_{k=0}^l (2k-l)^q\binom{l}{k}$$

$\binom{a}{a/2}$ is increasing, so it is maximised by taking $a$ as large as possible, that is, choosing $l$ as small as possible; so $l = 2$ ($0$ was eliminated above). Thus \begin{align*}\frac{\binom{2n-l}{n-l/2}}{\binom{2n}{n}} \sum_{k=0}^l (2k-l)^q\binom{l}{k} &\le \frac{\binom{2n-2}{n-1}}{\binom{2n}{n}} \sum_{k=0}^l (2k-l)^q\binom{l}{k} \\ &= \frac{n}{2n(2n-1)}\sum_{k=0}^l (2k-l)^q\binom{l}{k} \\ &= \frac{1}{2(2n-1)}\sum_{k=0}^l (2k-l)^q\binom{l}{k} \end{align*}

Thus, the approximation can't hold, since the left-hand side goes to $0$ as $n \to \infty$ while the right hand side doesn't.


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