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I've encountered the following problem that involves alternating sums of multinomial coefficients. Let $$f(k)=\sum_{i=0}^{n-k}(-1)^i\binom{n}{k,i,n-k-i}(k+i)^\alpha$$ where $\binom{n}{k,i,n-k-i}=\frac{n!}{k!i!(n-k-i)!}$ is the multinomial coefficient, and $\alpha\in\mathbb{R}$ is a constant that satisfies $-1\leqslant\alpha<0$

Problem: I'm trying to prove that $$\frac{f(k)}{f(k+1)}\leqslant\frac{k+1}{k}$$, and equality happens when $\alpha=-1$.

Background: this problem stems from my study of expected order statistics of Weibull distribution. When a Weibull distribution has shape parameter larger than 1 (assuming scale parameter equals 1), the expected difference between the $k$-th largest order statistic and the $k+1$-th is exactly $f(k)$. Numerical computations showed the results hold, but I cannot prove it analytically. Any suggestions or pointers are much appreciated!

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    $\begingroup$ Did you try the usual trick $x^{\alpha}=\Gamma(-\alpha)^{-1}\int e^{-tx} t^{-\alpha-1}dt$? $\endgroup$ – Abdelmalek Abdesselam Jun 2 '17 at 16:54
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The case $\alpha=-1$, i.e. we need to prove that $kf(k)=(k+1)f(k+1)$.

Let $m=n-k$ and $g(k)=kf(k)$ so that $g(k)=\sum_{j=0}^mF_k(m,j)$ where $$F_k(m,j):=(-1)^j\binom{m}j\binom{m+k}m\frac{k}{k+j}.$$ Define another function $$G_k(m,j):=(-1)^{j-1}\binom{m+1}j\binom{m+k}m\frac{kj}{(m+1)^2}.$$ Now, check this identity $F_k(m+1,j)-F_k(m,j)=G_k(m,j+1)-G_k(m,j)$, for instance routinely divide through by $F_k(m,j)$ and simplifying. As a next step, sum over all integers $j$: $$\sum_{j\in\mathbb{Z}}F_k(m+1,j)-\sum_{j\in\mathbb{Z}}F_k(m,k) =\sum_{j\in\mathbb{Z}}G_k(m,j+1)-\sum_{j\in\mathbb{Z}}G_k(m,j).$$ Since the RHS vanishes (equal sums), the outcome is: $g(k+1)-g(k)=0$. It follows that $g(k)=1$ because $g(0)=1$. Hence, $kf(k)=(k+1)f(k+1)$ as desired.

This approach uses the Wilf-Zeilberger method together with Zeilberger's algorithm.

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  • $\begingroup$ Thanks for this answer and pointers to the paper. I will try and see if this approach also works for the case of $-1<\alpha<0$. The case of $\alpha=-1$ actually reduces to order statistics of a standard exponential distribution, whereas the general case of Weibull distribution seems somewhat harder. $\endgroup$ – user3026001 Jun 2 '17 at 21:59
  • $\begingroup$ I expect the case $\alpha\neq-1$ to be harder. Warning: the above method does not apply when $\alpha\not\in\mathbb{Z}$. $\endgroup$ – T. Amdeberhan Jun 2 '17 at 22:01

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