Let \begin{align*} f(n, r, \pi, k) &= \sum_{z=0}^{n}\sum_{s=0}^{r}\binom{z}{s}\binom{n}{z}\binom{n-z}{r-s}(-1)^{r+s}\left(\frac{\pi}{1-\pi}\right)^{r/2-s}\pi^{z}(1-\pi)^{n-z}z^k \end{align*} I am postulating that $f$ is identically 0, based off some theoretical results from statistics. How would one go about in proving this? Are the above special cases of a class combinatorial identities?

Bonus: Another problem that came up in my research is \begin{align*} h(n, \pi) = \sum_{z=0}^{n}\frac{A(n, \pi, z)B(n, \pi, z)}{C(n, \pi, z)}\pi^{z}(1-\pi)^{n-z}\binom{n}{z} \end{align*} where \begin{align*} A(n, \pi, z) &= \sum_{r=2}^{n}\rho_r \sum_{s=0}^{r}\binom{z}{s}\binom{n-z}{r-s}(-1)^{r+s}\left(\frac{r}{2}-s\right)\left(\frac{\pi}{1-\pi}\right)^{r/2-s+1} \\ B(n, \pi, z) &= \sum_{r=3}^{n}c_r \sum_{s=0}^{r}\binom{z}{s}\binom{n-z}{r-s}(-1)^{r+s}\left(\frac{\pi}{1-\pi}\right)^{r/2-s} \\ C(n, \pi, z) &= 1+\sum_{r=2}^{n}\rho_r \sum_{s=0}^{r}\binom{z}{s}\binom{n-z}{r-s}(-1)^{r+s}\left(\frac{\pi}{1-\pi}\right)^{r/2-s} \end{align*} where $\rho_r, c_r$ are constants. Notice that in this expression, the $r$ is being marginalized out. This I have no idea what it equates to, and I'm frankly a bit pessimistic is finding a closed-form for this. But any help would be appreciated!

  • $\begingroup$ the sum on $s$ can probably be simplified, search in math.stackexchange.com/questions/tagged/binomial-coefficients the formulas for sum of products of binomial coefs $\endgroup$ – reuns May 26 '16 at 6:47
  • $\begingroup$ I did ask exactly that; in my previous post (math.stackexchange.com/questions/1793225/…), the best someone could do is an expression involving hypergeometric functions $\endgroup$ – Tom Chen May 26 '16 at 6:51
  • $\begingroup$ so try the sum on $z$ :) or try changing the triple product of binomial coefs to get a different sum on $s$. if it doesn't reduce to any formula on this forum, little chance that it can be simplified $\endgroup$ – reuns May 26 '16 at 6:53
  • $\begingroup$ That's what I've been struggling with. It's interesting, though, because most likely a sum on $z$ will involve another hypergeometric function, yet the sum on both $s, z$ will result in 0 (most likely, I confirmed this with several cases numerically). $\endgroup$ – Tom Chen May 26 '16 at 6:57
  • $\begingroup$ since you seem to know the contour integral / generating function method (as in math.stackexchange.com/a/1789628/276986 ) , can you write the whole double sum in term of contour integral of $(1+x/a)^z (1+x)^{n-z}$ etc., to see what it looks like ? $\endgroup$ – reuns May 26 '16 at 7:01

Hint: Something seems to be not correct, since the function can be written as \begin{align*} f(n, r, \pi, k) &= \sum_{z=0}^{n}\sum_{s=0}^{r}\binom{z}{s}\binom{n}{z}\binom{n-z}{r-s}(-1)^{r+s}\left(\frac{\pi}{1-\pi}\right)^{r/2-s}\pi^{z}(1-\pi)^{n-z}z^k\\ &=(-1)^r\pi^{\frac{r}{2}}(1-\pi)^{ n-\frac{r}{2}} \sum_{z=0}^n\binom{n}{z}\left(\frac{\pi}{1-\pi}\right)^zz^k \sum_{s=0}^r\binom{z}{s}\binom{n-z}{r-s}(-1)^s\left(\frac{1-\pi}{\pi}\right)^s \end{align*} So that \begin{align*} f(n=1,r=0,\pi,k)&=(1-\pi)\sum_{z=0}^1\binom{1}{z}\left(\frac{\pi}{1-\pi}\right)^zz^k\\ &=(1-\pi)\left(0+\frac{\pi}{1-\pi}\right)\\ &=\pi \end{align*} contrary to your assumption $f\equiv 0$.


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