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Solving some problem I have stumbled into the following sum :

$ \displaystyle \sum_{i=0}^{n-2} {e \choose i} (n-1-i) (1-p)^{e-i} p^i $

where $ 0 \leq e \leq {n \choose 2}$.

I am not very efficient with the evaluation of these sums so I would like to ask if there is any way to evaluate this sum or obtain a sharp lower bound for it?

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Did you get something out of a solution below? –  Did Apr 7 '11 at 7:53

1 Answer 1

Consider $$ S(k,e)=\sum_{i=0}^{k} {e \choose i} (1-p)^{e-i} p^i. $$ Then $S(k,e)=1$ if $k\ge e$, and your sum $S_{n,e}$ is $$ S_{n,e}=(n-1)S(n-2,e)-epS(n-3,e-1). $$ Hence $S_{n,e}=n-1-ep$ for every $n\ge e+2$.

If $n<e+2$, there might not exist so simple expressions of $S_{n,e}$ in all generality. Note however for small $n$ that $S(0,e)=(1-p)^e$, hence $S_{0,e}=S_{1,e}=0$ and $S_{2,e}=(1-p)^e$.

Edit According to E.C. Molina (Application to the Binomial Summation of a Laplacian Method for the Evaluation of Definite Integrals, Bell System Technical Journal, v8: i1 January 1929, 99-108, available here), for $k<e$, $$ S(k,e)=B_{k,e}(p)/B_{k,e}(0),\quad\mbox{where}\ B_{k,e}(p)=\int_p^1x^{k}(1-x)^{e-k-1}\mathrm{d}x. $$ The author then presents some approximations of $S(k,e)$.

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Didier, thanks for the link; I did not know until now that there is now a digital archive for The Bell System Technical Journal. –  J. M. May 10 '11 at 10:46
    
Yes. The treasure chest is here: alcatel-lucent.com/bstj –  Did May 10 '11 at 12:31

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