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When reading about ordered statistics, i've faced a sum, that I can't find:

$$\frac{d}{dp} \sum\limits_{i=k}^n C_{n}^{i} p^i (1 - p)^{n-i}$$ (where $\frac{d}{dp}$ stands for derivarive with respect to $p$, $ \quad C_{n}^{i}$ - binomial coefficient)

metamorphy
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Joitandr
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1 Answers1

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Take the derivative immediately, and use $i\binom{n}{i}=n\binom{n-1}{i-1}$, $(n-i)\binom{n}{i}=n\binom{n-1}{i}$ (where $\binom{n}{i}:=C_n^i$ is the modern notation). So, for $0<k\leqslant n$ [otherwise the result is trivially $0$], the result is the difference of $$\sum_{i=k}^{n}\binom{n}{i}ip^{i-1}(1-p)^{n-i}=n\sum_{i=k}^{n}\binom{n-1}{i-1}p^{i-1}(1-p)^{n-i}=n\sum_{i=k-1}^{n-1}\binom{n-1}{i}p^i(1-p)^{n-i-1}$$ (for the last step, $i$ is replaced by $i+1$ under the sum, and the range of summation is adjusted for that) and $$\sum_{i=k}^{\color{blue}{n-1}}\binom{n}{i}p^i(\color{blue}{n-i})(1-p)^{n-i-1}=n\sum_{i=k}^{n-1}\binom{n-1}{i}p^i(1-p)^{n-i-1};$$ now the summands match, hence the result is simply the summand at $i=k-1$, equal to $$n\binom{n-1}{k-1}p^{k-1}(1-p)^{n-k}.$$ By the way, this is yet another idea to prove the formula used in this answer of mine.

metamorphy
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