Sum over k of n Choose k by p^k by (1-p)^n-k by Absolute Value of k-np

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Theorem

Let $n \in \Z_{\ge 0}$ be a non-negative integer.


Then:

$\ds \sum_{k \mathop \in \Z} \dbinom n k p^k \paren {1 - p}^{n - k} \size {k - n p} = 2 \ceiling {n p} \dbinom n {\ceiling {n p} } p^{\ceiling {n p} } \paren {1 - p}^{n - 1 - \ceiling {n p} }$


Proof

Let $t_k = k \dbinom n k p^k \paren {1 - p}^{n + 1 - k}$.

Then:

$t_k - t_{k + 1} = \dbinom n k p^k \paren {1 - p}^{n - k} \paren {k - n p}$

Thus the stated summation is:

$\ds \sum_{k \mathop < \ceiling {n p} } \paren {t_{k + 1} - t_k} + \sum_{k \mathop \ge \ceiling {n p} } \paren {t_k - t_{k + 1} } = 2 t_{\ceiling {n p} }$

$\blacksquare$


Historical Note

This result was stated by Abraham de Moivre in his $1730$ work Miscellanea Analytica for the case where $n p$ is an integer.

The general case was proved by Henri Poincaré in his Calcul des Probabilités of $1896$.


Sources

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