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Let $p_1,\ldots,p_s$ be $s$ number in the unit interval such that $p_1+\ldots+p_s=1.$

Is it then true, that for every $n\geq 1$ we have $ \sum_{(k_1,\ldots,k_n)\in \{1,\ldots,s \}^n} p_{k_1}\cdot \ldots \cdot p_{k_n}=1 ?$

Checking it for $n=2$ indicates that it's true (but already for $n=3$ it isn't feasable to check it by hand)...

2 Answers 2

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Let $[s]=\{1,\dots,s\}$. For $n>1$ we have

$\begin{align*} \sum_{\langle k_1,\dots,k_n\rangle\in[s]^n}p_{k_1}\dots p_{k_n}&=\sum_{\langle k_1,\dots,k_{n-1}\rangle\in[s]^{n-1}}\left(p_{k_1}\dots p_{k_{n-1}}\sum_{k\in[s]}p_k\right)\\\\ &=\sum_{\langle k_1,\dots,k_{n-1}\rangle\in[s]^{n-1}}p_{k_1}\dots p_{k_{n-1}}\;, \end{align*}$

so the result follows by induction.

  • 0
    @user26698: Excellent. You’re welcome!2012-12-31
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$ \sum_{(k_1,\ldots,k_n)\in \{1,\ldots,s \}^n} p_{k_1}\cdot \ldots \cdot p_{k_n}=(p_1+p_2+\cdots+p_s)^n=1^n=1$