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Hoi there,

i'm trying to show something seemingly simple...

$ \left(\sum_{j=k}^n {n\choose j}p^j(1-p)^{n-j}\right)\left(\sum_{j=l}^n {n\choose j}p^j(1-p)^{n-j}\right)\geq \sum_{j=l+k}^n {n\choose j}p^j(1-p)^{n-j}$

Ofcourse $\sum_{j=0}^np^k(1-p)^{n-k}{n\choose j} = 1$ and $p\in (0,1)$

Let's write:

$a= \sum_{j=l+k}^na_j, \ \ \, \ \ \ b= \sum_{j=k}^{l+k-1}a_j, \ \ \ c = \sum_{j=l}^{l+k-1}a_j\\$

$a_j = {n\choose j}p^j(1-p)^{n-j}$

Then I want to show no matter how $l,k\leq n$ are chosen we have$(a+b)(a+c)\geq a$. But I'm 100% stuck on this...the only things we know are $a,b,c\leq 1$ and w.l.o.g. $b\geq c$.

Anyone maybe having an extreme clever insight? Some inequality we should be using?

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    Sorry, i dont quite understand how to proceed here: It amounts to showing $(1-A_{l-1}(p))(1-A_{k-1}(p)) \geq (1-A_{l+k-1(p)})$ But i don't see how the relation $A_k(p)=A_{n-k}(1-p)$ helps with this.2012-10-25

0 Answers 0