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What is an easy way to show that for positive integers $i,n$, a real $p \in (\frac12,1)$ and $\epsilon \in [0,p]$,

$$p^i(1-p)^{n-i} \geq (p-\epsilon)^i(1-(p-\epsilon))^{n-i}.$$

(I have a complicated way, where I first show that the left hand side is bigger when i = n/2 and then increasing i can only make the left hand side bigger. But is there some well known inequality which lets me formulate this shorter?)

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are there any restrictions? $n\in\mathbb{R}$,$p\in\mathbb{Z}$, etc . . . – Kate Apr 6 '11 at 3:16
Updated the restrictions, but my claim was wrong. Thanks for the quick response everyone. Now I understand why this thing caused me so much troubles to prove. :) – user915 Apr 6 '11 at 3:38
My error was that I assumed p(1-p) > (p-\epsilon)(1-(1-\epsilon)) despite knowing better... Math-overload, should take a break :) – user915 Apr 6 '11 at 3:47
up vote 4 down vote accepted

This inequality cannot be true. You need more conditions.
If $p$ is allowed to be larger than $1$ there will be sign problems since $n-i$ odd will make both sides negative. Even if we suppose $0<p<1$ and $0<e<p$ it does not work out.

Consider the case where $0.5n<i<0.6n<n\log_3 ( 2)$. Let $p=\frac{3}{4}$, and let $e=\frac{1}{4}$. Then the inequality becomes $$\left(\frac{3}{4}\right)^{i}\left(\frac{1}{4}\right)^{n-i}\geq\left(\frac{1}{2}\right)^{i}\left(\frac{1}{2}\right)^{n-i}$$ or equivalently $$\frac{3^{i}}{2^{n}}\geq1.$$ Since $i<n\log_{3}2$ we have that $3^{i}<2^{n}$ so that $$\frac{3^{i}}{2^{n}}<1$$ which is impossible.

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@Michael: If we have the conditions $0\leq p\leq \frac{1}{2}$, $0\leq e\leq p$, and $i\geq \frac{n}{2}$ then your inequality is true. (If $i\geq \frac{n}{2}$ then $p$ must be in $[0,\frac{1}{2}]$ rather than $[\frac{1}{2},1]$. If $i\leq \frac{n}{2}$ then it is the other way around) – Eric Naslund Apr 6 '11 at 14:43

It is not true. If $p=.8, n=5, i=3, \epsilon =.01, p^i(1-p)^{n-i}=.2048, (p-\epsilon)^i(1-(p-\epsilon))^{n-i}\approx .21743$ If you take the derivative $\frac{d(p^i(1-p)^{n-i})}{dp}=p^{i-1}(1-p)^{n-i-1}(i-np)\lt 0 \text { if } i-np \lt 0$, so if you decrease $p$ you increase the function.

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Let $F: (0,1) \rightarrow R, f(x)= x^i (1-x)^{n-i}$.

Then $f'$ is positive on $(\frac{i}{n}, 1)$ and negative on $(0,\frac{i}{n})$.

Thus your inequality holds if $p-\epsilon > \frac{i}{n}$, and is opposite if $p< \frac{i}{n}$. The left out case is much harder to stude.

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