Probability problem: upper bounds on binomial CDF

simba31415
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Homework Statement


Hi all, just a quick question here - the setup is as follows: X is a random variable, X \sim \operatorname{Bin}(m,p) where p=2^{-\sqrt{\log n}}(\log n)^2 and m \geq 2^{\sqrt{\log n}}c for constants c, n (n "large" here). I wish to show that \mathbb{P}(X < c) \leq e^{-(\log n)^2 c/3}. I've been told to use "Chernoff-esque bounds" here; however, after teaching myself a little about Chernoff bounds I haven't found a way to make this work - I can see that the multiplicative form could be useful but I haven't yet figured out how to translate the bounds which I've found online into a workable form for this problem.

I'm told observing the fact that \mu = \mathbb{E}(X) \geq (\log n)^2 c = \mu ' should also help, so I suspect maybe what we really need is to show is \mathbb{P}(X < c) = \mathbb{P}(X < \frac{\mu'}{(\log n) ^2}) < \mathbb{P}(X < \frac{\mu}{(\log n) ^2}) \leq ^{(*)} e^{- \mu / 3} \leq e^{-\mu ' /3} but as I said, no luck so far since I can't prove step (*) if indeed that is the way to do it. I suspect that this result only needs a few lines of work once you have the bound you require from Chernoff, so if anyone could show me how to do this I'd be very grateful! -S
 
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No thoughts anyone? Sorry to bump but I could really use some help with this, any thoughts at all please respond!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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