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Old 07-06-2013, 09:11 AM
BojanVujatovic BojanVujatovic is offline
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Default Hoeffding's Inequality vs. binomial distribution

Considering the bin case as described in the book and the lectures, we could write the probability of \mu being close to \nu:
P[|\mu - \nu| > \epsilon]
can be written excatly since \nu is binomaily distributed with parameters \mu and N. Now, the problem is that that expression has \mu in it which is unknown. But we can find the value of \mu for which P[|\mu - \nu| > \epsilon] has its maximum, and it turns out to be \mu = 0.5.
Now if we plug that in the expression we get the bound with same properties as Hoeffding's (valid for all \mu's, N's and \epsilon's).
Now, my view is that that bound is the tightest possible, tighter than the Hoeffding's. Am I correct? It could be (perhaps?) used in futher analysis, but the major donwside is that is is not so nice nor elegant to work with. Other opinions?

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Old 07-06-2013, 10:30 AM
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yaser yaser is offline
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Default Re: Hoeffding's Inequality vs. binomial distribution

Quote:
Originally Posted by BojanVujatovic View Post
Considering the bin case as described in the book and the lectures, we could write the probability of \mu being close to \nu:
P[|\mu - \nu| > \epsilon]
can be written excatly since \nu is binomaily distributed with parameters \mu and N. Now, the problem is that that expression has \mu in it which is unknown. But we can find the value of \mu for which P[|\mu - \nu| > \epsilon] has its maximum, and it turns out to be \mu = 0.5.
Now if we plug that in the expression we get the bound with same properties as Hoeffding's (valid for all \mu's, N's and \epsilon's).
Now, my view is that that bound is the tightest possible, tighter than the Hoeffding's. Am I correct? It could be (perhaps?) used in futher analysis, but the major donwside is that is is not so nice nor elegant to work with.
You are correct. Different approaches to similifying the expression for this bound can lead to different bounds of varying tightness and elegance.
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