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Old 05-20-2016, 11:29 PM
henry2015 henry2015 is offline
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Default Hoeffding Inequality With Probability > 1?

Given that the inequality shows an upper bound of 2*e^(-2*N*(epsilon)^2).

If we have N = 10, and epsilon = 0.1, then we have

2*e^(-2*10*(0.1^2)) = 1.637461506.

So what does it mean? I thought we would never have a probability > 1.
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Old 05-21-2016, 05:46 AM
ntvy95 ntvy95 is offline
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Default Re: Hoeffding Inequality With Probability > 1?

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Originally Posted by henry2015 View Post
Given that the inequality shows an upper bound of 2*e^(-2*N*(epsilon)^2).

If we have N = 10, and epsilon = 0.1, then we have

2*e^(-2*10*(0.1^2)) = 1.637461506.

So what does it mean? I thought we would never have a probability > 1.
I think you can check out the first question in the Q&A section of Lecture 02.
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Old 05-21-2016, 12:03 PM
henry2015 henry2015 is offline
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Default Re: Hoeffding Inequality With Probability > 1?

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Originally Posted by ntvy95 View Post
I think you can check out the first question in the Q&A section of Lecture 02.
Thanks for the link!
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