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 henry2015 05-21-2016 12:29 AM

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.

 ntvy95 05-21-2016 06:46 AM

Re: Hoeffding Inequality With Probability > 1?

Quote:
 Originally Posted by henry2015 (Post 12363) 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.

 henry2015 05-21-2016 01:03 PM

Re: Hoeffding Inequality With Probability > 1?

Quote:
 Originally Posted by ntvy95 (Post 12366) I think you can check out the first question in the Q&A section of Lecture 02.