LFD Book Forum Hoeffding Inequality With Probability > 1?
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#1
05-20-2016, 11:29 PM
 henry2015 Member Join Date: Aug 2015 Posts: 31
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.
#2
05-21-2016, 05:46 AM
 ntvy95 Member Join Date: Jan 2016 Posts: 37
Re: Hoeffding Inequality With Probability > 1?

Quote:
 Originally Posted by henry2015 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.
#3
05-21-2016, 12:03 PM
 henry2015 Member Join Date: Aug 2015 Posts: 31
Re: Hoeffding Inequality With Probability > 1?

Quote:
 Originally Posted by ntvy95 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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