Hoeffding's Inequality Probability $>1$
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in one of the courses the professor said that Hoeffding's Inequality equation is this
but for me that does not make any sense as it can be greater than $1.$
For example, assume
$$varepsilon = 0.00001$$
$$N = 1000$$
That will make the probability $leq 2( e^{-2cdot 0.0000000001cdot 1000}) = 2cdot 0.999999 = 1.99999.$
Is there something wrong here?
probability
edited Dec 23 '18 at 6:06
Gaby Alfonso
1,1901318
asked Dec 23 '18 at 2:48
asmgxasmgx
1587
$endgroup$
add a comment |
$begingroup$
in one of the courses the professor said that Hoeffding's Inequality equation is this
but for me that does not make any sense as it can be greater than $1.$
For example, assume
$$varepsilon = 0.00001$$
$$N = 1000$$
That will make the probability $leq 2( e^{-2cdot 0.0000000001cdot 1000}) = 2cdot 0.999999 = 1.99999.$
Is there something wrong here?
probability
edited Dec 23 '18 at 6:06
Gaby Alfonso
1,1901318
asked Dec 23 '18 at 2:48
asmgxasmgx
1587
$endgroup$
add a comment |
$begingroup$
in one of the courses the professor said that Hoeffding's Inequality equation is this
but for me that does not make any sense as it can be greater than $1.$
For example, assume
$$varepsilon = 0.00001$$
$$N = 1000$$
That will make the probability $leq 2( e^{-2cdot 0.0000000001cdot 1000}) = 2cdot 0.999999 = 1.99999.$
Is there something wrong here?
probability
edited Dec 23 '18 at 6:06
Gaby Alfonso
1,1901318
asked Dec 23 '18 at 2:48
asmgxasmgx
1587
$endgroup$
in one of the courses the professor said that Hoeffding's Inequality equation is this
but for me that does not make any sense as it can be greater than $1.$
For example, assume
$$varepsilon = 0.00001$$
$$N = 1000$$
That will make the probability $leq 2( e^{-2cdot 0.0000000001cdot 1000}) = 2cdot 0.999999 = 1.99999.$
Is there something wrong here?
probability
probability
edited Dec 23 '18 at 6:06
Gaby Alfonso
1,1901318
asked Dec 23 '18 at 2:48
asmgxasmgx
1587
edited Dec 23 '18 at 6:06
Gaby Alfonso
1,1901318
asked Dec 23 '18 at 2:48
asmgxasmgx
1587
edited Dec 23 '18 at 6:06
Gaby Alfonso
1,1901318
edited Dec 23 '18 at 6:06
Gaby Alfonso
1,1901318
edited Dec 23 '18 at 6:06
Gaby Alfonso
1,1901318
1,1901318
asked Dec 23 '18 at 2:48
asmgxasmgx
1587
asked Dec 23 '18 at 2:48
asmgxasmgx
1587
asked Dec 23 '18 at 2:48
asmgxasmgx
1587
1587
add a comment |
add a comment |
1 Answer
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Nothing is wrong - if the right hand side is greater than $1$, it just means that the inequality is pointless in that case, since all probabilities are less than or equal to $1$.
answered Dec 23 '18 at 2:53
pwerthpwerth
3,300417
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Nothing is wrong - if the right hand side is greater than $1$, it just means that the inequality is pointless in that case, since all probabilities are less than or equal to $1$.
answered Dec 23 '18 at 2:53
pwerthpwerth
3,300417
$endgroup$
add a comment |
$begingroup$
Nothing is wrong - if the right hand side is greater than $1$, it just means that the inequality is pointless in that case, since all probabilities are less than or equal to $1$.
answered Dec 23 '18 at 2:53
pwerthpwerth
3,300417
$endgroup$
add a comment |
$begingroup$
Nothing is wrong - if the right hand side is greater than $1$, it just means that the inequality is pointless in that case, since all probabilities are less than or equal to $1$.
answered Dec 23 '18 at 2:53
pwerthpwerth
3,300417
$endgroup$
Nothing is wrong - if the right hand side is greater than $1$, it just means that the inequality is pointless in that case, since all probabilities are less than or equal to $1$.
answered Dec 23 '18 at 2:53
pwerthpwerth
3,300417
answered Dec 23 '18 at 2:53
pwerthpwerth
3,300417
answered Dec 23 '18 at 2:53
pwerthpwerth
3,300417
answered Dec 23 '18 at 2:53
pwerthpwerth
3,300417
3,300417
add a comment |
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