If c <= (a_i)/(b_i) AA i, then c <= (sum_(i=1)^oo a_i)/(sum_(i=1)^oo b_i)

batystowy2b 2022-09-14 Answered
I'm trying to prove that the best moment tail bound is no worse than the best Chernoff bound. I have all pieces other than this little frustration here:
If c a i b i i, then c i = 1 a i i = 1 b i
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Answers (1)

Sanaa Holder
Answered 2022-09-15 Author has 20 answers
Assuming all numbers are positive and that the series are convergent, you have
c b i a i
for all i. Hence, summing all up, you have
c i b i i a i
which is equivalent to
c i a i i b i

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