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Dec 29, 2023 at 3:46 comment added john22445 thanks for your explanation, I would like to delete my post after this
Dec 28, 2023 at 16:18 comment added Iosif Pinelis @john22445 : I have added this detail as well.
Dec 28, 2023 at 16:17 history edited Iosif Pinelis CC BY-SA 4.0
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Dec 28, 2023 at 2:53 comment added john22445 I had fully understand your steps, thank you so much, but can i know the meaning of sup{r∈R:P(|Sn|≥r)=1} for n∈N? it is because from my understanding, this means the least upper bound which is actually infinity?
Dec 27, 2023 at 20:35 comment added Iosif Pinelis (i) I have added details on your first two comments. (ii) Why "pick negative for Sn-1?" -- Because it works. (iii) I am not really active on math.stackexchange. Most of my contributions there are spillovers from MathOverflow.
Dec 27, 2023 at 20:31 history edited Iosif Pinelis CC BY-SA 4.0
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Dec 27, 2023 at 19:47 comment added john22445 can you answer it here? Thanks in advance john22445 (math.stackexchange.com/users/1272887/john22445), Application on Lindeberg's condition, URL (version: 2023-12-27): math.stackexchange.com/q/4834304
Dec 27, 2023 at 19:44 comment added john22445 Then, why do you want to pick negative for Sn-1?
Dec 27, 2023 at 19:43 comment added john22445 For triangle inequality, if Sn - Sn-1 = Xn, then |P(Sn) - P(Sn-1)| <= |P(C)|, how do I get your inequality? it seems like you are using P(Xn intersect Sn-1)
Dec 27, 2023 at 19:27 comment added john22445 thanks for the answer. For P(|Sn−1|≤4n−an)=1, I know that it is because of P(|Sn|≥an), the problem is how to get P(|Sn|≥an)? Is it by our assumption that r>an?
Dec 27, 2023 at 17:50 history answered Iosif Pinelis CC BY-SA 4.0