Timeline for Estimate an expression about probability about Bernoulli random variables

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Jun 17, 2020 at 16:11	comment	added	Jianrong Li		thank you very much!
Jun 17, 2020 at 15:58	comment	added	Iosif Pinelis		@JianrongLi : Oops! I did indeed miss $3/4$; this is now corrected -- thank you for pointing this out. As for your detalization of the multi-line display before (3), what I had in mind was something like this, but a bit simpler: just using the fact that $\sum\limits_t P(X_M=t)=1$.
Jun 17, 2020 at 15:49	history	edited	Iosif Pinelis	CC BY-SA 4.0	deleted 2 characters in body
Jun 17, 2020 at 15:16	comment	added	Jianrong Li		sorry, I mean when $k \ge 1$, $D_k \le 3/4$.
Jun 17, 2020 at 10:34	comment	added	Jianrong Li		thank you very much for your great answer! I am trying to understand every step of your proof. In the formula before (3), it is said that $\sum_z P(X_{J \cap K}=z) \sum_y P(X_{K\backslash J}=y-z)=1$. I am trying to understand this step. I think that $$\sum_z P(X_{J \cap K}=z) \sum_y P(X_{K\backslash J}=y-z) \\=\sum_z P(X_{J \cap K}=z) \sum_y P(X_{K\backslash J}=y) = \sum_{y,z} P(X_{J \cap K}=z) P(X_{K\backslash J}=y) \\=\sum_{y,z}P(X_{K}=y+z)=\sum_y P(X_{K}=y)=1$$. Is my understanding correct? I checked that when $k=2$, $D_k=3/4$. So maybe for $k \ge 1$, $D_k \ge 3/4$?
Jun 17, 2020 at 7:53	vote	accept	Jianrong Li
Jun 17, 2020 at 0:12	history	edited	Iosif Pinelis	CC BY-SA 4.0	deleted 42 characters in body
Jun 17, 2020 at 0:02	history	answered	Iosif Pinelis	CC BY-SA 4.0