Timeline for Is the following inequality correct? How to prove it?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Aug 10, 2018 at 1:51 | vote | accept | Harry | ||
| Apr 19, 2017 at 14:25 | comment | added | Harry | @Manuel Fazio I am little confused about the correctness of $P(X \ge a_0 | Y \le b_0) \le f(y\le b_0)$. How can we arrive at this from the first inequality in the question? | |
| S Apr 19, 2017 at 14:16 | history | edited | overdisperse | CC BY-SA 3.0 |
change the answer for the current notations
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| S Apr 19, 2017 at 14:16 | history | suggested | Harry | CC BY-SA 3.0 |
change the answer for the current notations
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| Apr 19, 2017 at 14:12 | review | Suggested edits | |||
| S Apr 19, 2017 at 14:16 | |||||
| Apr 19, 2017 at 14:12 | comment | added | overdisperse | @ZHANGWei To be more precise, we should write $P(X \ge a_0 | Y = y) \le f(y), y \le b_0$. This way we avoid making it look like the value of $f$ is random. | |
| Apr 19, 2017 at 14:04 | comment | added | overdisperse | I always saw books made that exception when giving the formula for conditional probability. I never was too sure if it was just to keep zero out of the denominator or if allowing that possibility could have more dire implications that were merely glossed over. Given your comment, I'm assuming it was the former all along! | |
| Apr 19, 2017 at 13:59 | comment | added | Harry | Thank you very much for your answer. The first inequality in the question should be $\textbf{Pr}(X \ge a_0 | Y=y)\le f(y)$, which is more accurate. How can we obtain $\textbf{Pr}(X \ge a_0 | Y \le b_0) \le f(Y \le b_0)$? What does $f(Y \le b_0)$ mean? | |
| Apr 19, 2017 at 13:53 | comment | added | whuber♦ | This looks like a fine answer, regardless of whether $P(y\le b_0)$ is nonzero. | |
| Apr 19, 2017 at 13:40 | history | answered | overdisperse | CC BY-SA 3.0 |