Timeline for How to demonstrate a correlation inequality?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Dec 1, 2021 at 15:55 | comment | added | bdx77 | I don't think we have Cov(XZ, Y) = E(D(1-D))V(Y) with my construction. The covariance is positive (and equal to $m_1$ V(Y)) but small enough to ensure the final inequality. | |
| Dec 1, 2021 at 15:21 | comment | added | Mac Zhang | Hi, many thanks for your answer. I got the same conclusion but the computation process is different from you. As you have already defined X, Z. So we have XZ= [DY+(1-D)Y1][Dy2 + (1-D)Y]. Moreover, we have Cov(XZ, Y) = E(D(1-D))V(Y) = 0, so 0=ρ(XZ,Y)<ρ(X,Y)=p. | |
| Dec 1, 2021 at 14:49 | comment | added | bdx77 | Yes, sorry, I first misread your problem. I have completely modified my answer. | |
| S Dec 1, 2021 at 14:48 | review | First answers | |||
| Dec 1, 2021 at 15:12 | |||||
| S Dec 1, 2021 at 14:48 | history | edited | bdx77 | CC BY-SA 4.0 |
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| Dec 1, 2021 at 14:45 | comment | added | Mac Zhang | Based on your assumption, we have ρ(XZ,Y) >= 0 = ρ(X,Y). (When E(x) =0， ρ(XZ,Y)=ρ(X,Y)). The inequality still holds. | |
| S Dec 1, 2021 at 13:21 | review | First answers | |||
| Dec 1, 2021 at 13:32 | |||||
| S Dec 1, 2021 at 13:21 | history | edited | bdx77 | CC BY-SA 4.0 |
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| S Dec 1, 2021 at 12:15 | review | First answers | |||
| Dec 1, 2021 at 12:37 | |||||
| S Dec 1, 2021 at 12:15 | history | answered | bdx77 | CC BY-SA 4.0 |