Timeline for Justification of the use of residual plot
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jan 6, 2023 at 13:21 | comment | added | Iosif Pinelis | @Cheng-Yu : Thank you for your appreciation. | |
| Jan 6, 2023 at 6:13 | vote | accept | Cheng-Yu | ||
| Jan 6, 2023 at 6:12 | comment | added | Cheng-Yu | @losif Thank you very much! This answer is very helpful. I'm amazed by your problem solving skills. It's a shame that I don't have enough reputations to upvote your answer. | |
| Jan 6, 2023 at 4:52 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jan 6, 2023 at 4:28 | comment | added | Iosif Pinelis | @Cheng-Yu : I have added details on the $L^2$ thing. | |
| Jan 6, 2023 at 4:27 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jan 6, 2023 at 2:13 | comment | added | Cheng-Yu | And, would you mind elaborating how this step could be done? It seems not that easy to me. | |
| Jan 6, 2023 at 2:03 | comment | added | Cheng-Yu | Thanks for your clear explanation. It seems that the critical step to show the second result (converge in $L^2$), is showing that $\mathbb{E}[(g_n(X) - f(X))^2] < \eta \Longrightarrow |Cov[g_n(X), \epsilon_n]| < $ some_function($\eta$). So, if I define the goodness of approximation as $\mathbb{E}[(g_n(X) - f(X))^2] < \eta$, I can get a positive result? | |
| Jan 6, 2023 at 1:04 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jan 6, 2023 at 0:54 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |