Timeline for Why do you take the sqrt of 1/n for RMSE?
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17 events
| when toggle format | what | by | license | comment | |
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| Jun 3, 2022 at 18:27 | vote | accept | Circadian | ||
| Jun 2, 2022 at 22:20 | comment | added | Henry | $\sqrt{\frac{1}{n}\Sigma_{i=1}^{n}{\Big(\hat{y}_i -y_i\Big)^2}}$ is an measurement of how spread out the sample is and a reasonable estimator of how spread out the population is. If you took a sample with four times as many observations from the same population, you would typically get approximately the same result. $\frac{1}{n}\sqrt{\Sigma_{i=1}^{n}{\Big(\hat{y}_i -y_i\Big)^2}}$ multiplies the previous number by $\frac1{\sqrt n}$ and would tend to be smaller with a larger sample size, and is instead an estimator for the uncertainty in the sample mean. | |
| Jun 2, 2022 at 10:20 | answer | added | Corvus | timeline score: 5 | |
| Jun 2, 2022 at 7:16 | answer | added | Amin Shn | timeline score: 5 | |
| Jun 2, 2022 at 7:05 | comment | added | Nat | The expression for $`` MRSE "$ seems off: the mean-root-square-error would be$$\text{MRSE} = {\frac{1}{n}} \sum_{\forall i}{\sqrt{{\left(\hat{y}_i-y_i\right)}^{2}}} \,,$$with the thing being that the "mean" involves adding the stuff up and dividing through by the count. This would be least absolute deviations (LAD). | |
| Jun 2, 2022 at 6:00 | history | tweeted | twitter.com/StackStats/status/1532240596869447680 | ||
| Jun 2, 2022 at 4:42 | history | became hot network question | |||
| Jun 2, 2022 at 2:31 | answer | added | BehrouzB | timeline score: 0 | |
| Jun 1, 2022 at 23:51 | comment | added | Circadian | @whuber, I found an answer in your response to a similar question about the definition of standard deviation. For those interested: stats.stackexchange.com/questions/116342/… | |
| Jun 1, 2022 at 22:55 | comment | added | Circadian | @whuber, in the second example wouldn't multiplying the root by 1/n be considered taking the average of the root squared error? | |
| Jun 1, 2022 at 22:47 | history | edited | Circadian | CC BY-SA 4.0 |
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| Jun 1, 2022 at 22:05 | history | edited | Circadian | CC BY-SA 4.0 |
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| Jun 1, 2022 at 21:52 | comment | added | whuber♦ | You could do this. It would be a perfectly valid measure of residual size in any single instance. The problem is revealed when you consider what value to expect for its square. In the first case, you would be looking at the average squared residual. No matter what $n$ is, that expectation would be about the same. But in the second case you would be looking at $1/n$ times the average squared residual--and that gets really small as $n$ grows. Thus, it wouldn't be meaningful to compare the (modified) RMSEs of two datasets of different sizes. That wouldn't be terribly useful, would it? | |
| Jun 1, 2022 at 20:54 | answer | added | Demetri Pananos | timeline score: 18 | |
| Jun 1, 2022 at 20:51 | comment | added | Tylerr | Because it is the Root of the Mean Squared Error (thus RMSE) and MSE is defined as the stuff under the square root. | |
| S Jun 1, 2022 at 20:42 | review | First questions | |||
| Jun 1, 2022 at 20:57 | |||||
| S Jun 1, 2022 at 20:42 | history | asked | Circadian | CC BY-SA 4.0 |