Timeline for Linear regression forecast underestimation
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| May 8, 2012 at 5:28 | comment | added | Rob Hyndman | OK @whuber. I've edited my answer accordingly. | |
| May 8, 2012 at 5:28 | history | edited | Rob Hyndman | CC BY-SA 3.0 |
added 57 characters in body
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| May 7, 2012 at 16:50 | comment | added | whuber♦ | The edited question is fine, Robert. My latest comments were only an effort to clarify the logic behind Rob Hyndman's solution. | |
| May 7, 2012 at 16:30 | comment | added | Robert Kubrick | @whuber Actually the intercept is already in the question. Adding the simple mean of the fitted and observed values would just confuse more I believe. | |
| May 7, 2012 at 16:15 | comment | added | Robert Kubrick | @whuber I think I see your point now. If the intercept is large enough relative to the slope of the regression, then the mean of both the fitted values and observed values will tend to be close to the intercept and the difference between the two means relatively small. I'm going to add the intercept and mean values in the question, then the answer should maybe edited for clarity? | |
| May 7, 2012 at 14:57 | comment | added | whuber♦ | Alas, that seems not to be the case, @Rob. When the means are zero, then yes, the mean absolute value is a measure of dispersion. But consider what happens when the shift is large enough to make all values (actual and predicted) positive. Now the mean absolute values coincide with the means, giving us no information about the dispersions (absent special parametric assumptions). Notice that no analog of the decomposition of mean square into squared mean plus variance exists for the $L^1$ (or other $L^p$) measures of dispersion. | |
| May 5, 2012 at 13:31 | comment | added | Rob Hyndman | @whuber. The mean of the predicted values is presumably close to the mean of the original data (identical if the predictions are actually fitted values). The mean of the absolute data is a measure of their dispersion shifted by some constant that depends on mean. The mean of absolute predictions is a measure of their dispersion shifted by roughly the same constant due to having roughly the same mean. Hence I thought this was essentially a comparison of the two dispersions. | |
| May 4, 2012 at 15:24 | comment | added | whuber♦ |
I'm hung up on this phrase, Robert: "The average of the absolute predicted values is significantly lower than the absolute observed values mean". To see why this is puzzling, consider what would happen if, say, you were to add $100$ to every y-value. The fit would be unchanged--it would merely increase the intercept by $100$. But the mean absolute value would increase by almost $100$. Thus you seem to be making a comparison that reveals nothing about the quality of the fit. I think you get it right in the edit where you use mean(abs(mydata$Y - lm1.predict)) instead.
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| May 4, 2012 at 14:58 | comment | added | Robert Kubrick | @whuber In that formula we're looking at the MAE, that is the average absolute error of the regression. I only added that value because I was trying to understand why the predicted values are lower that the actual observations. I'm not clear where is the puzzling part. | |
| May 4, 2012 at 14:48 | comment | added | whuber♦ | Look a few lines above the Edit, Rob. What's puzzling me is not the use of the mean absolute value of residuals as a proxy for residual dispersion, but its (explicit) comparison to the mean absolute value of the data ("the actual Y absolute values"), which is not a proxy for its dispersion except when it's known the mean data value is close to zero. | |
| May 4, 2012 at 14:44 | comment | added | Robert Kubrick | @whuber By calculating the mean of the absolute values I'm effectively looking at a proxy for variance. mean(abs(values)) simply give us a measure of data dispersion. I'm not clear why you're referring to the simple mean of mydata$Y. I don't see it reported in the question. | |
| May 4, 2012 at 13:58 | comment | added | whuber♦ |
Rob, how did you get from mean(abs(mydata$Y)) to the "variance of the observations"? There seems to be a missing step in your reasoning. I think the edited information plays a crucial role here in revealing that the mean observation is close to zero.
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| May 3, 2012 at 23:36 | vote | accept | Robert Kubrick | ||
| May 3, 2012 at 23:36 | comment | added | Robert Kubrick | Of course! The variance of the regression line is always going to be smaller than the variance of the outcomes. | |
| May 3, 2012 at 22:14 | history | answered | Rob Hyndman | CC BY-SA 3.0 |