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Standard deviation of residuals different from manual calculation

Pablo Gil
Posted 1 year ago

Please can someone explain why pmSD1 and pmSD2 are different?

p = Predict[{1.2, 2, 3.5, 4, 5.2} -> {2.3, 1.2, 5.5, 7.7, 9.9}];
pm = PredictorMeasurements[p, {1.4 -> 2.5, 4.5 -> 8.6, 5.4 -> 10.2, 3.3 -> 3}];
pmSD1 = pm["StandardDeviation"]
pmSD2 = Sqrt[Total[pm["Residuals"]^2]/2]
Information[p, "Method"]
POSTED BY: Pablo Gil
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Pablo Gil
Pablo Gil
Posted 1 year ago

Many Thanks Jim, also can you tell how the baseline is computed?
POSTED BY: Pablo Gil
Jim Baldwin
Jim Baldwin
Posted 1 year ago

The documentation states

"StandardDeviationBaseline" standard deviation of test set values

And the formula used divides the sum of squares by the length of the test set values rather than the length of the test set values minus 1 (just as "StandardDeviation" does for a a PredictorMeasurementsObject).

The documentation isn't always very explicit but it is the first place to start and trying examples can sometimes help one figure out what formulas are being used.
POSTED BY: Jim Baldwin
Jim Baldwin
Jim Baldwin
Posted 1 year ago

pm["StandardDeviation"] uses the following formula:

Sqrt[Total[pm["Residuals"]^2]/Length[pm["Residuals"]]]

Here is a second example:

p = Predict[{1 -> 2, 3 -> 4.5, 5 -> 6, 7 -> 8.5}];
pm = PredictorMeasurements[p, {1.5 -> 2, 4 -> 5, 6 -> 5.5}];
pm["StandardDeviation"]
(* 1.12284 *)
Sqrt[Total[pm["Residuals"]^2]/Length[pm["Residuals"]]]
(* 1.12284 *)
POSTED BY: Jim Baldwin
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