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SinglePredictionErrors vs. SinglePredictionBands/2

Hi, all!

I cannot understand meanings of some LinearModelFit outputs.

Attachments:
LinearModelFit_problem.nb
POSTED BY: Vladimir Ivanov
6 Replies
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OK. Not perfectly clear, but I've collected some material to think about. Thank you.
POSTED BY: Vladimir Ivanov
Jim Baldwin
Jim Baldwin
Posted 4 years ago

Sorry, not true. What I was showing was that you could get the standard error for any value of x using fit["SinglePredictionBands"]. (You take the difference in the upper and lower values and divide by twice the t cut-off value.)

fit["SinglePredictionErrors"] only gets you the standard errors associated with the x values in your data. In other words, fit["SinglePredictionErrors"][[3]] gets you the standard error associated with the third element in your data. That just also happens to be x=3 for your data. But in general that certainly won't be the case.
POSTED BY: Jim Baldwin

It seems that you've made a misprint. In your example

fit["SinglePredictionErrors"][[3]]
(* 0.596900473 *)

and

Differences[fit["SinglePredictionBands"] /. x -> x0][[1]]/(2 t)
(* 0.596900473 *)

provide the same value. I guess in the last line must be

Differences[fit["SinglePredictionBands"] /. x -> x0][[1]]/2
(* 0.636611319 *)

If that is true, then I get the idea.
POSTED BY: Vladimir Ivanov
POSTED BY: Vladimir Ivanov
Jim Baldwin
Jim Baldwin
Posted 4 years ago

The standard error of prediction is not the same as the halfwidth of the 68.2689% confidence interval when the t-distribution is involved. One is $s$ and the other is $t \times s$. In your case the $t$-value is close to 1 so for large enough sample sizes, the values are similar.

If you had a huge sample size (i.e., you're dealing with a normal distribution rather than a t-distribution), then those two values are essentially the same.

If you just want the standard errors for the observed data points, then 

fit["SinglePredictionErrors"]

will get that for you. If you want an estimate of the standard error for a value of the predictor being x0, then the following will get that for you:

CLOneSigma = CDF[NormalDistribution[0, 1], 1] - CDF[NormalDistribution[0, 1], -1] // N;
fit = LinearModelFit[data, {1, x}, x, ConfidenceLevel -> CLOneSigma];
t = InverseCDF[StudentTDistribution[8], 1 - (1 - CLOneSigma)/2];
x0 = 3;
Differences[fit["SinglePredictionBands"] /. x -> x0][[1]]/(2 t)
POSTED BY: Jim Baldwin
Jim Baldwin
Jim Baldwin
Posted 4 years ago

That's because SinglePredictionErrors gives you the estimates of the standard error and the halfwidth of the confidence interval gives you the t-value times the estimate of the standard error.

fit["SinglePredictionBands"] /. x -> # & /@ data[[All, 1]]
(* {{0.455359, 1.81737}, {1.45006, 2.76207}, {2.43915, 3.71237}, {3.42211, 4.6688}, {4.39854, 5.63176}, 
{5.36824, 6.60146}, {6.3312, 7.57789}, {7.28763, 8.56085}, {8.23793, 9.54994}, {9.18263, 10.5446}}*)

fit["SinglePredictionConfidenceIntervals"]
(* {{0.455359, 1.81737}, {1.45006, 2.76207}, {2.43915, 3.71237}, {3.42211, 4.6688}, {4.39854, 5.63176}, 
{5.36824, 6.60146}, {6.3312, 7.57789}, {7.28763, 8.56085}, {8.23793, 9.54994}, {9.18263, 10.5446}}*)

t = InverseCDF[StudentTDistribution[8], 1 - (1 - CLOneSigma)/2];
fit["PredictedResponse"] + (# {-1, 1} & /@ (t*fit["SinglePredictionErrors"]))
(* {{0.455359, 1.81737}, {1.45006, 2.76207}, {2.43915, 3.71237}, {3.42211, 4.6688}, {4.39854, 5.63176}, 
{5.36824, 6.60146}, {6.3312, 7.57789}, {7.28763, 8.56085}, {8.23793, 9.54994}, {9.18263, 10.5446}}*)

or equivalently

spe1 = fit["SinglePredictionErrors"];
t = InverseCDF[StudentTDistribution[8], 1 - (1 - CLOneSigma)/2];
spe2 = Table[((#[[2]] - #[[1]])/(2 t)) &[fit["SinglePredictionBands"] /. x -> data[[i, 1]]], {i, Length[data]}];
spe2/spe1
(* {1., 1., 1., 1., 1., 1., 1., 1., 1., 1.} *)
POSTED BY: Jim Baldwin
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