Hi, all!
I cannot understand meanings of some LinearModelFit outputs.
LinearModelFit_problem.nb (11.9 KB)
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.} *)
Thank you for your answer, Jim.
But, frankly, I’m not a guru in statistic. I want to understand a simple thing: what value to use for estimate of the standard error of a prediction. Can I state it as follows: I need to use the halfwidth of the confidence interval, rather than the estimate of the standard error, since the former takes into account the finite size of my series, and the latter doesn’t?
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)
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.
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.
OK. Not perfectly clear, but I’ve collected some material to think about. Thank you.