The formulas are equivalent:
$$\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} =
\sqrt{\frac{\sigma_1^2}{\sqrt{n_1}^2} + \frac{\sigma_2^2}{\sqrt{n_2}^2}} =
\sqrt{\left(\frac{\sigma_1}{\sqrt{n_1}}\right)^2 + \left(\frac{\sigma_2}{\sqrt{n_2}}\right)^2} =
\sqrt{SE_1^2 + SE_2^2}$$
- First equality should be clear.
- Second equality is just pulling the square outside the fraction.
- Third equality is the formula for standard error of the mean. (also given in the question)
Hmm, so I think I figured out the issues I was having with segmented. It had to do with the weight statement (it doesn't work to weight the lm and the segmented model).
Segmented seems like the best bet for me.
It does a good job estimating breakpoints, even with my short datasets. It isn't terribly difficult to constrain the second slope to 0, and it has a built-in test for the significance of the change in slope. If anyone feels like explaining the difficulty with estimating error around a breakpoint estimate, I'm all ears!
library(segmented)
with second segment not constrained to 0
out.lm <- lm(y~x, data=dat)
o<-segmented(out.lm, seg.Z= ~x, weights=wt, psi=10)
with(dat, plot(x,y, ylim=c(0, max(y)), pch=16, cex=wt/(13), main="segmented"))
lines(x=dat$x, y=fitted(o), col="blue")
lines.segmented(o, col="blue")
fix from vito to fix slope to 0
o2<-lm(y~1)
xx<- -x
o3<-segmented(o2,seg.Z=~xx, weights=wt, psi=list(xx=-4))
points(x,fitted(o3),col="green")
ci<-confint(o3, rev.sgn=TRUE)
lines.segmented(o3, col="green", rev.sgn=TRUE, lwd=2)
test for slope change
davies.test(o3,~xx)
Best Answer
Caveat: I'm not an expert in this field.
If the theoretical distribution of sample impulse response function (IRF) is Gaussian (that is, at every time point the distribution of errors is Gaussian) then 1.96 standard errors covers about 95.0% of the distribution of the sample IRF, and 2 standard errors covers about 95.4% of the distribution of the sample IRF. As Michael Chernick pointed out in the comments, this is "close enough" that people use ±2 standard errors to mean "95% theoretical coverage". Note that the IRF for a typical Vector Autoregression model is asymptotically Gaussian [1].
If you imagine testing, at every time point, the hypothesis that the IRF is zero against the alternative that the IRF is nonzero, and if you assume that the test statistic has a Gaussian distribution, then a theoretical 95% confidence interval is equivalent to the 95% theoretical coverage interval described above.
So under a typical set of assumptions, they're more or less the same. But there are several ways to compute a confidence interval for an impulse response function [2], including simulation-based methods. These could in principle lead to a 95% confidence interval that does not line up with the theoretical Gaussian 95% coverage interval.
[1] Cf. lecture notes by Luca Gambetti, http://pareto.uab.es/lgambetti/VAR_Forecasting.pdf, citing: Hamilton, James D., 1994, Time Series Analysis, Princeton University Press.
[2] Griffiths, William and Lütkepohl, Helmut. (1990). Confidence intervals for impulse response functions: a comparison of asymptotic theory and simulation approaches. https://www.une.edu.au/__data/assets/pdf_file/0018/20484/emetwp42.pdf.