This may be a very simple question, but what does it mean "to fit a back-to-back Weibull distribution" to a residual series (or I suppose, to any arbitrary data)?

I think it means to fit a series like this:

$x_{t} = \sum_{k}\alpha_{k}\frac{k}{\lambda}\left(\frac{x}{\lambda}\right)^{k-1}e^{\left(\frac{x}{\lambda}\right)^{k}}$

but I don't really know.


From a problem in Cowpertwait's book "Introductory Time Series with R":

A hydrologist wishes to simulate monthly inflows to the Font Reservoir over the next 10-year period. Use the data in Font.dat (§2.3.3) to answer the following:

Plot a histogram of the residual errors of the fitted AR model, and comment on the plot. Fit a back-to-back Weibull distribution to the residual errors.


The residuals aren't symmetric around zero, and Cowpertwait next asks to simulate the distribution. Fitting a single distribution (and then randomly assigning each draw to be positive or negative) would generate a symmetric distribution around zero, which doesn't match the data. Instead, the intent may be to fit two Weibull distributions which are 'back-to-back' around zero. Split x into x1 > 0 and x2 < 0. Multiply x2 by -1. Then fit a separate Weibull to x1 and x2.

Example R code:

pos.resids <- resids[resids > 0]  # subset the positive residuals
neg.resids <- resids[resids < 0]  # subset the negative residuals
neg.resids <- neg.resids * -1  # transform the negative residuals
pos.Weibull <- fitdistr(pos.resids, "weibull")  # fit a Weibull to the positive residuals
neg.Weibull <- fitdistr(neg.resids, "weibull")  # fit a Weibull to the (transformed) negative reisduals

Since the residuals could be positive or negative, it could mean (just a guess) that instead of a density $f(x) =\frac{k}{\lambda}\left(\frac{x}{\lambda}\right)^{k-1}e^{\left(\frac{x}{\lambda}\right)^{k}}$ just for positive $x$, you might instead look at fitting a density for all real $x$ of $$f(x) =\frac{k}{2\lambda}\left(\frac{\left|x\right|}{\lambda}\right)^{k-1}e^{\left(\frac{\left|x\right|}{\lambda}\right)^{k}}.$$


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