Solved – What are the theoretical reasons for why extrapolation “less reliable” than interpolation

extrapolationinterpolation

Extrapolation is in general "unreliable". (See "What is wrong with extrapolation?")

But it is also commonly said that extrapolation is "less reliable" than interpolation.

But why should we generally assume that the model is "more reliable" between two known data points than to the right of the right-most data point (or the left of the left-most data point)?

From empirical examples, I can see that indeed, interpolation is often "more reliable" than extrapolation. But is there a more formal, theoretical justification for why this assertion is, in general, true?

Or is it just a purely empirical observation that interpolation tends to be "better"?

Best Answer

It is a theoretical result, at least for linear regression. Indeed, if one computes the so-called ''prediction error'' (see this link, slide 11), one can easily see that the further the independent variable $x$ is away from the sample average $\bar{x}$ (and for extrapolation one may be far away), the larger the prediction error. In the link that I referred to one can also see that in a graphical way.

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Solved – How to simplify a stretched exponential fit

The first thing to do, if possible, is to take care of the heteroscedasticity. Notice how the spread of the residuals consistently increases with the fit: in fact, the spread seems to increase almost quadratically with larger fit.

A standard cure is to return to the original response ($log(xy)$) and apply a strong transformation, such as a logarithm or even a reciprocal: something in that range is suggested by this pattern of heteroscedasticity. Then redo the fitting and recheck the residuals.

It's a good idea to fit lines by eye, using graphs of transformed $xy$ against $z$ (or $\log(z)$. This usually reveals more than any amount of manipulating a regression routine. Once you have a suitable model, you can finally use least squares (or robust regression) to produce a final fit.

In this instance you might also want to explore the relationships among $x$ and $z$ and $y$ and $z$ separately to see whether just one of $x$, $y$ is causing the sudden change in slope between 2.9 and 3.6. The change clearly is not quadratic: both "limbs" of the residual plot are linear. One way to model this change--if it persists after you have dealt with the heteroscedasticity--is with a changepoint model that posits one value of the slope $\beta_1$ for, say, $z \le 3.2$, and a different value for $z \gt 3.2$.

Finally, in Monte-Carlo simulations you have full control and you know exactly the mechanism that produces the responses. It can be useful to subject this to some analysis to find out just what the correct relationship among the triples ought to be.

Solved – Fourier/trigonometric interpolation

I am no expert on Fourier transforms, but...

Epstein's total sample range was 24 months with a monthly sample rate: 1/12 years. Your sample range is 835 weeks. If your goal is to estimate the average for one year with data from ~16 years based on daily data you need a sample rate of 1/365 years. So substitute 52 for 12, but first standardize units and expand your 835 weeks to 835*7 = 5845 days. However, if you only have weekly data points I suggest a sample rate of 52 with a bit depth of 16 or 17 for peak analysis, alternatively 32 or 33 for even/odd comparison. So the default input options include: 1) to use the weekly means (or the median absolute deviation, MAD, or something to that extent) or 2) to use the daily values, which provide a higher resolution.

Liebman et al. chose the cut-off point jmax = 2. Hence, Fig 3. contains fewer partials and is thus more symmetrical at the top of the sine compared to Fig 2. (A single partial at the base frequency would result in a pure sine wave.) If Epstein would have selected a higher resolution (e.g. jmax = 12) the transform would presumably only yield minor fluctuations with the additional components, or perhaps he lacked the computational power.

Through visual inspection of your data you appear to have 16-17 peaks. I would suggest you set jmax or the "bit depth" to either 6, 11, 16 or 17 (see figure) and compare the outputs. The higher the peaks, the more they contribute to the original complex waveform. So assuming a 17-band resolution or bit depth the 17th partial contributes minimally to the original waveform pattern compared to the 6th peak. However, with a 34 band-resolution you would detect a difference between even and odd peaks as suggested by the fairly constant valleys. The bit depth depends on your research question, whether you are interested in the peaks only or in both peaks and valleys, but also how exactly you wish to approximate the original series.

The Fourier analysis reduces your data points. If you were to inverse the function at a certain bit depth using a Fourier transform you could probably cross-check if the new mean estimates correspond to your original means. So, to answer your fourth question: the regression parameters you mentioned depend on the sensitivity and resolution that you require. If you do not wish for an exact fit, then by all means simply input the weekly means in the transform. However, beware that lower bit depth also reduces the data. For example, note how Epstein's harmonic overlay on Lieberman and colleagues' analysis misses the mid-point of the step function, with a skewed curve slightly to the right (i.e. temp. est. too high), in December in Figure 3.

Liebman and Colleagues' Parameters:

Bit Depth: 2

Epstein's Parameters:

Sample Rate: 12 [every month]
Sample Range: 24 months
Bit Depth: 6

Your Parameters:

Sample Rate: 365 [every day]
Sample Range: 5845 days

Exact Bit Depth Approach

Exact fit based on visual inspection. (If you have the power, just see what happens compared to lower bit-depths.)

Full Spectrum (peaks): 17
Full Spectrum (even/odd): 34

Variable Bit Depth Approach

This is probably what you wish to do:

Compare Peaks Only: 6, 11, 16, 17
Compare Even/Odd: 12, 22, 32, 34
Resynthesize and compare means

This approach would yield something similar to the comparison of Figures in Epstein if you inverse the transformation again, i.e. synthesise the partials into an approximation of the original time series. You could also compare the discrete points of the resynthesized curves to the mean values, perhaps even test for significant differences to indicate the sensitivity of your bit depth choice.

UPDATE 1:

Bit Depth

A bit - short for binary digit - is either 0 or 1. The bits 010101 would describe a square wave. The bit depth is 1 bit. To describe a saw wave you would need more bits: 0123210. The more complex a wave gets the more bits you need:

This is a somewhat simplified explanation, but the more complex a time series is, the more bits are required to model it. Actually, "1" is a sine wave component and not a square wave (a square wave is more like 3 2 1 0 - see attached figure). 0 bits would be a flat line. Information gets lost with reduction of bit depth. For example, CD-quality audio is usually 16 bit, but land-line phone quality audio is often around 8 bits.

Please read this image from left to right, focusing on the graphs:

FFT

You have actually just completed a power spectrum analysis (although at high resolution in your figure). Your next goal would be to figure out: How many components do I need in the power spectrum in order to accurately capture the means of the time series?

UPDATE 2

To Filter or not to Filter

I am not entirely sure how you would introduce the constraint in the regression as I am only familiar with interval constraints, but perhaps DSP is your solution. This is what I figured so far:

Step 1. Break down the series into sinus components through Fourier function on the complete data set (in days)
Step 2. Recreate the time series through an inverse Fourier transform, with the additional mean-constraint coupled to the original data: the deviations of the interpolations from the original means should cancel out each other (Harzallah, 1995).

My best guess is that you would have to introduce autoregression if I understand Harzallah (1995, Fig 2) correctly. So that would probably correspond to an infinite response filter (IIR)?

IIR http://paulbourke.net/miscellaneous/ar/

In summary:

Derive means from Raw data
Fourier Transform Raw data
Fourier Inverse Transform transformed Data.
Filter the result using IIR

Perhaps you could use an IIR filter without going through the Fourier analysis? The only advantage of the Fourier analysis as I see it is to isolate and determine which patterns are influential and how often they do reoccur (i.e. oscillate). You could then decide to filter out the ones that contribute less, for example using a narrow notch filter at the least contributing peak (or filter based on your own criteria). For starters, you could filter out the less contributing odd valleys that appear more like noise in the "signal". Noise is characterized by very few cases and no pattern. A comb filter at odd frequency components could reduce the noise - unless you find a pattern there.

Here's some arbitrary binning—for explanatory purposes only: Can you see the noise in the valleys?

Oops - There's an R Function for that!?

When searching for an IIR-filter I happen to discovered the R functions interpolate in the Signal package. Forget everything I said up to this point. The interpolations should work like Harzallah's: http://cran.r-project.org/web/packages/signal/signal.pdf

Play around with the functions. Should do the trick.

UPDATE 3

interp1 not interp

case.interp1 <- interp1(x=(ts.frame$no.influ.cases[!is.na(ts.frame$no.influ.case)]),y=ts.frame$yearday[!is.na(ts.frame$no.influ.case)],xi=mean(WEEKLYMEANSTABLE),method = c("cubic"))

Set xi to the original weekly means.

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