Fake fits

A few weeks ago I committed an act of deception. I needed a graph showing a few sets of data points, with curves fitted to them. This was strictly for show; neither the points nor the curves would mean anything; it just had to look good. So, out of laziness, I did it backwards: I drew a few arbitrary curves, chose some points on the curves, and then perturbed those points by small random displacements. The result is Figure 2 in Home Baked Graphics.

Apparently I have gotten away with this fakery, since no one has complained. But it nags at my conscience. The curves shown are not fitted to the given data points, but the other way around. And barring an extraordinary coincidence, we can’t expect the curves to give the best possible trend through the points.

I began to wonder: What if you iterate this process? In other words, you start out by generating a curve, say a simple quadratic:


You select a few points on the curve, in this case the five points with x coordinates equal to {1, 2, 3, 4, 5}. You perturb the y coordinates of these points by choosing random variates from a normal distribution with unit standard deviation:


Now you do a least-squares fit to the five selected points, generating a new quadratic:


At this stage you can start over, selecting five points on the new curve and randomly displacing them:


And a least-squares fit to those five purple points yields yet another quadratic curve:


So what happens if we continue in this way, repeatedly jiggling a few points on a curve and then fitting a new curve to the jiggled points? There’s a seeming hint of convergence in the three curves we’ve seen so far. They look like they might be approaching some fixed trajectory. Could that possibly be true?

I’ll post an update in a few days to say what more I’ve been able to learn about this question. If you just can’t wait, click here for a sneak peek.

The promised update (2010-04-02): You can’t fool anybody in this crowd, not even circa April 1. Of course there is no convergence to a limiting curve.


The butterfly above shows 100 quadratic curves, selected from 100,000 iterations of the least-squares fitting process, with that process applied at points whose x values are in the set {–5, –4, –3, –2, –1, 0, 1, 2, 3, 4, 5}. Clearly enough, the coefficients of the quadratic curves are wandering widely; in particular note that the x2 term can be either positive or negative, yielding curves that are concave upward or downward.

On the other hand, these are not just any old quadratic curves. They all have their apex (the point where f′(x) = 0) fairly close to the origin. Here’s a closer view:


(When Ros glanced at this graph on the screen, her comment was: “Bad hair day.”) And here’s a tracing of the wandering of the apex (taken from a different, longer, run of the program):


The path has the look of a classic random walk, but note that it’s confined to a narrow range of x values. What attractive force holds the apex in this region? It’s the decision to fit the curves to points in the interval –5 ≤ x ≤ 5. Here’s what happens if we instead fit to points in the range 100 ≤ x ≤ 110:


And note that without the symmetry-breaking x1 term, the curve would always be a parabola with its apex on the y axis; only the curve’s shape (or scale) and the height of the y intercept could vary.

This whole process of repeatedly fitting curves to data and then deriving new data from the fitted curves is weird and useless enough that it seems unlikely to appear anywhere in the universe outside my own idle mind. And yet…. Think of the Dow-Jones index, which is calculated as a weighted average of the prices of 30 selected stocks. They are not always the same 30 stocks. Indeed, recent years have seen quite a lot of turnover in the components of the Dow: AIG was replaced by Kraft Foods; General Motors was dropped in favor of Cisco Systems; Honeywell lost out to Chevron. Would it be too utterly cynical of me to suggest that what’s going on here resembles a process of fitting a curve to randomly chosen points, then choosing more points that lie near the curve?

This entry was posted in mathematics, statistics.

6 Responses to Fake fits

  1. Barry Cipra says:

    OK, I admit to cheating, I went straight to the sneak peek before thinking about the question. But it seems clear (which means I’m likely wrong…) that you’re setting yourself up for some sort of random walk in 3 dimensions (the 3 coefficients of your quadratic curves). I would expect the coefficients to slowly “diffuse” into larger and larger values, with the rate of diffusion inversely proportional to something like the number of (equi-spaced) jiggled points. It’s certainly true that if you jiggle one point on a constant curve (aka, horizontal line) and re-fit the “curve” to the jiggled point, all you’re doing is taking a random walk in 1-d.

  2. ivansml says:

    Interesting problem (and a nice opportunity to procrastinate a little…):

    I also think it should be a random walk (though with correlated increments): say that at t-th round, you have vector of coefficients b_t, and generate vector of y-coordinates as y = X * b_t + e, where X is your data matrix (so in your case, first column would be vector of 1′s, second column would be vector (1,2,3,4,5) and the third column (1,4,9,16,25)), and e is vector of random errors. Then, OLS estimate of coefficients will be:

    b_{t+1} = (X’ * X)^{-1} * X’ * y = (X’ * X)^{-1} * X’ * (X * b_t + e) = b_t + (X’ * X)^{-1} * X’ * e = b_t + u,

    where u is a transformed error term: u = (X’ * X)^{-1} * X’ * e (so in general, if elements of e were independent, elements of u will be correlated). In your example, you pick the same x-coordinates, so X would stays constant in all rounds - if you randomized those as well, I think the result would be the same, only we would have X_t with time index, the transformation e -> u would be different in each round and thus the covariance matrix of errors u would be changing randomly at each round as well.

  3. Iain says:

    Using the same notation as the last comment, one can also write down the progression of the y’s:

    y_{t+1} = A * y_t + e_{t+1}, where A = X * (X’ * X) * X’.

    This is a “vector autoregression model”, specificially a VAR(1) model, http://en.wikipedia.org/wiki/Vector_autoregression . For those interested, a better starting point might be http://en.wikipedia.org/wiki/Autoregressive_model which involves less linear algebra.

  4. Iain says:

    Another comment: the observation of growing coefficients (apparent from the process identified in ivansml’s post) might provide some intuition as to why regularization or shrinkage can make sense.

  5. Iain says:

    Correction, with apologies for multiple comments. I missed out a superscript: A = X * (X’ * X)^{-1} * X’

  6. unekdoud says:

    What happens if the coefficients of the quadratic are perturbed, rather than the y-coordinates of points on it?
    Another variation: What if the y-coordinates are randomly scaled up or down by small factors (say about 5%), rather than simply adding a unit standard normal? (I believe the effect would be different for different choice of X)