Re: Curve-fitting/data analysis question...

• To: mathgroup at smc.vnet.net
• Subject: [mg73403] Re: Curve-fitting/data analysis question...
• From: "sherifffruitfly" <sherifffruitfly at gmail.com>
• Date: Thu, 15 Feb 2007 05:06:34 -0500 (EST)
• References: <200702110615.BAA11578@smc.vnet.net><equnhn\$ife\$1@smc.vnet.net>

```On Feb 14, 2:16 am, "Robert Dodier" <robert.dod... at gmail.com> wrote:
> sherifffruitfly wrote:
> > For each point p in my interval, I constructed two models - a linear
> > one using the points to the LEFT of p, and an exponential one using
> > the points to the RIGHT of p. I ran chi-squared to see which model p
> > belongs to. At the beginning, LEFT wins pretty much every time. When
> > RIGHT started winning is where I called the cut-off.
>
> Better still would be to choose the cut-off point to minimize the sum
> of the goodness of fit (i.e., chi-square) on both sides.
> (It's not a question of which side fits better, but rather how to get
> better fit overall.) Maybe you're already doing that; I can't tell for
> sure.

Good idea. At best, I was doing a very ghetto version of your idea.

> I would be interested to hear whether the total chi-square is
> sensitive
> to the placement of the cut-off point. Do you find that there is one
> point which is much better than the rest, or is there a range of
> points which are all more or less OK?

I haven't noticed that - but that doesn't imply that it isn't there to
be noticed.

I *have* however noticed the following. For decreasing x, it's
difficult to distinguish an exponential from a constant function. So
really the bulk of the goodness-of-fit work is being done for
increasing x of the linear models - linear models fail dramatically
when the data becomes "exponential-ish".

Is it reasonable to apply the above thought to your suggestion of
minimizing the total goodness of fit by way of somehow weighting the
linear part more heavily than the exponential, and then minimizing? If
that is reasonable, are there any guiding heuristics? (I've never had
to answer this sort of problem before - or even seen one.)

> HTH
> Robert Dodier

Thanks!

cdj

```

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