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POFinding the best trade-off point on a curve
 

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  1. POFinding the best trade-off point on a curve
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    <p>Say I had some data, for which I want to fit a parametrized model over it. My goal is to find the best value for this model parameter.</p> <p>I'm doing model selection using a <a href="http://en.wikipedia.org/wiki/Akaike_information_criterion" rel="noreferrer">AIC</a>/<a href="http://en.wikipedia.org/wiki/Bayesian_information_criterion" rel="noreferrer">BIC</a>/<a href="http://en.wikipedia.org/wiki/Minimum_description_length" rel="noreferrer">MDL</a> type of criterion which rewards models with low error but also penalizes models with high complexity (we're seeking the simplest yet most convincing explanation for this data so to speak, a la <a href="http://en.wikipedia.org/wiki/Occam%27s_Razor" rel="noreferrer">Occam's razor</a>).</p> <p>Following the above, this is an example of the sort of things I get for three different criteria (two are to be minimized, and one to be maximized):</p> <p><img src="https://i.stack.imgur.com/iBgo3.png" alt="aic-bic"> <img src="https://i.stack.imgur.com/AG52Y.png" alt="fit"></p> <p>Visually you can easily see the elbow shape and you would pick a value for the parameter somewhere in that region. The problem is that I'm doing do this for large number of experiments and I need a way to find this value without intervention.</p> <p>My first intuition was to try to draw a line at 45 degrees angle from the corner and keep moving it until it intersect the curve, but that's easier said than done :) Also it can miss the region of interest if the curve is somewhat skewed.</p> <p>Any thoughts on how to implement this, or better ideas?</p> <p>Here's the samples needed to reproduce one of the plots above:</p> <pre><code>curve = [8.4663 8.3457 5.4507 5.3275 4.8305 4.7895 4.6889 4.6833 4.6819 4.6542 4.6501 4.6287 4.6162 4.585 4.5535 4.5134 4.474 4.4089 4.3797 4.3494 4.3268 4.3218 4.3206 4.3206 4.3203 4.2975 4.2864 4.2821 4.2544 4.2288 4.2281 4.2265 4.2226 4.2206 4.2146 4.2144 4.2114 4.1923 4.19 4.1894 4.1785 4.178 4.1694 4.1694 4.1694 4.1556 4.1498 4.1498 4.1357 4.1222 4.1222 4.1217 4.1192 4.1178 4.1139 4.1135 4.1125 4.1035 4.1025 4.1023 4.0971 4.0969 4.0915 4.0915 4.0914 4.0836 4.0804 4.0803 4.0722 4.065 4.065 4.0649 4.0644 4.0637 4.0616 4.0616 4.061 4.0572 4.0563 4.056 4.0545 4.0545 4.0522 4.0519 4.0514 4.0484 4.0467 4.0463 4.0422 4.0392 4.0388 4.0385 4.0385 4.0383 4.038 4.0379 4.0375 4.0364 4.0353 4.0344]; plot(1:100, curve) </code></pre> <hr> <h2>EDIT</h2> <p>I accepted the solution given by <a href="https://stackoverflow.com/questions/2018178/finding-the-best-trade-off-point-on-a-curve/2022348#2022348">Jonas</a>. Basically, for each point <code>p</code> on the curve, we find the one with the maximum distance <code>d</code> given by:</p> <p><img src="https://i.stack.imgur.com/IoRbN.png" alt="point-line-distance"></p>
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    1. COI was going to say draw a 45 degree line too :\
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    2. COHow much do you expect your graphs to deviate from the general shapes in the above examples? In other words, do you expect that the "elbow" of the graph will always be near the same corner of the graph?
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    3. CO@Amro: Is it OK to have only 6 points out the 100 in the upper part curve? Try `plot(1:100,curve,'.')`
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