Timeline for Fitting a curve to the edge of a distribution
Current License: CC BY-SA 3.0
13 events
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| Jan 18, 2012 at 0:45 | comment | added | whuber♦ | On the contrary, I noted these are bivariate data, as your description confirms. It is really strange that the variance of the measurements would increase as the measurements themselves decrease. This suggests that both $x$ and $y$ are actually differences relative to a large value (perhaps around $100$ or so). It would be more productive and informative to work with the original $(x,y)$ data. | |
| Jan 3, 2012 at 9:13 | comment | added | xboxrob | Hi all, Firstly thank you for your responses. The y-axis is a measurement of difference between two measurements. The x-axis is the minimum value of the two measurements. As you have noted this is univariate data and hence when x is small there is much larger differences by random chance. The idea therefore is that points that are far away from the edge of the distribution (roughly the red line) are actually important values as the difference between the two measurements should be more believable | |
| Dec 19, 2011 at 2:05 | comment | added | whuber♦ | xboxrob, it is unclear what you are asking for, and consequently the replies you are getting may or may not be applicable; at any rate, future readers will not be able to gauge whether they are appropriate solutions. The K-S test does not seem applicable--you display bivariate data and it's for univariate data only--and finding outliers does not seem related to your red curve. What exactly do you want to accomplish? | |
| Dec 18, 2011 at 23:01 | answer | added | EDi | timeline score: 1 | |
| Dec 18, 2011 at 7:49 | answer | added | athula herath | timeline score: 0 | |
| Dec 17, 2011 at 8:23 | history | tweeted | twitter.com/#!/StackStats/status/147955180678283264 | ||
| Dec 17, 2011 at 6:59 | comment | added | Xi'an | Yes, this is a weird one: if your points are random, their boundary is not a density function... There is a large literature on non-parametric support estimation, see e.g. [Devroye 1980](www.jstor.org/stable/2100656) or Tsybakov 1997. | |
| Dec 16, 2011 at 16:21 | comment | added | onestop | Might help if you say why you need to find outliers. By eye, I wouldn't say there are any particularly severe outliers, considering the size of the data set. The only obvious anomaly is the little line of values with y=0 for x=1,2,3,4.. with a gap just above it. | |
| Dec 16, 2011 at 16:12 | comment | added | Wayne | +1 whuber. @xboxrob: Aren't you, by fitting a curve, already making assumptions about what might be an outlier? I'm not sure. | |
| Dec 16, 2011 at 15:14 | history | edited | Andy W | CC BY-SA 3.0 |
added image directly in post
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| Dec 16, 2011 at 15:14 | comment | added | whuber♦ | You will find it illuminating to plot these bivariate data on log-log axes. (To include the zero values you will need to add a small constant to all values first.) Among other things, you will find that the exponential decay is not quite right; the relationship may be better approximated with a power law. | |
| Dec 16, 2011 at 14:48 | history | edited | whuber♦ |
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| Dec 16, 2011 at 14:29 | history | asked | xboxrob | CC BY-SA 3.0 |