Comments on: The Banff Challenge [Eqn]
http://hea-www.harvard.edu/AstroStat/slog/2008/eotw-banff-challenge/
Weaving together Astronomy+Statistics+Computer Science+Engineering+Intrumentation, far beyond the growing bordersFri, 01 Jun 2012 18:47:52 +0000hourly1http://wordpress.org/?v=3.4By: hlee
http://hea-www.harvard.edu/AstroStat/slog/2008/eotw-banff-challenge/comment-page-1/#comment-320
hleeWed, 03 Sep 2008 06:04:24 +0000http://hea-www.harvard.edu/AstroStat/slog/?p=357#comment-320A fun video clip about LHC from Youtube, Large Hadron Rap.
<object width="425" height="344"><param name="movie" value="http://www.youtube.com/v/j50ZssEojtM&hl=en&fs=1"></param><param name="allowFullScreen" value="true"></param><embed src="http://www.youtube.com/v/j50ZssEojtM&hl=en&fs=1" type="application/x-shockwave-flash" allowfullscreen="true" width="425" height="344"></embed></object>A fun video clip about LHC from Youtube, Large Hadron Rap.
]]>By: Paul B
http://hea-www.harvard.edu/AstroStat/slog/2008/eotw-banff-challenge/comment-page-1/#comment-311
Paul BMon, 28 Jul 2008 22:17:50 +0000http://hea-www.harvard.edu/AstroStat/slog/?p=357#comment-311Hmmm, indeed this is a tricky problem! N,Y and Z are <b>conditionally</b> independent, given the parameters, but the real problem is that epsilon is a multiplicative factor on the interest parameter. That makes inference very sensitive to the efficiency of the detector -- and is the main reason why no method works perfectly for a wide range of source and efficiency values.
Finding the various types of MLE (profile, modified profile etc) is not too bad, the real problem is providing reliable estimates of the uncertainty. Finding a 95% confidence interval that actually provides 95% confidence is far easier said than done (see the references for some explanations). This is also the primary criterion for many of the physicists involved, so it leaves plenty of work to be done.
It still bothers me that there isn't a decent Bayesian solution yet (that I've seen anyway), but I'm sure one is out there somewhere...Hmmm, indeed this is a tricky problem! N,Y and Z are conditionally independent, given the parameters, but the real problem is that epsilon is a multiplicative factor on the interest parameter. That makes inference very sensitive to the efficiency of the detector — and is the main reason why no method works perfectly for a wide range of source and efficiency values.
Finding the various types of MLE (profile, modified profile etc) is not too bad, the real problem is providing reliable estimates of the uncertainty. Finding a 95% confidence interval that actually provides 95% confidence is far easier said than done (see the references for some explanations). This is also the primary criterion for many of the physicists involved, so it leaves plenty of work to be done.
It still bothers me that there isn’t a decent Bayesian solution yet (that I’ve seen anyway), but I’m sure one is out there somewhere…
]]>By: vlk
http://hea-www.harvard.edu/AstroStat/slog/2008/eotw-banff-challenge/comment-page-1/#comment-310
vlkMon, 28 Jul 2008 21:16:54 +0000http://hea-www.harvard.edu/AstroStat/slog/?p=357#comment-310Yeah, N, Y, and Z are independent. With just the first two equations, the problem is very simple, and has been solved analytically numerous times. But including the third one throws it for a loop. The hardest part is to make sure that the frequency coverage on lambda_S is correct at the low counts (N~few) level. See these two for example for how complicated it can become:
Edlefson, P. (JSM2007) - <a href="http://www.amstat.org/meetings/jsm/2007/onlineprogram/index.cfm?fuseaction=abstract_details&abstractid=310238" rel="nofollow">A Dempster-Shafer Bayesian Solution to the Banff A1 Challenge</a>
Baines, P. (JSM2007) - <a href="http://www.amstat.org/meetings/jsm/2007/onlineprogram/index.cfm?fuseaction=abstract_details&abstractid=308563" rel="nofollow">Upper Limits for Source Detection in the Three-Poisson Model</a>Yeah, N, Y, and Z are independent. With just the first two equations, the problem is very simple, and has been solved analytically numerous times. But including the third one throws it for a loop. The hardest part is to make sure that the frequency coverage on lambda_S is correct at the low counts (N~few) level. See these two for example for how complicated it can become:
Edlefson, P. (JSM2007) – A Dempster-Shafer Bayesian Solution to the Banff A1 Challenge
Baines, P. (JSM2007) – Upper Limits for Source Detection in the Three-Poisson Model
]]>By: brianISU
http://hea-www.harvard.edu/AstroStat/slog/2008/eotw-banff-challenge/comment-page-1/#comment-309
brianISUMon, 28 Jul 2008 01:18:57 +0000http://hea-www.harvard.edu/AstroStat/slog/?p=357#comment-309Being completely ignorant on the physics of the problem, what are the assumptions that are reasonable to make on the data? e.g., are N, Y, and Z independent random variables? That might begin to make things easier. Then, with a derived likelihood, maybe maximize out the nuisance parameters and just be left with the parameter of interest. Or one could try numerical maximizing techniques. This problem also seems to be screaming a Bayesian approach and look at the posterior of the parameter of interest, especially since there are experts in this field that could give reliable prior information. Of course everything I am saying is just a fun thought process and I am not claiming I would really know what to do.Being completely ignorant on the physics of the problem, what are the assumptions that are reasonable to make on the data? e.g., are N, Y, and Z independent random variables? That might begin to make things easier. Then, with a derived likelihood, maybe maximize out the nuisance parameters and just be left with the parameter of interest. Or one could try numerical maximizing techniques. This problem also seems to be screaming a Bayesian approach and look at the posterior of the parameter of interest, especially since there are experts in this field that could give reliable prior information. Of course everything I am saying is just a fun thought process and I am not claiming I would really know what to do.
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