Learning Parameters When Some Data is Missing - Sprinkler Example
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2012년 4월 6일 금요일 오후 5:09
Noob question please :) In the sprinkler example packaged with Infer.NET, the 2nd part of the program attempts to learn parameters of the model given data C, S, R, W. What if I cannot observe C, but would still like to infer ProbCloudyPosterior?
Thanks a lot in advance
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2012년 4월 7일 토요일 오전 10:37
In LearnParameters you will see the line
Cloudy.ObservedValue = cloudy;
If you change this to:Cloudy.ClearObservedValue();
Then Cloudy will be left unobserved. See ProbRain for an example of this.
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2012년 4월 9일 월요일 오후 4:41Thanks a lot Tom!
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2012년 4월 17일 화요일 오후 10:34
Hi Tom,
It seems that if I do not provide any observed values to Cloudy (i.e. Cloudy.ClearObservedValue()), the model has a very hard time learning the probabilities. The learned probabilities, even when I generate 10,000 sample data points, are:
Prob. Cloudy: Ground truth: 0.30, Inferred: 0.50
Prob. Sprinkler | Cloudy: Ground truth: 0.10, Inferred: 0.38
Prob. Sprinkler | Not Cloudy: Ground truth: 0.50, Inferred: 0.38
Prob. Rain | Cloudy: Ground truth: 0.80, Inferred: 0.38
Prob. Rain | Not Cloudy: Ground truth: 0.20, Inferred: 0.38
Prob. Wet Grass | Sprinkler, Rain: Ground truth: 0.99, Inferred: 0.99
Prob. Wet Grass | Sprinkler, Not Rain Ground truth: 0.90, Inferred: 0.89
Prob. Wet Grass | Not Sprinkler, Rain: Ground truth: 0.90, Inferred: 0.90
Prob. Wet Grass | Not Sprinkler, Not Rain: Ground truth: 0.00, Inferred: 0.00(note: I changed the actual prob. cloudy from (0.5, 0.5) in the original code to (0.3, 0.7)).
Any help is greatly appreciated.
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2012년 4월 18일 수요일 오전 10:46
By the looks of it, the parameters have come out symmetric. When you don't observe Cloudy, the model effectively becomes a mixture model. So you have to be careful about breaking symmetry as explained in the Mixture of Gaussians example.
- 편집됨 Tom MinkaMicrosoft Employee 2012년 4월 18일 수요일 오전 10:47
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2012년 4월 18일 수요일 오후 6:54
Thanks so much Tom. Did as you suggested, and now, P(S | C) and P(R | C) are better calculated. But it seems P(C) still can't be inferred.
Prob. Cloudy: Ground truth: 0.30, Inferred: 0.50
Prob. Sprinkler | Cloudy: Ground truth: 0.10, Inferred: 0.16
Prob. Sprinkler | Not Cloudy: Ground truth: 0.50, Inferred: 0.60
Prob. Rain | Cloudy: Ground truth: 0.80, Inferred: 0.60
Prob. Rain | Not Cloudy: Ground truth: 0.20, Inferred: 0.16
Prob. Wet Grass | Sprinkler, Rain: Ground truth: 0.99, Inferred: 0.99
Prob. Wet Grass | Sprinkler, Not Rain Ground truth: 0.90, Inferred: 0.90
Prob. Wet Grass | Not Sprinkler, Rain: Ground truth: 0.90, Inferred: 0.90
Prob. Wet Grass | Not Sprinkler, Not Rain: Ground truth: 0.00, Inferred: 0.00
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2012년 4월 19일 목요일 오후 12:47Yes, it isn't possible to recover the true parameters exactly when cloudy is not observed. This is because the model has too many parameters. There is 1 for cloudy, 2 for sprinkler, 2 for rain, totalling 5 for p(sprinkler,rain) alone. But you only need 3 parameters to describe any joint distribution of (sprinkler,rain).
- 편집됨 Tom MinkaMicrosoft Employee 2012년 4월 19일 목요일 오후 12:47
