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  #151  
Old 10-30-2019, 10:23 AM
viewchinovision viewchinovision is offline
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I agree that they should accept both answers for #38. When building the tree, it's true that the number of predictors m used are fewer than the total predictors available p. However, the final tree could very well be comprised of all p predictors. It is very ambiguously phrased indeed, which is why it may hard to argue for just one answer over another.
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  #152  
Old 10-30-2019, 12:01 PM
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Originally Posted by daaaave View Post
Since you know that the total area under the curve is 1, you could try to visually estimate something by dividing the region under the curve into rectangles whose relative sizes you can estimate. But one thing that you know is that the total area under the curve is 1, the total range of the parameter is also 1, so the average posterior density is 1. That means that the peak values of the density are greater than 1, and 0.3 = integral (posterior density over HPDI). The HPDI is going to be a region where the posterior is big, so here greater than 1, so to make that integral 0.3, the range we are integrating over must be less than 0.3. That is, the HPDI is shorter than 0.3. Only choice D is both shorter than 0.3 and includes the mode.

You can improve on that argument slightly: almost all the area under the curve is when the parameter is between 0.3 and 0.9, so the average on (0.3, 0.9) is on the order of 1/0.6 = 1.67, and from that you can conclude that our HPDI is shorter than 0.18 in length.
Hey Dave, do you have an idea when the written solutions for the exam will be up? Any discrepancies with the preliminary key on a first pass?
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  #153  
Old 10-30-2019, 12:22 PM
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I'm still working on written solutions, I'm hoping tomorrow. My only issues so far with the preliminary answer key are that the given information in #35 is impossible so I'm not sure yet how to proceed and I agree that the wording of I in #38 is bad and I can see both answers being reasonable.
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  #154  
Old 10-30-2019, 02:37 PM
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Originally Posted by daaaave View Post
I'm still working on written solutions, I'm hoping tomorrow. My only issues so far with the preliminary answer key are that the given information in #35 is impossible so I'm not sure yet how to proceed and I agree that the wording of I in #38 is bad and I can see both answers being reasonable.
Thanks for all your work!
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  #155  
Old 10-30-2019, 04:10 PM
RentLaVBoM RentLaVBoM is offline
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Quote:
Originally Posted by daaaave View Post
I'm still working on written solutions, I'm hoping tomorrow. My only issues so far with the preliminary answer key are that the given information in #35 is impossible so I'm not sure yet how to proceed and I agree that the wording of I in #38 is bad and I can see both answers being reasonable.
Yes, thank you for your work on the answer key! Given your comment above, does that mean you think #4 was fair game? It previously sounded like there was agreement that the binomial distribution was outside of the sections cited on the syllabus. I think I am going to protest that one, but wanted to make sure I'm not missing something. Thank you!
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  #156  
Old 10-30-2019, 04:25 PM
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Since it is only asking about credibility for frequency and not aggregate it’s probably ok. The derivation in the text starts generally for frequency then only adds the Poisson assumption at the last step. And you can argue that a strong candidate could redo the derivation on the fly.

That doesn’t mean you shouldn’t appeal, just that it seems like a long shot.
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  #157  
Old 10-31-2019, 02:56 PM
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I have to go pick up the kids so they can get into their Halloween costumes, here's the draft solutions that I have.

Of the controversial questions, I think #35 should be E. The given information in the problem isn't possible (if the sample mean is 2 and the prior is exponential, the posterior mode is below 2 and in particular not 2.1), so maybe the question will be thrown out. But choices II and III definitely have a lower posterior mode than the original prior, and choice I has a higher posterior mode than the original prior so long as the sample mean is below 7 (which it is), so I only, or E is still the best answer.

For #38, I would have answer said I is false because it talked about individual trees, not splits. I don't know how the CAS will rule on that.

And to add a new one to the mix, on #34, I think the answer can be anything from 0 to 0.5. I think the intention of the problem is that we have run for long enough to be roughly stationary, and our stationary distribution is symmetric, so the next accepted step of the chain is 50-50 to be right or left. I agree with that part. But if sigma is extremely small, there is a high probability that the next proposal will be rejected, and the next iteration will remain where it is, making a small probability of moving right (or left)

I'll work on cleaning these up some more tomorrow afternoon, but since people might be appealing things felt an early draft would be useful.
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