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| Jan Schellenberger
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28
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02-13-2005 05:34 AM ET (US)
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/m27There is a trick. You can use the CDF of the uniform (P(z<Z)) to correspond to some CDF of the radii (P(r>R)). Once you have that, the PDF is the derivative of the CDF.
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| Stephen Krotosky
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27
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02-13-2005 01:55 AM ET (US)
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For problem 4,
I've figured out what the distribution looks like, and I believe I can define it mathematically, but am having trouble doing so. Basically what I would like to do is transform a uniform distribution to the desired one. Here is my logic.
If we look at a cross section of an organelle of radius r, the cross section can be thought of as cutting the organelle at a point x that is generated uniformly from -r..r. This corresponds to the front and back of the sphere relative to the cutting plane.
If we know where the sphere is cut, we can compute the observed radius, Z = sqrt(r^2 - x^2). My questions is how can I use the function for Z and the fact that x is uniform to find the pdf (likelihood function) for Z.
I know this is possible but am having difficulty actually doing it. Thanks.
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Charles Elkan
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26
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02-12-2005 03:47 PM ET (US)
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/m25 answer: Yes, you may assume that the number of cross-sections n is fixed.
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| Hyun Min Kang
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25
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02-12-2005 12:13 PM ET (US)
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/m21 /m24 Thanks. I understand what you mean. One more question. Can we assume that the number of cross-section is always n? If the organelles are distributed randomly, the number of cross-section will actually vary. So, the distribution should involve the probability distribution of n also. But since there is no description about the density of the organelles, I cannot compute the exact distribution. (Also, the problem becomes much more complicated).
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Charles Elkan
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24
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02-12-2005 01:07 AM ET (US)
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Charles Elkan
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23
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02-12-2005 01:02 AM ET (US)
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Charles Elkan
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22
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02-12-2005 12:56 AM ET (US)
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| Ryan Kelley
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21
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02-11-2005 11:40 PM ET (US)
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\m20 Just as a random variable has a distribution, so does some function of that variable. Since an estimator is just a function of the data, it will some distribution as well. In your example, if x1,...,xn are iid gaussian(\mu,\sigma^2), then avg(x1,...,xn) will follow a gaussian distribution with mean \mu and variance \frac{\sigma^2}{n}
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| Hyun Min Kang
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20
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02-11-2005 09:44 PM ET (US)
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Edited by author 02-11-2005 09:44 PM
In problem 4, I don't get the last sentence. What did you mean by "distribution of estimate"? For example, if avg(x1,..,xn) is the MLE of \mu in Gaussian, what is the distribution of the estimate?
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| Hyun Min Kang
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19
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02-11-2005 12:37 PM ET (US)
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/m17 Thanks for the information. I found dirichlet there, but Zipf distribution is not stated clearly, and there is no power law distribtution. Where can I get these info?
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| Hyun Min Kang
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18
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02-11-2005 12:33 PM ET (US)
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In part 3(c), "Explain carefully whether or not the theorem relies on the exponential family being described using its natural parameters." I don't get that part. What does that mean, and what I am supposed to explain?
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| Banu Dost
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17
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02-11-2005 05:19 AM ET (US)
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from wikipedia.com Banu
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| Hyun Min Kang
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16
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02-11-2005 03:03 AM ET (US)
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Where can I get the formal definition of Dirichlet distributions, power law distirbutions, and Zipf distributions?
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Charles Elkan
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15
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02-10-2005 11:55 AM ET (US)
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/m10, /m11 answer: Yes, the binomial assumes N is large. You may assume this; I should have mentioned it in the problem statement. No need to do the more difficult hypergeometric calculations.
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Charles Elkan
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14
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02-10-2005 11:52 AM ET (US)
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/m10 answer: I think the reasoning with the binomial is correct, and the hypergeometric is not needed for this problem. Be careful with this claim: "if the actual N is less than the claimed N, we would expect to see more tagged animals." I'm not saying it's false (or true) just that it requires careful thought to be sure.
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Charles Elkan
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13
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02-10-2005 11:45 AM ET (US)
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/m7 answer: I discussed this with Ryan, and I think he is correct. This does not mean that the Gaussian-based answer is incorrect, just that it may be unreliable in the real world. It points out the need to investigate which distribution(s) model real-world returns well.
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