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Messages 39-38 deleted by topic administrator 07-22-2008 05:11 AM |
goldstonesoft
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07-10-2008 01:25 AM ET (US)
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Edited by author 07-10-2008 02:05 AM
Easy Way to Convert Video Files to MP4 with Cucusoft Ultimate Video Converter Convert to MP4
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| kalison
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36
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07-07-2008 05:29 AM ET (US)
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Messages 35-31 deleted by topic administrator between 07-03-2008 02:42 AM and 02-25-2008 11:11 AM |
| best@gmail.com
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30
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07-30-2007 04:40 PM ET (US)
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gmc trucks here www.gmctruck.fora.pl
gmc from america www.gmctruck.fora.pl
real gmc www.gmctruck.fora.pl
and www.emeraldring.fora.pl rings
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| Goldshield
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29
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11-18-2006 02:17 PM ET (US)
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Can anyone tell me what is the probability of two people writing a 50 word paragraph to be exactly the same?
Gold
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Messages 28-27 deleted by topic administrator between 07-22-2006 09:30 AM and 07-21-2006 09:00 AM |
| Charles Elkan
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02-01-2005 02:54 PM ET (US)
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\m25 answer: Your approach to part b is sound. For part a, can you start with the definition of conditional probability as a fraction, and see what cancels?
Hint: look for an "n choose k" coefficient; what is the other name for "n choose k"?
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| Stephen Krotosky
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02-01-2005 03:01 AM ET (US)
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For problem 4, can you give some more intuition on how to approach this?
For part a, I can see that the distibution associated with Z = \sum x_i is also a Poisson Distribition, but I can't figure out how to use that to get the Conditional distribution.
For part b, I'm assuming that \sum x_i is the sufficient statistic, since we were supposed to prove that in part a. I'm guessing that we should use an initial estimator of
P(x_i = 0 event) = 1 for x1 = 0, 0 for x1 != 0.
This would be an unbiased estimator, and then we could find what P(x_i = 0 event| \sum x_i = A) which should give the MVUE.
I think I have outlined the idea of how to do it, but I can't seem to do the specific computations to get the results. Could you please offer some ideas on how to actually compute these values. Thanks.
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| Charles Elkan
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01-30-2005 10:53 PM ET (US)
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/m21 answer: Constant variance does not imply zero covariance. However, you may assume that all estimators are independent here. In other words, each estimator has a different argument x where all the x are independent of each other.
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