m
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Ken
 
Ken
 
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this is also <math>n \cdot \left(\frac{1}{n} + \frac{1}{n-2} + \cdots + \frac{1}{1}\right) </math>. The <math>\left(\frac{1}{n} + \frac{1}{n-2} + \cdots + \frac{1}{1}\right) </math> portion is the harmonic number. I do not believe there is a closed form to this number
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 +
Writing it as: <br>
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  <math> n \cdot \sum_{i=n}^1\frac{1}{i}\<\math>  
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<br> can help you see that the summation is the Harmonic number. I do not believe there is a closed form to this number.
  
 
AJ
 
AJ
 
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Revision as of 16:33, 6 October 2008

E as opposed to P


I am not entirely certain, but since $ \frac{n - i + 1}{n}\! $ is the probability of getting a different coupon for each one, souldn't the expected value be:


$ \sum_{i=1}^n\frac{n}{n - i + 1}\! $


The sum of the individual expected values should be the expected value of the sum, right? Just a thought. I am not sure, if that is right.


Perhaps, someone else could expand on this idea with more word-for-word, verbatim note-copying.


Virgil, you are right. The question is, can this be simplified to a closed form (non-summation) equation? Still trying to determine this.

Ken


Writing it as:

$  n \cdot \sum_{i=n}^1\frac{1}{i}\<\math>  <br> can help you see that the summation is the Harmonic number. I do not believe there is a closed form to this number.   AJ ------- $

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