m (2a Ken Pesyna moved to 4.2a Ken Pesyna: Improperly named originally)
 
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The E[x] equation you come up with in this problem can be simplified (rid the summation term) by using a differentiated for of the commonly used geometric series. <math>\sum_{n=0}^\infty r^n = 1/(1-r)</math>
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The E[x] equation you come up with in this problem can be simplified (rid the summation term) by using a differentiated form of the commonly used geometric series equation: <math>\sum_{n=0}^\infty r^n = 1/(1-r)</math>
  
 
Now take the derivative with respect to r and you get:
 
Now take the derivative with respect to r and you get:
  
<math>\sum_{n=0}^\infty n*r^n = 1/(1-r)^2</math>
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<math>\sum_{n=0}^\infty nr^{n-1} = 1/(1-r)^2</math>
  
 
You can use this equation to simplify your expected value.
 
You can use this equation to simplify your expected value.

Latest revision as of 06:43, 15 October 2008

The E[x] equation you come up with in this problem can be simplified (rid the summation term) by using a differentiated form of the commonly used geometric series equation: $ \sum_{n=0}^\infty r^n = 1/(1-r) $

Now take the derivative with respect to r and you get:

$ \sum_{n=0}^\infty nr^{n-1} = 1/(1-r)^2 $

You can use this equation to simplify your expected value.

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang