(New page: Regarding Seraj's solution, i believe the solution would be <math>\sqrt(A) cos(''2pi*''drand48())</math>)
 
 
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Regarding Seraj's solution, i believe the solution would be
 
Regarding Seraj's solution, i believe the solution would be
  
<math>\sqrt(A) cos(''2pi*''drand48())</math>
+
<math>\sqrt(A) cos(2pi*drand48())</math>
 +
 
 +
where A is still the exponential RV calculated in 1b.
 +
 
 +
 
 +
- Yes you are correct, I forgot the 2pi factor to change the range of the uniform random variable from [0-1] to [1-2pi].
 +
 
 +
thank you for pointing it out.
 +
 
 +
Seraj

Latest revision as of 15:07, 21 October 2008

Regarding Seraj's solution, i believe the solution would be

$ \sqrt(A) cos(2pi*drand48()) $

where A is still the exponential RV calculated in 1b.


- Yes you are correct, I forgot the 2pi factor to change the range of the uniform random variable from [0-1] to [1-2pi].

thank you for pointing it out.

Seraj

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