Thanks for the forward. John Maddock wrote:
Nelis Franken wrote:
The Wikipedia reference: http://en.wikipedia.org/wiki/Exponential_distribution
The Boost implementation (line 54 of exponential_distribution.hpp) -result_type(1) / _lambda * log(result_type(1)-eng());
This is -1/lambda * log(1-eng()) This matches what wikipedia shows.
The first, less critical point: If eng() is uniform, then so is (1 - eng()), which allows for the simplification of the statement passed to log() from: log(result_type(1)-eng()) to log(eng())
Not quite. Our engines return values in the range [0,1) (lower bound inclusive, upper bound exclusive). log(0) is undefined, thus we avoid that by using (1-eng()). Wikipedia states "Moreover, if U is uniform on (0;1), then so is 1 − U. " Note the interval that is open on both ends in the precondition. I believe the boost implementation is correct here. Thanks, Jens