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Nov 4, 2012 at 13:39 answer added wolfies timeline score: 5
Nov 2, 2012 at 9:51 comment added chris @Rojo old is beautiful ;-)
Nov 2, 2012 at 9:49 comment added chris @J.M. I have: mathematica.stackexchange.com/questions/14009/…
Nov 2, 2012 at 1:43 history tweeted twitter.com/#!/StackMma/status/264180971115524097
Nov 2, 2012 at 1:35 comment added J. M.'s missing motivation @chris, sounds like a new question...
Nov 2, 2012 at 1:32 comment added J. M.'s missing motivation @Sasha, think of Release[] as what Mathematica had before they split up that function into ReleaseHold[] and Evaluate[]. A chimera, if you will...
Nov 2, 2012 at 0:48 comment added Rojo @Sasha I think you just learned something old
Nov 1, 2012 at 21:14 comment added chris @Sasha ah! that's just because I learned mathematica 1.2 which had Release which has been depreciated since
Nov 1, 2012 at 20:56 answer added Sjoerd C. de Vries timeline score: 12
Nov 1, 2012 at 20:46 comment added Sasha @chris I did not know of existence of Release. The current way is to use Evaluate. The kudos was because I learned something new.
Nov 1, 2012 at 19:51 comment added Sasha @chris Kudos for Release!
Nov 1, 2012 at 19:40 comment added chris @J.M. and more importantly why is it wrong?? Show[Plot[PDF[\[ScriptCapitalD], x], {x, -5, 5}], RandomVariate[\[ScriptCapitalD], 1500]//Histogram[#, Automatic, "Probability"]&]
Nov 1, 2012 at 19:35 comment added chris @J.M. why does RandomVariate[\[ScriptCapitalD], 15] from reference.wolfram.com/mathematica/ref/… works though?
Nov 1, 2012 at 18:39 history edited J. M.'s missing motivation CC BY-SA 3.0
deleted 37 characters in body
Nov 1, 2012 at 16:41 comment added J. M.'s missing motivation FWIW: it looks to me (after some experimentation) that RandomVariate[] is not equipped to handle arbitrary discrete probability distributions; in general, you might have to roll your own algorithm.
Nov 1, 2012 at 16:41 review First posts
Nov 1, 2012 at 19:17
Nov 1, 2012 at 16:37 answer added chris timeline score: 19
Nov 1, 2012 at 16:32 comment added J. M.'s missing motivation "the distribution I want is a derived mixture of two Poisson distributions" - could you maybe specify how the two Poisson distributions are mixed? I presume you've tried MixtureDistribution[] already?
Nov 1, 2012 at 16:28 comment added chris It would work if for instance your new distribution is derived from a known one via TransformedDistribution as in RandomVariate[ TransformedDistribution[x - 1, x \[Distributed] NormalDistribution[]] // Release, 15]
Nov 1, 2012 at 16:24 history asked Rafal CC BY-SA 3.0