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Mar 31, 2014 at 18:39 comment added jaradniemi Yes. That is correct.
Mar 31, 2014 at 4:23 vote accept Heisenberg
Mar 31, 2014 at 1:53 comment added Heisenberg I see. And what you wrote is true because $f(X|\theta) = f(X|\theta, a, b)$, right?
Mar 31, 2014 at 1:01 history answered jaradniemi CC BY-SA 3.0