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ordered vector abs #48
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Why can't we just Because of the joint density of ordered statistics? Take The Jacobian in this case is exactly the density chosen. We don't need to compute |
I'm finding that using the absolute value to increment is more stable than$\exp(\cdot)$ . The jacobian adjustment
is constant.
I have to put a prior on x or else it's improper. When I compare to the built-in I notice the built-in has 1 or 2 divergences with a normal prior (N(0, 10)) and this one doesn't.
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