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Jul 31, 2015 at 15:35 vote accept maksay
Jul 31, 2015 at 15:22 answer added ziggystar timeline score: 6
Jul 31, 2015 at 15:14 comment added maksay I understand that. However, when I marginalize over $x_{i+2},\cdots $, I can't seem to get rid of an expression that grows exponentially with the size of $n-i$. I guess what I was really trying to ask is if there is a form of $P(x_i|x_{i-1})$ that does not involve later terms. (I.e. $P(x_i|X_{i-1})=\psi(x_{i-1},x_i)\phi(x_i)$. Which would get what I need, but, unfortunately, is not a valid distribution)
Jul 31, 2015 at 15:03 comment added ziggystar You have to marginalize over $x_{i+2},\ldots,x_n$, and you can drop the factors involving $x_1,\ldots,x_{i-1}$ because of conditional independence.
Jul 31, 2015 at 15:02 history edited maksay CC BY-SA 3.0
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Jul 31, 2015 at 14:54 history asked maksay CC BY-SA 3.0