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Dec 30, 2019 at 17:38 vote accept Noah Stebbins
May 7, 2019 at 5:47 history edited dnqxt CC BY-SA 4.0
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May 7, 2019 at 0:42 comment added dnqxt Where do you see independence in that equality?
May 7, 2019 at 0:25 comment added Noah Stebbins Hey @dnqxt, I'm not quite sure I follow. If the $x_i$-s are not assumed conditionally independent, then how does $p(x_{1:n}|\theta)p(x_{n+1}|x_{1:n}, \theta) = p(x_{1:n+1}|\theta)$ in the second approach?
May 7, 2019 at 0:09 comment added dnqxt That's a good question, but the answer is no, $x_i$-s are not assumed conditionally independent given the initial $\theta$. $\theta$ changes at each stage if one adopts the second (step by step or online) approach above. That's why it's necessary to include all $x_i$-s in the "batch" likelihood in the first approach according to the product rule. As for conditional independence of $x_i$-s given $\theta$ that indeed shows up in a popular Naive Bayes model, but that's another topic.
May 6, 2019 at 23:48 comment added Noah Stebbins Hi @dnqxt, quick follow-up question: how did you go from $p(x_{1:n}|\theta)p(x_{n+1}|x_{1:n}, \theta)$ to $p(x_{1:n+1}|\theta)$? Did you assume conditional independence between $x_{1:n}$ and $x_{n+1}$ ergo $p(x_{n+1}|x_{1:n}, \theta) = p(x_{n+1}|\theta)$ followed by using conditional independence between $x_{1:n}$ and $x_{n+1}$ again, ergo $p(x_{1:n+1}|\theta) = p(x_{1:n}|\theta)p(x_{n+1}|\theta)$? Thus, is conditional independence between $x_{1:n}$ and $x_{n+1}$ necessary to yield the same posterior distribution?
May 6, 2019 at 2:28 history edited dnqxt CC BY-SA 4.0
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May 5, 2019 at 21:46 history edited dnqxt CC BY-SA 4.0
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May 5, 2019 at 21:38 history edited dnqxt CC BY-SA 4.0
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May 5, 2019 at 21:04 history edited dnqxt CC BY-SA 4.0
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May 5, 2019 at 20:36 history edited dnqxt CC BY-SA 4.0
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May 5, 2019 at 20:28 history edited dnqxt CC BY-SA 4.0
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May 5, 2019 at 20:15 history answered dnqxt CC BY-SA 4.0