Say we have B and C. B ~ N(0,1) and C ~ N(0,4). We have A =

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Say we have B and C. B ~ N(0,1) and C ~ N(0,4). We have A = B+C. So A ~ N(0,1). B and C are independent, but of course A and B are not independent, because B carries information about A. What is the MAP estimate of B given that A takes value 2? Which of the Bayes rules should we apply

Posted Date: May 19, 2024 03:03 PM

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Say we have B and C. B ~ N(0,1) and C ~ N(0,4). We have A =

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