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I am reading about bayes estimator and it says that the parameter for MAP (maximum a posteriori) estimation is equal to Bayes estimation. However, there is no proof whatsoever and I cannot seem to find proof of this online.

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In Bayesian estimation one views the parameter as a random variable and bases estimates on its posterior distribution given the data. However, there is no single, "correct" way to get an estimate from a posterior distribution, so one common approach is to take the mode of the posterior distribution. This is sort of the Bayesian analog of maximum likelihood. Another approach is to take the mean of the posterior, as one does when using the squared error loss function.

So to answer your question, the idea of taking the most likely value of the parameter is just one approach to doing Bayesian estimation and is not necessarily something that needs proving.

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  • $\begingroup$ There is actually a proof here: cs.ubc.ca/~murphyk/Papers/bayesGauss.pdf but I cannot follow equation (17) $\endgroup$
    – Eyess1982
    Oct 10, 2016 at 12:15
  • $\begingroup$ The author is not actually deriving the MAP estimator here, just writing down a posterior. This appears just to be some algebra, but the author is writing in a very confusing way, using notation that he hasn't even defined yet. Try substituting $\sigma_n^{2} = \left ( 1 / \sigma^2_0 + n / \sigma^2 \right )^{-1}$ into (17) and see if you get (16). $\endgroup$
    – dsaxton
    Oct 10, 2016 at 14:28
  • $\begingroup$ By the way, in the Gaussian case the MAP estimator is just the posterior mean, which actually can be derived as a Bayes estimator under squared error loss. $\endgroup$
    – dsaxton
    Oct 10, 2016 at 14:30

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