MAP estimation

01Understand the idea#

Maximum a posteriori, or MAP, estimation selects the parameter candidate with the greatest posterior probability. It combines the likelihood of the observed data with a prior preference over candidates. The candidate favored by data alone need not win after that prior is included.

The prior favors A, and likelihood favors C. Multiply their matching values: weights are 0.1,0.15,0.12. B has the largest posterior weight.

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Every posterior mass has the same positive evidence denominator. For candidate $i$, multiply its likelihood $L_i$ by prior $\pi_i$. Choose every candidate tied at the largest weight.

For candidates A, B, C with prior $(0.5,0.3,0.2)$ and likelihood $(0.2,0.5,0.6)$, multiply matching pairs to get $(0.1,0.15,0.12)$. B has the greatest weight. All posterior masses would share the same positive normalization constant, so normalization is unnecessary if you only need the winner.

02For prior (0.6,0.4) and likelihood (0.2,0.5), weights are (0.12,0.2)#

For prior $(0.6,0.4)$ and likelihood $(0.2,0.5)$, weights are $(0.12,0.2)$. The second candidate wins even though its prior is smaller. Comparing just one input can change the answer.

03Keep all candidates tied at the largest weight#

Keep all candidates tied at the largest weight. A zero prior produces zero posterior weight when evidence is positive. Equal nonzero priors preserve likelihood ordering; unequal priors can change it.

04MAP reports a best candidate, not the full uncertainty over candidates#

MAP reports a best candidate, not the full uncertainty over candidates. Preserve all tied maxima when the task asks for every posterior mode. To answer a probability question about several candidate settings, retain and normalize the full grid instead.

Further questions